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
lV215V2?hf,1_2?(?d?Kc)2?(0.025?0.1?0.5)2?2.13V2Z1?
Substitute the known variables equation a
V239630??2.13V2 9.81×6.66=21000the velocity is V =3.13m/s
the flow rate of water is
??Vh?3600?d2V?3600??0.12?3.13?88.5m3/h
44 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 ?4.81V2
input 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，
p3.512?2?26.2 9.81×6.66=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 ??2d22?hf2
In the equation ?2?0.03
l2??le2?10m
d2?0.053
u2?3600?2.72?4?0.343m/s?0.0532input the data into equation c
?hf2100.3432?0.03???0.333J/kg
0.0532The energy loss in the main pipe is
?hf1??hf2l1u12??1?0.333
d12So u1?0.333?0.3?2?2.36m/s
0.018?2The water discharge of main pipe is
?Vh1?3600??0.32?2.36?601m3/h
4Total water discharge is
Vh?601?2.72?603.7m3/h
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 2.5R 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 meter
p1poVo2V12??z1g???z2g?hf 1 ?2?2rearranging the equation above, and set (z2-z1)=x
Figure for problem 1.16
p1?po?Vo2?V12??xg?hf 2
222from continuity equation
?d1Vo?V1??d?o??5?????V1?6.25V1 3 ??2??substituting equation 3 for Vo into equation 2 gives
p1?po?39.06V12?V12??xg?hf?19.03V12?hf?19.032.5R2?118.94R?xg?hf??2?xg?hf 4
from the hydrostatic equilibrium for manometer
p1?po?R(?Hg??)g?x?g 5
substituting equation 5 for pressure difference into equation 4 obtains
R(?Hg??)g?x?g?R(?Hg??)g?118.94R?xg?hf 6
rearranging equation 6
hf???118.94R?123.61R?118.94R?4.67R?2.288J/kg
1.17.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/m3
Solution: a)
D010??0.2 D150p1?p2?R(?Hg??)g?0.1(13600?1300)?9.81? V2?
Co?D01???D?1????42?p1?p2???0.611?0.242?0.1(13600?1300)?9.811300?0.6118.56?0.61?4.31?2.63m/s
44b) approximate pressure drop m??D02V2????0.012?2.63?1300?0.268kg/s
p1?p2?R(?Hg??)g?0.1(13600?1300)?9.81?12066.3Pa
pressure difference due to increase of velocity in passing through the orifice
22?Do???V?V22?22D1?V2?V12.632(1?0.24)??p1?p2?????1300?4488.8Pa
222 pressure drop caused by friction loss
4?pf?12066.3?4488.8?7577.5Pa
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 26m3/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. Solution:
Write the mechanical energy balance equation between the suction connection and discharge connection
2u12p1u2p Z1???H?Z2??2?Hf,1_2
2g?g2g?gwhere
Z2?Z1?0.4m
p1??2.47?104Pa(gaugepressure）p2?1.52?105Pa(gaugepressure）u1?u2Hf,1_2?0