7.44: PROBLEM DEFINITION
Situation:
Apumpfills a tank with water from a river.
Dtank =5m,Dpipe =5cm.
hL=10V2
2/2g,hp=20−4×104Q2.
Find:
Time required to fill tank to depth of 10 m.
Assumptions:
α=1.0.
SOLUTION
Energy equation (locate 1 on the surface of the river, locate 2 on the surface of the
water in the tank).
where V2/2gistheheadlossduetotheabruptexpansion.Then
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But Q=ATdh/dt where AT=tank area, so
But t=0when h=0so const. =−2(18)0.5.Then
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7.45: PROBLEM DEFINITION
Situation:
A pump transfers SAE-30 oil between two tanks.
Dtank =12m,Dpipe =20cm.
hL=20
V2
2g,hp=60m.
zA=20m,zB=1m.
Find:
Time required to transfer oil (h).
PLAN
Apply the energy equation between the top of the fluid in tank A to that in tank B.
SOLUTION
Energy equation
Solve for velocity
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Continuity equation
Separate variables
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7.46: PROBLEM DEFINITION
Situation:
A pump is used to pressurize a tank.
Dtank =2m,Dpipe =4cm.
hL=10
V2
2g,hp=50m.
zA=20m,zB=1m.
pT=3
4−ztp0,p0=0kPagage =100kPa.
Find:
Write a computer program to show how the pressure varies with time.
Time to pressurize tank to 300 kPa ( s) .
PLAN
Apply the energy equation between the water surface at the intake and the water
surface inside the tank.
SOLUTION
Energy equation
Expressing the head loss in terms of the velocity allows one to solve for the velocity
in the form
Substituting in values
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A computer program can be written taking time intervals and finding the fluid level
and pressure in the tank at each time step. The time to reach a pressure of 300 kPa
300
350
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7.47: PROBLEM DEFINITION
Situation:
Water is flowing in a horizontal pipe.
D=0.15 m,L=60m.
V=2m/s.h
L=2m.
Find:
Pressure drop (Pa).
Pumping power (W).
Properties:
Water (15 ◦C), Table A.5: γ=9800N/m3.
PLAN
1. Find pressure drop using the energy equation.
2. Find power using the power equation.
SOLUTION
1. Energy equation:
2. Power equation:
˙
Wp=γQhp=˙mghp
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REVIEW The pump would need to supply about 0.9 hp to the water.
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7.48: PROBLEM DEFINITION
Situation:
A pump supplies energy to a flowing fluid
Q=3ft
3/s,α=1.0.
DA=1.0ft,p
A=5psig.
DB=0.5ft,p
B=55psig.
Find:
Horsepower delivered by pump (hp).
PLAN
Apply the flow rate equation, then the energy equation from A to B. Then apply the
power equation.
SOLUTION
Flow rate equation:
Energy equation:
Power equation:
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7.49: PROBLEM DEFINITION
Situation:
A pump moves water from a tank through a pipe.
Q=8m
3/s,hL=7V2/2g.
D=1m.
Find:
Power supplied to flow (MW).
Assumptions:
α=1.0.
PLAN
Find power using the power equation. The steps are
1. Find velocity in the pipe using the flow rate equation.
2. Find head of the pump using the energy equation.
3. Calculate power.
SOLUTION
1. Flow rate equation
2. Energy equation (locate 1 on the reservoir surface; locate 2 at the out of the pipe).
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3. Power equation
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7.50: PROBLEM DEFINITION
Situation:
A subsonic wind tunnel is being designed.
A=4m
2,V=60m/s.
hL=0.025V2
T
2g.
Find:
Power required (kW).
Assumptions:
α=1.0.
Properties:
ρ=1.2kg/m3.
PLAN
To find power, apply the power equation. The steps are
1. Find the head of the pump by applying theenergyequationandthecontinuity
equation together.
2. Calculate power.
SOLUTION
1. Finding head of the pump
•Energy equation
•Continuity principle
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•Combining previous two equations
2. Power equation
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7.51: PROBLEM DEFINITION
Situation:
A pumping system delivers water.
zA=117ft,zB=154ft.
zC=110ft,zD=90ft.
A=0.1ft
2.
Find:
Power delivered by pump (hp).
SOLUTION
Power equation
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7.52: PROBLEM DEFINITION
Situation:
A pumping system delivers water.
zA=40m,zB=65m.
zC=35m,zD=30m.
A=25cm
2.
Find:
Power delivered by pump (kW).
PLAN
Apply the energy equation from the reservoir water surface to point B. Then apply
the power equation.
SOLUTION
Energy equation
Flow rate equation
Power equation
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7.53: PROBLEM DEFINITION
Situation:
Water is pumped from a reservoir through a pipe.
Q=0.35 m3/s,hL=2
V2
2
2g.
z1=6m,z2=10m.
D1=40cm,D2=10cm.
p2=100kPa.
Find:
Power (kW) that the pump must supply.
Assumptions:
α=1.0.
Properties:
Water (10 ◦C), Table A.5: γ=9810N/m3.
PLAN
Apply the flow rate equation, then the energy equation from reservoir surface to the
10 m elevation. Then apply the power equation.
SOLUTION
Flow rate equation:
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Power equation:
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7.54: PROBLEM DEFINITION
Situation:
Oil is pumped through a pipe attached to a manometer filled with mercury.
Q=6ft
3/s,∆h=46in.
D1=12in,D2=6in.
Find:
Horsepower pump supplies ( hp).
Assumptions:
α=1.0.
Properties:
Table A.5: Smercury =13.55.
Soil =0.88
PLAN
Apply the flow rate equation, then the energy equation. Then apply the power
equation.
SOLUTION
Flow rate equation
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Energy equation
Power equation
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