1.1 What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m
below the surface? ρwater = 1000 kg/m 3, and P atmosphere = 101kN/m 2.
Solution:
Rearranging the equation 1.1-4
gh p p a b ρ+=
Set the pressure of atmosphere to be zero, then the gauge pressure at depth 12m below the surface
is
kPa gh p p a b 72.1171281.910000=⨯⨯+=+=ρ
Absolute pressure of water at depth 12m
kPa Pa gh p p a b 72.2182187201281.91000101000==⨯⨯+=+=ρ
1.3 A differential manometer as shown in Fig. is sometimes ud to measure small pressure
dopc
过犹不及在线阅读difference. When the reading is zero, the levels in two rervoirs are equal. Assume that fluid B is
methane （甲烷）, that liquid C in the rervoirs is kerone (specific gravity = 0.815), and that
liquid A in the U tube is water. The inside diameters of the rervoirs and U tube are 51mm and
6.5mm , respectively. If the reading of the manometer is145mm., what is the pressure difference
over the instrument In meters of water, (a) when the change in the level in the rervoirs is
neglected, (b) when the change in the levels in the rervoirs is taken into account? What is the
nasdaqpercent 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=0.145m
When the pressure difference between two rervoirs is incread, the volumetric changes in the rervoirs and U tubes
R d x D 224
4
ππ
= (1) so R D d x 2⎪⎭
上海洋泾中学
⎫ ⎝⎛= (2) and hydrostatic equilibrium gives following relationship
g R g x p g R p A c c ρρρ++=+21 (3)
袼褙so
g R g x p p c A c )(21ρρρ-+=- (4)
maskedsubstituting 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 rervoirs 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 rervoirs 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 gaous 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
Figure for problem 1.4
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
fire flyc 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
1.5 Water discharges from the rervoir 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 rervoir 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 rervoir, tank A and the exit of
drainpipe are all open to air.
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
2
donatella2u Hg =
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
Solving for H
H=1.39m
1.6 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-ctional area A 1 m 2 at a pressure p 1 N/m 2, and then it pass to a ction of the pipe in which the area is reduced gradually to A 2 m 2 and the pressure is p
iu怎么读2. Assuming no friction loss, 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,
222
200u u p =+ρ
()＝＝＝144.281.92100081.910002125.11112442-⨯⨯⨯--⎪⎭
⎫ ⎝⎛==ρρg h d D u Hg
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,
2)1(212121
21-=-A A V p p ρ ⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛-12
221211A A p p V ρ＝
北京德语培训班Performing the same derivation but in terms of V 2,
⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛--21221212
A A p p V ρ＝
1.7 A liquid who 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 2
32d V μ. Oil of viscosity 0.05 Pas flows through a pipe of diameter 0.1m with a average velocity of 0.6m/s.
