FLUID MECHANICS:FUND.+APPL.(LL)>CUSTOM<
FLUID MECHANICS:FUND.+APPL.(LL)>CUSTOM<
3rd Edition
ISBN: 9781260244342
Author: CENGEL
Publisher: MCG CUSTOM
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Chapter 2, Problem 55CP
To determine

The percentage error caused by Boussinesq approximation.

Expert Solution & Answer
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Answer to Problem 55CP

The percentage error caused by Boussinesq approximation is 6%.

Explanation of Solution

Given information:

The pressure of the air is 95kPa, the initial temperature is 20°C, the final temperature is 60°C, and the midway temperature is 40°C.

Write the expression for the Boussineq approximation for 20°C.

  ρ20=ρ0[1β(TT0)]   ..... (I)

Here, the density of fluid at 20°C is ρ20, the density of air is ρ0, the temperature is T, the midway temperature is T0 and the coefficient of volume expansion is β.

Write the expression for the Boussineq approximation for 60°C.

  ρ60=ρ0[1β(TT0)]   ..... (II)

Here, the density of fluid at 60°C is ρ60.

Write the expression for the pressure at 20°C using ideal gas law.

  P=(ρ 20)aRT   ..... (III)

Here, the pressure is P the actual density at 20°C is (ρ 20)a.and the universal gas constant is R.

Write the expression for the pressure at 60°C using ideal gas law.

  P=(ρ 60)aRT   ..... (IV)

Here, the actual density at 60°C is (ρ 60)a.

Write the expression the percentage error at 20°C.

  %error=(ρ 20 ( ρ 20 )a ( ρ 20 )a)×100   ..... (V)

Write the expression the percentage error at 60°C.

  %error=(ρ 60 ( ρ 60 )a ( ρ 60 )a)×100   ..... (VI)

Calculation:

Convert the value of midway temperature in Kelvin.

  T0=40°C=(40°C+273.15)K=313.15K

Refer Boussinesq approximation data at 313.5K to obtain the values of ρo as 1.127kg/m3 and β as 3.2×103/K.

Substitute 1.127kg/m3 for ρ0, 3.2×103/K for β, 20°C for T and 313.5K for T0 in Equation (I).

  ρ20=(1.127kg/ m 3)[1( 3.2× 10 3 /K)(20°C313.15K)]=(1.127kg/ m 3)[1( 3.2× 10 3 /K)(( 20°C+273.15)K313.15K)]=1.199kg/m3

Substitute 1.127kg/m3 for ρ0, 3.2×103/K for β, 60°C for T and 313.5K for T0 in Equation (II).

  ρ60=(1.127kg/ m 3)[1( 3.2× 10 3 /K)(60°C313.15K)]=(1.127kg/ m 3)[1( 3.2× 10 3 /K)(( 60°C+273.15)K313.15K)]=1.054kg/m3

Substitute 95kPa for P, 0.287×103J/kgK for R and 20°C for T in Equation (III).

  95kPa=( ρ 20)a×(0.287× 103J/kgK)×(20°C)( ρ 20)a=( 95kPa× 1000N/ m 2 1kPa )( ( 0.287× 10 3 J/ kgK )× 1Nm 1J )×( 20°C+273.15)K( ρ 20)a=1.129kg/m3

Substitute 95kPa for P, 0.287×103J/kgK for R and 60°C for T in Equation (IV).

  95kPa=( ρ 60)a×(0.287× 103J/kgK)×(60°C)( ρ 60)a=( 95kPa× 1000N/ m 2 1kPa )( ( 0.287× 10 3 J/ kgK )× 1Nm 1J )×( 60°C+273.15)K( ρ 60)a=0.993kg/m3

Substitute 1.199kg/m3 for ρ20 and 1.129kg/m3 for (ρ 20)a in Equation (V).

  %error=( 1.199 kg/ m 3 1.129 kg/ m 3 1.129 kg/ m 3 )×100=0.06×100=6%

Substitute 1.054kg/m3 for ρ60 and 0.993kg/m3 for (ρ 60)a in Equation (VI).

  %error=( 1.054 kg/ m 3 0.993 kg/ m 3 0.993 kg/ m 3 )×100=0.06×100=6%

Conclusion:

The percentage error caused by Boussinesq approximation is 6%.

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Chapter 2 Solutions

FLUID MECHANICS:FUND.+APPL.(LL)>CUSTOM<

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