MONASH UNIVERSITY DEPARTMENT OF CHEMICAL ENGINEERING
CHE3166 Problem Set: Pumping Liquids
Question 1:
It is required to pump cooling water from a pond to a condenser at a height of 15m above the level of the pond through 100m of pipe of internal diameter 100mm. The pressure at the exit of the condenser should be 30kPa gauge. The pump characteristics are given below:
Discharge 25 50 75 100 (m3/h)
head (m) 23.5 22.5 19.8 15.2
The head loss in the condenser is equivalent to 18 velocity heads based on the pipe diameter. Taking the friction factor to be constant at 0.005 over the range of flow rates used, determine the flow rate, the head developed and the power drawn by the pump assuming it has an overall efficiency of 50%.
Solution:
Applying Equation 9,
éPumphead ù=-ws =H -H +åh +åh êë r e q u i r e d p e r k g o f f l u i d úû g 2 1 f i
H 2 - H 1 = ( p 2 - p 1 ) + ( v 2 2 - v 12 ) + ( z 2 - z 1 ) rg 2g
p =p ;p =p +30´103 (p =atmos.pressure) 1A2A A
f r
v1 = 0 (pond surface is stationary); Assume v2 = v = velocity in pipe. z1 = 0, z2 = 15m
Hence, H2 -H1 =18.06+0.051v2 (metres)
Pipefrictionloss, åhfr = 2fv2L = 2´0.005´v2 ´100 =1.0194v2
gD 9.81´ 0.10
æv2 ö
Loss in fittings (in condenser), åhf i = 18 ´ ç 2g ÷ = 0.9174v2
èø
Therefore, pump head required = - ws = 18.06 + 0.051v2 + 1.0194v2 + 0.9174v2
g
= 18.06 + 1.988v2
CHE3166 Problem Set: Pumping Liquids p. 1
&2
If V& is the volume flow rate in m3/h, then v2 = V and so:
-3 & 2 Pump head required = 18.06 + 2.4867 ´10 V
This expression is plotted on the same coordinates as the pump characteristic curve. The point of intersection of the two curves gives the "duty point", the head and flow rate delivered by the pump operating with this system.
\ -ws =23´9.81=225.6 J/kg
Power delivered by pump = 225.6 x 12.08 W
= 2725 W Powerinputtopump=édeliveredpowerù=2725=5450 W
êë e f f i c i e n c y úû 0 . 5
Duty point
Flow rate (m3/h)
Head (m)
Head (m) Reqd head (m)
Pump head required = 23m; Volume flow rate delivered = 43.5 m3/h Mass flow rate = 12.08 kg/s (cooling water density = 1000 kg/m3)
Pump head required = - w s , where ws is the work done per kg of fluid. g
CHE3166 Problem Set: Pumping Liquids
p. 2
continued.......
Question 2:
Ethanol at 35 ̊C is to be pumped at a rate of 2kg/s from a storage tank, held at atmospheric pressure, through a heat exchanger to a reactor at an elevation of 5m above the discharge level of the pump. The liquid depth in the tank is held constant at 0.2m. The pipe is schedule 40, 1.5 inch nominal size with a roughness of 0.08 mm. The pressure required at the inlet of the reactor is 1.5 bar gauge. The pressure drop across the heat exchanger may be assumed to be 10 velocity heads based on the velocity of fluid in the pipe. Further details are given in the diagram below:
liquid level
Heat exchanger
reactor
5m
0.2m
2m
liquid level storage tank
pump
5m
50m
Select a suitable pump from the performance curves shown on next page. Estimate the pump power consumption. Determine the minimum height of the liquid level in the storage tank above the suction point of the pump required to avoid cavitation.
40mm ID smooth pipe
open gate valve
For ethanol at 35 ̊C: Viscosity
Density
Vapour pressure
9.5 x 10-4 Pas 800 kg/m3 93,300 Pa
CHE3166 Problem Set: Pumping Liquids p. 3
Solution:
Applying Equation 9,
éPumpheadrequiredù -w (p -p) (v2-v2)
= s=2 1+2 1+(z-z)+h+h
ê ú 2 1 åfi åfr ë per kg fluid û g rg 2g
Take position 1 as the level of the liquid in the tank Take position 2 as the point of entry to the reactor.
p =Patm,p=Patm +1.5x105 Pa.Therefore ép2 -p1ù=19.113m 1 2 êë rg úû
Assume that the level in the tank is constant, then: v1 = zero.
Assume velocity of the ethanol entering the reactor is the pipe velocity. Then v2 = v
(z2 -z1)=5-2=3m Friction loss in pipes, å hfr
Total length of straight pipe = (2 - 0.2) + 5 + 50 + 5 = 61.8 m.
Inside diameter of schedule 40 1.5 pipe: D = 40.9 mm (from pipe data table)
velocity of liquid in the pipes = 2
æ p0.0409 2
= 1.903 m/s 4 ÷ø
ö
Reynolds number = 1.903 ´ 800 ´ 0.0409 = 65543 (hence turbulent)
8 0 0 ´ çè 9.5´10-4
From the friction factor chart (with e/D = 0.08/40.9 = 0.002), f = 0.0058 hence friction head loss in straight pipes, å hfr = 2 fv 2 L = 6.43m
Friction loss in fittings/equipment, åhfi
2 x 90 degree bends: (2 @ 0.75) = 1.5 velocity heads
gD
1 open gate valve:
1 tank exit (smooth): Heat exchanger:
(1 @ 0.25) = 0.25 velocity heads (1 @ 0.50) = 0.5 velocity heads
(1@10) = 10 velocity heads
Total loss in fittings/equipment , åhfi = 12.25 velocity heads based on pipe diameter
=12.25´v2 =2.26m 2g
Totalpumpheadrequiredperkgfluid= -ws =19.11+0.185+3+6.43+2.26= g
31.62 m
Thus, we need to select a pump which will develop a differential head of at least 32m at a flow rate of 9 m3/h (2 kg/s) of ethanol.
Pump D will deliver a head of 33m at 9 m3/h. We will therefore select pump D.
CHE3166 Problem Set: Pumping Liquids p. 4
3 flow rate (m /h)
Efficiency
Pump D
Pump C
Pump B
Pump A
Head developed (m)
NSPH(R) (m)
Performance curves for pumps A, B, C and D
Power requirement:
Pump power requirement per kg fluid = - ws
where, ws is the work done per kg of the fluid.
The pump will actually deliver a head of 33m at the required flowrate. The extra head would be dissipated by using a valve to trim the flow rate.
therefore, -ws = 33 x 9.81 = 323.7 J/kg
\ delivered power = 323.7 x 2 = 647.5 W (mass flow rate = 2 kg/s)
From the performance curve, the pump efficiency is about 72%, so the actual power
consumed by the pump = 647.5 = 899.3 W 0.72
Cavitation:
To avoid cavitation:
Total head at suction point ³ NPSH(R) + vapour pressure head
CHE3166 Problem Set: Pumping Liquids p. 5
g
Total head at suction point = total head at the liquid level in the supply tank less the friction losses in pipes and fittings between the supply tank and the suction point.
i.e.
p1 +v12 +z-åhfr -åhfit ³NPSH(R)+Pv rg 2g rg
Where Pv is the liquid vapour pressure.
Assuming that the level in the supply tank is constant, v1 = 0. Hence:
p1 +z-åhfr -åhfit ³NPSH(R)+Pv rg rg
p1 =1.013x105 Pa, Pv =93,300Pa
Friction losses in the line to the pump ( åhfr , åhfit ): Length of straight pipe = 5 + (z – 0.2) = (4.8 + z) m
From first part of the problem, velocity = 1.903 m/s, Re = 65543 and f = 0.0058 Hence, åhfr = 2fv2L = 0.1047(4.8 + z) m
gD
Friction loss in fittings, åhfi
1 x 90 degree bends: (1 @ 0.75) = 0.75 velocity heads
1 tank exit (smooth): (1 @ 0.50) = 0.5 velocity heads
Total loss in fittings/equipment , åhfi = 1.25 velocity heads based on pipe diameter
=1.25´v2 =0.185m 2g
From the pump performance curve, NPSH(R) = 1.9 m Hence:
p1 +z-åhfr -åhfit ³NPSH(R)+Pv becomes: rg rg
1.013 ´105 + z - 0.1047(4.8 + z)- 0.185 ³1.9 + 93300 800 ´ 9.81 800 ´ 9.81
Andso: z³1.80m
The minimum height of the liquid level above the suction point of the pump is 1.8m
(less than the specified height of 2m, so the present design is acceptable).
CHE3166 Problem Set: Pumping Liquids p. 6
Question 3: Consider this pump curve (from
Part A: The pump in question has this pump curve, and is operating at about 1750 RPM.
The fluid being pumped is an organic chemical solvent with SG of about 900 kg/m3 The flowmeter shows 140 cubic metres per hour.
The inlet pressure gauge shows 3.2 barg
The discharge pressure gauge shows 4.5 barg Each pressure gauge has an accuracy of +/- 0.1 bar
1. What impellor is being used in this pump? (8)
Solution: ΔP=ρgh → h=(4.5-3.2)*105/(900*9.81)=14.7 m
From graph intersection of head=14.7m, Q=140m3/h gives impeller size = 8 inch
2. How much brake power is being delivered to the pump? (About 8Hp or about 6kW)
Solution: Ws=14.7*9.81=141.26 kJ/kg → Power= Ws*m·/efficiency= 141.26 kJ/kg *35kg/s /0.82~ 6kW
3. Assuming 95% efficiency, how much power is leaving the motor? (8Hp/0.95 = 8.4 Hp or 6.3 kW)
4. Based on the following table, and assuming the motor is a 4-pole motor, and applying the expected motor efficiencies from the table, what is the smallest
CHE3166 Problem Set: Pumping Liquids p. 7
motor that will work for this application? (Table from
(Item 9)
Power= 6.3kW so we have to select item 9 which deliver 7.5 kW
Item
Powe r (HP)
Power (kW)
Energy Efficiency Standard (Percentage)
Enclosed
2 Pole
4 Pole
6 Pole
8 Pole
Part B: Because operations overnight. is offered for auction
problem with the board of directors,
Within the week, all equipment within the now closed company
at very low prices.
of a
the company ceases
CHE3166 Problem Set: Pumping Liquids p. 8
For your project, you buy the pump (but not the motor) for $1. You have to go collect and remove the pump yourself. You have a project where you will use the pump
· Fluid
· Suction pressure
· Flowrate
Non-corrosive brine, SG = 1.3 465 kPag
200 cubic metres /hour
For this application, you will need to buy a 2-pole motor, which will run at nominally 3500 RPM. Again, assume the connection between the pump and the motor operates at 95% efficiency. What is the smallest motor you can buy that will work? (item 14)
Solution: Affinity law
We can use the above graph if we change 1750 rpm to 3500 rpm
So
n1/n2=Q1/Q2 → 200 m3/h in 3500 rpm is corresponding to 100 m3/h in 1750 rpm from graph: Q=100 m3/h , 8 Inch impeller size → ~ 17 m
3500/1750=(x/17 )1/2→ x=68 m
From graph → P= 7.8 HP, normally power curves (shaft power) are based on water density (ρ=1000 kg/m3). We need to correct the shaft power with the density of brine solution (ρ=1300 kg/m3). Actual shaft power needed for brine solution =7.8×1300/1000=10.14 HP
3500/1750=(x/10.14 )1/3 → x~81.1 HP motor efficiency=94.5 → REAL POWER= 81.1/0.945 = 85.8 HP item 18
Part C: The motor we buy does not run at EXACTLY 3500 RPM. The nameplate says 3390 RPM. If we try for the same flowrate as part 2, what is the expected discharge pressure for this pump/motor combination?
(n2/n1)2= h2/h1 = (3390/3500)2=h2/68 → h2=63 m
ΔP=ρgh= 1300*9.81*63=803.44 Kpa
ΔP= Pdischarge-Psuction → = Pdischarge=803.44 + 465 = 1268.44 kpag
���1 ���1 h1 1/2 ���1 1/3
���2 =���2 =(h2) =(���2)
CHE3166 Problem Set: Pumping Liquids p. 9
