Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 12, Problem 59P
To determine

The kinetic energy correction factor for this flow.

Expert Solution & Answer
Check Mark

Explanation of Solution

Given:

The velocity profile for turbulent flow in a circular pipe u(r) is umax(1rR)1/n

The value of (n)  is 9.

Calculation:

Draw velocity profile for turbulent flow in a circular pipe as in Figure 1.

Fundamentals of Thermal-Fluid Sciences, Chapter 12, Problem 59P

Show the area of circular pipe (A) using the formula;

  A=πr2

Differentiate the above equation with respect to r.

  dAdr=2πrdA=2πrdr

Determine the correction factor (α)  using the relation;

  α=1AA(u(r)Vavg)3dA=1AVavg3A(u(r)3)dA

  α=1AA(u(r)Vavg)3dA=1AVavg3A(u(r)3)dA

Substitute πR2 for A and umax(1rR)1/n for u(r).

  α=1πR2Vavg3A(umax(1rR)1/n)3dA=umax3πR2Vavg3r=0R((1rR)3/n)dA

Substitute 2πrdr for dA.

  α=umax3πR2Vavg3r=0R((1rR)3/n)2πrdr

  α=2umax3R2Vavg3r=0R((1rR)3/n)rdr        (1)

Calculate the average velocity (Vavg) using the relation;

  Vavg=1AAu(r)dA=1πR2r=0R(umax(1rR)1/n)2πrdr

  Vavg=2umaxR2r=0R(1rR)1/nrdr        (2)

Using the integral table formula.

  (a+bx)nxdx=(ax+bx)n+2b2(n+2)a(a+bx)n+1b2(n+1)

Calculate the value of r=0Ru(r)rdr using integral table formula;

  r=0Ru(r)rdr=r=0R(1rR)1nrdr=[(1rR)1n+21R2(1n+2)(1rR)1n+11R2(1n+1)]r=0R=n2R2(n+1)(2n+1)

Substitute n2R2(n+1)(2n+1) for r=0R(1rR)1/nrdr in the equation (2).

  Vavg=2umaxR2n2R2(n+1)(2n+1)=2umaxn2(n+1)(2n+1)

Substitute 9 for n.

  Vavg=2umax(9)2(9+1)(2(9)+1)=0.8167umax

Calculate the value of r=0Ru(r)3rdr using integral table formula;

  r=0Ru(r)3rdr=r=0R(1rR)3nrdr=[(1rR)3n+21R2(3n+2)(1rR)3n+11R2(3n+1)]r=0R=n2R2(n+3)(2n+3)

Substitute n2R2(n+3)(2n+3) for r=0R((1rR)3/n)rdr in the equation (1).

  α=2umax3R2Vavg3n2R2(n+3)(2n+3)=2umax3R2(2n2R2(n+1)(2n+1))3n2R2(n+3)(2n+3)=(n+1)3(2n+1)34n4(n+3)(2n+3)=(9+1)3(2(9)+1)34(9)4(9+3)(2(9)+3)=1.04

Thus, the kinetic energy correction factor for this flow is 1.04_.

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Chapter 12 Solutions

Fundamentals of Thermal-Fluid Sciences

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