流体力学与传热习题参考解答(英文)

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流体力学与传热：1-3 Basic Equations of Fluid Flow

Since mass carries with it associated energy due to its position, motion, or physical state, we will find that each of these types of energy will appear in the energy balance.
It is usual to take α to 1 in the calculation.
1.3.4 Overall Mechanical Energy Balance for Steady-state Flow System
The total energy balance, Eq. (4) is not often used when appreciable enthalpy changes occur or appreciable heat is added (or subtracted), since the kinetic- and potentialenergy terms are usually small and can be neglected.
a
b
At steady state the mass flow in equals the mass flow out,
m Q VaSaa VbSbb const
(1.3-6 )
The equation is also called the equation of continuity.
For an incompressible fluids
a b
If the fluid flows through a channels of circular cross section ,then the volumetric flow rate is

流体力学与传热习题：problems and solutions forchapter1,2

1.3 A differential manometer as shown in Fig. is sometimes used to measure small pressuredifference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B ismethane （甲烷）, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and thatliquid A in the U tube is water. The inside diameters of the reservoirs and U tube are 51mm and6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure differenceover the instrument In meters of water, (a) when the change in the level in the reservoirs isneglected, (b) when the change in the levels in the reservoirs is taken into account? What is thepercent error in the answer to the part (a)?Solution ：p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesR d x D 2244ππ= (1) so R D d x 2⎪⎭⎫ ⎝⎛= (2) and hydrostatic equilibrium gives following relationshipg R g x p g R p A c c ρρρ++=+21 (3)sog R g x p p c A c )(21ρρρ-+=- (4)substituting the equation (2) for x into equation (4) givesg R g R D d p p c A c )(221ρρρ-+⎪⎭⎫ ⎝⎛=- (5) （a ）when the change in the level in the reservoirs is neglected,()Pa g R g R g R D d p p c A c A c 26381.98151000145.0)()(221=⨯-=-≈-+⎪⎭⎫ ⎝⎛=-ρρρρρ（b ）when the change in the levels in the reservoirs is taken into account()Pa g R g R D d g R g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22221=⨯-+⨯⨯⨯⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-ρρρρρρ error=％＝7.68.2812638.281-1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. The readings of two U-tube manometers are R 1=400mm ，R 2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R 3=50mm. Try to calculate the pressure at point A and B .Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point A is given by hydrostatic equilibriumg R R g R g R p g Hg O H A )(32232+-+=ρρρg ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplifiedc A p p ≈=232gR gR Hg O H ρρ+=1000×9.81×0.05+13600×9.81×0.05=7161N/m²1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/mFigure for problem 1.41.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank A and the exit ofdrainpipe are all open to air.Solution: Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: 2222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:22⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭⎫ ⎝⎛=d D u u o 2 Bernoulli equation is written between the throat and the station 2-23 Combining equation 1,2,and 3 gives222u Hg =222200u u p =+ρSolving for HH=1.39m1.6 A liquid with a constant density ρ kg/m 3 is flowing at an unknown velocity V 1 m/s through a horizontal pipe of cross-sectional area A 1 m 2 at a pressure p 1 N/m 2, and then it passes to a section of the pipe in which the area is reduced gradually to A 2 m 2 and the pressure is p2. Assuming no friction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured. Solution :In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,2112A A V V = For the items in the Bernoulli equation , for a horizontal pipe,z 1=z 2=0Then Bernoulli equation becomes, after substituting 2112A A V V = for V 2, ρρ22121211212020p A A V p V ++=++ ()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u HgRearranging,2)1(21212121-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-12221211A A p p V ρ＝Performing the same derivation but in terms of V 2,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--21221212A A p p V ρ＝1.7 A liquid whose coefficient of viscosity is µ flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity V . Show that the pressure loss in a length of pipe L p ∆ is 232dV μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.Solution :The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2 substituting equation 2 for u into equation 1 and integrating3 rearranging equation 3 gives ⎰⎰==R R rdr u R udA A V 020211ππ⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--=22014R r R L p p u L μ2032D L p p V L μ-=1.8. In a vertical pipe carrying water, pressure gauges areinserted at points A and B where the pipe diameters are0.15m and 0.075m respectively. The point B is 2.5m belowA and when the flow rate down the pipe is 0.02 m 3/s, thepressure at B is 14715 N/m 2 greater than that at A.Assuming the losses in the pipe between A and B can beexpressed as g V k 22where V is the velocity at A, find the value of k . If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres.Solution:d A =0.15m; d B =0.075mz A -z B =l =2.5mQ =0.02 m 3/s,p B -p A =14715 N/m 2s m d QV V d Q A A AA /132.115.0785.002.044222=⨯===ππs m d QV V d Q B B BB /529.4075.0785.002.044222=⨯===ππWhen the fluid flows down, writing mechanical balance equation222222A B B B A A A V k V g z p V g z p +++=++ρρ 213.1253.4100014715213.181.95.2222k ++=+⨯ Figure for problem 1.8 232d V L p μ=∆Pa d VL p 115201.01206.005.0323222=⨯⨯⨯==∆μk 638.0260.10715.14638.0525.24++=+=k 0.295making the static equilibriumgR g x g l p g R g x p Hg A B ρρρρρ+∆++=+∆+()()mm g g l p p R g H A B 7981.91260081.910005.214715-=⨯⨯⨯-=---=ρρρ1.9．The liquid vertically flows down through the tube from thestation a to the station b , then horizontally through the tube fromthe station c to the station d , as shown in figure. Two segments ofthe tube, both ab and cd ，have the same length, the diameter androughness.Find:（1）the expressions of g p ab ρ∆, h fab , g p cd ρ∆ and h fcd , respectively. （2）the relationship between readings R 1and R 2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station dFigure for problem 1.922V d l h fab λ=22V d l h fcd λ=fcdfab h h =fab b a h p p +=+ρρlg fab b a h p p =+-lg ρfcd d c h p p +=ρρhence3From static equationp a -p b =R 1（ρˊ-ρ）g -l ρg 4p c -p d =R 2（ρˊ-ρ）g 5Substituting equation 4 in equation 2 ，thentherefore6Substituting equation 5 in equation 3 ，then7ThusR 1=R 21.10 Water passes through a pipe of diameter d i=0.004 m with the average velocity 0.4 m/s, as shown in Figure.1) What is the pressure drop –∆P when water flows through the pipe length L =2 m, in m H 2O column?2) Find the maximum velocity and point r at which it occurs.3) Find the point r at which the average velocityequals the local velocity. 4）if kerosene flows through this pipe ，how do thevariables above change ？（the viscosity and density of Water are 0.001 Pasand 1000 kg/m 3，respectively ；and the viscosityand density of kerosene are 0.003 Pas and 800kg/m 3，respectively ）solution:1）1600001.01000004.04.0Re =⨯⨯==μρud fcd d c h p p =-ρfab h g l g R =+--'lg 1ρρρρ）（g R h fab ρρρ-'=1g R h fcd ρρρ-'=2Figure for problem 1.10from Hagen-Poiseuille equation1600004.0001.024.0323222=⨯⨯⨯==∆d uL P μ m g p h 163.081.910001600=⨯=∆=ρ 2）maximum velocity occurs at the center of pipe, from equation 1.4-19max 0.5V u = so u max =0.4×2=0.8m3）when u=V=0.4m/s Eq. 1.4-172max 1⎪⎪⎭⎫ ⎝⎛-=wr r u u 5.0004.01max2=⎪⎭⎫ ⎝⎛-u V r ＝ m r 00284.071.0004.05.0004.0=⨯== 4) kerosene:427003.0800004.04.0Re =⨯⨯==μρud Pa p p 4800001.0003.01600=='∆='∆μμ m g p h 611.081.98004800=⨯=''∆='ρ1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entranceis 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m³/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m²). l e /d ≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.Solution ：(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’ and the section of pressure point 2—2’，and take the center of section 2—2’ as the referring plane, then ∑+++=++21,2222121122—f h p u gZ p u gZ ρρ （a ） In the equation 01=p (the gauge pressure)222/396304.181.910004.081.913600m N gh gR p O H Hg =⨯⨯-⨯⨯=-=ρρ0021=≈Z uWhen the gate valve is fully closed, the height of water level in the reservoir can be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).gR h Z g Hg O H ρρ=+)(12 （b ）where h=1.5mR=0.6mSubstitute the known variables into equation b 2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600V V V K d l h m Z c f =+⨯=+==-⨯=∑λ Substitute the known variables equation a9.81×6.66=2213.21000396302V V ++ the velocity is V =3.13m/s Figure for problem 1.12the flow rate of water is h m V d V h /5.8813.312.0436004360032=⨯⨯⨯=⨯=ππ2） the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 1—1’ and 3-3´，then∑+++=++31,3233121122—f h p V gZ p V gZ ρρ （c ） since m Z 66.61=311300p p u Z =≈=2223_1,81.4 2]5.0)151.035(025.0[ 2)(V V V K d l l h c e f =++=++=∑λ input the above data into equation c ，9.8122V 81.4266.6+=⨯V the velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’ and 2——2’, for the same situation of water level ∑+++=++21,2222121122—f h p V gZ p V gZ ρρ （d ）since m Z 66.61=212103.51/0(page pressure Z u u m s p =≈≈=）kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λ input the above data into equationd ， 9.81×6.66=2.261000251.322++pthe pressure is: 329702=p1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60⨯3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution ： The variables of main pipe are denoted by a subscript 1, and branch pipe by subscript 2.The friction loss for parallel pipelines is2121S S s f f V V V h h +==∑∑The energy loss in the branch pipe is 22222222u d l l h e f ∑∑+=λ In the equation 03.02=λs m u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation ckg J h f /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is 333.022111121===∑∑u d l h h f f λ So s m u /36.22018.023.0333.01=⨯⨯⨯= The water discharge of main pipe is h m V h /60136.23.043600321=⨯⨯⨯=π Total water discharge ish m V h /7.60372.26013≈+=1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be R 5.2 m/s, where R is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution: Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meterf o o hg z V p g z V p +++=++22121122ρρ 1 rearranging the equation above, and set (z 2-z 1)=xf o oh xg V V p p ++-=-22121ρ 2 from continuity equation 11221125.625V V d d V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()f f f f oh xg R h xg R h V h xg V V p p ++=++=+=++-=-94.1185.203.1903.19206.3922121211ρ 4from the hydrostatic equilibrium for manometerg x g R p p Hg o ρρρ+-=-)(1 5substituting equation 5 for pressure difference into equation 4 obtainsf Hgh xg R gx g R ++=+-94.118)(ρρρρ 6 rearranging equation 6 kg J R R R R g R h Hg f /288.267.494.11861.12394.118)(==-=--=ρρρFigure for problem 1.161.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m 3Solution: a)2.0501010==D D =⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3pas kg V D m /268.0130063.201.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3Pa pressure difference due to increase of velocity in passing through the orificePa D D V V V V p p o 8.44882)2.01(63.213002242412222212221=-=⎪⎪⎭⎫ ⎝⎛-=-=-ρρ pressure drop caused by friction lossPa p f 5.75778.44883.12066=-=∆2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow rate is 26m 3/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition. ()s m p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρSolution:Write the mechanical energy balance equation between the suction connection and discharge connection 2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρ wherem Z Z 4.012=-(Pa 1052.1(Pa 1047.22_1,215241≈=⨯=⨯-=f H u u pressure gauge p pressure gauge p ））total heads of pump is m H 41.1881.9100010247.01052.14.055=⨯⨯+⨯+= efficiency of pump is N N e /=ηsince kW g QH N e 3.1360081.9100041.18263600=⨯⨯⨯==ρ N=2.45kWThen mechanical efficiency %1.53%10045.23.1=⨯=η The performance of pump is Flow rate ，m³/h26 Total heads ，m18.41 Shaft power ，kW2.45 Efficiency ，%53.12.2 Water is transported by apump from reactor, which has200 mm Hg vacuum, to thetank, in which the gaugepressure is 0.5 kgf/cm 2, asshown in Fig. The totalequivalent length of pipe is200 m including all localfrictional loss. The pipeline isφ57×3.5 mm , the orificecoefficient of C o and orificediameter d o are 0.62 and 25mm, respectively. Frictionalcoefficient λ is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution:Equation(1.6-9)Mass flow rates kg S V m o o /02.21000025.0414.312.42=⨯⨯⨯==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2f H z gp p H +∆+-=ρ12 ∆z=10mPa p 7570710013.17602001081.95.054=⨯⨯+⨯⨯=∆ ∆p/ρg=7.7mThe relation between the hole velocity and velocity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the netpositive suction head must be at least 3m above the cavitation vaporpressure of 710mm mercury vacuum. If losses in the suction pipeaccounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet?Solution :From an energy balance,s m Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫ ⎝⎛-=ρρρ）（s m D d V V /12112.42200=⎪⎭⎫ ⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l f H f 1.581.92105.0200025.02422=⨯⨯==NPSH H gp p H f v o g ---=ρWhereP o =760-640=120mmHgP v =760-710=50mmHgUse of the equation will give the minimum height H g as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?• Density of acid 1840kg/m 3• Viscosity of acid 25×10-3 PasSolution: Velocity of acid in the pipe:s m d m d mpipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04sec 222=⨯⨯===-=ρπρReynolds number:6109102532.31840025.0Re 3=⨯⨯⨯==-μρud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9kg J u d l f ph f /450232.3025.0600085.042422=⨯==∆=ρ kPa p 5.8271840450=⨯=∆ or friction factor is calculated from equation1.4-25kg J u d l u d l f ph f /426232.3025.0606109046.042Re 046.042422.022.02=⨯⨯⨯==∆=--＝ρkPa p 84.7831840426=⨯=∆ if the pressure drop falls to 783.84/2=391.92kPam NPSH H g p p H f v o g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=（ρ8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u d l u d l p p =⎪⎭⎫ ⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ＝ so s m u /27.236.489..1079012.03919208.18.1==⨯= new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m 3Viscosity of acid 25×10-3 Pa Friction factor 32.0Re500.00056.0+=f for hydraulically smooth pipe Solution: Write energy balance equation:f hg u z g p H g u z g p +++=+++2222222111ρρ gu d l g p h H f 22λρ=∆== 342=ρπu ds m d u /32.31840025.014.3124322=⨯⨯=⨯=ρπ 611510251840025.032.3Re 3=⨯⨯⨯=- 0087.061155.00056.0Re 500.00056.032.032.0=+=+=f 92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρ Δp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38⨯2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o ＝0.63. the permanent loss in pressure is3.5×104N/m 2, the friction coefficient λ=0.024. find:(1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W, 60%efficiency? (The density of fluid is 870kg/m 3 )solution ：∑+++=+++f A A A A AA h u p g z w u p g z 2222ρρ ρλρ022p u d l h p p f BA ∆+==-∑ 247.0334.162=⎪⎭⎫ ⎝⎛=A A o ()()s m gR C u /5.8870870136006.081.9297.063.02247.01200=-⨯⨯=''--=ρρρ ∴u = (16.4/33)2×8.5=2.1m/s∴242/76855105.321.2033.030870024.0m N h p p f B A =⨯+⨯==-∑ρ (2)W u d p Wm 1381.2033.0785.0768554Ne 22=⨯⨯⨯=∆==ρπρ sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid ％％＝461006.0500138⨯⨯。

流体力学与传热(英文版)chapter3a

The total drag on the body is the sum of the integrals of these quantities, each evaluated over the entire surface of the body in contact with the fluid.
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In addition, when the wall of the body is
vertical to the direction of flow, shown in Fig.1-11(b), p.33, drag force is due entirely to the pressure component.

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The
key points to be discussed in this chapter: The situation where the solid is immersed in, and surrounded by fluid is the subject of this chapter.
P cos α dA (form drag)
Direction Of flow
α
Tw sin α dA (wall drag)
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流体力学英文版课后习题答案

1.1 What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? ρwater = 1000 kg/m3, and P atmosphere = 101kN/m 2. Solution:Rearranging the equation 1.1-4ghp p abSet the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surface iskPaghp p ab72.1171281.91000Absolute pressure of water at depth 12mkPaPaghp p ab72.2182187201281.910001010001.3 A differential manometer as shown in Fig. is sometimes used to measure small pressure difference. When the reading is zero, the levels in two reservoirs are equal. Assume that fluid B is methane （甲烷）, that liquid C in the reservoirs is kerosene (specific gravity = 0.815), and that liquid A in the U tube is water. The inside diameters of the reservoirs and U tube are51mm and6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference over the instrument In meters of water, (a) when the change in the level in the reservoirs is neglected, (b) when the change in the levels in the reservoirs is taken into account? What is thepercent error in the answer to the part (a)? Solution ：p a =1000kg/m3p c =815kg/m3p b =0.77kg/m3D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesRd x D 2244(1)soR Dd x2(2)and hydrostatic equilibrium gives following relationshipg Rg xp g Rp Acc21(3)sog R g xp p cA c)(21(4)substituting the equation (2) for x into equation (4) gives gR g RDd p p cA c)(221(5)（a ）when the change in the level in the reservoirs is neglected,Pag R g R g RDd p p cA cA c26381.98151000145.0)()(221（b ）when the change in the levels in the reservoirs is taken into accountPagR g RD d gR g RD d p p cA ccA c8.28181.98151000145.081.9815145.0515.6)()(22221error=％＝7.68.2812638.2811.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in the figure. Thereadings of two U-tube manometers are R 1=400mm ，R 2=50mm, respectively. The indicating liquid is mercury. The top of the manometer is filled with the water to prevent from the mercury vapor diffusing into the air, and the height R 3=50mm. Try to calculate the pressure at point A and B.Solution: There is a gaseous mixture in the U-tube manometer meter. The densities of fluids are denoted by HgOH g,,2, respectively. The pressure at point A is given by hydrostaticequilibriumFigure for problem 1.4gR R gR gR p gHgOH A)(32232gis small and negligible in comparison withHgand ρH2O , equation above can besimplifiedc Ap p =232gR gR HgOH =1000×9.81×0.05+13600×9.81×0.05 =7161N/m21gR pp p HgADB=7161+13600×9.81×0.4=60527N/m1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. Theratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume that fluid flow is a potential flow. The reservoir, tank Aand the exit ofdrainpipe are all open to air.Solution:Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: 2222222111u gz p u gz p Where p 1=0, p 2=0, and u 1=0, simplification of the equation Ddp ap aHhAFigure for problem 1.5222u Hg1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation: 22Dd u u o22dD u u o 2Bernoulli equation is written between the throat and the station 2-23Combining equation 1,2,and 3 gives Solving for H H=1.39m1.6 A liquid with a constant density ρ kg/m 3is flowing at an unknown velocityV 1 m/s through ahorizontal pipe of cross-sectional area A 1 m 2at a pressure p 1 N/m 2, and then it passes to a section of the pipe in which the area is reduced gradually toA 2 m 2and the pressure is p 2. Assuming nofriction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured.Solution: In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,222200uu p ＝＝＝144.281.92100081.910002125.11112442g h dD uHg2112A A V V For the items in the Bernoulli equation , for a horizontal pipe, z 1=z 2=0Then Bernoulli equation becomes, after substituting 2112A A V V for V 2,221212112122p A A V p V Rearranging,2)1(21212121A A V p p 12221211A A p p V ＝Performing the same derivation but in terms of V 2,21221212A A p p V ＝1.7 A liquid whose coefficient of viscosity isμ flows below the critical veloc ity for laminar flowin a circular pipe of diameter d and with mean velocity V. Show that the pressure loss in a length of pipeLp is232dV.Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s.Calculate the loss of pressure in a length of 120m. Solution :The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2substituting equation 2 for u into equation 1 and integrating 3rearranging equation 3 gives1.8. In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is2.5m below A and when the flow rate down the pipe is 0.02 m 3/s, the pressure at B is 14715 N/m 2greater than that at A.Assuming the losses in the pipe between A and B can be expressed as gVk 22where V is the velocity at A, find thevalue of k.If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. Solution:d A =0.15m; d B =0.075m z A -z B =l =2.5m Q=0.02 m 3/s, p B -p A =14715 N/m2Figure for problem 1.8RRrdru RudAAV 022112214Rr R Lp p u L2032DLp p VL 232dVLp Pad VLp115201.01206.005.0323222sm d QV V d Q AAA A /132.115.0785.002.044222sm d QV V d Q BBB B /529.4075.0785.002.044222When the fluid flows down, writing mechanical balance equation 222222A B B BA A AV kV gz p V gz p 213.1253.4100014715213.181.95.2222kk638.0260.10715.14638.0525.24k0.295making the static equilibriumgRg x g l pgR gx p HgABmmgg l p p RgH AB 7981.91260081.910005.2147151.9．The liquid vertically flows down through the tube from the station a to the station b, then horizontally through the tube from the station c to the station d, as shown in figure. Two segments of the tube, both ab and cd ，have the same length, the diameter and roughness. Find:（1）the expressions ofg p ab , h fab,gp cd and h fcd , respectively.（2）the relationship between readings R 1and R 2 in the U tube. Solution:(1) From Fanning equationFigure for problem 1.922V dl h faband soFluid flows from station a to station b, mechanical energy conservation gives hence 2from station c to station d hence 3From static equation p a -p b =R 1（ρˊ-ρ）g -l ρg 4p c -p d =R 2（ρˊ-ρ）g 5Substituting equation 4 in equation 2 ，then therefore 6Substituting equation 5 in equation 3 ，then7Thus R 1=R 222V dl h fcd fcdfabh h fabbah p p lgfabbah p p lgfcddch p p fcddch p p fabh gl gR lg1）（gR h fab1gR h fcd 21.10 Water passes t hrough a pipe of diameter di=0.004 m with the average velocity 0.4 m/s, as shown in Figure. 1)What is the pressure drop –P when water flows through the pipe lengthL=2 m, in m H 2Ocolumn?2) Find the maximum velocity and point r at which it occurs.3) Find the point r at which the average velocity equals the local velocity.4）if kerosene flows through this pipe ，how do the variables above change ？（the viscosity and density of Water are 0.001 Pas and 1000 kg/m 3，respectively ；and the viscosity and density of kerosene are 0.003 Pas and 800kg/m 3，respectively ）solution: 1）1600001.01000004.04.0Reudfrom Hagen-Poiseuille equation1600004.0001.024.0323222duL Pmgp h163.081.9100016002）maximum veloc ity occurs at the center of pipe, from equation 1.4-19m ax0.5V u so u max =0.4×2=0.8m3）when u=V=0.4m/s Eq. 1.4-172max1wr r u u 5.0004.01max2u V r ＝mr 00284.071.0004.05.0004.04) kerosene: 427003.0800004.04.0ReudPap p 4800001.0003.01600LrFigure for problem 1.10mgp h611.081.980048001.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a)When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly,R=400mm, h=1400mm. The friction coefficientλ is0.025, and the loss coefficient of the entrance is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m3/h) b)When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure,N/m2). l e /d ≈15 when the gate valve is widely open, and the friction coefficientλ is still 0.025.Solution ：(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’　and the section of pressure point 2—2’，and take the center of section 2—2’　as the referring plane, then21,2222121122—f hpu gZp u gZ（a ）In the equation 01p (the gauge pressure)222/396304.181.910004.081.913600mN ghgRp OH Hg Figure for problem 1.12021Z u When the gate valve is fully closed, the height of water level in the reservoir can be relatedto h (the distance between the center of pipe and the meniscus of left arm of U tube).gRh Z g HgOH )(12（b ）where h=1.5mR=0.6mSubstitute the known variables into equation b2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600VVVK dl h mZ c fSubstitute the known variables equation a9.81×6.66=2213.21000396302VVthe velocity isV =3.13m/sthe flow rate of water ishmVd V h/5.8813.312.04360043600322）the pressure of the point where pressure is measured when the gate valve is wide-open. Write mechanical energy balance equation between the stations 1—1’　and 3-3′，then31,3233121122—fh p V gZp V gZ（c ）since m Z 66.61311300p p u Z 2223_1,81.42]5.0)151.035(025.0[2)(VVVK d l l h c ef input the above data into equation c ，9.8122V81.4266.6Vthe velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’　and 2——2’, for the same situation of water level21,2222121122—f hp V gZp V gZ（d ）since m Z 66.612121003.51/0(page pressure Z u u m sp ）kgJ V K dl h c f /2.26251.3)5.01.015025.0(2)(222_1,input the above data into equation d ，9.81×6.66=2.261000251.322p the pressure is: 329702p 1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long.There is a attached-pipe (Φ603.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the reading of the rotameter is 2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution ：The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2.The friction loss for parallel pipelines is2121S S sf f V V V h h The energy loss in the branch pipe is22222222u d l l h e fIn the equation03.02sm u dml l e /343.0053.04360072.2053.01022222input the data into equation ckgJ h f/333.02343.0053.01003.022The energy loss in the main pipe is333.022111121u d l hhf f So sm u /36.22018.023.0333.01The water discharge of main pipe ishmV h /60136.23.043600321Total water discharge ishmV h/7.60372.260131.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to beR5.2m/s, where R is the manometer reading in metres of mercury. Determine the loss of headbetween inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution:Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meter fo o h g z V p gz V p 221211221rearranging the equation above, and set (z 2-z 1)=x f ooh xg V V p p 221212from continuity equationFigure for problem 1.1611221125.625V V d d V V oo 3substituting equation 3 for V o into equation 2 gives fff f oh xgRh xg R h V h xg V V p p 94.1185.203.1903.19206.39221212114from the hydrostatic equilibrium for manometergx g R p p Hgo)(15substituting equation 5 for pressure difference into equation 4 obtains fHgh xgR gx g R 94.118)(6rearranging equation 6 kgJ R R R R gR h Hgf/288.267.494.11861.12394.118)(1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid,calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercury as 13.6, and the density of water as 1000 kg/m 3Solution: a)2.0501010D D 81.9)130013600(1.0)(21g R p p Hgsm p p D D C V o/63.231.461.056.1861.0130081.9)130013600(1.022.0161.0214214102skg V D m/268.0130063.201.0442220b)approximate pressure drop81.9)130013600(1.0)(21g R p p Hg12066.3Papressure difference due to increase of velocity in passing through the orificePaD D V V V V p p o 8.44882)2.01(63.213002242412222212221pressure drop caused by friction lossPapf5.75778.44883.120662.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suction connection of the pump is 24.7 kPa as the flow rate is 26m 3/h.The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same.Calculate the mechanical efficiency of pump and list the performance of the pump under thisoperating condition. Solution:Write the mechanical energy balance equation between the suction connection and discharge connection2_1,2222121122f Hgpg uZH gp g uZwheremZ Z 4.012(Pa 1052.1(Pa 1047.22_1,215241f Hu u pressure gaugep pressure gaugep ））total heads of pump is mH 41.1881.9100010247.01052.14.055efficiency of pump is NN e/sincekWg QHN e3.1360081.9100041.18263600N=2.45kWThen mechanical efficiency%1.53%10045.23.1The performance of pump isFlow rate ，m3/h 26 Total heads ，m 18.41 Shaft power ，kW 2.45 Efficiency，%53.12.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm 2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is 57×3.5 mm , the orifice coefficient of C o and orifice diameter d o are 0.62 and 25 mm, respectively. Frictional coefficient is 0.025.Calculate: Developed head H of pump, in m (the reading R of U pressuregauge in orifice meter is 168 mm Hg)Solution: Equation(1.6-9)Mass flow rateskg S V moo /02.21000025.0414.312.422) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2fHzg p p H12z=10m10m21sm Rg Dd C V f/12.444.69375.062.01000)100013600(81.9168.025025162.02144000）（Pap 7570710013.17602001081.95.054p/g=7.7mThe relation between the hole velocity and veloc ity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet? Solution :From an energy balance, WhereP o =760-640=120mmHg P v =760-710=50mmHgUse of the equation will give the minimum height Hg as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ??Density of acid 1840kg/m3?Viscosity of acid25×10-3PasSolution:Velocity of acid in the pipe:H gsm Dd V V /12112.422mg ud l fHf1.581.92105.0200025.02422NPSHHgp p Hfv ogmNPSHHgp p Hfv og55.335.181.9100081.913600)05.012.0（sm dm dmpipeof area tionalcrossflowrate volumetric u/32.3025.01840785.03785.04sec 222Reynolds number: 6109102532.31840025.0Re3ud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9 kgJ ud l fph f/450232.3025.0600085.042422kPap5.8271840450or friction factor is calculated from equation1.4-25 kgJ ud l ud l fph f/426232.3025.0606109046.042Re046.042422.022.02＝kPap 84.7831840426if the pressure drop falls to 783.84/2=391.92kPa8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u uudlu dl p p＝so sm u/27.236.489..1079012.03919208.18.1new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m3Viscosity of acid 25×10-3Pa Friction factor 32.0Re500.00056.0f for hydraulically smooth pipeSolution:Write energy balance equation: fh gu z g p Hg u z gp 2222222111gud l gp h Hf22342ud sm du/32.31840025.014.3124322611510251840025.032.3Re30087.061155.00056.0Re500.00056.032.032.0f 92.4681.9232.3025.0600087.04222gud l gp h HfΔp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38 2.5mmdiameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o ＝0.63. the permanent loss in pressure is 3.5×104N/m 2, the friction coefficient λ=0.024. find: (1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when theshaft work is 500W , 60%efficiency? (The density of fluid is 870kg/m3)solution ：fA A A A AA h u p gz wu p gz 222222p ud l h p p fBA247.0334.162AA osm gR C u /5.8870870136006.081.9297.063.02247.01200∴u= (16.4/33)2×8.5=2.1m/s ∴242/76855105.321.2033.030870024.0mN h p p fB A (2) Wud p Wm1381.2033.0785.0768554Ne 22sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid％％＝461006.0500138。

07流体力学与传热期末考试题(含答案).

诚信应考,考试作弊将带来严重后果！华南理工大学期末考试2007《流体力学与传热英文》试卷注意事项：1. 考前请将密封线内填写清楚；2. 所有答案请直接答在试卷上(或答题纸上)；3．考试形式：开（闭）卷；一．(38 points) True or false or fill the black or choose the correct answer(1)In SI system, the dimension of viscosity is ( M/Lq ), and its unit is ( kg/ms ).(2) ( T ) If the volume flow rate Q increases, the developed head H of centrifugal pumps will decrease.(3) ( F )When the gas temperature increases, its viscosity will decrease.(4) ( F ) Water need not to be added during the start of centrifugal pump because it has priming function.(5) If water with density r=1000kg/m3, viscosity µ=0.001Pa.s and u=1m/s flows through a pipe with di=0.001m, its Re=(1000), f (Fanning(maximum velocity)=( 2 m/s); If the pipe factor)=( 0.016), umaxlength is 2m, its pressure drop Dp=(128000 Pa).(6) A single spherical solid particle settles freely in static liquid, in which the settling obeys Stoke's Law (laminar flow). If the diameter of particle increases, the settling velocity uwill ( be up ); when thetwill ( be down ).temperature decreases, ut(7) The revolution n of centrifugal pump increase, its volumetric flow rate Q will increases , its developed head H will increase .(8) Orifice meter measures ( C ) to get mass flow rate of fluid ina pipe, rotameter measures ( A ) to get mass flow rate of fluid in a pipe. The former ( E ) keeps constant, and the later ( D ) keeps unchanged.A Volume flow rate;B Local velocity;C Average velocity;D Pressure drop;E cross section area(9) The pressure of 10 m H2O is equal to 735.7 mmHg. Ifatmospheric pressure is 1 atm and the reading of centrifugal pump outlet gauge is 5000Pa(N/m2)= 106300 N/m2 (absolute pressure).(10) When the flow rate is fixed, the maximum suction lift Za will increase if the diameter of suction pipe increases. Whenthe flow rate is fixed, Zawill decreases if a valve and some fittings are set up in the suction pipe.(11) The expression of Newton's Law of cooling is ( hADt ).(12)If the temperature drop DT o between the saturated vapor and pipe wall is double in a vertical tube, the condensation heat transfer film coefficient h will increase to ( A ) of original value.A) 2-1/4 ; B) 21/4 ; C) 23/4 ; D) 2(13) Based on the temperature distribution as shown in figure, which has largest temperature drop A , and which has the largest thermal conductivity C if the thickness of each layer is the same?A) layer A; B) layer B; C) layer C(14) The removal of uncondensable gas in the condensing vapor will ( B )A increase condensing film coefficient h;B decrease condensing film coefficient h;C do not change condensing film coefficient h;D be uncertain.(15) A flat wall thickness b=0.1m and heat transfer area A=2m2 consists of some material with conductivity k=0.1 w/m×K, heat transfer rate q will be 200 (w) when the temperature difference between both sides of wall is 100 K. If heat resistance is double and temperature drop between both sides of wall keeps constant, heat transfer rate q will be 100 (w).(16) If hi is much smaller than hoand wall heat resistance is ignored,overall heat transfer coefficient U will be approximately equal to ( A ); in order to enhance heat transfer, the best way is to increase ( ).A hi ; B ho; C average of hiand ho; D uncertain(17) If the side of a square is 0.5m, its equivalent diameter is (0.5 m).(18) Saturated vapor is cooled by air in a shell-tube exchanger. When air flow rate increases, the outlet temperature of air will ( B )A increase;B decrease;C be the same;D be uncertain(19) Water flows fully turbulently through a pipe and the velocity changes from 1.5m/s to 3m/s. If inlet and outlet temperatures of water keep the same, heat transfer coefficient of water hi is ( A )A．1.74 of origin; B．2 of origin; C．4 of origin; D. 1/2 of origin2 (12 points) A slurry is filtered by a 1 m2 filter press at constant pressure. The cake is incompressible. The filter basic equation is as follows:(V + 1)2 = 30 (t + 2)where V is filtrate volume, liter; t is filtration time, minFind 1) how much filtrate is obtained after 28 min, in liter?2) if the pressure drop decreases to 50% of original value, how much filtrate can be got after 28 min, in liter?3) The filtration rate in the beginning and at 28 min in case 1, in liter/minSolution: 1) V=29 L; 2) V=20.2 L3) dV/dt=KA2/2(V+Ve)=30/(2*1)=15 l/min; and dV/dt=KA2/2(V+Ve)=30/2(29+1)=0.5 l/min3 (25points) Water (r=1000kg/m3) flows from the top tank to the low one with a velocity 2m/s, as shown in Fig. There are one orifice and a gate valve in the pipeline with f57´3mm. The total pipe length including all local frictional equivalent length is 60 m. If the frictional coefficient l=4f is 0.036 and Kv(local loss coefficient of valve) is 7.5 (wide open), Find(1) mass flow rate in pipeline, in kg/s.(2) liquid surface height difference DZ between top and low tanks.(3)pressure drop through the valve and reading R2of U-shape pressure gauge.(4) if the valve is closed a little bit, there is 20% decreases of readingR 1for U-shape pressure meter. Assume orifice coefficient Cokeepsunchanged, what is the mass flow rate in the pipeline, in kg/s?Hint: velocity equation in orifice uo =Co[(2R1g(ro-r)/r)1/2 , and rois13600kg/m3.Solution: (1) m=2*1000*0.785*(0.051)2=4kg/s (2) DZ=4fL*u2/2dg=0.036*60*22/2*0.051*9.81=8.63m(3) Dp=R2(ro-r)g and Dp/r=Ku2/2=7.5*22/2=15Dp=15000= R2(13600-1000)*9.81 R2=0.12m(4) m'/m= Co [(2R'g(ro-r)/r)1/2 / Co[(2Rg(ro-r)/r)1/2 =(R'/R)1/2=0.81/2=0.894m'=0.894*m=0.894*4=3.58kg/s4. (25points) Heated oil (heat capacity Cp=3.35kJ/kg·℃) is cooled by water (heat capacity Cp=4.187kJ/kg·℃) and they flow countercurrently through a double pipe exchanger with inner pipe dimension φ180×10mm. Water at 15℃ of inlet temperature goes through the pipe and leaves at 55℃. Oil passes through the annulus with mass flow rate 500kg/h and inlet and outlet temperatures of oil are 40℃ and 90℃, respectively. The heattransfer coefficients hi and hofor water and oil are 1000W/m2·℃ and299W/m2·℃，and heat resistances of pipe wall and fouling as well as heat loss can be ignored. Find(1) water flow rate, in kg/h.(2) overall heat transfer coefficient Uobased on outer surface of pipe.(3) LMTD Dtmand pipe length of pipe, in m.(4) if inlet temperature of water becomes 20℃, what happens to that?(5) in order to enhance heat transfer, what ways can be employed? Solution: (1) w=500*3.35*50/(4.187*40)=500kg/h(2) Uo=1/(180/1000*160+1/299)=223.7w/m2K(3) LMTD=75-15/ln75/15=37.3oCL=q/Uo pdoLMTD=500*3.35*50/(3600*p*0.18*37.3)=4.9m4.oil outlet temperature decreases5.increase oil velocity and change 1-1 pass to 1-2 pass, decrease oilheat resistance.。

流体力学思考题双语

1 if the inlet and outlet temperatures of fluids are given the LMTD of countercurrent is always larger than that of parallel—current flow?2．For countercurrent flow，the outlet temperature of cold fluid can be higher than that of hot one．3．The value of overall heat transfer coeff．U is closed to that of larger heat transfer film coeff．4．Dirty overall heat transfer coef．is smaller than clean overall heat transfer coef．5．If hi and ho are 10 and 1000 w／(m2.℃)，respectively，we should try to increase ho in order to elevate overall heat transfer coefficient U．6．For no phase change，∆T of 1-2 pass shell—tube exchanger is smaller than LMTD of countercurrent flow．7．Explain simply the advantages of countercurrent flow over parallel—current flow and in what situations，parallel flow should be used.8．Dimensional analysis can directly produce the numerical results without experimental data．9．The thicker the insulating layer，the smaller the heat loss20．For heat transfer through a series of layers at steady-state，the smaller the temperature drop at a cerain layer，the larger the heat resistance at the same layer．21．At steady-state heat transfer conduction through a pipe wall，q／A is a constant．22．For heat transfer through two layers of insulating materials having the same thickness but different conductivities at a flat wall．Temperatures of both sides of insulating materials keep constant．If two 1avers ofinsulating materials change their places how does heat loss change? Expiain why?23．For the same conditions as the above question，but heat transfer through a cylinder，how does heat loss change?Explain why?24·Which can develop more total head H, one pump or two same pump which work in series?25．The dimension for viscosity in SI system is____________, and what about the unit for it?26．What is the relationship between the gauge and absolute pressure，the vacuum and absolute pressure? If the reading at entrance of pump is 0.029 MPa(vacuum)，what are vacuum in mmHg and gauge pressure in mnmHg? if reading at the exit of pump is 0.67Mpa(gauge),what is absolute pressure (atmosphere-pressure 0.1MPa)?27．The total (developed) Head of centrifugal punlp Hmeans__________________and maximum suctionlift implies ____________________and net positive suction head (NPSH)___________________.28 .Spherical particle settles in air and obeys the stock’s law, if the temperature changes from 25 to 50℃, the terminal velocity will .29. A single spherical particle settling freely in the fluid and it is laminar flow, when the particle diameter increases, the terminal velocity u will ; when the viscosity of fluid increases, u will ; if the fluid is a gas, what happens to u if the temperature increase?30.For laminar flow, the average velocity in a pipe is ____________ maximum velocity, for turbulent flow average velocity is______________ maximum velocity.31 For laminar flow in a pipe, the flow rate is constant, when d i increases, the friction coefficientwill______________,and friction loss will____________.32.When the pipe installation changes from horizontal to vertical position and the velocity keeps unchanged, the frictional loss will be .A）. increased; B）. decreased; C）. the same; D）. uncertain34. In a Newtonian fluid ，the relationship between the shear tress and shear rate is（）35.Fluid flows through the pipe at low flow rates, when the velocity rise in double time, the pressure drop in the fluid will( ).36.Fluid flows through the pipe at high flow rates, when the velocity rise in double time, the pressure drop in the fluid will( ).37.For Newtonian fluids，if the type of flow through the pipe is laminar flow，then its characteristics are:__________________________________；______________________;________________________.38.For given dust system the capacity of gravity settling chamber depends on ( )A) The height and length of chamberB) Height of chamberC) The bottom area of chamberD) Volume of chamber39.For constant pressure filtration, if the filtering medium (septum) resistance is ignored, the filtering time is double and other operating conditions keep unchanged, the filtrate volume will be （）（A）double; (B) half of original; (C) 21/2 of original; (D) uncertain40.The critical diameter implies for the performance of cyclone.(A)minimum diameter of particle removed completely by cyclone(B)diameter of cyclone for maximum separation efficiency(C)maximum diameter of particle to maintain laminar flow(D)minimum diameter of particle leaving the cyclone41.A centrifugal pump operates on air, then, ___________ liquid from an initially empty suction line.42.Positive-displacement(reciprocating往复) pumps ______________) air binding.43 ( )One pump can develop more H than that two same pumps which work in series.44. The methods of adjusting volume flow rate of centrifugal pump are (1) _________________________;(2)____________________;(3) _____________________; The commonest way in practice is ( ).45.When the temperature of fluid to be transported increases, the maximum lift of suction will ( ). If the velocity of suction pipe decreases, the maximum suction lift will ( ). If the flow rate remains constantbut the diameter of suction pipe decreases, the maximum suction lift will ( ).A）increase; B ）decrease; C）keep the same D）be uncertain46.Air flows along the tube and saturated vapor passes through the shell in a shell-tube exchanger. In order to enhance heat transfer, which way is feasible in practice as follows.（）A）increase vapor velocity; B) employ superheated vaporC) increase air velocity. D) set up the baffle in the shell.47.When fouling (scale) and pipe wall resistances are ignored and the difference between hi and ho is very huge, overall heat transfer coefficient U is close to( ); wall temperature is close to ( )A) larger h; B) smaller h; C) average value between h i and h o ; D) uncertain48. What is cavitation? At what situations,thecavitation will occur?49.if temperture of fluids increases，what happen to viscosities of liquid and gas?50..When the pipe changes from horizontal to vertical position and velocity keeps the same．what happens to energy loss?51.The solid dust is removed from gases in a gravity-settling chamber．If settling is within the laminar region．cornpare the productivity at 20 and 200℃ , which is larger?52．The filter basic equation is derived based on_____________________________________．53．What is equivalent diameter for 0.5m square?54．Orifice and venturi meters measure the __________ velocity of the entire stream of fluid.The pitot tube measures the __________ velocity of fluid.Rotametermust be________ installed in the pipe line, and the fluid must flow ______through the tapered tube and suspends freely a float.55.How do you adjust the flow rate of centrifugal and reciprocating pumps?56．The reading of vacuum gauge in the equipment is 100mmHg,try to calculate the absolute pressure and the gauge pressure, respectively. Given that the atmospheric pressure in this area is 740mmHg.。

流体力学与传热习题参考解答(英文).

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+ U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U U Z g+W Z g+h 22ρρP P ++=++ 10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15= 422.67x101.97W 15x9.81120246.88J /kg 12002=-+++= N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, finda. water velocity at section A-A'.b. water flow rate, in m 3 /h.22112212f U U Z g+Z g+h 22ρρP P +=++ 1U 0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+ U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m /s == 22a a b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z 22a b b a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tankand delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑ 31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u d Re 2.78x10ρμ== f 0.063= 2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be takenas 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have 22112212f 1-2u u gZ gZ h 22ρρP P ++=+++ 2212Hg H o 0 g(R h)39630N/m ρρP =P =-= 2212f1-2c u u u 0 Z =0 h 4f +k 2.13u d 22===l We can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++ 311Z 0 Z 6.66m u =0== 22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++ 112120 Z 6.66 Z 0 u 0 u 3.51P ===== 22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N 32970mρP =++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe line fA fB total A B22A fA A 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg = 22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /h π∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+ b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==>12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr= ()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=. ()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k ==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====> ()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=. (2) 12w 2w = 4421Re Re /2=2x1010=> 0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭ 0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k == (3) 44333u d 2000x0.01Re 2x10100.001ρμ===> 0.81/3Nu 0.023Re Pr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease dThe first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes throughthe pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21()20h 642.9w/m k =12t +t LMTD=702∆∆℃= Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln 14080-=--℃ 1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=- C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃ 212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T )80/()20ln(20802---=∆h h m T T T Th=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P 1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+ q=240 V=qA=240*0.1=24(2) K 2k =∆P K'2k '=∆P'2∆P =∆P K'2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。

流体力学与传热习题参考解答(英文)(完整资料).doc

【最新整理，下载后即可编辑】1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=012P =P10Z =W=60j/kg f h 30/kg = 2U =1m/s 2(60300.5)/g 3m Z =--= 21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P =5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*.1Z 0=2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation: h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++1U 0=12P =P1Z 6m =2Z 0=2f h 6.5U=22U 6x9.81 6.5U 2=+U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure thepressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A =2b U 2.5*(33/47)1.23m /s ==22aa b b a b f U U Z g+Z g+h 22ρρP P +=++a b Z =Z22abb a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocalU U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ=300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u dRe 2.78x10ρμ==f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken as 0.61, what is the flow rate of water?0o o2gR()u c ρρρ-=20o 02gx0.6x(136001000)V u s 0.61x /4x0.0001x1000π-==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f 1-2u u gZ gZ h 22ρρP P ++=+++2212Hg H o 0 g(R h)39630N/m ρρP =P =-=2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81ud 20.01l l ++=+==229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+=22229.81x6.66 3.15/226.2N 32970m ρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B 22A fAA 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a.2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d10x0.02x1.06 Re=1.06x10100.02x10ρμ-==> 12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr0.027x 1.06x10x 0.69239.66==.=.()2i ii h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1)444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2)12w 2w =4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023Re Pr =()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find: (1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturatedvapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a)0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 ()20h 642.9w/m k =12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca)300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m (c) 8020LMTD=86.514020ln14080-=--℃1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)oi h h Uo 111+= (1)o i o h h U 1'1'1+= (2) (1)-(2)=0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=-C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet andoutlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆ ()()h h 12c c 21W C T -T 'W C t 't =- 2150100401515080t 15--=-- 2t 50=℃212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =-()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭ We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24 (2) K2k=∆P=∆P K'2k'∆P=∆P K'2K500'2==()()2q'+10500x249.60.4=+q’=343.6 v=34.36。

流体力学与传热(英文版)chapter1b

the shear stress is proportional to the shear rate du (1.2-3)

dy
(the proportionality constant μ is called the viscosity, unit of μ is kg/(m.s) ) Eq. (1.2-3) is called Newton’s law of viscosity. It is good for Newton’s fluids and laminar flow. Gases and most liquids belong to Newtonian fluids
Chapter 1
1.2 Fluid-flow phenomena
• Ideal fluid
which has zero viscosity. Without friction or resistance when fluid flows through the channel.
• Potential flow
The velocity field
• If the wall is at rest in the reference frame, the velocity of the fluid at the interface is zero, but there’re variations in velocity from point to point in the flowing stream at distances away from the solid . • The velocity field exists in the space occu pied by the fluid.

传热学简答题中英文版

1.对流换热表面传热系数的定义，与什么有关。

牛顿冷却公式----------- 对流换热包括热对流和热传导，它与流动状态是层流还是紊流，流体性质，流动性质（自然对流还是强制对流），表面材料形状，面积大小和性质等有关，还与换热是否相变有关。

Newton's cooling equation ----------- include convection heat convection and heat conduction, and its flow is laminar or turbulent state of fluid properties, flow properties (natural convection or forced convection), the surface shape of the material , size and nature, but also on whether the phase change heat related.2.角系数一个表面投射到另一个表面的能量占总投射能的份额。

性质：可换性，完整性，可加性。

纯几何因子条件：物体表面是漫反射表面且等强辐射，定向辐射强度与方向无关。

Onto the surface to a surface energy of another share of the total energy of the projection.Nature: interchangeable, integrity, additive.Pure geometric factor conditions: the surface is diffuse surface and other strong radiation, the radiation intensity and direction of orientation-independent.3.边界层中温度变化率的绝对值何处最大，对于一定换热温差的同一流体，为何能用温度梯度的绝对值大小来判断表面传热系数的大小。

流体力学英文版课后习题问题详解

1.1 What will be the (a) the gauge pressure and (b) the absolute pressure of waterat depth 12m below the surface? ρwater = 1000 kg/m 3, and P atmosphere = 101kN/m 2.Solution:Rearranging the equation 1.1-4gh p p a b ρ+=Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m belowthe surface iskPa gh p p a b 72.1171281.910000=⨯⨯+=+=ρAbsolute pressure of water at depth 12mkPa Pa gh p p a b 72.2182187201281.91000101000==⨯⨯+=+=ρ1.3 A differential manometer as shown in Fig. is sometimes used to measure smallpressure difference. When the reading is zero, the levels in two reservoirs are equal.Assume that fluid B is methane （甲烷）, that liquid C in the reservoirs is kerosene(specific gravity = 0.815), and that liquid A in the U tube is water. The insidediameters of the reservoirs and U tube are 51mm and 6.5mm , respectively. If thereading of the manometer is145mm., what is the pressure difference over theinstrument In meters of water, (a) when the change in the level in the reservoirsis neglected, (b) when the change in the levels in the reservoirs is taken intoaccount? What is the percent error in the answer to the part (a)?Solution ：p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145mWhen the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubesR d x D 2244ππ= (1) so R D d x 2⎪⎭⎫ ⎝⎛= (2) and hydrostaticequilibrium gives followingrelationship g R g x p g R p A c c ρρρ++=+21 (3)sog R g x p p c A c )(21ρρρ-+=- (4)substituting the equation (2) for x into equation (4) givesg R g R D d p p c A c )(221ρρρ-+⎪⎭⎫ ⎝⎛=- (5)（a ）when the change in the level in the reservoirs is neglected,()Pa g R g R g R D d p p c A c A c 26381.98151000145.0)()(221=⨯-=-≈-+⎪⎭⎫ ⎝⎛=-ρρρρρ（b ）when the change in the levels in the reservoirs is taken into account ()Pa g R g R D d g R g R D d p p c A c c A c 8.28181.98151000145.081.9815145.0515.6)()(22221=⨯-+⨯⨯⨯⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-+⎪⎭⎫ ⎝⎛=-ρρρρρρ error=％＝7.68.2812638.281- 1.4 There are two U-tube manometers fixed on the fluid bed reactor, as shown in thefigure. The readings of two U-tube manometers are R 1=400mm ，R 2=50mm, respectively.The indicating liquid is mercury. The top of the manometer is filled with the waterto prevent from the mercury vapor diffusing into the air, and the height R 3=50mm.Try to calculate the pressure at point A and B .Solution: There is a gaseous mixture in the U-tube manometer meter. The densitiesof fluids are denoted by Hg O H g ρρρ,,2, respectively. The pressure at point AFigure for problem 1.4is given by hydrostatic equilibriumg R R g R g R p g Hg O H A )(32232+-+=ρρρg ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplifiedc A p p ≈=232gR gR Hg O H ρρ+=1000×9.81×0.05+13600×9.81×0.05=7161N/m ²1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/m1.5 Water discharges from the reservoir through the drainpipe, which the throat diameter is d. The ratio of D to d equals 1.25. The vertical distance h between the tank A and axis of the drainpipe is 2m. What height H from the centerline of the drainpipe to the water level in reservoir is required for drawing the water from the tank A to the throat of the pipe? Assume thatfluid flow is a potentialflow. The reservoir, tankA and the exit of drainpipeare all open to air.Solution:Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane:2222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation1The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:22⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭⎫ ⎝⎛=d D u u o 2 Bernoulli equation is written between the throat and the station 2-2 3Combining equation 1,2,and 3 givesSolving for HH=1.39m1.6 A liquid with a constant density ρ kg/m 3is flowing at an unknown velocity V 1m/s through a horizontal pipe of cross-sectional area A 1 m 2 at a pressure p 1 N/m 2,and then it passes to a section of the pipe in which the area is reduced graduallyto A 2 m 2 and the pressure is p 2. Assuming no friction losses, calculate the velocities V 1 and V 2 if the pressure difference (p 1 - p 2) is measured. Solution :In Fig1.6, the flow diagram is shown with pressure taps to measure p 1 and p 2. From the mass-balance continuity equation , for constant ρ where ρ1 = ρ2 = ρ,222u Hg =222200u u p =+ρ()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u Hg2112A A V V = For the items in the Bernoulli equation , for a horizontal pipe,z 1=z 2=0 Then Bernoulli equation becomes, after substituting 2112A A V V = for V 2, ρρ22121211212020p A A V p V ++=++ Rearranging,2)1(21212121-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-12221211A A p p V ρ＝Performing the same derivation but in terms of V 2,⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--21221212A A p p V ρ＝1.7 A liquid whose coefficient of viscosity is µ flows below the critical velocity for laminar flow in a circular pipe of diameter d and with mean velocity V . Show that the pressure loss in a length of pipe L p ∆ is 232d V μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s. Calculate the loss of pressure in a length of 120m.Solution : The average velocity V for a cross section is found by summing up all the velocities over the cross section and dividing by the cross-sectional area1From velocity profile equation for laminar flow2 substituting equation 2 for u into equation 1 and integrating3rearranging equation 3 gives1.8. In a vertical pipe carrying water, pressuregauges are inserted at points A and B where the pipediameters are 0.15m and 0.075m respectively. Thepoint B is 2.5m below A and when the flow rate downthe pipe is 0.02 m 3/s, the pressure at B is 14715N/m 2 greater than that at A. Figure for problem 1.8⎰⎰==R R rdr u R udA A V 020211ππ⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛--=22014R r R L p p u L μ2032D L p p V L μ-=232d V L p μ=∆Pa d VL p 115201.01206.005.0323222=⨯⨯⨯==∆μAssuming the losses in the pipe between A and B can be expressed as g V k 22where V is the velocity at A, find the value of k .If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres.Solution:d A =0.15m; d B =0.075mz A -z B =l =2.5mQ =0.02 m 3/s,p B -p A =14715 N/m 2s m d QV V d Q A A AA /132.115.0785.002.044222=⨯===ππ s m d QV V d Q B B BB /529.4075.0785.002.044222=⨯===ππWhen the fluid flows down, writing mechanical balance equation222222A B B B A A A V k V g z p V g z p +++=++ρρ 213.1253.4100014715213.181.95.2222k ++=+⨯ k 638.0260.10715.14638.0525.24++=+=k 0.295making the static equilibriumgR g x g l p g R g x p Hg A B ρρρρρ+∆++=+∆+()()mm g g l p p R g H A B 7981.91260081.910005.214715-=⨯⨯⨯-=---=ρρρ1.9．The liquid vertically flows down through the tube fromthe station a to the station b , then horizontally throughthe tube from the station c to the station d, as shown inFigure for problem 1.9figure. Two segments of the tube, both ab and cd ，have the same length, the diameter and roughness.Find:（1）the expressions of gp ab ρ∆, h fab , g p cd ρ∆ and h fcd , respectively. （2）the relationship between readings R 1and R 2 in the U tube.Solution:(1) From Fanning equationandsoFluid flows from station a to station b , mechanical energy conservation giveshence2from station c to station dhence3From static equationp a -p b =R 1（ρˊ-ρ）g -l ρg 4p c -p d =R 2（ρˊ-ρ）g 5Substituting equation 4 in equation 2 ，thentherefore6 22V d l h fab λ=22V d l h fcd λ=fcdfab h h =fab b a h p p +=+ρρlg fab b a h p p =+-lg ρfcd d c h p p +=ρρfcd d c h p p =-ρfab h g l g R =+--'lg 1ρρρρ）（g R h fab ρρρ-'=1Substituting equation 5 in equation 3 ，then7ThusR 1=R 21.10 Water passes through a pipe of diameter d i=0.004 m with the average velocity 0.4 m/s, as shown in Figure.1) What is the pressure drop –∆P when water flows through the pipe length L =2 m, in m H 2O column?2) Find the maximum velocity and point r at which it occurs.3) Find the point r at which the averagevelocity equals the local velocity. 4）if kerosene flows through this pipe ，howdo the variables above change ？（the viscosity and density of Water are0.001 Pas and 1000 kg/m 3，respectively ；andthe viscosity and density of kerosene are0.003 Pas and 800 kg/m 3，respectively ）solution:1）1600001.01000004.04.0Re =⨯⨯==μρud from Hagen-Poiseuille equation1600004.0001.024.0323222=⨯⨯⨯==∆d uL P μ m g p h 163.081.910001600=⨯=∆=ρ 2）maximum velocity occurs at the center of pipe, from equation 1.4-19 max 0.5V u =so u max =0.4×2=0.8m3）when u=V=0.4m/s Eq. 1.4-172max 1⎪⎪⎭⎫ ⎝⎛-=w r r u u g R h fcd ρρρ-'=2Figure for problem 1.105.0004.01max2=⎪⎭⎫ ⎝⎛-u V r ＝ m r 00284.071.0004.05.0004.0=⨯== 4) kerosene: 427003.0800004.04.0Re =⨯⨯==μρud Pa p p 4800001.0003.01600=='∆='∆μμ m g p h 611.081.98004800=⨯=''∆='ρ1.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entrance is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m ³/h)b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m ²). l e /d ≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.Solution ：(1) When the gate valve is opened partially, the water discharge isSet up Bernoulli equation between the surface of reservoir 1—1’and the sectionof pressure point 2—2’，and take the center of section 2—2’as the referring plane, then∑+++=++21,2222121122—f h p u gZ p u gZ ρρ （a ）In the equation 01=p (the gauge pressure)222/396304.181.910004.081.913600m N gh gR p O H Hg =⨯⨯-⨯⨯=-=ρρ 0021=≈Z uWhen the gate valve is fully closed, the height of water level in the reservoircan be related to h (the distance between the center of pipe and the meniscus of left arm of U tube).gR h Z g Hg O H ρρ=+)(12（b ）where h=1.5mR=0.6mSubstitute the known variables into equation b2222_1,113.22)5.01.015025.0(2)(66.65.110006.013600V VV K d l h mZ cf =+⨯=+==-⨯=∑λ Substitute the known variables equation a9.81×6.66=2213.21000396302V V ++ the velocity is V =3.13m/sthe flow rate of water isFigure for problem 1.12h m V d V h /5.8813.312.0436004360032=⨯⨯⨯=⨯=ππ2） the pressure of the point where pressure is measured when the gate valve is wide-open.Write mechanical energy balance equation between the stations 1—1’and 3-3´，then∑+++=++31,3233121122—f h p V gZ p V gZ ρρ （c ）since m Z 66.61=311300p p u Z =≈= 2223_1,81.4 2]5.0)151.035(025.0[ 2)(V V V K d l l h c e f =++=++=∑λ input the above data into equation c ，9.8122V 81.4266.6+=⨯V the velocity is: V =3.51 m/sWrite mechanical energy balance equation between thestations 1—1’and 2——2’, for the same situation of water level∑+++=++21,2222121122—f h p V gZ p V gZ ρρ（d ）since m Z 66.61=2121003.51/0(page pressure Z u u m sp =≈≈=）kg J V K d l hc f /2.26251.3)5.01.015025.0(2)(222_1,=+⨯=+=∑λinput the above data into equation d ，9.81×6.66=2.261000251.322++p the pressure is: 329702=p1.14 Water at 20℃ passes through a steel pipe with an inside diameter of 300mm and 2m long. There is a attached-pipe (Φ60⨯3.5mm) which is parallel with the main pipe. The total length including the equivalent length of all form losses of the attached-pipe is 10m. A rotameter is installed in the branch pipe. When the readingof the rotameter is 2.72m 3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively.Solution ： The variables of main pipe are denoted by a subscript1, and branch pipeby subscript 2. The friction loss for parallel pipelines is 2121S S s f f V V V h h+==∑∑The energy loss in the branch pipe is22222222u d l l he f ∑∑+=λ In the equation 03.02=λsm u d ml l e /343.0053.04360072.2053.01022222=⨯⨯===+∑πinput the data into equation c kg J hf /333.02343.0053.01003.022=⨯⨯=∑The energy loss in the main pipe is333.022111121===∑∑u d l h hf f λSo s m u /36.22018.023.0333.01=⨯⨯⨯=The water discharge of main pipe ish m V h /60136.23.043600321=⨯⨯⨯=πTotal water discharge ish m V h /7.60372.26013≈+=1.16 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury U-tube manometer. The velocity of flow along the pipe is found to be R 5.2 m/s, where R is the manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the Venturi when R is 0.49m. (Relative density of mercury is 13.6). Solution: Writing mechanical energy balance equation between the inlet 1 and throat o for Venturi meterf o o hg z V p g z V p +++=++22121122ρρ 1 rearranging the equation above, and set (z 2-z 1)=xf o oh xg V V p p ++-=-22121ρ2from continuity equation11221125.625V V dd V V o o =⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛= 3 substituting equation 3 for V o into equation 2 gives()fff f oh xg R h xg Rh V h xg V V p p ++=++=+=++-=-94.1185.203.1903.19206.3922121211ρ4from the hydrostatic equilibrium for manometerg x g R p p Hg o ρρρ+-=-)(1 5substituting equation 5 for pressure difference into equation 4 obtainsf Hgh xg R gx g R ++=+-94.118)(ρρρρ 6rearranging equation 6Figure for problem 1.16kg J R R R R gR h Hg f /288.267.494.11861.12394.118)(==-=--=ρρρ1.17.Sulphuric acid of specific gravity 1.3 is flowing through a pipe of 50 mm internal diameter. A thin-lipped orifice, 10mm, is fitted in the pipe and the differential pressure shown by a mercury manometer is 10cm. Assuming that the leads to the manometer are filled with the acid, calculate (a)the weight of acid flowing per second, and (b) the approximate friction loss in pressure caused by the orifice.The coefficient of the orifice may be taken as 0.61, the specific gravity of mercuryas 13.6, and the density of water as 1000 kg/m 3Solution:a)2.0501010==D D =⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρs kg V D m /268.0130063.201.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)130013600(1.0)(21g R p p Hg ρρ12066.3Papressure difference due to increase of velocity in passing through the orificePa D D V V V V p p o8.44882)2.01(63.213002242412222212221=-=⎪⎪⎭⎫⎝⎛-=-=-ρρpressure drop caused by friction lossPa p f 5.75778.44883.12066=-=∆()s m p p D D C V o /63.231.461.056.1861.0130081.9)130013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρ2.1 Water is used to test for the performances of pump. The gauge pressure at the discharge connection is 152 kPa and the reading of vacuum gauge at the suctionconnection of the pump is 24.7 kPa as the flow rate is 26m 3/h. The shaft power is 2.45kw while the centrifugal pump operates at the speed of 2900r/min. If the vertical distance between the suction connection and discharge connection is 0.4m, the diameters of both the suction and discharge line are the same. Calculate the mechanical efficiency of pump and list the performance of the pump under this operating condition. Solution:Write the mechanical energy balance equation between the suction connection and discharge connection2_1,2222121122f H gp g u Z H g p g u Z +++=+++ρρwherem Z Z 4.012=-(Pa 1052.1(Pa 1047.22_1,215241≈=⨯=⨯-=f H u u pressure gauge p pressure gauge p ））total heads of pump is m H 41.1881.9100010247.01052.14.055=⨯⨯+⨯+= efficiency of pump is N N e /=η since kW g QH N e 3.1360081.9100041.18263600=⨯⨯⨯==ρ N=2.45kWThen mechanical efficiency%1.53%10045.23.1=⨯=η The performance of pump isFlow rate ，m ³/h26Total heads ，m 18.41Shaftpower ，kW 2.45Efficiency，% 53.12.2 Water is transportedby a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gaugepressure is 0.5 kgf/cm 2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is φ57×3.5 mm , the orifice coefficient of C o and orifice diameter d o are 0.62 and 25 mm, respectively. Frictional coefficient λ is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)Solution:Equation(1.6-9)Mass flow rates kg S V m o o /02.21000025.0414.312.42=⨯⨯⨯==ρ 2) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V 1=V 2f H z gp p H +∆+-=ρ12 ∆z=10mPa p 7570710013.17602001081.95.054=⨯⨯+⨯⨯=∆ ∆p/ρg=7.7mThe relation between the hole velocity and velocity of pipeFriction losssoH=7.7+10+5.1=22.8m2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum.Ifs m Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫⎝⎛-=-⎪⎭⎫ ⎝⎛-=ρρρ）（s m D d V V /12112.42200=⎪⎭⎫⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l f H f 1.581.92105.0200025.02422=⨯⨯==losses in the suction pipe accounted for a head of 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet? Solution :From an energy balance,WhereP o =760-640=120mmHg P v =760-710=50mmHgUse of the equation will give the minimum height H g as2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?• Density of acid 1840kg/m 3• Viscosity of acid 25×10-3PasSolution:Velocity of acid in the pipe:sm d m d mpipe of area tional cross flowrate volumetric u /32.3025.01840785.03785.04sec 222=⨯⨯===-=ρπρReynolds number:6109102532.31840025.0Re 3=⨯⨯⨯==-μρud from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9kg J u d l f ph f /450232.3025.0600085.042422=⨯==∆=ρkPa p 5.8271840450=⨯=∆or friction factor is calculated from equation1.4-25NPSH H gp p H f vo g ---=ρmNPSH H gp p H f vo g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=（ρkgJ u d l u d l f ph f /426232.3025.0606109046.042Re 046.042422.022.02=⨯⨯⨯==∆=--＝ρkPa p 84.7831840426=⨯=∆if the pressure drop falls to 783.84/2=391.92kPa8.18.12.12.038.12.12.022.0012.089.1079`2025.060102518401840046.042046.042Re 046.043919202u u u dl u d l p p =⎪⎭⎫ ⎝⎛⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯==∆='∆----ρμρρ＝sos m u /27.236.489..1079012.03919208.18.1==⨯=new mass flowrate=0.785d 2u ρ=0.785×0.0252×2.27×1840=2.05kg/s2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?Density of acid 1840kg/m 3Viscosity of acid 25×10-3Pa Friction factor 32.0Re500.00056.0+=f for hydraulically smooth pipe Solution:Write energy balance equation:f h gu z g p H g u z g p +++=+++2222222111ρρ gu d l g p h H f 22λρ=∆==342=ρπu ds m d u /32.31840025.014.3124322=⨯⨯=⨯=ρπ 611510251840025.032.3Re 3=⨯⨯⨯=-0087.061155.00056.0Re 500.00056.032.032.0=+=+=f92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρΔp=46.92×1840×9.81=847.0kpa2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38⨯2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient C o ＝0.63.the permanent loss in pressure is 3.5×104N/m 2, the friction coefficient λ=0.024. find:(1) What is the pressure drop along the pipe AB?(2)What is the ratio of power obliterated in pipe AB to total power supplied to thefluid when the shaft work is 500W, 60%efficiency? (The density of fluid is 870kg/m 3)solution ：∑+++=+++f AA A A AA h u p g z w u p g z 2222ρρρλρ022p u d l h p p f BA ∆+==-∑247.0334.162=⎪⎭⎫⎝⎛=A A o ()()s m gR C u /5.8870870136006.081.9297.063.02247.01200=-⨯⨯=''--=ρρρ∴u = (16.4/33)2×8.5=2.1m/s∴242/76855105.321.2033.030870024.0m N h p p f B A =⨯+⨯==-∑ρ (2)W u d p Wm 1381.2033.0785.0768554Ne 22=⨯⨯⨯=∆==ρπρ sothe ratio of power obliterated in friction losses in AB to total power supplied to the fluid％％＝461006.0500138⨯⨯。

流体力学与传热(英文版)Chapter4(d)4.5

t tLstw Nhomakorabeas
D
t
w
(a)
(b)
(c )
• （a) and (b) film type, (c) dropwise
In film condensation
the liquid condensate forms a film, or continuous layer, of liquid that flows over the surface of the tube under the action of gravity.
• Friction losses in a condenser are normally small, so that condensation is essentially a constant-pressure process.
• The condensing temperature of a single pure substance depends only on the pressure, and therefore the process of condensation of a pure substance is isothermal.
4.5.1.1 Dropwise and film-type condensation A vapor may condense on a cold surface in one of two ways, which are well described by the terms dropwise and film-type.
Phase change covers condensation, boiling, melting, crystallization and sublimation (升华, 凝华)

流体力学和传热习题参考解答(英文)

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U UZ g+W Z g+h 22ρρP P ++=++10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15=422.67x101.97W 15x9.81120246.88J /kg 12002=-+++=N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation: h f =6.5U 2where U is the velocity in the pipe, find a. water velocity at section A-A'. b. water flow rate, in m 3 /h.22112212f U UZ g+Z g+h 22ρρP P +=++ 1U0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m /s ==22aa b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z22abb a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ= k=0.025mm k/d=0.025/25=0.001 c l r k =0.4 k =1 k =2x0.07=0.14 el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u d Re 2.78x10ρμ== f 0.063=2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meterand orifice coefficient can be taken as 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalentlength of the valve is 1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3,ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have22112212f 1-2u u gZ gZ h 22ρρP P ++=+++ 2212Hg H o 0 g(R h)39630N/m ρρP =P =-= 2212f1-2c u u u 0 Z =0 h 4f +k 2.13ud 22===lWe can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have22331113f1-3u u gZ gZ h 22ρρP P ++=+++ 311Z 0 Z 6.66m u =0==22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++112120 Z 6.66 Z 0 u 0 u 3.51P =====22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N32970mρP=++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe linefA fB total A B22A fAA 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg =22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /hπ∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C. a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied?a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2) x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==> 12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr=()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=.()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find: a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain? Hint: for laminar flow, Nu=1.86[Re Pr]1/3 for turbulent flow Nu=0.023Re 0.8 Pr 1/3(1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====>()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=.(2) 12w 2w = 4421Re Re /2=2x1010=>0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k ==(3) 44333u d 2000x0.01Re 2x10100.001ρμ===>0.81/3Nu 0.023RePr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease d The first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ? (2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged? (4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?(a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 ()20h 642.9w/m k = 12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln14080-=--℃1122L t70/86.5L t ∆==∆ 2L 0.81L1 4.4m ==(d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface areaof pipe) is 2115 w/(m 2o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uois 2660 w/( m 2o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored) o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)=0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=-C=2859io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturatedvapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T)80/()20ln(20802---=∆h h m T T TTh=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p 1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatureschange from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2o C) and 1.7kw/(m 2o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -==()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1) 10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3 660 9.10⨯10-33.50 17.1 2.27⨯10-3 233 9.10⨯10-3 Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm21-s K=2k ∆P 1-s K'=2k '∆P1sK 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E eV KA 2V+V d d θ⎛⎫=⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows: (q+10)2 = 250(t+ 0.4) where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+ q=240 V=qA=240*0.1=24 (2) K 2k =∆P K'2k '=∆P '2∆P =∆P K'2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。

2004级流体力学与传热期中考试题(双语班)

Midterm Examination for Fluid Mechanics and Heat Transfer Grade 2004Name Student No. Class Score1.(10 points) True or false(1)( ) When a fluid passes through a pipe, the total mechanical energy in section1 (inlet) is always equal to that in section2 (outlet).(2)( ) When the developed head H is the same, one pump can not develop morevolume flow rate than that two same pumps which work in parallel.(3)( ) In a laminar flow, the frictional loss is proportional to velocity square.(4) ( ) A reciprocating pump is not subject to air binding.(5) ( ) There are three types of flow, that is, laminar, transitional and turbulent flow.2.(30 points) Fill in the blanks or choose the correct answer(1) The orifice meter belongs to , flow rate meter; while the rotameter is , flow rate meter.A) constant pressure drop; B) constant area; C) variable pressure drop; D) variable area(2) ( ) In order to avoid the cavitation, which of the following methods is feasible?A)increase the operating temperature;B)decrease the diameter of suction pipe;C)install the valve or fittings in the suction pipeline;D)reduce the frictional loss in the suction pipeline(3) The orifice meter is used to measure ; the rotameter is utilized tomeasure ; the Pitot tube is used to measure .A.average velocity in the orifice;B.volume flow rate in the orifice;C.average velocity in annular space between the float and tube wall;D.volume flow rate in annular space between the float and tube wall;E.local velocity in the pipe;F.average velocity in the pipe(4) The methods of adjusting volume flow rate of centrifugal pump are1) ; 2) ;(3) .The most common used way in practice is ( ). (5) To avoid cavitation, the pressure at the centrifugal pump inlet must exceed the vapor pressure by a certain value called the .(6) 1atm= mH2O= N/m2；1cp= P= N·s/m2.(7) When the temperature of fluid to be transported increases, the maximum suction lift will ( ). If the velocity of suction pipe decreases, the maximum suction lift will ( ).(8) The viscosity of a gas with temperature increasing, and viscosity of a liquid usually significantly when the temperature is raised.(9) In the boundary layer, there exist . Outside the boundary layer, potential flow survives.(10) When the pipe installation changes from horizontal to vertical position and the velocity keeps unchanged, the frictional loss will be .A. increased;B. decreased;C. the same;D. uncertain3. (30 points) A liquid with density 900 kg/m3 and viscosity μ=30 cp flows from tank 1 to tank 2 through the pipeline of inner diameter 40 mm. The levels of liquid surfaces in two tanks keep constant. When the valve is closed, the readings of gauges A and B are 8.82⨯104 N/m2 and4.41⨯104 N/m2, respectively. When the valve is open, the total length of pipeline is 100 m including all local frictional equivalent length, find:(1)the volumetric flow rate in the pipe, in m3/h?(2)when the valve is open, what happens to the readings of gauges A and B,explain simply why?Hint: for laminar flow, f=16/Re or λ=4f=64/ReFor turbulent flow, λ=0.3164/Re0.254. (30 points)The characteristics of a centrifugal pump are measured under certain rotational speed. The inside diameters of the suction and discharge pipelines are 70mm and 50mm respectively. When the volumetric flow rate of water is 30m3/h, the readings of the vacuum gauge at the suction station and the pressure gauge at the discharge station of the pump are 40kPa and 215kpa respectively. The vertical distance between the pressure tap (pressure measurement point) at the suction station and the pressure tap at the discharge station is 0.4m. The shaft power is 3.45kW. Calculate the developed head and efficiency of the pump.。

流体力学与传热(英文版)Chapter4(f)4.7

Chapter 4
4.7 heat-exchange equipment
In industrial processes heat energy is transferred by a variety of methods. Heat exchangers are of importance for heat energy transport. There are lots of exchangers. Including conduction-convection in exchangers, boilers, and condensers; radiation in furnaces and radiant heat dryer.
Multipass construction increases the fluid velocity, with a corresponding increase in the heat-transfer coefficient.
An even number of tube-side passes are used in multipass exchangers. The shell side may be either single-pass or multipass.
Flash p343
The disadvantages of multipass are that
(1) the exchanger is slightly more complicated ; (2) Some sections in the exchanger have parallel flow, which limits the temperature approach; (3) the friction loss rough the equipment is increased because of the larger velocities and multiplication (增大） of exit and entrance losses.

流体力学思考题双语

流体力学与传热习题参考解答英文

1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, finda. water velocity at section A-A'.b. water flow rate, in m 3 /h.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tank and delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be taken as 0.61, what is the flow rate of water?7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we haveWe can get Z1 from the valve closed(2) when the valve opens fully, for section 1-1’ and 3-3’, we haveFor section 1-1’ and 2-2’8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe line10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied?a.2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+ b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====> (2) 12w 2w = 4421Re Re /2=2x1010=> (3) 44333u d 2000x0.01Re 2x10100.001ρμ===> (4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )π (5) there methods : increase u or hi or decrease dThe first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes through the pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why?(a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21 12t +t LMTD=702∆∆℃=Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln 14080-=--℃ (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=- C=2859 io h Uo h 111-= ho=8127W/(m2K) 17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected)18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored)Th=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h 22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 2We can see K=0.015 qe=0.026For p=3.5Kg/cm 2 23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+q=240 V=qA=240*0.1=24(2) K 2k =∆P K'2k '=∆Pq ’=343.6 v=34.36。

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1. Water is pumped at a constant velocity 1m/s from large reservoir resting on the floor to the open top of an absorption tower. The point of discharge is 4 meter above the floor, and the friction losses from the reservoir to the tower amount to 30 J/kg. At what height in the reservoir must the water level be kept if the pump can develop only60 J/kg?2222112f 1U P U P w=Z g+h (Z g+)22ρρ++-+ U 1=0 12P =P 10Z = W=60j/kg f h 30/kg =2U =1m/s 2(60300.5)/g 3m Z =--=21Z Z Z 431m ∆=-=-=2. The fluid (density 1200 kg/m 3 ) is pumped at a constant rate 20 m 3 /h from the large reservoir to the evaporator. The pressure above the reservoir maintains atmosphere pressure and the pressure of the evaporator keeps 200 mmHg (vacuum). The distance between the level of liquid in the reservoir and the exit of evaporator is 15 meter and frictional loss in the pipe is 120 J/kg not including the exit of evaporator, what is the pump effective work and power if the diameter of pipe is 60 mm?22112212f U U Z g+W Z g+h 22ρρP P ++=++ 10P = 5422200P x1.013x10 2.67x10N /m 760=-=- 31200Kg /m ρ= 1U 0= f h 120J /kg =22V 20U 1.97m /s A 3600*4006===π/*. 1Z 0= 2Z 15= 422.67x101.97W 15x9.81120246.88J /kg 12002=-+++= N W Q 246.88x1200x20/3600=1646W ρ==3. Water comes out of the pipe (Φ108x4 mm), as shown in Fig. The friction loss of the pipeline which does not cover the loss at the exit of pipe can be calculated by the following equation:h f =6.5U 2where U is the velocity in the pipe, finda. water velocity at section A-A'.b. water flow rate, in m 3 /h. 22112212f U U Z g+Z g+h 22ρρP P +=++ 1U 0= 12P =P 1Z 6m = 2Z 0=2f h 6.5U = 22U 6x9.81 6.5U 2=+ U 2.9m/s = 23V=UA=2.94x01x360082m /h =π/.4. Water passes through the variable pipe. The velocity in the small pipe is 2.5 m/s. The vertical glass tubes are inserted respectively at the section A and B to measure the pressure (see fig.) If the friction loss between two section is 15 J/kg, what is the water column difference between two glass tubes? By the way, draw the relative liquid column height of two tubes in the Fig.a ab b U A U A = 2b U 2.5*(33/47)1.23m /s == 22a a b b a b f U U Z g+Z g+h 22ρρP P +=++ a b Z =Z 22a b b a f U U h 22ρρP P -=-+221.23/2 2.5/21512.63=-+= a b P P R g ρ-=∆ 3312.63R=1.29x10m 9.8x10-∆=5. A centrifugal pump takes brine (density 1180 kg/m 3 , viscosity 1.2 cp) from the bottom of a supply tankand delivers it into another tank. The line between the tanks is 300 m of 25 mm diameter pipe (inner diameter). The flow rate is 2 m 3 /h. In this line, there are two gate valves, four elbows (90o ) and one return bend, what is the friction loss if the roughness of pipe is 0.025 mm?22f fst flocal U U h h h 4f k d 22l ∑=+=+∑ 31180kg /m ρ= 300m, d=0.025m l =3-3v 2m /h =1.2cp=1.2x10Pa.s μ=k=0.025mm k/d=0.025/25=0.001c l r k =0.4 k =1 k =2x0.07=0.14el re k 4x0.75 3 k 1.5-2.2===2u v /A 2/(3600x /4x0.025)1.13m /s π===4u d Re 2.78x10ρμ== f 0.063= 2f 2h 4x0.0063x300/0.025x1.13/2+(0.4+1+2x0.07+4x0.7+1.5)x1.13/2 =197.86J/kg∑=6. The orifice meter (diameter of orifice 0.0001 m) is installed for measuring the flow rate. The indicating liquid of orifice is mercury if U shape pressure gauge reading is 0.6 meter and orifice coefficient can be takenas 0.61, what is the flow rate of water?o u c =20o 0V u s 0.61x /4x0.0001π==835.8x10m /s -=7. Water flows through a pipe with a diameter di 100 mm as shown in figure.a. when the valve is closed, R is 600 mm and h equals 1500 mm. While the valve opens partially, R=400 mm and h=1400 mm, f=0.00625 (Finning factor) and k c =0.5 (contraction coefficient), what is the flow rate of water, in m 3 /h?b. If the valve opens fully, what is the pressure of section 2-2', in N/m 2 ? The equivalent length of the valve is1.5 m and the Fanning factor f keeps the same?(ρH2O =1000kg/m 3, ρHg =13600kg/m 3)(1) the valve opens partially ,for selection 1-1’ and 2-2’ , we have 22112212f 1-2u u gZ gZ h 22ρρP P ++=+++ 2212Hg H o 0 g(R h)39630N/m ρρP =P =-= 2212f1-2c u u u 0 Z =0 h 4f +k 2.13u d 22===l We can get Z1 from the valve closed21Hg H O h=1.5m R=0.6m Z gR/h 6.66m ρρ=-=229.81x6.66u /2 2.13u 39630/1000=++23h u=3.13m/s V 3600x /4x0.1x3.1388.5m /h π==(2) when the valve opens fully, for section 1-1’ and 3-3’, we have 22331113f1-3u u gZ gZ h 22ρρP P ++=+++ 311Z 0 Z 6.66m u =0== 22e f1-3c u 3.1.5h (4f k )(4x0.00625x +0.5) 4.81u d 20.01l l ++=+== 229.81x6.66u /2 4.81u =+ u 3.51m/s =For section 1-1’ and 2-2’22112212f1-2u u gZ gZ h 22ρρP P ++=+++ 112120 Z 6.66 Z 0 u 0 u 3.51P ===== 22f1-2c l u h (4f k )(4x0.00625x15/0.10.5)3.51/226.2J /kg d 2=+=+= 22229.81x6.66 3.15/226.2N 32970mρP =++P =8. The rotameter is installed to measure the water flow rate, as shown in figure. If the total length including equivalent length of pipeline A is 10 m and the reading of rotameter is 2.72 m 3 /h, what is the flow rate for pipeline B? (f A =0.0075, f B =0.0045)For parallel pipe line fA fB total A B22A fA A 2A h h V V +V u (l+le) 2.72h 4f 4x0.0075x10/0.053/2()d 23600x /4x0.053π∑=∑=∑∑== 0.333J /kg = 22B fB B B B 23B B B B u (l+le)h 4f 4x0.0045x2/0.3/2xu 0.333d 2u 2.36m /s V =u A 2.36x /4x0.23600m /h π∑∑======10. A flat furnace wall is constructed of 120 mm layer of sil-o-cel brick, with a thermal conductivity 0.08 w/(m o C), backed by a 150 mm of common brick, of conductivity 0.8 w/(m o C), the temperature of inner face of the wall is 1400 o , and that of the outer face is 200o C.a. What is the heat loss through the wall in w per square meter.b. To reduce the heat loss to 600 w/m 2 by adding a layer of cork with k 0.2 w/(m o C) on the outside of common brick, how many meters of cork are requied? a. 2Q t 1400200711N /m 11L R 0.080.80.120.15∑∆-===∑+ b. 600=(1400-200)/(0.12/0.08+0.15/0.8+x/0.2)x=0.0625m13. Air at the normal pressure passes through the pipe (d i 20 mm) and is heated from 20o C to 100o C. What is the film heat transfer coefficient between the air and pipe wall if the average velocity of air is 10 m/s? The properties of air at 60 o C are as follows:density 1.06 kg/m 3 , viscosity 0.02 cp, conductivity 0.0289 w/(m o C), and heat capacity 1 kJ/kg-K443u d 10x0.02x1.06 Re=1.06x10100.02x10ρμ-==>12T +T 20100T=6022+==℃ 0.141ωμμ⎛⎫= ⎪⎝⎭10000.020.0010.6920.0289p c x x k μ==Pr= ()()0.81/3081/34Nu 0027Re Pr 0.027x 1.06x10x 0.69239.66==.=. ()2i i i h d 39.66 h 39.66x0.0289/0.02=57.22w/m .k k ==14. A hot fluid with a mass flow rate 2250 kg/h passes through a ∅25x2.5 mm tube. The physical properties of fluid are as follows:k=0.5 w/(m o C), C p =4 kJ/kg-K, viscosity 10-3 N-s/m 2 , density 1000 kg/m 3 Find:a. Heat transfer film coefficient h i , in w/(m 2 -K).b. If the flow rate decreases to 1125 kg/h and other conditions are the same, what is the h i ?c. If the diameter of tube (inside diameter) decreases to 10 mm, and the velocity u keeps the same as that of case a, calculate h i .d. When the average temperature of fluid and quantity of heat flow per meter of tube are 40 o C and 400 w/m, respectively, what is the average temperature of pipe wall for case a?e. From this problem, in order to increase the heat transfer film coefficient and enhance heat transfer, what kinds of methods can you use and which is better, explain?Hint: for laminar flow, Nu=1.86[Re Pr]1/3for turbulent flow Nu=0.023Re 0.8 Pr 1/3 (1) 444N 2250x4u d Gd d 3600x x0.02Re 3.98x10100.001ρππμμμ=====> ()()1/30.8081/3424Nu 0023Re Pr 0.023x 3.98x10220.10.5Nuk 220.1x0.5hi 5500w /m k d 0.02⎛⎫== ⎪⎝⎭===.=. (2) 12w 2w = 4421Re Re /2=2x1010=> 0.80.82211Nu Re 0.5Nu Re ⎛⎫== ⎪⎝⎭ 0.8i2i1h 0.5h = ()0.82i2h 5500x0.53159w /m k == (3) 44333u d 2000x0.01Re 2x10100.001ρμ===> 0.81/3Nu 0.023Re Pr = ()2hi=6347w/m k(4)i i w w Q=h A (t-t )=400=500x2x0.02(t-t )πw t=40t 39.41=℃ ℃(5) there methods : increase u or hi or decrease dThe first is better15. In a double pipe exchange (Φ23x2 mm), the cold fluid (Cp=1 kJ/kg, flow rate 500 kg/h) passes throughthe pipe and the hot fluid goes through the outside. The inlet and outlet temperatures of cold fluid are 20 and 80 o , and the inlet and outlet temperatures of hot fluid are 150 and 90o , respectively. The h i (film coefficient inside pipe) is 700 w/(m 2 o C)and overall heat transfer coefficient U o (based on the outside surface of pipe) is 300w/(m 2 o C), respectively. If the heat loss is ignored and the conductivity of pipe wall (steel) is taken as 45 w/(m o C), find:(1) heat transfer film coefficient outside the pipe h o ?(2) the pipe length required for counter flow, in m?(3) what is the pipe length required if the heating medium changes to saturated vapor(140 o C) and it condenses to saturated liquid and other conditions keep unchanged?(4) When the exchanger is used for a year, it is found that it cannot meet the need of production (the outlet temperature of cold fluid cannot reach 80 o C), explain why? (a) 0m o 0i i m d l d 111230.002x23h Vo h d kd 300700x1945x21⎛⎫=-+=-- ⎪⎝⎭ 1/h0=1/U0-(do/hidi+bdo/kdm)=1/300-23/700*19-0.002*23/45*21()20h 642.9w/m k =12t +t LMTD=702∆∆℃= Q=UoAo ∆Tm=mcCp(Tcb-Tca) 300*2π*0.023*70L=500/3600*1000*(80-20)L=5.4m(c) 8020LMTD=86.514020ln 14080-=--℃ 1122L t 70/86.5L t ∆==∆ 2L 0.81L1 4.4m == (d) scale is formed on the outside ,V 0 is decreased16. Water flows turbulently in the pipe of Φ25x2.5 mm shell tube exchanger. When the velocity of water u is 1 m/s, overall heat transfer coefficient Uo (based on the outer surface area of pipe) is 2115 w/(m 2 o C). If the u becomes 1.5 m/s and other conditions keep unchanged, Uo is 2660 w/( m 2 o C ). What is the film coefficient ho outside the pipe? (Heat resistances of pipe wall and scale are ignored)o i h h Uo 111+= (1) oi o h h U 1'1'1+= (2) (1)-(2)= 0.80.80.80.81211111121152660u C u C 1C 1.5C-=-=- C=2859 io h Uo h 111-= ho=8127W/(m2K)17. Water and oil pass parallelly through an exchanger which is 1 m long. The inlet and outlet temperatures of water are 15 and 40 o C, and those of oil are 150 and 100 o C, respectively. If the outlet temperature of oil decreases to 80 o C, and the flow rates and physical properties and inlet temperatures of water and oil maintain the same, what is the pipe length of new exchanger? (Heat loss and pipe wall resistance are neglected) ()()h h 12c c 21m Q W C T -T W C t t VA t ==-=∆()()h h 12c c 21W C T -T 'W C t 't =-2150100401515080t 15--=-- 2t 50=℃ 212m1112m2L T T 't 1508092.51.85L T T t 15010069.8-∆-===-∆- 2m1m2L 1.85m L1=1m t 92.5 t 69.8=∆=∆=18. Air which passes through the pipe in turbulent flow is heated from 20 to 80 o C. The saturated vapor at 116.3 o C condenses to saturated water outside the pipe. If air flow rate increases to 120% of the origin and inlet and outlet temperatures of air stay constant, what kind of method can you employ in order to do that? (Heat resistance of pipe wall and scale can be ignored))(111ca cb pc c m i i T T C m T A h -=∆=1Q)'(2212ca cb pc c m i T T C m T A h -=∆=2Q 128.012112i22.12.1h m m c c m i m T T m m T h T ∆∆===∆∆ )803.116/()203.116ln(20801---=∆m T )80/()20ln(20802---=∆h h m T T T Th=118.5oC19. Water flows through the pipe of a Φ25x2.5 mm shell-tube exchanger from 20 to 50 o C. The hot fluid (C p1.9 kJ/kg o C, flow rate 1.25 kg/s) goes along the shell and the temperatures change from 80 to 30 o C. Film coefficients of water and hot fluid are 0.85kw/(m 2 o C) and 1.7 kw/(m 2 o C). What is the overall heat transfer coefficient Uo and heat transfer area if the scale resistance can be ignored? (the conductivity of steel is 45w/(m o C).W=1.25Kg/s Cp=1.9Kj/kg ℃()()2h p 12Q W C T T 1.25x1.9x 80-30119Kw =-==m 3010t 30ln 10-∆= ()200m 00i i m 1V 472w/m k d l d 1h h d kd ++==32i 0m Q 119x10A 13.9m V t 472x18.2===∆20. A spherical particle (density 2650 kg/m 3) settles freely in air at 20 o C (density of air 1.205 kg/m 3 , viscosity 1x10-5 Pa.s). Calculate the maximum diameter of particle if the settle obeys the Stoke s’ Law?Re ≤1 ()2p t p D g U 18D ρρμμρP -== ()23p 18D g μρρρP =- ()1/3-10p 18x10D 1.205x9.81x 2650-1.205⎛⎫= ⎪ ⎪⎝⎭=3.85x10-521. A filter press(A=0.1 m 2 ) is used for filtering slurry. The vacuum inside the filter is 500 mm Hg. One liter filtrate can be got after filtering of 5 min and 0.6 more liter filtrate is obtained after 5 more min. How much filtrate will be got after filtering of 5 more min?for filter press 22e V 2VV =KA θ+5 min 22e 12V 0.1x5K +=(1)10min 22e 1.62x1.6V 0.1x10K +=(2)From (1) (2),we can see Ve=0.7 K=4815 min 22V 2x0.7V=48x0.1x15+ V=2.07m 3/h22. The following data are obtained for a filter press (A=0.0093 m 2) in a lab.------------------------------------------------------------------------------------------------pressure difference (kg f /cm 2 ) filtering time (s) filtrate volume (m 3 )1.05 502.27⨯10-3660 9.10⨯10-33.50 17.1 2.27⨯10-3233 9.10⨯10-3Find1) filtering constant K, q e , t e at pressure difference 1.05 kg f /cm 2 ?2) if the frame of filter is filled with the cake at 660 s, what is the end filtering rate (dV/dt)E at P 1.05 kg f /cm 2 ?3) compressible constant of cake s?For p=1.05Kg/cm 22e 2e 2e q 2qq K 0.002270.0002272x q 50K 0.00930.000930.000910.000912x q 660K 0.000930.00093θ+=⎛⎫+= ⎪⎝⎭⎛⎫+= ⎪⎝⎭We can see K=0.015 qe=0.026For p=3.5Kg/cm 21-s K=2k ∆P 1-s K'=2k '∆P 1s K 'K '-∆P ⎛⎫= ⎪∆P ⎝⎭ ()2E e V KA 2V+V d d θ⎛⎫= ⎪⎝⎭23. A slurry is filtered by a 0.1 m 2 filter press at constant pressure if the cake is incompressible. The filter basic equation is as follows:(q+10)2 = 250(t+ 0.4)where q---l/m 2 t----minfind (1) how much filtrate is got after 249.6 min?(2) if the pressure difference is double and the resistance of cake is constant, how much filtrate can be obtained after 249.6 min? (cake is imcompressible)（1）let θ=249.6 ()()2q+10250x 249.60.4=+ q=240 V=qA=240*0.1=24(2) K 2k =∆P K'2k '=∆P'2∆P =∆P K'2K 500== ()()2q'+10500x 249.60.4=+ q ’=343.6 v=34.36。

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