Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 7, Problem 62P

An incompressible fluid of density ρ and viscosity μ flows at average speed V through a long, horizontal section of round pipe of length L, inner diameter D, and inner wall roughness height ε (Fig. P7-62). The pipe is long enough that the flow is fully developed, meaning that the velocity profile does not change down the pipe. Pressure decreases (linearly) down the pipe in order to “push” the fluid through the pipe to overcome friction. Using the method of repeating variables, develop a nondimensional relationship between pressure drop Δ P = P 1 P 2 and the other in the problem. Be sure to modify your Π groups as necessary to achieve established nondimensional parameters, and name them. (Hint: For consistency, choose D rather than L or ε as one of your repeating parameters.)
Answer: Eu = f (Re, ε /D, UD)

Expert Solution & Answer
Check Mark
To determine

The non -dimensional relationship parameters.

The non-dimensional for first pi terms.

The non-dimensional for second pi terms.

The non- dimensional for third pi terms.

The non-dimensional for fourth pi terms.

Answer to Problem 62P

The non -dimensional parameter for first pi terms is Euler number.

The non- dimensional parameter for second pi terms is Reynolds number.

The non -dimensional parameter for third pi terms is aspect ratio.

The non -dimensional parameter for fourth pi terms is roughness ratio.

The non-dimensional relationship is Eu=f(Re,LD,εD).

Explanation of Solution

Given information:

A homogenous wire with a mass per unit length is 0.056kg/m.

Write the expression for the moment of inertia of the link 3.

  (Iy)3=(Iy)3+ml2   .....(I)

Here, the moment of inertia of the link 3 is (Iy)3, the centroidal component of the moment of inertia of link 3 is (Iy)3, the mass is m and the length of link 3 is l.

Write the expression for the moment of inertia of the link 4.

  (Iy)4=(Iy)4+m(x¯2+z¯2)   .....(II)

Here, the moment of inertia of the link 4 is (Iy)4, the centroidal component of the moment of inertia of link 4 is (Iy)4, the mass is m and the length of link 4 is l.

Write the expression for the centroidal component.

  (Iy)4=ml212

Write the expression for the moment of inertia of the link 5.

  (Iy)5=(Iy)5+m(x¯2+z¯2)   .....(III)

Here, the moment of inertia of the link 5 is (Iy)5, the centroidal component of the moment of inertia of link 5 is (Iy)5, the mass is m and the length of link 5 is l.

Write the dimension of the diameter of the pipe in FLT unit system.

  D=[L]

Here, the dimension for diameter of the pipe is [L].

Write the dimension of the length of pipe in FLT unit system.

  L=[L]

Here, the dimensions for length of the pipe is [L].

Write the dimension of the height of pipe in FLT unit system.

  ε=[L]

Here, the dimension for the height of the pipe is [L].

Write the expressions for the density.

  ρ=mv   .....(I)

Here, the mass is m, the volume is v and the density is ρ.

Substitute [M] for m and [L3] for v in Equation (I).

  ρ=[M][L3]=[ML3]

Write the expression for the pressure.

  ΔP=FA   .....(II)

Here, the pressure is ΔP, the force is F and the area is A.

Substitute [MLT2] for F and [L2] for A in Equation (II).

  ΔP=[MLT 2][L2]=[ML1T2]

Write the dimension for the viscosity.

  μ=[ML1T1]

Write the dimension for the velocity.

  V=[LT1]

Write the expression for the number of pi-terms.

  n=kr   .....(III)

Here, the number of variable is k, the number of dimensions is r and the number of pi terms is n.

Write the expression for first pi terms.

  Π1=ΔPDaρbVc    .....(IV)

Here, the constant are a, c and b.

Write the dimension for pi term.

  Π=L0T0

Write the expression for second pi terms.

  Π2=μDaρbVc    .....(V)

Write the expression for third pi terms.

  Π3=LDaρbVc    .....(VI)

Write the expression for fourth pi terms.

  Π4=εDaρbVc    .....(VII)

Write the expression for relation between the pi terms.

  Π1=f(Π2,Π3,Π4)   .....(VIII)

Calculation:

The number of variables are 7 and the number of dimensions is 3.

Substitute 7 for k and 3 for r in Equation (III).

  n=73=4

Substitute [L] for D, [M0L0T0] for Π1, [ML1T2] for ΔP, [ML3] for ρ and [LT1] for V in Equation (IV).

  [M0L0T0]=[ML1T2][L]a[ML3]b[LT1]c[M0L0T0]=[ML1T2][La][MbL3b][LcTc][M0L0T0]=[M1+bL1+a3b+cT2c]    .....(IX)

Compare the coefficients of M in Equation (IX).

  1+b=0b=1

Compare the coefficients of T in Equation (IX).

  2c=0c=2

Compare the coefficients of L in Equation (IX).

  1+a3b+c=0   .....(X)

Substitute 2 for c and 1 for b in Equation (XI).

  1+a3(1)2=01+a+32=0a+22=0a=0

Substitute 2 for c and 1 for b and 0 for a in Equation (IV).

  Π1=ΔPD0ρ1V2=ΔPρ1V2=ΔPρV2

The non-dimensional for first pi terms is Euler number.

Substitute [L] for D, [M0L0T0] for Π2, [ML1T1] for μ, [ML3] for ρ and [LT1] for V in Equation (V).

  [M0L0T0]=[ML1T1][L]a[ML3]b[LT1]c[M0L0T0]=[ML1T1][La][MbL3b][LcTc][M0L0T0]=[M1+bL1+a3b+cT1c]    .....(XI)

Compare the coefficients of M in Equation (XI).

  1+b=0b=1

Compare the coefficients of T in Equation (XI).

  1c=0c=1

Compare the coefficients of L in Equation (XI).

  1+a3b+c=0   .....(XII)

Substitute 1 for c and 1 for b in Equation (XII).

  1+a3(1)1=01+a+31=0a+21=0a=1

Substitute 1 for c and 1 for b and 1 for a in Equation (V).

  Π2=μD1ρ1V1=μDρV

The non-dimensional for second pi terms is Reynolds number.

Substitute [L] for D, [M0L0T0] for Π3, [L] for L μ, [ML3] for ρ and [LT1] for V in Equation (VI).

  [M0L0T0]=[L][L]a[ML3]b[LT1]c[M0L0T0]=[L][La][MbL3b][LcTc][M0L0T0]=[MbL1+a3b+cTc]   .....(XIII)

Compare the coefficients of M in Equation (XIII).

  b=0

Compare the coefficients of T in Equation (XIII).

  c=0

Compare the coefficients of L in Equation (XIII).

  1+a3b+c=0......... (XIV)

Substitute 0 for c and 0 for b in Equation (XIV).

  1+a3(0)0=01+a=0a=1

Substitute 0 for c and 0 for b and 1 for a in Equation (V).

  Π3=LD1ρ0V0=LD

The non-dimensional for third pi terms is aspect ratio.

Substitute [L] for D, [M0L0T0] for Π4, [L] for ε, [ML3] for ρ and [LT1] for V in Equation (VII).

  [M0L0T0]=[L][L]a[ML3]b[LT1]c[M0L0T0]=[L][La][MbL3b][LcTc][M0L0T0]=[MbL1+a3b+cTc]    .....(XV)

Compare the coefficients of M in Equation (XV).

  b=0

Compare the coefficients of T in Equation (XV).

  c=0

Compare the coefficients of L in Equation (XV).

  1+a3b+c=0   .....(XVI)

Substitute 0 for c and 0 for b in Equation (XVI).

  1+a3(0)0=01+a=0a=1

Substitute 0 for c and 0 for b and 1 for a in Equation (V).

  Π4=εD1ρ0V0=εD1=εD

The non-dimensional for fourth pi terms is roughness ratio.

Substitute εD for Π4, LD for Π3, Re for Π2 and Eu for Π1 in Equation (VIII).

  Eu=f(Re,LD,εD)

Conclusion:

The non -dimensional parameter for first pi terms is Euler number.

The non- dimensional parameter for second pi terms is Reynolds number.

The non -dimensional parameter for third pi terms is aspect ratio.

The non -dimensional parameter for fourth pi terms is roughness ratio.

The non-dimensional relationship is Eu=f(Re,LD,εD).

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Chapter 7 Solutions

Fluid Mechanics: Fundamentals and Applications

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