1.52?105?0.247?105?18.41m total heads of pump is H?0.4?1000?9.81efficiency of pump is ??Ne/N
QH?g26?18.41?1000?9.81??1.3kW 36003600N=2.45kW
Then mechanical efficiency
1.3?100%?53.1% ??2.45The performance of pump is
Flow rate ，m3/h 26 Total heads，m 18.41 Shaft power ，kW 2.45 Efficiency ，% 53.1
2.2 Water is transported by a pump from reactor, which has 200 mm Hg vacuum, to the tank, in
2which the gauge
pressure is 0.5 kgf/cm2,
10mas shown in Fig. The total equivalent length
1of 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 coefficient ? 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/s ?0.9375
Mass flow rate
since Ne?
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 m?VoSo??4.12?H?p2?p1??z?Hf ?g200?1.013?105?75707Pa 760?z=10m
?p?0.5?9.81?104??p/?g=7.7m
The relation between the hole velocity and velocity of pipe
22 ?d0??1?V?V0???4.12????1m/s ?2??D?Friction loss
lu220012?0.025??5.1m Hf?4fd2g0.052?9.81 so
H=7.7+10+5.1=22.8m
2.3 . A centrifugal pump is to be used to extract water from a condenser in which the vacuum is 640 mm of mercury, as shown in figure. At the rated discharge, the net positive suction head must be at least 3m above the cavitation vapor pressure of 710mm mercury vacuum. If losses in the suction pipe accounted for a head of Hg 1.5m. What must be the least height of the liquid level in the condenser above the pump inlet? Solution:
From an energy balance,
p?pv?Hf?NPSH Hg?o?g
Where
Po=760-640=120mmHg Pv=760-710=50mmHg
Use of the equation will give the minimum height Hg as
po?pv?Hf?NPSH Hg??g
（0.12?0.05)?13600?9.81 ??1.5?3??3.55m1000?9.81
2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half, what will the new flowrate be ?
? Density of acid 1840kg/m3 ? Viscosity of acid 25×10-3 Pas
Solution:
Velocity of acid in the pipe:
mu?volumetricflowratem3?????3.32m/s22?cross?sectionalareaofpipe0.785?d0.785?1840?0.025d24d?u0.025?1840?3.32?6109
25?10?3Reynolds number:
Re???from Fig.1.22 for a smooth pipe when Re=6109, f=0.0085 pressure drop is calculated from equation 1.4-9
lu2603.322hf??4f?4?0.0085?450J/kg
?d20.0252?p?p?450?1840?827.5kPa
or friction factor is calculated from equation1.4-25
2lu2603.322?0.2lu?0.2hf??4f?4?0.046Re＝4?0.046?6109?426J/kg?d2d20.0252?p?p?426?1840?783.84kPa
if the pressure drop falls to 783.84/2=391.92kPa
????plu2?0.2?p???391920?4?0.046Re?＝4?0.046??????2d2???1840??4?0.046?1840??3??25?10??0.21.8?0.2lu1.8?1.22d
60u1079.891.8?`u1.220.0120.025so
u?1.8391920?0.0121.8?4.36?2.27m/s
1079..89new mass flowrate=0.785d2uρ=0.785×0.0252×2.27×1840=2.05kg/s
2.4 Sulphuric acid is pumped at 3 kg/s through a 60m length of smooth 25 mm pipe. Calculate the drop in pressure. If the pressure drop falls by one half on assumption that the change of friction factor is negligible, what will the new flowrate be ?
Density of acid 1840kg/m3 Viscosity of acid 25×10-3 Pa Friction factorf?0.0056?0.500 for hydraulically smooth pipe Re0.32Solution:
Write energy balance equation:
2p1u12p2u2?z1??H??z2??hf ?g2g?g2g?plu2 H?hf????gd2g?4d2u??3
u?3?412??3.32m/s 22?d?3.14?0.025?18403.32?0.025?1840?6115 ?325?10Re?f?0.0056?0.5000.5?0.0056??0.0087 0.320.32Re6115?plu2603.322H?hf????4?0.0087?46.92
?gd2g0.0252?9.81Δp=46.92×1840×9.81=847.0kpa
2.6 The fluid is pumped through the horizontal pipe from section A to B with the φ38?2.5mm diameter and length of 30 meters, shown as figure. The orifice meter of 16.4mm diameter is used to measure the flow rate. Orifice coefficient Co＝0.63. the permanent loss in pressure is 3.5×104N/m2, the friction coefficient λ=0.024. find:
(1) What is the pressure drop along the pipe AB?
(2)What is the ratio of power obliterated in pipe AB to total power supplied to the fluid when the shaft work is 500W, 60?ficiency? (The density of fluid is 870kg/m3 )
solution：
22uApAuAzAg???w?zAg????hf
?2?2pApA?pB?lu2?p0 ??hf???d2?2Ao?16.4?????0.247 A?33?u0?C01?0.24722gR??????0.632?9.81?0.6?13600?870???8.5m/s
??0.97870∴u= (16.4/33)2×8.5=2.1m/s
302.12?3.5?104?76855N/m2 ∴pA?pB???hf?0.024?8700.0332(2)
Ne?Wm??p?2du??76855?0.785?0.0332?2.1?138W ?4so
the ratio of power obliterated in friction losses in AB to total power supplied to the fluid 138?100％＝46％
500?0.6
3.2 A spherical quartzose particle（颗粒） with a density of 2650 kg/m3 settles（沉淀） freely in the 20℃ air, try to calculate the maximum diameter obeying Stocks’ law and the minimum diameter obeying Newton’s law.
Solution:
The gravity settling is followed Stocks’ law, so maximum diameter of particle settled can be calculated from Re that is set to 1
Ret?dcut???1, then
ut?
? dc?