Chapter 12, Problem 102P

Air enters a 5-cm-diameter, 4-m-long adiabatic duct with inlet conditions of ${\text{Ma}}_{\text{1}}=\text{ 2}.\text{8,}{T}_{\text{1}}=\text{ 38}0\text{ K}$ , and $P_{\text{1}}=\text{8}0\text{ kPa}$ . It is observed that a normal shock occurs at a location 3 m from the inlet. Taking the average friction factor to be 0.007, determine the velocity, temperature, and pressure at the duct exit

FIGURE P12-102

To determine

Velocity, temperature and pressure at the duct exit.

Answer to Problem 102P

  ${\text{T}}_{\text{4}}\text{=813K}$

  ${\text{P}}_{\text{4}}\text{=328 kPa}$

  ${\text{V}}_{\text{4}}\text{=572 m/s}$

Explanation of Solution

Given:

The properties of air to be,

  $\begin{array}{l}\text{ k = 1}\text{.4}\\ \text{ cp = 1}\text{.005 kJ/kg}\text{.K}\\ \text{ R = 0}\text{.287 kJ/kg}\text{.K}\text{.}\end{array}$

  $\text{M}{\text{a}}_{\text{1}}\text{=2}\text{.8}$

Temperature ${\text{T}}_{\text{1}}\text{=380K}$

Pressure at Inlet ${\text{P}}_1\text{=80kPa}$

The average friction factor is given to be $\text{f = 0}\text{.007}$

Calculation:

The Fanno flow functions corresponding to the inlet Mach number of $2.8$ are, from $\text{Table A-16}$,

  $\begin{array}{l}\text{M}{\text{a}}_1\text{=2}\text{.8}\\ {\left(\frac{\text{fL*}}{\text{Dh}}\right)}_1\text{=0}\text{.4898}\\ \frac{{\text{T}}_1}{\text{T*}}\text{=0}\text{.4673}\end{array}$

First we check that the flow in everywhere shock at upstream is supersonic. The required length of duct from the inlet $L_1*$ for the flow to reach sonic conditions is given by,

  $\begin{array}{l}{\text{L}}_1\text{*=0}\text{.4898}\frac{\text{D}}{\text{f}}\\ {\text{L}}_1\text{*=0}\text{.4898}\frac{\text{0}\text{.05m}}{\text{0}\text{.007}}\\ {\text{L}}_1\text{*=3}\text{.50m}\end{array}$

This is higher than the actual length of $\text{3 m}$.

Therefore, the flow is indeed supersonic when the normal shock appears at the indicated location. Then, using the actual duct length ${\text{L}}_1$,

  $\begin{array}{l}\frac{\text{f}{\text{L}}_{\text{1}}}{{\text{D}}_{\text{h}}}\text{=}\frac{\text{0}\text{.007×3}}{\text{0}\text{.05}}\\ \frac{\text{f}{\text{L}}_{\text{1}}}{{\text{D}}_{\text{h}}}\text{=0}\text{.4200}\end{array}$

Taking the value ${\text{L}}_{\text{1}}={\text{L}}_{\text{1}}*-{\text{L}}_2*,$ the function $\frac{\text{fL*}}{{\text{D}}_{\text{h}}}$ at the exit state and the corresponding Mach number are given by,

  $\begin{array}{l}{\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=}{\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{1}}\text{-}\frac{\text{f}{\text{L}}_{\text{1}}}{{\text{D}}_{\text{h}}}\\ {\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=0}\text{.4898-0}\text{.4200}\\ {\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=0}\text{.0698}\\ \text{M}{\text{a}}_{\text{2}}\text{=1}\text{.315}\end{array}$

From $\text{Table A-16, at Ma2 =1}\text{.315}$,

  $\begin{array}{l}{\text{T}}_{\text{2}}\text{/T \&*\#x00A0;= 0}\text{.8918 }\\ {\text{P}}_{\text{2}}\text{/P \&*\#x00A0;= 0}\text{.7183}\end{array}$

Calculating the temperature, pressure, and velocity before the shock,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{T}}_{\text{2}}\text{/T *}}{{\text{T}}_{\text{1}}\text{/T *}}\text{=}\frac{\text{0}\text{.8918}}{\text{0}\text{.4673}}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=1}\text{.9084}\\ {\text{T}}_{\text{2}}\text{=1}\text{.9084}{\text{T}}_{\text{1}}\\ {\text{T}}_{\text{2}}\text{=1}\text{.9084×380K}\\ {\text{T}}_{\text{2}}\text{=725}\text{.2K}\end{array}$

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}\text{/P *}}{{\text{P}}_{\text{1}}\text{/P *}}\text{=}\frac{\text{0}\text{.7183}}{\text{0}\text{.2441}}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=2}\text{.9426}\\ {\text{P}}_{\text{2}}\text{=2}\text{.9426}{\text{P}}_{\text{1}}\\ {\text{P}}_{\text{2}}\text{=2}\text{.9426×80kPa}\\ {\text{P}}_{\text{2}}\text{=235}\text{.4kPa}\end{array}$

The normal shock functions corresponding to Mach number $1.315$ are, from $\text{Table A-14}$,

  $\begin{array}{l}\text{M}{\text{a}}_{\text{2}}\text{=1}\text{.315}\\ \text{M}{\text{a}}_{\text{3}}\text{=0}\text{.7786}\\ \frac{{\text{T}}_{\text{3}}}{{\text{T}}_{\text{2}}}\text{=1}\text{.2001}\\ \frac{{\text{P}}_{\text{3}}}{{\text{P}}_{\text{2}}}\text{=1}\text{.8495}\end{array}$

Then the temperature after the shock will be,

  $\begin{array}{l}{\text{T}}_{\text{3}}\text{=1}\text{.2001}{\text{T}}_{\text{2}}\\ {\text{T}}_{\text{3}}\text{=1}\text{.2001×725}\text{.2K}\\ {\text{T}}_{\text{3}}\text{=870}\text{.3K}\end{array}$

Pressure after the shock will be,

  $\begin{array}{l}{\text{P}}_{\text{3}}\text{=1}\text{.8495}{\text{P}}_{\text{2}}\\ {\text{P}}_{\text{3}}\text{=1}\text{.8495×235}\text{.4kPa}\\ {\text{P}}_{\text{3}}\text{=435}\text{.4kPa}\end{array}$

Sonic conditions exist at the exit of duct, and the flow downstream the shock will be Fanno flow. From $\text{Table A-16}$,

  $\begin{array}{l}\text{M}{\text{a}}_{\text{3}}\text{=0}\text{.7786}\\ \frac{{\text{T}}_{\text{3}}}{\text{T*}}\text{=1}\text{.0702}\\ \frac{{\text{P}}_{\text{3}}}{\text{P*}}\text{=1}\text{.3286}\\ \text{M}{\text{a}}_{\text{4}}\text{=1;}\\ \frac{{\text{T}}_{\text{4}}}{\text{T*}}\text{=1}\\ \frac{{\text{P}}_{\text{4}}}{\text{P*}}\text{=1}\end{array}$

Calculating the temperature at the duct exit is given by,

  $\begin{array}{l}\frac{{\text{T}}_{\text{4}}}{{\text{T}}_{\text{3}}}\text{=}\frac{{\text{T}}_{\text{4}}\text{/T*}}{{\text{T}}_{\text{3}}\text{/T*}}\text{=}\frac{\text{1}}{\text{1}\text{.0702}}\\ {\text{T}}_{\text{4}}\text{=}\frac{{\text{T}}_{\text{3}}}{\text{1}\text{.0702}}\\ {\text{T}}_{\text{4}}\text{=}\frac{\text{870}\text{.3K}}{\text{1}\text{.0702}}\\ {\text{T}}_{\text{4}}\text{=813K}\end{array}$

Therefore, the temperature at the duct exit is ${\text{T}}_{\text{4}}\text{=813K}$.

Calculating the pressure at the duct exit is given by,

  $\begin{array}{l}\frac{{\text{P}}_{\text{4}}}{{\text{P}}_{\text{3}}}\text{=}\frac{{\text{P}}_{\text{4}}\text{/P*}}{{\text{P}}_{\text{3}}\text{/P*}}\text{=}\frac{\text{1}}{\text{1}\text{.3286}}\\ {\text{P}}_{\text{4}}\text{=}\frac{{\text{P}}_{\text{3}}}{\text{1}\text{.3286}}\\ {\text{P}}_{\text{4}}\text{=}\frac{\text{435}\text{.4kPa}}{\text{1}\text{.3286}}\\ {\text{P}}_{\text{4}}\text{=328 kPa}\end{array}$

Therefore, the pressure at the duct exit is ${\text{P}}_{\text{4}}\text{=328 kPa}$.

Calculating the velocity at the duct exit is given by,

  $\begin{array}{l}{\text{V}}_{\text{4}}\text{=M}{\text{a}}_{\text{4}}{\text{c}}_{\text{4}}\\ {\text{V}}_{\text{4}}\text{=1}\sqrt{\text{kR}{\text{T}}_{\text{4}}}\\ {\text{V}}_{\text{4}}\text{=1}\sqrt{\text{1}\text{.4×0}\text{.287 kJ/kg}\text{.K×813 K}\left(\frac{\text{1000 }{\text{m}}^{\text{2}}\text{/}{\text{s}}^{\text{2}}}{\text{1 kJ/kg}}\right)}\\ {\text{V}}_{\text{4}}\text{=572 m/s}\end{array}$

Therefore, the velocity at the duct exit is ${\text{V}}_{\text{4}}\text{=572 m/s}$.