流体力学与传热习题：solutions

1.3

Solution ：

p a =1000kg/m 3 p c =815kg/m 3 p b =0.77kg/m 3 D/d=8 R=0.145m

When the pressure difference between two reservoirs is increased, the volumetric changes in the reservoirs and U tubes

R d x D 2244ππ

= (1) so R D d x 2⎪⎭

⎫ ⎝⎛= (2) and hydrostatic equilibrium gives following relationship

g R g x p g R p A c c ρρρ++=+21 (3)

so

g R g x p p c A c )(21ρρρ-+=- (4)

substituting the equation (2) for x into equation (4) gives

g 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

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 equilibrium

g R R g R g R p g Hg O H A )(32232+-+=ρρρ

g ρis small and negligible in comparison with Hg ρand ρH2O , equation above can be simplified

c A p p ≈=232gR gR Hg O H ρρ+

=1000×9.81×0.05+13600×9.81×0.05

=7161N/m²

1gR p p p Hg A D B ρ+=≈=7161+13600×9.81×0.4=60527N/m

Figure for problem 1.4

1.5

Solution: Bernoulli equation is written between stations 1-1 and 2-2, with station 2-2 being reference plane: 2

222222111u gz p u gz p ++=++ρρ Where p 1=0, p 2=0, and u 1=0, simplification of the equation

1

The relationship between the velocity at outlet and velocity u o at throat can be derived by the continuity equation:

22⎪⎭

⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛D d u u o 22⎪⎭

⎫ ⎝⎛=d D u u o 2 Bernoulli equation is written between the throat and the station 2-2

3 Combining equation 1,2,and 3 gives

222u Hg =222

200u u p =+ρ

Solving for H

H=1.39m

1.6

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,

2

112A A V V = For the items in the Bernoulli equation , for a horizontal pipe,

z 1=z 2=0

Then Bernoulli equation becomes, after substituting 2

112A A V V = for V 2, ρρ2212121

1212020p A A V p V ++=++ Rearranging,

()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭

⎫ ⎝⎛==ρρg h d D u Hg