l2??le2?10m
d2?0.053
u2?3600?2.72?4?0.343m/s?0.0532input the data into equation c
?h?hf2100.3432?0.03???0.333J/kg
0.0532l1u12??1?0.333
d12The energy loss in the main pipe is
f1??hf2So u1?0.333?0.3?2?2.36m/s
0.018?2The water discharge of main pipe is
Vh1?3600?
?4?0.32?2.36?601m3/h
Total 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
p1?po?Vo2?V12??xg?hf 2
2Figure for problem 1.16
from continuity equation
?d1Vo?V1??d?o??5?????V1?6.25V1 3 ??2??22substituting 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?
Co2?p1?p2?0.612?0.1(13600?1300)?9.81V?? 24?13001?0.24??D0 1???D?? ?1? ?0.6118.56?0.61?4.31?2.63m/s
m??42D0V2???4?0.012?2.63?1300?0.268kg/s
b) approximate pressure drop
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
V22?V12p1?p2??2?DoV22?V22??D?1??2???24??13002.63(1?0.2)?4488.8Pa
24 pressure drop caused by friction loss
?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??H?Z2??2?Hf,1_2 Z1?2g?g2g?gwhere
Z2?Z1?0.4m
p1??2.47?104Pa(gaugepressure）p2?1.52?105Pa(gaugepressure）u1?u2Hf,1_2?01.52?105?0.247?105?18.41m total heads of pump is H?0.4?1000?9.81efficiency of pump is ??Ne/N since Ne?
QH?g26?18.41?1000?9.81??1.3kW 36003600N=2.45kW
Then mechanical efficiency
??1.3?100%?53.1% 2.45
26 18.41 2.45 53.1
The performance of pump is
Flow rate ，m3/h Total heads，m Shaft power ，kW Efficiency ，%
2.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in which the gauge pressure is 0.5 kgf/cm2, as shown in Fig. The total equivalent length of pipe is 200 m including all local frictional loss. The pipeline is ?57×3.5 mm , the orifice coefficient of Co and orifice diameter do are 0.62 and 25 mm, respectively. Frictional
210m1coefficient ? is 0.025. Calculate: Developed head H of pump, in m (the reading R of U pressure gauge in orifice meter is 168 mm Hg)
Solution:
Equation(1.6-9)
2Rg（?f??）C00.622?0.168?9.81(13600?1000) V??044?1000 ?d0??25?1???1???
?50??D?
0.62 ??6.44?4.12m/s0.9375
Mass flow rate
m?VoSo??4.12?3.14?0.0252?1000?2.02kg/s 42) Fluid flow through the pipe from the reactor to tank, the Bernoulli equation is as follows for V1=V2
H?p2?p1??z?Hf ?g?z=10m