Chapter 6, Problem 6.1P

Interpretation Introduction

Interpretation: The maximum length of the pipe for different inlet conditions is to be determined.

Concept introduction: The length between the starting and the end point of a pipe is known as the maximum length of the pipe. The maximum length for a given Mach number is given as,

  $\frac{{\text{<mover accent="true"><mo>f </mo><mo>¯</mo></mover> L}}_{\text{max}}}{{\text{r}}_{\text{H}}}\text{=}\frac{\text{1}}{\text{γ}}\left(\frac{\text{1}}{{\text{Ma}}_{\text{a}}{}^{\text{2}}}\text{-}\text{1}\text{-}\frac{\text{γ+1}}{\text{2}}\text{ln}\frac{\text{2}\left\{\text{1+}\left[\text{(γ-1)/2}\right]{\text{Ma}}_{\text{a}}{}^{\text{2}}\right\}}{{\text{Ma}}_{\text{a}}{}^{\text{2}}\text{(γ+1)}}\right)$ ....... (1)

The notations used here are,

  ${\text{Ma}}_{\text{a}}$ = Mach number at the entrance of the pipe

  ${\text{L}}_{\max }$ = Maximum length of the pipe

  ${\text{r}}_{\text{H}}$ = Hydraulic radius of pipe = $\frac{\text{D}}{\text{4}}$

  $\text{γ}$ = Heat capacity ratio

  $\text{<mover accent="true"><mo>f </mo><mo>¯</mo></mover>}$ = Average friction factor

The Reynolds number is given as,

  $\text{Re = }\frac{\text{GD}}{\text{μ}}$ ....... (2)

The notations used here are,

G = Gas Mass velocity

D = Diameter of pipe

  $\text{μ}$ = Viscosity of the fluid

The Mach number is given as,

  $\text{Ma}\text{=}\frac{\text{u}}{\text{a}}\text{=}\frac{\text{Speed}\text{of}\text{fluid}}{\text{Speed}\text{of}\text{sound}}$ ....... (3)

The speed of sound is given as,

  $\text{a}\text{=}\sqrt{\frac{\text{γRT}}{\text{M}}}$ ....... (4)

M = Molecular weight of the air

R = Universal gas constant

T = Temperature of the air

Answer to Problem 6.1P

The maximum length for the inlet conditions of example 6.2 is 2383 ft and the maximum length for the inlet conditions of example 6.3 is 83.17 m.

Explanation of Solution

The data from example 6.2 are,

  $\text{G}\text{=}\text{61}{\text{.61 lb/ft}}^{\text{2}}\text{.s}$ .

  $\frac{{\text{<mover accent="true"><mo>f </mo><mo>¯</mo></mover> L}}_{\text{max}}}{{\text{r}}_{\text{H}}}\text{=}\text{280}$ ....... (5)

The data for air at $\text{540°F}$

  $\text{μ}\text{=}\text{2}\text{.8}\text{×}{\text{10}}^{\text{-5}}$ Pa-s = 0.028 cP

  $\begin{array}{l}\text{D}\text{=}\text{2 inch}\\ \text{1 inch = }\frac{\text{1}}{\text{12}}\text{ft}\\ \text{D}\text{=}\frac{\text{2}}{\text{12}}\text{ft}\text{=}\text{0}\text{.1667}\text{ft}\end{array}$

Substitute the value in equation (2),

  $\text{Re}\text{=}\frac{\left(\text{61}\text{.61}{\text{lb/ft}}^{\text{2}}\text{.s}\right)\text{×(0}\text{.1667}\text{ft)}}{\text{(0}\text{.028}\text{cP)×(6}\text{.72}\text{×}{\text{10}}^{\text{-4}}\text{)}}\text{=}\text{5}\text{.46}\text{×}{\text{10}}^{\text{5}}$

Corresponding to the given Reynolds number the value of f = 0.0049 from chart of Re vs f.

f = $\text{<mover accent="true"><mo>f </mo><mo>¯</mo></mover>}$ = 0.0049

r_H = $\frac{\text{0}\text{.1667}\text{ft}}{\text{4}}\text{=}\text{0}\text{.0417}\text{ft}$

Substitute these values in equation (5),

  $\begin{array}{l}\frac{\text{(0}\text{.0049)}\text{×}{\text{L}}_{\text{max}}}{\text{(0}\text{.0417}\text{ft)}}\text{=}\text{280}\\ {\text{L}}_{\text{max}}\text{=}\frac{\text{280}\text{×0}\text{.0417}\text{ft}}{\text{0}\text{.0049}}\text{=}\text{2383}\text{ft}\end{array}$

The data for the air is,

  $\begin{array}{l}\text{M}\text{=}\text{0}\text{.029}\text{kg/mol}\\ \text{R}\text{=}\text{8}\text{.314}\text{J/mol}\text{.K}\\ \text{T}\text{=}\text{288}\text{K}\\ \text{γ}\text{=}\text{1}\text{.4}\end{array}$

Substitute these values in equation (4),

  $\begin{array}{l}\text{a}\text{=}\sqrt{\frac{\text{(1}\text{.4)}\text{×}\text{8}\text{.314}\text{J/mol}\text{.K}\text{×}\text{288K}}{\text{0}\text{.029}\text{kg/mol}}}\\ \text{a}\text{=}\text{340}\text{m/s}\end{array}$

From example 6.3, the data useful is,

  $\begin{array}{l}\text{q}\text{=}\text{0}\text{.265}{\text{m}}^{\text{3}}\text{/s}\\ \text{A}\text{=}\text{0}\text{.00442}{\text{m}}^{\text{2}}\\ \text{u}\text{=}\frac{\text{q}}{\text{A}}\text{=}\frac{\text{0}\text{.265}{\text{m}}^{\text{3}}\text{/s}}{\text{0}\text{.00442}{\text{m}}^{\text{2}}}\text{=}\text{60}\text{m/s}\end{array}$

Substitute the value of u and a in equation (3),

  $\text{Ma}\text{=}\frac{\text{60}\text{m/s}}{\text{340}\text{m/s}}\text{=}\text{0}\text{.176}$

Other data from example 6.3,

  $\begin{array}{l}\text{f}\text{=}\text{<mover accent="true"><mo>f </mo><mo>¯</mo></mover>}\text{=}\text{0}\text{.0044}\\ {\text{r}}_{\text{H}}\text{=}\text{0}\text{.01875}\text{m}\end{array}$

Substitute the data in equation (1),

  $\begin{array}{l}\frac{\text{(0}{\text{.0044)L}}_{\text{max}}}{\text{(0}\text{.01875}\text{m)}}\text{=}\frac{\text{1}}{\text{1}\text{.4}}\left(\frac{\text{1}}{{\text{(0}\text{.176)}}^{\text{2}}}\text{-}\text{1}\text{-}\frac{\text{1}\text{.4+1}}{\text{2}}\text{ln}\frac{\text{2}\left\{\text{1+}\left[\text{(1}\text{.4}\text{-}\text{1)/2}\right]{\text{(0}\text{.176)}}^{\text{2}}\right\}}{{\text{(0}\text{.176)}}^{\text{2}}\text{(1}\text{.4+1)}}\right)\\ \frac{\text{(0}{\text{.0044)L}}_{\text{max}}}{\text{(0}\text{.01875}\text{m)}}\text{=}\text{0}\text{.7143}\left(\text{32}\text{.283}\text{-1}\text{-}\text{1}\text{.2ln}\frac{\text{2}\text{.0124}}{\text{0}\text{.0743}}\right)\\ \frac{\text{(0}{\text{.0044)L}}_{\text{max}}}{\text{(0}\text{.01875}\text{m)}}\text{=}\text{19}\text{.517}\\ {\text{L}}_{\text{max}}\text{=}\frac{\text{(0}\text{.01875}\text{m)}}{\text{(0}\text{.0044)}}\text{×19}\text{.517}\\ {\text{L}}_{\text{max}}\text{=}\text{83}\text{.17}\text{m}\end{array}$

Conclusion

The maximum length for the inlet conditions of example 6.2 is 2383 ft and the maximum length for the inlet conditions of example 6.3 is 83.17 m.