流体力学与传热习题：solutions
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⨯⨯。
流体力学与传热作业-(含答案)
1、当地大气压为745mmHg 测得一容器内的绝对压强为350mmHg ，则真空度为 mmHg 。
测得另一容器内的表压强为1360 mmHg ，则其绝对压强为 mmHg 。
2、为测量腐蚀性液体贮槽中的存液量，采用图示的装置。
测量时通入压缩空气，控制调节阀使空气缓慢地鼓泡通过观察瓶。
今测得U 形压差计读数为R=130mm ，通气管距贮槽底面h=20cm ，贮槽直径为2m ，液体密度为980kg/m 3。
试求贮槽内液体的贮存量为多少吨？解：由题意得：R=130mm ，h=20cm ，D=2m ，=ρ980kg/3m ，=Hg ρ3/13600m kg 。
（1）管道内空气缓慢鼓泡u=0，可用静力学原理求解。
（2）空气的ρ很小，忽略空气柱的影响。
g R g H Hg ρρ=∴ m R H Hg 8.113.098013600.=⨯==ρρ 吨）(15.6980)2.08.1(2785.0)(4122=⨯+⨯⨯=+=∴ρπh H D W 3、测量气体的微小压强差，可用附图所示的双液杯式微差压计。
两杯中放有密度为1ρ的液体，U 形管下部指示液密度为2ρ，管与杯的直径之比d/D 。
试证气罐中的压强B p 可用下式计算：22112)(Dd hg hg p p a B ρρρ---=证明： 作1-1等压面，由静力学方程得：g h g h P g h P B a 211ρρρ+∆+=+ （1） 22144d hD h ππ=221Dd h h =∴代入（1）式得：g h g D d h P g h P B a 21221ρρρ++=+即22112)(Dd hg g hg P P a B ρρρ---=3、查阅资料，写出比重计（密度计）的设计原理4、本题附图所示的开口容器内盛有油和水。
油层高度h 1=0.7m 、密度ρ1=800kg/m 3，水层高度h 2=0.6m 、密度ρ2=1000kg/m 3。
（1）判断下列两关系是否成立，即 p A =p'A p B =p'B（2）计算水在玻璃管内的高度h 。
流体力学与热传递在工程设计中的应用考核试卷
1.流体在圆形管道中做层流运动时，流速分布呈_______状。
（）
2.热对流的基本方程是_______方程。
（）
3.在热辐射中，黑体的辐射强度与温度的_______成正比。
（）
4.流体的粘度与温度的关系是_______。
（）
D.流体粘性很小，雷诺数小于2000
4.在热传递中，热对流的主要影响因素是：（）
A.流体的热导率
B.流体的速度
C.流体的比热容
D.流体与固体表面的温差
5.对于热辐射，以下哪种说法是正确的：（）
A.热辐射不需要介质就可以传播
B.热辐射的强度与物体温度成反比
C.热辐射在真空中不能传播
D.热辐射的强度只与物体颜色有关
A.热边界层是温度梯度的过渡层
B.热边界层厚度随流体速度增加而减小
C.热边界层厚度与流体的热导率有关
D.热边界层厚度与流体与固体表面的温差有关
20.在流体力学中，以下哪些方程用于描述流体的运动：（）
A.纳维-斯托克斯方程
B.连续性方程
C.能量方程
D.动量方程
（以上为多选题部分，以下是其他题型，请根据需要自行补充）
4.热对流是依靠流体的宏观运动进行热量传递的。（）
5.热辐射的强度与物体的表面积成正比。（）
6.在热交换器中，增加流体流速会降低传热效率。（）
7.热力学第二定律表明，热能总是自发地从高温物体传递到低温物体。（）
8.湍流流动的流线是平行的，层流流动的流线是弯曲的。（）
9.热边界层的厚度与流体的热导率成反比。（）
四、判断题
1. ×
2. √
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。
(完整版)流体力学练习题及答案
(完整版)流体力学练习题及答案-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN流体力学练习题及答案一、单项选择题1、下列各力中，不属于表面力的是（ ）。
A ．惯性力B ．粘滞力C ．压力D ．表面张力2、下列关于流体粘性的说法中，不准确的说法是（ ）。
A ．粘性是实际流体的物性之一B ．构成流体粘性的因素是流体分子间的吸引力C ．流体粘性具有阻碍流体流动的能力D ．流体运动粘度的国际单位制单位是m 2/s3、在流体研究的欧拉法中，流体质点的加速度包括当地加速度和迁移加速度，迁移加速度反映（ ）。
A ．由于流体质点运动改变了空间位置而引起的速度变化率B ．流体速度场的不稳定性C ．流体质点在流场某一固定空间位置上的速度变化率D ．流体的膨胀性4、重力场中平衡流体的势函数为（ ）。
A ．gz -=πB ．gz =πC ．z ρπ-=D ．z ρπ=5、无旋流动是指（ ）流动。
A ．平行B ．不可压缩流体平面C ．旋涡强度为零的D ．流线是直线的6、流体内摩擦力的量纲[]F 是（ ）。
A . []1-MLtB . []21--t MLC . []11--t ML D . []2-MLt 7、已知不可压缩流体的流速场为xyj zi x 2V 2+= ，则流动属于（ ）。
A ．三向稳定流动B ．二维非稳定流动C ．三维稳定流动D ．二维稳定流动8、动量方程 的不适用于(??? ??) 的流场。
A ．理想流体作定常流动B ．粘性流体作定常流动C ．不可压缩流体作定常流动D ．流体作非定常流动9、不可压缩实际流体在重力场中的水平等径管道内作稳定流动时，以下陈述错误的是：沿流动方向 ( ) 。
A ．流量逐渐减少B ．阻力损失量与流经的长度成正比C ．压强逐渐下降D ．雷诺数维持不变10、串联管道系统中,其各支管内单位质量流体的能量损失（ ）。
A ．一定不相等B ．之和为单位质量流体的总能量损失C ．一定相等D ．相等与否取决于支管长度是否相等11、边界层的基本特征之一是（ ）。
流体力学与传热学考试题目
流体力学与传热学考试题目1-1 下图所示的两个U 形管压差计中,同一水平面上的两点A 、B 或C 、D 的压强是否相等？答：在图1—1所示的倒U 形管压差计顶部划出一微小空气柱。
空气柱静止不动，说明两侧的压强相等，设为P 。
由流体静力学基本方程式： 11gh gh p p A 水空气ρρ++=11gh gh p p B 空气空气ρρ++=空气水ρρ>∴BA p p >即A 、B 两点压强不等。
而 1gh p p C 空气ρ+=1gh p p D 空气ρ+=也就是说，Cp 、D p 都等于顶部的压强p 加上1h 高空气柱所引起的压强，所以C 、D 两点压强相等。
同理，左侧U 形管压差计中，B A p p ≠ 而DC p p =。
分析：等压面成立的条件—静止、等高、连通着的同一种流体。
两个U 形管压差计的A 、B 两点虽然在静止流体的同一水平面上，但终因不满足连通着的同一种流体的条件而非等压。
1-2 容器中的水静止不动。
为了测量A 、B 两水平面的压差，安装一U 形管压差计。
图示这种测量方法是否可行？为什么？ 答：如图1—2，取1—1/为等压面。
由1'1p p =可知：)(2H R g p O H B ++ρ=gRH h g p Hg O H A ρρ+++)(2ghp p O H A B 2ρ+=将其代入上式，整理得 0)(2=-gR O H Hg ρρ∵2≠-OHHg ρρ ∴0=RR 等于零，即压差计无读数，所以图示这种测量方法不可行。
分析：为什么压差计的读数为零？难道A 、B 两个截面间没有压差存在吗？显然这不符合事实。
A 、B 两个截面间确有压差存在，即h 高的水柱所引起的压强。
问题出在这种测量方法上，是由于导管内充满了被测流体的缘故。
连接A 平面测压口的导管中的水在下行过程中，位能不断地转化为静压能。
此时，U 型管压差计所测得的并非单独压差，而是包括位能影响在内的“虚拟压强”之差。
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。
(完整版)流体力学与传热学试题及答案
流体力学与传热学试题及参考答案一、填空题：（每空1分）1、对流传热总是概括地着眼于壁面和流体主体之间的热传递，也就是将边界层的 和边界层外的 合并考虑，并命名为给热。
答案：热传导；对流传热2、在工程计算中，对两侧温度分别为t1,t2的固体，通常采用平均导热系数进行热传导计算。
平均导热系数的两种表示方法是 或 。
答案；221λλλ+=-；221t t +=-λ3、图3-2表示固定管板式换热器的两块管板。
由图可知，此换热器为 管程，管程流体的走向为 或 。
1 2 3图3-2 3-18 附图答案：4；2→4 →1→5→3；3→5→1→4→24、黑体的表面温度从300℃升至600℃，其辐射能力增大到原来的 倍. 答案: 5.39分析: 斯蒂芬-波尔兹曼定律表明黑体的辐射能力与绝对温度的4次方成正比, 而非摄氏温度,即4273300273600⎪⎭⎫⎝⎛++=5.39。
5、3-24 用0.1Mpa 的饱和水蒸气在套管换热器中加热空气。
空气走管内，由20℃升至60℃，则管内壁的温度约为 。
答案：100℃6、热油和水在一套管换热器中换热，水由20℃升至75℃。
若冷流体为最小值流体，传热效率0.65，则油的入口温度为 。
答案：104℃ 分析： ε=2020751--T =0.65 ∴1T =104℃1 2 37、因次分析法的基础是 ，又称因次的和谐性。
答案：因次的一致性8、粘度的物理意义是促使流体产生单位速度梯度的_____________。
答案：剪应力9、如果管内流体流量增大1倍以后，仍处于滞流状态，则流动阻力增大到原来的 倍。
答案：210、在滞流区，若总流量不变，规格相同的两根管子串联时的压降为并联时的 倍。
答案：411、流体沿壁面流动时，在边界层内垂直于流动方向上存在着显著的_______________，即使____________很小，____________仍然很大，不容忽视。
答案：速度梯度；粘度；内摩擦应力 12、雷诺数的物理意义实际上就是与阻力有关的两个作用力的比值，即流体流动时的______ 与__ ____ 之比。
流体力学与传热模拟试题全套-new.
流体力学与传热模拟试题（一）一、选择题1 离心泵的调节阀开大时，A 吸入管路阻力损失不变B 泵出口的压力减小C 泵入口的真空度减小D 泵工作点的扬程升高2在完全湍流时（阻力平方区），粗糙管的摩擦系数λ数值A 与光滑管一样B 只取决于ReC 取决于相对粗糙度D 与粗3 离心泵开动之前必须充满被输送的流体是为了防止发生＿＿＿＿。
A 汽化现象B 汽蚀现象C 气缚现象D 气浮现象二填空题1在滞流区，颗粒的沉降速度与颗粒直径的＿＿次方成正比，在湍流区颗粒的沉降速度与颗粒直径的＿＿次方成正比。
2流体在圆形直管中作滞流流动时，其速度分布是型曲线，其管中心最大流速为平均流速的倍，摩擦系数λ与Re的关系为。
3 离心泵的安装高度超过允许安装高度时会发生现象4导热系数的单位为，对流传热系数的单位为，总传热系数的单位为。
5流体流动的连续性方程是＿＿＿＿＿＿＿＿＿＿＿＿；适用于圆形直管的不可压缩流体流动的连续性方程为＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿。
三计算题1 在一单程逆流列管换热器中用水冷却空气，两流体的进口温度分别为20℃和110℃。
在换热器使用的初期，冷却水及空气的出口温度分别为45℃和40℃，使用一年后，由于污垢热阻的影响，在冷热流体的流量和进口温度不变的情况下，冷却水出口温度降至38℃，试求：（1）空气出口温度为多少？（2）总传热系数为原来的多少倍？（3）若使冷却水加大一倍，空气流量及两流体进口温度不变，冷热流体的出口温度各为多少？（α水>>α空气）（4）冷却水流量加大后，换热器的传热速率有何变化？变为多少？2 如图所示，用泵将水从贮槽送至敞口路尺寸为φ83×3.5mm，泵的进出口管道上分别安装有真空表和压力表，真空表安装位置离贮槽的水面高度H1为4.8m，压力表安装位置离贮槽的水面高度H2为5m。
当输水量为36m3/h时，进水管道全部阻力损失为1.96J/kg，出水管道全部阻力损失为 4.9J/kg，压力表读数为2.452×105Pa，泵的效率为70%，水的密度ρ为1000kg/m3，试求：（1）两槽液面的高度差H为多少？（2）泵所需的实际功率为多少kW？（3）真空表的读数为多少kgf/cm2？3）一卧式列管冷凝器，钢质换热管长为6m，直径为φ25×2mm。
华南理工大学考研真题—流体力学与传热
7．写出恒压过滤方程式，并且标出各个物理量的单位。（5分）
8．传热有哪三种方式？指出它们的主要区别。（5分）
9．从液膜的分布形式，蒸汽冷凝可分为哪两种类型？（5分）
10．空气天然对流换热系数的范围是多少？（5分）
二、计算题（20分）
若用压力表测得输送水、油（密度880kg/m3）、98%硫酸（密度1830kg/m3）的某段水平等直径管路的压力降均为49kPa,问三者的压头损失的数值是否相等？各为多少米液柱？
三、计算题（30）
有一高位槽输水系统，管径为Φ57mm×3.5mm。已知水在管路中流动的机械能损失为∑hf= 4.5 u2∕2（u为管内流速）。试求水的流量。欲将水的流量增强20%，应将高位槽水面升高多少米？（水密度1000kg/m3）
四、计算题（20分）
某离心泵用15℃的清水举行性能实验，水的体积流量为540m3/h，泵出口压力表读数350 kPa,泵入口真空表读数30kPa,若压力表与真空表测压截面间的垂直距离为350mm，吸入管与压出管内径分离为350 mm和310 mm，试求泵的扬程。（水密度1000kg/m3）
一、问答题（50分）
1．牛顿型流体与非牛顿型流体有什么区别？，画出它们的流变图。（5分）
时，倘若两截面之间的管道有散热损失，如何考虑？（5分）
4．离心泵中，闭式叶轮后盖板上开有平衡孔，其作用是什么？（5分）
5．描述离心泵性能主要有哪五个参数？（5分）
五、计算题（30分）
有一台运转中的单程逆流列管式换热器，热空气在壳程由120℃降到80℃，其对流换热系数α1=50W/m2·K。管程的冷却水从15℃升到90℃，其对流换热系数α2=2000W/m2·K。管壁热阻与污垢热阻皆可不计，管壁当平壁处理。当冷却水流量增强一倍时，试求：（1）水和空气的出口温度，忽略流体物性随温度的变化（20分）。（2）传热速率比本来增强多少？（10分）
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。
传热学习题及答案
传热学习题及答案传热学习题及答案传热学是热力学的一个重要分支，研究热量如何在物质之间传递的过程。
在工程和科学领域中，传热学的知识对于理解和解决各种热传递问题至关重要。
下面，我们将提供一些传热学的学习题及其答案，帮助读者巩固对该领域的理解。
1. 对流传热是指什么？请列举几个常见的对流传热的例子。
答案：对流传热是指通过流体（气体或液体）的运动而传递热量的过程。
常见的对流传热例子包括：自然对流（如烟囱中的烟气上升）、强迫对流（如风扇吹过热食物）、冷却系统中的冷却液循环等。
2. 传热过程中的三种传热方式是什么？请分别解释它们。
答案：传热过程中的三种传热方式是导热、对流和辐射。
- 导热是指通过物质内部的分子振动和传递热量的方式。
导热通常发生在固体和液体中，如铁棒两端的温度差会导致热量从高温端传递到低温端。
- 对流是指通过流体的运动传递热量的方式。
对流通常发生在气体和液体中，如热水从底部加热，底部的热水上升并与顶部的冷水交换热量。
- 辐射是指通过电磁波传递热量的方式。
辐射传热不需要介质，可以在真空中传递热量。
例如，太阳辐射的热量可以穿过空气和云层到达地球表面。
3. 请解释传热中的热传导方程。
答案：热传导方程是描述导热过程的数学方程。
它可以用来计算热量在物质中的传递速率。
热传导方程的一般形式为：q = -kA(dT/dx)其中，q表示单位时间内通过物质传递的热量，k是物质的热导率，A是传热面积，dT/dx是温度梯度（温度变化率）。
4. 请解释传热中的对流换热系数。
答案：对流换热系数是描述对流传热过程中热量传递速率的参数。
它表示单位面积上的热量传递速率与温度差之间的比值。
对流换热系数取决于流体的性质、流体的速度、流体与固体表面的接触情况等。
通常，对流换热系数越大，热量传递速率越快。
5. 请解释传热中的辐射换热系数。
答案：辐射换热系数是描述辐射传热过程中热量传递速率的参数。
它表示单位面积上的热量传递速率与温度差之间的比值。
流体力学与传热练习题2
一、填空题1．相对压强的起量点为；绝对压强的起量点为。
2．当理想流体在水平变径管路中作稳定的连续流动时，在管子直径缩小的地方，其质量通量；势能。
3．当流量及其他条件一定，离心泵的吸入管径增加，则泵的允许安装高度，输送流体的温度提高，则离心泵的允许安装高度。
4．调节往复泵流量的方法有：，等。
5．滤饼过滤是指：的操作。
6．对恒压过滤，介质阻力可以忽略时，过滤面积增大一倍，则过滤速率为原来的。
过滤时间增加一倍，过滤液量增加为原来的倍。
7．列举两种间壁式换热器：、等。
（任填两种）8．为了减少保温瓶的热损失，在瓶胆的夹层中抽真空是为了减少形式的热损失；在瓶胆的夹层中镀水银是为了减少形式的热损失。
二、选择题1．当管子由水平放置改为垂直放置，而流速不变，其能量损失( ) 。
A. 增大B. 减小C. 不变D. 不定2．流体静压强P的作用方向为（）。
A．指向受压面B．垂直指向受压面C．垂直受压面D．平行受压面3．流体在圆形直管中流动时，若流动处于层流区，则下列四种论述中不正确的是( )。
A．摩擦系数与速度成反比，阻力损失与速度成正比B．管内速度分布呈抛物体状C．阻力损失与流过的管长成正比，与管径的平方成反比D．摩擦系数与速度成正比，阻力损失与速度成正比4．重力沉降和离心沉降的理论依据是：（）A．颗粒与流体的密度差B．含尘气体中颗粒的浓度C．颗粒自身的重量D．颗粒的直径大小5．为提高离心泵的允许安装高度，以下哪种措施是不当的？（）A．提高流体的温度B．增大离心泵吸入管的管径C．缩短离心泵吸入管的管径D．减少离心泵吸入管路上的管件6．以下哪一措施不利于提高转筒真空过滤机的生产能力？（）A．提高转速B．增加浸没度C．降低悬浮液的温度D．提高真空度7．有一套管换热器，在内管中空气从20C︒被加热到50C︒，环隙内有119.6C︒的水蒸汽︒。
冷凝，管壁温度接近（）CA．35 B．119.6 C．77.3 D．无法判断8．翅片管加热器一般用于（）。
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。
流体力学课后习题与解答
流体力学课后习题与解答1.1 按连续介质的概念，流体质点是指：（）（a ）流体的分子；（b ）流体的固体颗粒；（c ）几何的点；（d ）几何尺寸同流动空间相比是极小量，又含有大量分子的微元体。
1.2 作用于流体的质量力包括：（）（a ）压力；（b ）摩擦阻力；（c ）重力；（d ）表面力。
1.3 单位质量力的国际单位是：（）（a ）N ；（b ）Pa ；（c ）kg N /；（d ）2/s m 。
1.4 与牛顿摩擦定律直接有关的因素是：（）（a ）剪应力和压强（b ）剪应力和剪应变率（c ）剪应力和剪应变（d ）剪应力和流速 1.5 水的动力黏度μ随温度的升高：（）（a ）增大；（b ）减小；（c ）不变；（d ）不定。
1.6 流体运动黏度ν的国际单位是：（）（a ）2/s m ；（b ）2/m N ；（c ）m kg /；（d ）2/m s N ?。
1.7 无黏性流体的特征是：（）（a ）黏度是常数；（b ）不可压缩；（c ）无黏性；（d ）符合RT p=ρ。
1.8 当水的压强增加1个大气压时，水的密度增大约为：（）（a ）1/20000；（b ）1/10000；（c ）1/4000；（d ）1/2000。
2.1 静止流体中存在：（）（a ）压应力；（b ）压应力和拉应力；（c ）压应力和剪应力；（d ）压应力、拉应力和剪应力。
2.2 相对压强的起算基准是：（）（a ）绝对真空；（b ）1个标准大气压；（c ）当地大气压；（d ）液面压强。
2.3 金属压力表的读值是：（）（a ）绝对压强（b ）相对压强（c ）绝对压强加当地大气压（d ）相对压强加当地大气压2.4 某点的真空度为65000Pa ，当地大气压为0.1MPa,该点的绝对压强为：（）（a ）65000Pa ；（b ）55000Pa ；（c ）35000Pa ；（d ）165000Pa 。
2.5 绝对压强abs p 与相对压强p 、真空度V p 、当地大气压a p 之间的关系是：（）（a ）abs p =p +V p ；（b ）p =abs p +a p ；（c ）V p =a p -abs p ；（d ）p =V p +V p 。
流体力学习题及答案
题2－40图题2－41图
2－41油库侧壁有一半球形盖，直径为d＝0.6m，半球中心在液面下的淹没深度H＝2.0m，测压管中液面高出油库中液面的高度h＝0.6m，石油重度为6867N/m3，试求液体作用在半球盖上的水平分力及铅垂分力。
2－18U形水银压差计中，已知h1＝0.3m，h2＝0.2m，h3＝0.25m。A点的相对压力为pA＝24.5kPa，酒精的比重为0.8，试求B点空气的相对压力。
题2－18图题2－19图
2－19一直立的煤气管，在底部的测压管中读数为h1＝100mmH2O，在H＝20m高处测得h2＝115mmH2O。管外空气的重度γa＝12.64N/m3，求管中静止煤气的重度。
1－8体积为5m3的水在温度不变的条件下，压力从1大气压增加到5大气压，体积减小了1L，求水的体积压缩系数和弹性系数值。
1－9某液体的动力粘度为0.045Pa·s，其比重为0.85，试求其运动粘度。
1－10某气体的重度为11.75N/m3，运动粘度为0.157cm2/s，试求其动力粘度。
1－11温度为20℃的空气在直径为2.5cm的管道中流动。在距管壁1mm处空气流速为3cm/s，试求：(1)管壁处的切应力；(2)单位管长的粘性阻力。
2－20图示封闭容器中有空气、油和水三种流体，压力表A读数为－1.47N/cm2。(1)试绘出容器侧壁上的静压力分布图；(2)求水银测压计中水银柱高度差。
题2－20图题2－21图
2－21三个U形水银测压计，其初始水银面如图A所示。当它们装在同一水箱底部时，使其顶边依次低下的距离为a＝1m，水银的比重为13.6，试问三个测压计中的读数h1、h2、h3各为多少？
Solns to 2005级流体力学与传热考试题(双语班)(A卷)
(2) There are many industrial filters, give three of them:1) Cake filters (pressure filters, vacuum filters) ; 2) Clarifying filters ;3) crossflow filters (ultrafilter) .(3) The orifice meter belongs to B , C flow meter; while the rotameter is A , D flow meter.A) constant pressure drop; B) constant area; C) variable pressure drop; D) variable area(4) When the filter cake is incompressible, a cake resistance is not influenced by ( D ).A) the cake layer thicknessB) the specific cake resistanceC) the porosity (void fraction)D) the filtration rate(5) 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? ( C )A) increase vapor velocity; B) employ superheated vapor;C) increase air velocity; D) set up the baffles in the shell.(6) The removal of non-condensable gas in the condensing vapor will ( A )A) increase condensing film coefficient.B) decrease condensing film coefficient.C) not change condensing film coefficient.D) be uncertain.(7) Based on the temperature distribution as shown in theattached figure, which layer has the largest heat resistance?A , and which has the largest thermal conductivity?B , if the thickness of each layer is the same.A) layer A; B) layer B; C) layer C(8) When fouling (scale) and pipe wall resistances are ignored and the difference between h i and h o is very huge, overall heat transfer coefficient U is close to ( B ); and wall temperature is close to ( A ) side.A) larger h ; B) smaller h ; C) average value between h i and h o ; D) uncertain (9) For laminar flow, the average velocity in a pipe is 0.5 times the maximum velocity, for turbulent flow average velocity is 0.82~0.87 times the maximum velocity. (10) For laminar flow in a pipe, if the flow rate is constant, when diameter D i increases, the friction coefficient will increase , and friction loss will decrease .3. (15 points) It is proposed to pump 10000 kg/h of toluene at 114℃and 1.1 atm abs pressure from the reboiler of a distillation tower to a second distillation unit without cooling the toluene before it enters the pump. If the friction loss in the line between the reboiler and pump is 7 kN/m 2 and the density of toluene is 866 kg/m 3, how far above the pump must the liquid level in the reboiler be maintained to give a net positive suction head of 2.5 m? Solution.Use the equation for the suction lift H g .m NPSH g h g p p H fs v a g 32.35.280665.9803.8-=--=---'=ρ4. (20 points) A centrifugal pump takes brine from the bottom of a supply tank and delivers it into the bottom of another tank. The brine level in the discharge tank is 45 m above that in the supply tank, and both brine levels are open to the atmosphere. The line between the tanks is 180 m long of 100 mm inner diameter pipe. The flow rate is 90 m 3/h. In the line are 2 gate valves, 4 standard tees, and 4 elbows. What is the power of the motor required for running this pump? The density of brine is 1180 kg/m 3, and the viscosity of brine is 1.2 cP. The overall efficiency of pump and motor is 60 percent.Hint: Loss coefficient for gate valve is 0.17, for elbow is 0.75, for tee is 1.0 For laminar flow, f=16/Re or λ=4f=64/Re For turbulent flow, λ=0.3145/Re 0.25Solution.Bernoulli equation between station a and b becomesb fa b b b p a a ah Vgz p W V gz p -+++=+++2222ρηρ According to the given conditions,2)4()81.945(6.01)(1%60,,0452V K K K D L f h h h g z W p p V V m z z z f e c bfa b fa b fa p b a b a a b ⨯∑+++⨯=+⨯=+⋅∆=∴=====-=∆---ηη Here, from the given conditions,53221013.3102.11180183.31.0Re /183.336001.04904,1.0,180⨯=⨯⨯⨯==∴=⨯⨯⨯====-μρππV D s m D q V m D m LSo it is turbulent flow, and0133.0)1013.3(3145.0425.05=⨯==λf34.70.1475.0417.020.1,4.0=⨯+⨯+⨯=∑==f e c K K KkgJ V K K K D L f h f e c bfa /5.1652183.3)34.70.14.01.01800133.0(2)4(22=⨯+++⨯=⨯∑+++⨯=- kg J h h g z W b fa b fa p /1012)5.16581.945(6.01)81.945(6.01)(1=+⨯=+⨯=+⋅∆=∴--ηPower of motor W W mP p B 3109.2910123600118090⨯=⨯⨯==5. (20 points) In the shell-and-tube heat exchanger with the parallel-current flow, water is used to cool the oil. The temperature of entering water and leaving water are 15 o C and 40 o C, respectively. While the temperature of entering and leaving oil are 150 o C and 100 o C, respectively. Because of production condition changes, the leaving temperature of the oil is to be decreased to 80 o C. Suppose that the mass flow rate, the entering temperatures and the physical properties of fluids are not changed, if the tube length of the original heat exchanger is 1m, how long has the tube to be increased to meet the production requirement? The heat loss of the heat exchanger can be ignored. Solution.The logarithmic mean temperature difference is(15015)(10040)92.513560m t Cn---∆==︒ From the heat balance2112()40150.5()150100h Ph C Pc W C t t W C T T --===-- When the temperature of the oil drops to 80℃, from the heat balance2(15080)(15)h Ph c pc Q W C W C t '=-=-or2150.515080h PhC Pct W C W C '-==-From which 250t C '=︒And (15015)(8050)7013530m t C n ---∆==︒From the heat transfer rate equation, The former heat exchanger:(150100)(92.5)h ph m W C KS t Kn dL π-=∆= The new heat exchanger:(15080)(70)h ph m W C KS t Kn dL π'''-=∆=7092.5, 1.855070hence L m '=⨯=6. (15 points) A slurry is filtered by a filter press of 0.1m 2 filtering area at constant pressure, the equation for a constant pressure filtration is as follows)4.0(250)10(2+=+t qwhere q=filtrate volume per unit filtering area in l/m 2, t= filtering time, in min calculate:(1) how much filtrate will be gotten after 249.6min?(2) If the pressure difference is doubled and both the resistances of the filtration medium and cake are constant, how much filtrate will be obtained after 249.6min?solution:(1) From 22250)4.06.249(250)4.0(250)10(=+=+=+t q We can get 2/240m l q = So filtrate volumel qA V 241.0240=⨯==(2) ∵P rcPK ∆∝∆=μ2 ∴5002==∆'∆='K K pp K And t cons t K Kt q mm m tan 1004.02502=''==⨯== ∴[]min 2.0500100100=='='K t m∴()()m mm t t K t t K q t K q +'≈'+'=+'=+'22)10( ∴222502)10(⨯≈+'q∴2/6.343102250m l q =-=' ∴l A q V 36.341.06.343=⨯='='。
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。
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1.3Solution ：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=％＝7.68.2812638.281-1.4Solution: 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 1.41.5Solution: 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=1.39m1.6Solution :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 ++=++ Rearranging,()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭⎫ ⎝⎛==ρρg h d D u Hg2)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＝1.7Solution :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 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-=232d V L p 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 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．.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 ，then Figure for problem 1.9 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 =-ρtherefore6Substituting equation 5 in equation 3 ，then7ThusR 1=R 21.10solution: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 1.4-19max 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=⨯==fab h g l g R =+--'lg 1ρρρρ）（g R h fab ρρρ-'=1g R h fcd ρρρ-'=2Figure for problem 1.104) kerosene:427003.0800004.04.0Re =⨯⨯==μρud Pa p p 4800001.0003.01600=='∆='∆μμ m g p h 611.081.98004800=⨯=''∆='ρSolution ：(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 bFigure for problem 1.122222_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/sthe 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=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 Solution ： The variables of main pipe are denoted by a subscript1, and branch pipe by subscript2.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 Solution:Writing mechanical energy balance equation between the inlet 1 and throat of o o h V p V p ++=+222211ρρ 1 rearranging the equation abovef o oh 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()ff f f oh R h Rh V h V V p p +=+=+=+-=-94.1185.203.1903.19206.3922121211ρ4from the hydrostatic equilibriumg R p p O H Hg o )(21ρρ-=- 5substituting equation 5 for pressure difference into equation 4 obtainsf OH O H Hg h R gR +=-94.118)(22ρρρ 6rearranging equation 6kg J R R R R gR h OH O H Hg f /288.267.494.11861.12394.118)(22==-=--=ρρρ1.17.Solution: a)2.0501010==D D =⨯-=-=-81.9)1300013600(1.0)(21g R p p Hg ρρs kg V D m /187.013000183.001.0442220=⨯⨯⨯==πρπb) approximate pressure drop=⨯-=-=-81.9)1300013600(1.0)(21g R p p Hg ρρ588.6Pa2.1Solution: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=-Figure for problem 1.17()s m p p D D C V o /183.030.061.00906.061.01300081.9)1300013600(1.022.0161.021*******=⨯=≈⨯-⨯-=-⎪⎪⎭⎫ ⎝⎛-=ρ(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 ，m³/h 26 Total heads ，m 18.41 Shaft power ，kW 2.45 Efficiency ，%53.12.2Solution:Equation(1.6-9)sm Rg D d C V f /12.444.69375.062.01000)100013600(81.9168.025025162.02144000=⨯=-⨯⨯⎪⎭⎫ ⎝⎛-=-⎪⎭⎫⎝⎛-=ρρρ）（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 .. Solution :From an energy balance,WhereP 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 ?s m D d V V /12112.42200=⎪⎭⎫⎝⎛⨯=⎪⎭⎫ ⎝⎛=m g u d l fH f 1.581.92105.0200025.02422=⨯⨯==NPSH H gp p H f vog ---=ρmNPSH H gp p H f vo g 55.335.181.9100081.913600)05.012.0-=--⨯⨯⨯-=---=（ρ• 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=⨯⨯⨯==-μρudfrom 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 equation1.4-25kgJ 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 25×10-3 Pa Friction factor 32.0Re 500.00056.0+=f for hydraulically smooth pipeSolution: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=+=+=f 92.4681.9232.3025.0600087.04222=⨯⨯==∆==g u d l g p h H f λρΔp=46.92×1840×9.81=847.0kpa2.6solution ：∑+++=+++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⨯⨯3.2Solution:The gravity settling is followed Stocks ’ law, so maximum diameter of particle settled can be calculated from Re that is set to 11Re ==μρt c t u d , thenρμc t d u =equation 3.2-16 for the terminal velocityμρρρμ18)(2g d d S c c -=solving for critical diameter32)(224.1gd S c ρρμ-=Check up the appendixThe density of 20℃ air ρ=1.205 kg/m³ and viscosity µ=1.81×10-3N ·s/m 2mm d c μ3.571073.5205.1)205.12650()1081.1(224.15323=⨯=⨯-⨯=--when Reynolds number ≥1000, the flow pattern follows Newton ’s law and terminal velocity can be calculated by equation 3.2-19()ρρρ-=p p t gd u 75.1 1critical Reynolds number is 1000Re ='=μρt ct u d ， 2rearranging the equation 2 givesρμct d u '=1000 3 combination of equation 1 with equation 3ρρρρμg d d S c c )(74.11000-'=' solving for critical diameter32)(3.32μρρρ-='S cdumm d c151210512.1205.1)205.12650()1081.1(3.323323=⨯=⨯-⨯='--3.3solution:to calculate terminal velocity from the equation 3.2-16()μρρ182g d u p pt -=The density of 21℃ air ρ=1.205 kg/m³ and viscosity µ=1.81×10-5N ·s/m 2()s m g d u p pt /181.01081.11881.9)205.12403()1050(185262=⨯⨯⨯-⨯-=--＝μρρt BLu Q = so286.20181.0605.226m u Q BL t =⨯==1 from equation3.3-4tu H u L = the maximum permissible velocity of the air is 3m/s181.03HL =H L 58.16= 2set B to be 3m , then from equation 1 L =7m And H =0.42m3.4 Solution: D =0.4m B =D /4=0.1m h =D /2=0.2ms m hB Q u i /9.131.02.036001000=⨯⨯==According to the equation3.3-12 for N=5：m m u N B d i p c μπρρπμ81089.13)02300(5)1.0)(106.3(9)(965=⨯=⨯-⨯≈-=--3.6Solution:The equation for the constant-pressure filtrationt KA VV V m 222=+5min .1l 51.02122⨯=+K V m 10min .6.1l 101.06.126.122⨯⨯=⨯+K V msolving the equations above for V m and K7.0=m V and K =48For min 15=t 151.0487.0222⨯⨯=⨯⨯+V VSolving for V =2.073 l3.7 The following data are obtained for a filter press of 0.0093 m 2 filtering area in the test(1) filtration constant K , V m at the pressure difference of 1.05(2) if the frame of the filter is filled with the cake at 660s, what is the final rate of filtration Edt dV ⎪⎭⎫ ⎝⎛ (3) and what is the compressible constant of cake n ?solution:①from equation 3.4-19aKt qq q m =+22 For pressure difference 05.1=p ㎏/㎝2500093.01027.220093.01027.2323⨯=⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 1 6600093.0101.920093.0101.9323⨯=⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 2 solving the equations 1 and 2 gives s m K mm q m /1056.10379.02323-⨯==② )(22m E V V KA dt dV +=⎪⎭⎫ ⎝⎛=s m /1014.736-⨯For pressure difference 5.3=p ㎏/㎝271.10093.01027.220093.01027.2323⨯'='⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 3 2330093.0101.920093.0101.9323⨯'='⨯⨯+⎪⎪⎭⎫ ⎝⎛⨯--K q m 4 solving the equations 3 and 4 gives s m K m m q m /1037.40309.02323-⨯='='n p p -⎪⎪⎭⎫ ⎝⎛∆'∆=K K '1 then n ---⎪⎭⎫ ⎝⎛=⨯⨯13305.15.31056.11037.433.3ln 8.2ln 1=-n solving for n =0.1423.8solution:(1)22250)4.06.249(250)4.0(250)10(=+=+=+t qq=240 l/m 2V=24 l (2) the pressure difference is double2=∆'∆='pp K K soK ´=500210102=⎪⎪⎭⎫ ⎝⎛++'q q 2/5.3421025041.110)10(2m l q q =-⨯=-+='V=34.25 lSolution:Area of filtration: A =2×0.32×20=3.6m 2Δp =248.7kN/m 2t 1=300s, V 1=1/4×0.7=0.175m 3t 2-t 1=1800s, ΔV=0.7-0.175=0.525m 3or t 2 =2100s, V 2=0.7m 3KV andKt KA V m m 21006.37.027.03006.3175.02175.022222⨯=⨯+⨯==⨯+V m =0.2627m 3 And 522101517.3210096.122627.04.149.021006.34.17.0-⨯=⨯⨯+=⨯+=m V K capacitys m Q /10692.250021007.034-⨯=+=for the rotary drum filterArea of filtration: A =πdL =3.14×2.2×1.5=10.362m 2Operating pressure:Δp =70kN/m 2Andthe drum submerged : φ=25%For rotary drum filter the filtration coefficient: keep V m unchanged, K changes with changing in pressure difference6510871.87.24870101517.3--⨯=⨯=∆'∆='p p K K the capacity of rotary drum filter is equal to that of press filter⎪⎪⎭⎫ ⎝⎛-+⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛-+'==⨯--2627.02627.025.0362.1010871.810692.2226224n n V V n A K n nV m m ϕSolving for n from the equation above by trial and error⎪⎪⎭⎫ ⎝⎛-+=⎪⎪⎭⎫ ⎝⎛-+'==⨯-2627.0069.00002381.010692.2224n n V V n A K n nV m m ϕ n=0.0088=0.048rpm(转/min)Solution:From equation4.2-11 (a) heat loss:22211/89.68838.1229.0138.0114.06.76760m W k B k B T R T A q i i =+-=+∆=∆=∑∑ (b) the temperature at interface:211/89.688138.0114.0760m W t k B T A q =-=∆= t=191℃(c) contact resistance is 0.088°C·m 2/W, heat loss:22211/4.683088.038.1229.0138.0114.06.76760088.0m W k B k B T R T A q i i =++-=++∆=∆=∑∑4.4.Solution ：Similar to compound resistances in series through the flat wall, the total resistance across insulating layer is as followsLL kA B R m 2357.0)426.0/278.1ln(/)426.0278.1(615.0426.0=-⨯==π Heat loss per meter tube through the wall ism W R T L q /7.5892357.038177=-=∆= and temperature distribution is()m W r r t R T L q /7.589213.0ln 213.02615.0177=-⨯-=∆=π()213.0ln213.05.2277177r r t --=4.7. Solution:106001002.006.11002.0Re 3=⨯⨯⨯==-μρdu 692.00289.01002.01000Pr 3=⨯⨯==-k c p μ 6.47863.05.166003324.0692.01060002.00289.0023.0Pr Re 023.04.08.04.08.0=⨯⨯=⨯⨯==i i d k h 4.8. Methyl alcohol flowing in the inner pipe of a double-pipe exchanger is cooled with water flowing in the jacket. The inner pipe is made from 25-mm Schedule 40 steel pipe. The thermalconductivity of steel is 45 W/(m·o C). The individual coefficients and fouling factors are given in the following Table. What is the overall coefficient, based on the outside area of the inner pipe? TABLE Data for Problem 8Coefficient/ W/(m 2·°C)Alcohol coefficient h i 1020Water coefficient h o 1700Inside fouling factor h di 5680Outside fouling factor h do 2840Solution: From equation (4.3-37)od o m o i i o di i o o h h d d k b h d d h d d U 111++++= from appendix 3, the wall thickness is 3.25mm and outside diameter is 33.5mm for nominal diameter 25mm Schedule 40 standard steel pipemm d m 25.302275.33=+=Cm W U o o 244534/4.40710521.310882.510998.710216.110184.21284011700125.305.334500325.01020275.335680275.331=⨯+⨯+⨯+⨯+⨯=+++⨯+⨯=-----4.9.Solution:Outside diameter of inner tube d o =19mm, inside diameter of outer tube Di=32mmThe equivalent diameter of annular space of double tube de ismm d D d D d D d o i o i o i e 131932)()(4422=-=-=+-=ππfrom equation4.08.0023.0⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=k c u d d k h p e e μμρ 1 C t o m 5022080=+= the properties of benzene: µ=0.45×10-3Pas ，c p =1.81kJ/kg.℃，k=0.138W/m ℃，ρ= 850kg/m 3 at average temperature of 50℃andaverage velocity of benzene flowing through the annular space of double tube is calculated as followu d D mo i )(422-=πρ()s m d D m u o i /13.1019.0032.0850360018004)(42222=-⨯⨯⨯=-=ππρ Substituting the variables into equation 1 givesC)·W/(m 17809.527748244.0138.01045.01081.11045.085013.1013.0013.0138.0023.0023.024.08.04.0338.034.08.0︒⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯⎪⎭⎫ ⎝⎛⨯⨯⨯⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫ ⎝⎛=--＝＝k c u d d kh p e e μμρ4.13Solution:From the appendix 6, we obtain the latent heat of the saturated vapor, 2256.7kJ/kg at 100 o C.The average temperature of the condensate film=(100+92)/2=96℃The properties of the saturated water at 96 o C are as follows:µ=0.282×10-3Pas(from appendix 8); ρ=958kg/m 3; k=0.68W/m.K(from appendix5)from equation (4.5-12)()())/(5471101327.1943.010282.05.292100107.225681.995868.0943.0943.0241154133234123C m W L T g k h o f o f f ⋅=⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯⨯-⨯⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛∆=-μλρ()h kg s kg t t hA m w s /8.20/005785.0107.2256921005.2038.05471)(3==⨯-⨯⨯⨯=-=πλ Check the flow pattern:)/(04848.0038.014.3005785.0ms kg d m o =⨯==Γπ 68810282.004848.044Re 3=⨯⨯=Γ=-μ(laminar flow) If the tube is placed horizontally, then from equation (4.5-14)4123729.0⎪⎪⎭⎫ ⎝⎛∆='f o o f f d T g k h μλρ then2.2848.2773.0038.05.2773.0943.0729.04141=⨯=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛='od L h h h´=2.2h=2.2×5471=12036.2W/m 2℃hence,hh m m '=' m´=2.2×20.8=45.8kg/h4.14.Solution: (1)From equation (4.3-37)3001212345002.070011923111=++=++=o o m o i i o o h h d d k b h d d U 300110868.4001729.015=+⨯+-ohh o =642.8 W/m 2℃(2) write energy balance equationm o o m o o p t L d U t A U t t mc ∆=∆=-π)(12 1logarithmic mean temperature difference for countercurrent flow∆t 1=150-80=70 o C ; ∆t 2=90-20=70 o C so ∆t m =70 o Ccalculate the length of double tube from equation 1m t d U t t mc L m o o p 5.570023.03003600)2080(1000500)(12=⨯⨯⨯-⨯=∆-=ππ (3)∆t 1=140-80=60 o C ; ∆t 2=140-20=120 o C so2121ln t t t t t m ∆∆-=∆ m t t d U mc t t t t d U t t mc t d U t t mc L o o p o o p m o o p 4.412060ln 023.030036001000500ln ln )()(2121211212=⨯⨯⨯-=∆∆=∆∆--=∆-='ππππ(4) the scale depositing on the surfaces of the tube to give additional thermal resistances4.15.Solution:From equation (4.3-37), neglecting thermal resistance of tube wall giveso i o h h U 111+=when the velocity of water is 1m/s, the individual coefficient of water to wall of tubes is h i , overall coefficient2115111=+=o i o h h U 1when the velocity of water is 1.5m/s, the individual coefficient of water to wall of tubes is h ´i , and h ´i related to h i by equation(4.4-25)383.15.18.08.0==⎪⎭⎫ ⎝⎛'='u u h h i iand overall coefficient is26601383.111111=+=+'=o i o i o h h h h U 2solving equation 1 and 2 for h i and h o h i =2858 W/( m 2⋅ o C ); h o =8135 W/( m 2⋅ o C )4.22Solution :(1)Write energy balance equation211212ln )(t T t T t t A U t A U t t mc o o m o o p ---=∆=- sopo o mc A U t T t T =--21ln set U o =h i because the thermal resistance from saturated vapor to wall of tube and resistance of wall are very small and negligible 608.0100025.1203817030170ln 2=⨯⨯=--t 837.1100025.12038170301702=⨯⨯=--t t 2=93.8 o Cthe rate of condensate for one condenser ()s kg t t mc m p h /0388.010002054308.93100025.1)(121=⨯-⨯=-=λ sum of condensate for two condensers in parallel m h =2m h1=2×0.0388=0.0776kg/s(2) two condensers operate in series heat transfer coefficient between the air and wall of tubes 74.128.08.0==⎪⎭⎫ ⎝⎛'=⎪⎪⎭⎫ ⎝⎛'='u u h h U U i i o oU ´o =1.74U o =1.74×38=66.12 W/m 2⋅K. po o mc A U t T t T 2ln21⨯'='-- so 058.110005.24012.6617030170ln 2=⨯⨯='--t 88.2170301702='--t t´2=121.4 o C()s kg t t c m m p h /111.010*********.12110005.2)(12=⨯-⨯=-''=λ。