?p?0.5?9.81?104?200?1.013?105?75707Pa 760?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 1.5m. What must be the least height of the liquid Hg level in the condenser above the pump inlet? Solution:
From an energy balance,
p?pv Hg?o?Hf?NPSH?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/scross?sectionalareaofpipe?20.785?d20.785?1840?0.0252d4d?uReynolds number:
Re???0.025?1840?3.32?6109 ?325?10from 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
?0.2????plu2?0.2??p???391920?4?0.046Re?＝4?0.046????2d2????1840??4?0.046?1840??3??25?10?so
?0.2lu1.8?1.22d
60u1.81079.891.8?`u0.0120.0251.22u?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?Solution:
0.500 for hydraulically smooth pipe
Re0.32Write energy balance equation:
2p1u12p2u2?z1??H??z2??hf ?g2g?g2g?plu2 H?hf????gd2g?4d2u??3
3?412??3.32m/s ?d2?3.14?0.0252?1840u?Re?3.32?0.025?1840?6115 ?325?10f?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
4
3.5×10N/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??????0.63???0.972?9.81?0.6?13600?870??8.5m/s
870∴u= (16.4/33)2×8.5=2.1m/s
302.12∴pA?pB???hf?0.024?870?3.5?104?76855N/m2
0.0332(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?equation 3.2-16 for the terminal velocity
2dc(?S??)g?? dc?18?solving for critical diameter
dc?1.2243?2(?S??)g
Check up the appendix
The density of 20℃ air ρ=1.205 kg/m3 and viscosity μ=1.81×10-3N·s/m2