From velocity profile equation for laminar flow 2??p?pr??0 L 2 u ? 2 ?R1??????4?L ??R??substituting equation 2 for u into equation 1 and integrating p0?pL2V?D 32 ? L 3
rearranging equation 3 gives ?p32?V? Ld2
32?VL32?0.05?0.6?120 ?p???11520Pa22 d0.1
1.8. In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m below A and when the flow rate down the pipe is 0.02 m3/s, the pressure at B is 14715 N/m2 greater than that at A.
Assuming the losses in the pipe between A and B
V2can be expressed as kwhere V is the velocity
2gFigure for problem 1.8
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:
dA=0.15m; dB=0.075m zA-zB=l=2.5m Q=0.02 m3/s,
pB-pA=14715 N/m2
Q??42dAVAVA?Q?4?2dA 0.02?1.132m/s0.785?0.152
Q??42dBVBVB?Q?42dB 0.02??4.529m/s0.785?0.0752When the fluid flows down, writing mechanical balance equation
pA22VApBVB2VA ?zAg???zBg??k?2?221.132147154.5321.1322.5?9.81????k
210002224.525?0.638?14.715?10.260?0.638k
k?0.295
making the static equilibrium
pB??x?g?R?g?pA?l?g??x?g?R?HggR??pB?pA??l?g??Hg??g??14715?2.5?1000?9.81??79mm
12600?9.81
1.9．The liquid vertically flows down through the tube from the station a to the station b, then horizontally through the tube from the station c to the station d, as shown in figure. Two segments of the tube, both ab and cd，have the same length, the diameter and roughness. Find:
（1）the expressions of
?pab?pcd, hfab, and hfcd,
?g?gFigure for problem 1.9
respectively.
（2）the relationship between readings R1and R2 in the U tube.
Solution:
(1) From Fanning equation
lV2 hfab??d2
and
lV2 hfcd??d2
so
hfab?hfcdFluid flows from station a to station b, mechanical energy conservation gives papb? h ? lg ? fab
??
hence
p?p a b ? lg ? h fab 2 ?
from station c to station d pp c?d?hfcd??
hence p?pcd? h fcd 3 ?
From static equation pa-pb=R1（ρˊ-ρ）g -lρg 4 pc-pd=R2（ρˊ-ρ）g 5 Substituting equation 4 in equation 2 ，then R?（1???）g?l?g?lg?hfab
?
therefore
???? h fab ? R 1 g 6
?
Substituting equation 5 in equation 3 ，then
???? ? 7 h fcdR 2 g?
Thus R1=R2
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 it occurs.
3) Find the point r at which the average velocity equals the local velocity. 4）if kerosene flows through this pipe，how do the variables above change？
（the viscosity and density of Water are 0.001 Pas and 1000 kg/m3，respectively；and the viscosity and density of kerosene are 0.003 Pas and 800 kg/m3，respectively）
solution: 1）Re?ud?L r Figure for problem 1.10
??0.4?0.004?1000?1600
0.001from Hagen-Poiseuille equation
32uL?32?0.4?2?0.001?P???1600 22d0.004h??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
uumax?r??1???r??
?w?22V?0.5umaxV?r?1???0.5 ?＝?0.004?umaxr?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?9.81
1.12 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