Chapter 12, Problem 138P

To determine

The velocity, temperature, and pressure at the duct exit.

Answer to Problem 138P

  ${\text{T}}_{\text{4}}\text{=727K}$

  ${\text{P}}_{\text{4}}\text{=430 kPa}$

  ${\text{V}}_{\text{4}}\text{=520 m/s}$

Explanation of Solution

Given:

The properties of air:

  $\text{R =0}\text{.280 kJ/kg}\text{.K}$

  $c_p\text{=1}\text{.005 kJ/kg}\text{.K}$

  $\text{k=1}\text{.33}$

Inlet Temperature $T_1=510K$

Inlet Pressure $P_1=180kPa$

Mach number $M{a}_1=2.0$

Shock is located distance of $\text{2m}$ from inlet.

Diameter $D=0.1m$

Average friction factor $f=0.010$

Calculation:

Fanno flow equations related to the Mach number of inlet are from $\text{Table A-16}$.

  $\begin{array}{l}\text{M}{\text{a}}_{\text{1}}\text{=2}\\ {\left(\frac{\text{fL*}}{\text{Dh}}\right)}_{\text{1}}\text{=0}\text{.3402}\\ \frac{{\text{T}}_{\text{1}}}{\text{T*}}\text{=0}\text{.7018}\\ \frac{{\text{P}}_{\text{1}}}{\text{P*}}\text{=0}\text{.4189}\\ \frac{{\text{V}}_{\text{1}}}{\text{V*}}\text{=0}\text{.2405}\end{array}$

  $\begin{array}{l}{\text{L}}_{\text{1}}\text{=}\frac{\text{0}\text{.3402D}}{\text{f}}\\ {\text{L}}_{\text{1}}\text{=}\frac{\text{0}\text{.3402×0}\text{.1}}{\text{0}\text{.01}}\\ {\text{L}}_{\text{1}}\text{=3}\text{.402 m}\end{array}$

Actual duct length is given by,

  $\begin{array}{l}\left(\frac{\text{f}{\text{L}}_1}{\text{Dh}}\right)=\frac{\text{0}\text{.010}\times 2}{\text{0}\text{.10}}\\ \left(\frac{\text{f}{\text{L}}_1}{\text{Dh}}\right)=0.200\end{array}$

We know,

  ${\text{L}}_{\text{1}}\text{=}{\text{L}}_{\text{1}}\text{*-}{\text{L}}_{\text{2}}\text{*}$

Then,

  $\begin{array}{l}{\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=}{\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{1}}\text{-}\frac{\text{f}{\text{L}}_{\text{1}}}{{\text{D}}_{\text{h}}}\\ {\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=0}\text{.3402-0}\text{.2000}\\ {\left(\frac{\text{fL*}}{{\text{D}}_{\text{h}}}\right)}_{\text{2}}\text{=0}\text{.1402}\end{array}$

From $\text{Table A-16}$

  $\text{M}{\text{a}}_{\text{2}}\text{=1}\text{.476}$ we have,

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{\text{T*}}\text{=0}\text{.8568}\\ \frac{{\text{P}}_{\text{2}}}{\text{P*}}\text{=0}\text{.6720}\end{array}$

Temperature before the shock is given by

  $\begin{array}{l}\frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{{\text{T}}_{\text{2}}\text{/T*}}{{\text{T}}_{\text{1}}\text{/T*}}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=}\frac{\text{0}\text{.8568}}{\text{0}\text{.7018}}\\ \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}\text{=1}\text{.2209}\\ {\text{T}}_{\text{2}}\text{=1}\text{.2209}{\text{T}}_{\text{1}}\\ {\text{T}}_{\text{2}}\text{=1}\text{.2209×510}\\ {\text{T}}_{\text{2}}\text{=622}\text{.7K}\end{array}$

Pressure before the shock is given by

  $\begin{array}{l}\frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}\text{/P*}}{{\text{P}}_{\text{1}}\text{/P*}}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=}\frac{\text{0}\text{.6270}}{\text{0}\text{.4189}}\\ \frac{{\text{P}}_{\text{2}}}{{\text{P}}_{\text{1}}}\text{=1}\text{.4968}\\ {\text{P}}_{\text{2}}\text{=1}\text{.2209}{\text{P}}_{\text{1}}\\ {\text{P}}_{\text{2}}\text{=1}\text{.2209×180}\\ {\text{P}}_{\text{2}}\text{=269}\text{.4 kPa}\end{array}$

From normal shocks $\text{Table A-16}$ we have

  $\begin{array}{l}\text{M}{\text{a}}_{\text{2}}\text{=1}\text{.476}\\ \text{M}{\text{a}}_{\text{3}}\text{=0}\text{.7052}\\ \frac{{\text{T}}_{\text{3}}}{{\text{T}}_{\text{2}}}\text{=1}\text{.2565}\\ \frac{{\text{P}}_{\text{3}}}{{\text{P}}_{\text{2}}}\text{=2}\text{.3466}\end{array}$

Temperature after the shock is given by

  $\begin{array}{l}\frac{{\text{T}}_{\text{3}}}{{\text{T}}_{\text{2}}}\text{=1}\text{.2565}\\ {\text{T}}_{\text{3}}\text{=1}\text{.2565×622}\text{.7}\\ {\text{T}}_{\text{3}}\text{=782}\text{.4K}\end{array}$

Pressure after the shock is given by

  $\begin{array}{l}\frac{{\text{P}}_{\text{3}}}{{\text{P}}_{\text{2}}}\text{=2}\text{.3466}\\ {\text{P}}_{\text{3}}=2.3466\times 269.4\\ {\text{P}}_{\text{3}}=632.3\text{ kPa}\end{array}$

Then,

  $\begin{array}{l}\text{M}{\text{a}}_{\text{3}}\text{=0}\text{.7052}\\ \frac{{\text{T}}_{\text{3}}}{\text{T*}}\text{=1}\text{.0767}\\ \frac{{\text{P}}_{\text{3}}}{\text{P*}}\text{=1}\text{.4713}\\ \text{M}{\text{a}}_4\text{=1}\\ \frac{{\text{T}}_4}{\text{T*}}\text{=1}\\ \frac{{\text{P}}_4}{\text{P*}}\text{=1}\end{array}$

Temperature at duct exit is given by

  $\begin{array}{l}\frac{{\text{T}}_{\text{4}}}{{\text{T}}_{\text{3}}}\text{=}\frac{{\text{T}}_{\text{4}}\text{/T*}}{{\text{T}}_{\text{3}}\text{/T*}}\\ \frac{{\text{T}}_{\text{4}}}{{\text{T}}_{\text{3}}}\text{=}\frac{\text{1}}{\text{1}\text{.0767}}\\ {\text{T}}_{\text{4}}\text{=}\frac{\text{782}\text{.4}}{\text{1}\text{.0767}}\\ {\text{T}}_{\text{4}}\text{=727K}\end{array}$

Pressure at duct exit is given by

  $\begin{array}{l}\frac{{\text{P}}_{\text{4}}}{{\text{P}}_{\text{3}}}\text{=}\frac{{\text{P}}_{\text{4}}\text{/P*}}{{\text{P}}_{\text{3}}\text{/P*}}\\ \frac{{\text{P}}_{\text{4}}}{{\text{P}}_{\text{3}}}\text{=}\frac{\text{1}}{\text{1}\text{.4713}}\\ {\text{P}}_{\text{4}}\text{=}\frac{\text{632}\text{.3}}{\text{1}\text{.4713}}\\ {\text{P}}_{\text{4}}\text{=430 kPa}\end{array}$

Velocity at duct exit is given by

  $\begin{array}{l}{\text{V}}_{\text{4}}\text{=M}{\text{a}}_{\text{4}}{\text{c}}_{\text{4}}\text{=1}{\sqrt{\text{kRT}}}_{\text{4}}\\ {\text{V}}_{\text{4}}\text{=}\sqrt{\text{1}\text{.33×0}\text{.280}\left(\text{727}\right)\left(\frac{\text{1000 J/kg}\text{.K}}{\text{1}\text{kJ/kg}\text{.K}}\right)}\\ {\text{V}}_{\text{4}}\text{=520 m/s}\end{array}$

Conclusion:

Therefore, it is given ${\text{L}}_{\text{3}}\text{ \&*\#x00A0;= 2}\text{.13 m}$, and for the duct the total length is $\text{4}\text{.13 m}$. If the duct is going to be further enlarged, then normal shock is going to move upstream further and eventually to the inlet of the duct.