FLUID MECHANICS FUND. (LL)-W/ACCESS
FLUID MECHANICS FUND. (LL)-W/ACCESS
4th Edition
ISBN: 9781266016042
Author: CENGEL
Publisher: MCG CUSTOM
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Textbook Question
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Chapter 8, Problem 94P

Repeat Prob. 8-93 for cast lion pipes of the same diameter.

Expert Solution
Check Mark
To determine

(a)

The electric power consumption of the system for pumping of water.

Answer to Problem 94P

The electric power consumption of the system for pumping of water is 8126.25kW.

Explanation of Solution

Given information:

The temperature of the water is 110°C, density of the water is 950.6kg/m3, dynamic viscosity of the water is 0.255×103kg/ms, diameter of the pipe is 60cm, roughness of the cast iron pipe surfaces is 2.6×104m, specific heat capacity of water is 4.229kJ/kg°C, pump-motor efficiency is 80%, length of the pipe is 12km and volume flow rate of the water is 1.5m3/s.

Write the expression for the Reynolds number.

  Re=ρvDμ........... (I)

Here, Reynolds number is Re, density of the water is ρ, velocity of the water is v, diameter of the pipe is D and dynamic viscosity of the water is μ.

Write the expression for the volume flow rate of the water.

  Q˙=π4D2v........... (II)

Here, the volume flow rate of the water is Q˙.

Write the expression for the friction factor for turbulent flow.

  1f=2log(εD3.7+2.51Ref)........... (III)

Here, friction factor for turbulent flow is f and roughness of the cast iron pipe surfaces is ε.

Write the expression for the head loss through the pipe.

  hf=fLv22gD........... (IV)

Here, length of the pipe is L, gravitational acceleration is g and the head loss through the pipe is hf.

Write the expression for the pressure drop through the pipe.

  ΔP=ρghf........... (V)

Here, the pressure drop through the pipe is ΔP.

Write the expression for the electric power consumption of the system for pumping of water.

  W˙pump=Q˙ΔP........... (VI)

Here, electric power consumption of the system for pumping of water is W˙pump.

Write the expression for the Bernoulli Equation.

  P1ρg+v122g+Z1+hpump=P2ρg+v222g+Z2+hf........... (VII)

Here, pressure at point 1 is P1, velocity at point 1 in v1, density of the fluid is ρ, gravitational acceleration is g, elevation of point 1 is Z1, pressure at point 2 is P2, elevation of point 2 is Z2, velocity at point 2 in v2 and head loss in pump is hpump.

Write the expression for the motor-pump efficiency.

  ηpump=W˙pumpW˙electric........... (VIII)

Here, the motor-pump efficiency is ηpump.

Calculation:

Substitute 1.5m3/s for Q˙ and 60cm for D in Equation (II).

  (1.5 m 3/s)=π4(60cm)2v(1.5 m 3/s)=π4(( 60cm)[ 1m 100cm ])2vv=5.305m/s

Substitute 950.6kg/m3 for ρ, 5.305m/s for v, 0.255×103kg/ms for μ and 60cm for D in Equation (I).

  Re=( 950.6 kg/ m 3 )( 5.305m/s )( 60cm)( 0.255× 10 3 kg/ ms )=( 950.6 kg/ m 3 )( 5.305m/s )( 60cm)[ 1m 100cm]( 0.255× 10 3 kg/ ms )=1.187×107

Since value of the Reynolds number is greater than 4000, therefore flow is turbulent.

Substitute 1.187×107 for Re, 2×106m for ε and 60cm for D in Equation (III).

  1f=2log( ( 2.6× 10 4 m ) ( 60cm ) 3.7+ 2.51 ( 1.187× 10 7 ) f )12f=log( ( 2.6× 10 4 m ) ( 60cm )[ 1m 100cm ] 3.7+ 2.51 ( 1.187× 10 7 ) f )f=0.0162

Substitute 0.0162 for f, 12km for L, 5.305m/s for v, 9.81m/s2 for g and 60cm for D in Equation (IV).

  hf=( 0.0162)( 12km) ( 5.305m/s )22( 9.81m/ s 2 )( 60cm)=( 0.0162)( 12km)[ 1000m 1km] ( 5.305m/s )22( 9.81m/ s 2 )( 60cm)[ 1m 100cm]=465m

Substitute 465m for hf, 950.6kg/m3 for ρ and 9.81m/s2 for g in Equation (V).

  ΔP=(950.6kg/ m 3)(9.81m/ s 2)(465m)=(4334× 103kgm/ s 2)(1m 2)[1N1 kgm/ s 2 ][1Pa1N/ m 2 ][1kPa 10 3Pa]=4334kPa

Substitute 1.5m3/s for Q˙ and 4334kPa for ΔP in Equation (VII).

  W˙pump=(1.5 m 3/s)(4334kPa)=(1.5 m 3/s)(4334kPa)[ 10 3Pa1kPa][1N/ m 2 1Pa][1J1Nm][1W1J/s]=(6501× 103W)[1kW1000W]=6501kW

Substitute 6501kW for W˙pump and 80% for ηpump in Equation (VIII).

  80%=6501kW W ˙ electric(80%)[ 10 21%]=6501kW W ˙ electricW˙electric=8126.25kW

Conclusion:

The electric power consumption of the system for pumping is 8126.25kW.

Expert Solution
Check Mark
To determine

(b)

The daily cost of the power consumption of the system.

Answer to Problem 94P

The daily cost of the power consumption of the system is $11701.8day1.

Explanation of Solution

Calculation:

The unit cost of the electricity is $0.06kWh1.

Write the expression for the amount of daily cost of the power consumption.

  A=ctW˙electric........... (IX)

Here, amount of daily cost of the power consumption is A, unit cost is c and time is t.

Substitute $0.06kWh1 for c, 8126.25kW for W˙electric and 24hr/day for t in Equation (IX).

  A=($0.06 kWh 1)(24hr/day)(8126.25kW)=$11701.8day1

Conclusion:

The daily cost of the power consumption of the system is $11701.8day1.

Expert Solution
Check Mark
To determine

(c)

The frictional heating during flow will make up drop in temperature or not.

Answer to Problem 94P

The frictional heating during flow will make up drop in temperature.

Explanation of Solution

Given information:

The drop-in temperature of geothermal water during flow is 0.5°C.

Write the expression for the rise in temperature of geothermal water.

  W˙pump=ρQ˙CPΔT........... (X)

Here, rise in temperature of geothermal water during flow is ΔT.

Calculation:

Substitute 950.6kg/m3 for ρ, 6501kW for W˙pump, 1.5m3/s for Q˙ and 4.229kJ/kg°C for CP in Equation (X).

  6501kW=(950.6kg/ m 3)(1.5 m 3/s)(4.229kJ/kg°C)ΔT(6501kW)[1 kJ/s1kW]=(950.6kg/ m 3)(1.5 m 3/s)(4.229kJ/kg°C)ΔTΔT=1.078°C

Since rise in temperature of geothermal water due to frictional heating is greater than the drop in temperature of geothermal water during flow, therefore the frictional heating during flow will make up drop in temperature.

Conclusion:

The frictional heating during flow will make up drop in temperature.

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Chapter 8 Solutions

FLUID MECHANICS FUND. (LL)-W/ACCESS

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