1.10 Water passes through a pipe of diameter di=0.004 m with the average velocity 0.4 m/s, as shown in Figure.

1) What is the pressure drop –?P when water flows through the pipe length L=2 m, in m H2O column?

2) Find the maximum velocity and point r at which

L it occurs.

3) Find the point r at which the average velocity equals the local velocity.

r 4）if kerosene flows through this pipe，how do the variables above change？

（the viscosity and density of Water are 0.001 Pas and 1000 kg/m3，respectively；and the viscosity

Figure for problem 1.10 and density of kerosene are 0.003 Pas and 800

kg/m3，respectively）

solution: 1）Re?ud???0.4?0.004?1000?1600

0.001from Hagen-Poiseuille equation

?P?h?32uL?32?0.4?2?0.001??1600 d20.0042?p1600??0.163m ?g1000?9.812）maximum velocity occurs at the center of pipe, from equation 1.4-19

so umax=0.4×2=0.8m

3）when u=V=0.4m/s Eq. 1.4-17

V?0.5umaxuumax?r?1???r?w2??? ?2V?r?1???0.5 ?＝0.004u??maxr?0.0040.5?0.004?0.71?0.00284m

4) kerosene:

Re?ud???0.4?0.004?800?427

0.003?p???p??0.003?1600?4800Pa ?0.001h??

?p?4800??0.611m ??g800?9.811.12 As shown in the figure, the water level in the reservoir keeps constant. A steel drainpipe (with the inside diameter of 100mm) is connected to the bottom of the reservoir. One arm of the U-tube manometer is connected to the drainpipe at the position 15m away from the bottom of the reservoir, and the other is opened to the air, the U tube is filled with mercury and the left-side arm of the U tube above the mercury is filled with water. The distance between the upstream tap and the outlet of the pipeline is 20m.

a) When the gate valve is closed, R=600mm, h=1500mm; when the gate valve is opened partly, R=400mm, h=1400mm. The friction coefficient λ is 0.025, and the loss coefficient of the entrance is 0.5. Calculate the flow rate of water when the gate valve is opened partly. (in m3/h)

b) When the gate valve is widely open, calculate the static pressure at the tap (in gauge pressure, N/m2). le/d≈15 when the gate valve is widely open, and the friction coefficient λ is still 0.025.

Figure for problem 1.12

Solution：

(1) When the gate valve is opened partially, the water discharge is Set 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

2u12p1u2pgZ1???gZ2??2??hf,1—2 （a）

2?2?In the equation p1?0(the gauge pressure)

p2??HggR??H2Ogh?13600?9.81?0.4?1000?9.81?1.4?39630N/m2

u1?0Z2?0

When 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).

?HOg(Z1?h)??HggR

2 （b）

where h=1.5m

R=0.6m

Substitute the known variables into equation b

13600?0.6?1.5?6.66m1000 22lV15V?hf,1_2?(?d?Kc)2?(0.025?0.1?0.5)2?2.13V2Z1?Substitute the known variables equation a

V2396309.81×6.66=??2.13V2

21000the velocity is V =3.13m/s

the flow rate of water is

Vh?3600??4d2V?3600??4?0.12?3.13?88.5m3/h

2） 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

V32p3V12p1gZ1???gZ3????hf,1—3 （c）

2?2?since Z1?6.66m

Z3?0

u1?0 p1?p3l?leV2?hf,1_3?(?d?Kc)235V2 ?[0.025(?15)?0.5]

0.12 ?4.81V2input the above data into equation c，

V2?4.81V2 9.81?6.66?2the velocity is: V=3.51 m/s

Write mechanical energy balance equation between thestations 1—1’ and 2——2’, for the same situation of water level

V12p1V22p2gZ1???gZ2????hf,1—2

2?2?（d）

since Z1?6.66m

Z2?0

u1?0u2?3.51m/s

p1?0(page pressure）?h

f,1_2lV2153.512?(??Kc)?(0.025??0.5)?26.2J/kg

d20.12input the above data into equation d，

p3.5129.81×6.66=?2?26.2

21000the pressure is: p2?32970

1.14 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 is 2.72m3/h, try to calculate the flow rate in the main pipe and the total flow rate, respectively. The frictional coefficient of the main pipe and the attached-pipe is 0.018 and 0.03, respectively. Solution： The variables of main pipe are denoted by a subscript1, and branch pipe by subscript 2.

The friction loss for parallel pipelines is

?hf1??hf2Vs?VS1?VS2

The energy loss in the branch pipe is

2l2??le2u2??2

d22?hf2In the equation ?2?0.03