Chapter 12, Problem 126EP

Helium expands in a nozzle from 220 psia, 740 R, and negligible velocity to 15 psia. Calculate the throat and exit areas for a mass flow rate of 0.2 Ibm's, assuming the nozzle is isentropic. Why must this nozzle be converging-diverging?

To determine

The throat and exit areas of the nozzle for given mass flow rates and also why the nozzle must be of converging-diverging type.

Answer to Problem 126EP

The throat area of the nozzle is $\text{A*=8}\text{.19×1}{\text{0}}^{\text{-4}}\text{f}{\text{t}}^{\text{2}}.$

The exit area of the nozzle is ${\text{A}}_{\text{2}}\text{=0}\text{.00164 f}{\text{t}}^{\text{2}}.$

The exit Mach number comes out to be greater than the 1, hence the nozzle must be of converging-diverging type.

Explanation of Solution

Given:

Stagnation Temperature ${\text{T}}_1\text{= 740 R}$

Stagnation Pressure ${\text{P}}_1\text{=220 psia}$

Exit pressure at nozzle is ${\text{P}}_{\text{2}}\text{ = 15 psia}$

Mass flow rate $\stackrel{\text{.}}{\text{m}}\text{ = 0}\text{.2}$

Assuming flow is isentropic.

Calculation:

Properties of helium,

  $\text{R=0}\text{.4961 Btu/lbm}\text{.R=2}\text{.6809 psia/f}{\text{t}}^{\text{3}}$

  $c_p\text{=1}\text{.25 Btu/lbm}\text{.R}$

  $\text{k=1}\text{.667}$

R is gas constant, $c_p$ is specific heat, k is the specific heat ratio.

Mach number is given by

  $\begin{array}{l}\frac{{\text{P}}_{\text{0}}}{{\text{P}}_{\text{3}}}\text{ = }{\left(\text{1+}\frac{\text{k-1}}{\text{2}}\text{M}{\text{a}}_{\text{3}}{}^{\text{2}}\right)}^{\text{k/k-1}}\\ \text{Initial pressure of helium is }{\text{P}}_{\text{0}}\text{,index of helium at adiabatic is k and}\\ \text{final pressure of helium is }{\text{P}}_{\text{3}}\text{.}\\ \frac{\text{220}}{\text{15}}\text{ = }{\left(\text{1+}\frac{\text{1}\text{.667-1}}{\text{2}}\text{M}{\text{a}}_{\text{3}}{}^{\text{2}}\right)}^{\text{1}\text{.667/1}\text{.667-1}}\\ \text{M}{\text{a}}_{\text{3}}\text{=2}\text{.405>1 nozzle is converging-diverging}\text{.}\end{array}$

The flow is imagined to be isentropic, so the stagnation temperature and stagnation pressure remain sameoverall the nozzle.

  ${\text{T}}_{02}\text{=}{\text{T}}_{01}\text{= 740 R}$

  ${\text{P}}_{02}\text{=}{\text{P}}_{01}\text{=220 psia}$

Throat temperature is given by

  $\begin{array}{l}\text{T*=}{\text{T}}_{\text{0}}\left(\frac{\text{2}}{\text{k+1}}\right)\\ \text{Intial temperature of helium }{\text{T}}_{\text{0}}.\\ \text{T*=740}\left(\frac{\text{2}}{\text{1}\text{.667+1}}\right)\\ \text{T*=554}\text{.93 R}\end{array}$

Throat pressure is given by

  $\begin{array}{l}\text{P*=}{\text{P}}_{\text{0}}{\left(\frac{\text{2}}{\text{k+1}}\right)}^{k/k-1}\\ \text{P*=220 psia}{\left(\frac{\text{2}}{\text{1}\text{.667+1}}\right)}^{1.667/1.667-1}\\ \text{P*=107}\text{.16 psia}\end{array}$

Exit area of throat is given by

  $\text{A*=}\frac{\stackrel{\text{.}}{\text{m}}}{\text{ρ*V*}}\mathrm{.......}(1)$

Throat density is given by

  $\begin{array}{l}\text{ρ*=}\frac{\text{P*}}{\text{RT*}}\\ \text{ρ*=}\frac{\text{107}\text{.2}}{\text{2}\text{.6809×554}\text{.9}}\\ \text{ρ*=0}\text{.07203 lbm/f}{\text{t}}^{\text{3}}\end{array}$

Inlet velocity is given by

  $\begin{array}{l}\text{V*=c*=}\sqrt{\text{kRT*}}\\ \text{V*=}\sqrt{\text{1}\text{.667×0}\text{.4961}\left(\text{554}\text{.9}\right)\left(\frac{\text{25,037 f}{\text{t}}^{\text{2}}\text{/}{\text{s}}^{\text{2}}}{\text{1Btu/lbm}}\right)}\\ \text{V*=3390 ft/s}\end{array}$

Substitute $\text{ρ*=0}\text{.07203 lbm/f}{\text{t}}^{\text{3}}$ and $\text{V*=3390 ft/s}$ in equation $\text{(1)}$ is given by $\begin{array}{l}\text{A*=}\frac{\stackrel{\text{.}}{\text{m}}}{\text{ρ*V*}}\\ \text{A*=}\frac{\text{0}\text{.2}}{\text{0}\text{.07203×3390}}\\ \text{A*=8}\text{.19×1}{\text{0}}^{\text{-4}}\text{f}{\text{t}}^{\text{2}}\end{array}$

Therefore, throat area is given by $\text{A*=8}\text{.19×1}{\text{0}}^{\text{-4}}\text{f}{\text{t}}^{\text{2}}$.

Pressure at the exit of nozzle is given as ${\text{P}}_{\text{2}}\text{ = 15 psia}\text{.}$

Pressure at nozzle exit is given by

  $\begin{array}{l}{\text{P}}_{\text{3}}\text{ =}{\text{P}}_{\text{0}}{\left(\text{1+}\frac{\left(\text{k-1}\right)}{\text{2}}\text{M}{\text{a}}_{\text{3}}{}^{\text{2}}\right)}^{\text{-k/k-1}}\\ {\text{P}}_{\text{3}}\text{ =220 }{\left(\text{1+}\frac{\left(\text{1}\text{.667-1}\right)}{\text{2}}\text{2}\text{.40}{\text{5}}^{\text{2}}\right)}^{\text{-1}\text{.667/1}\text{.667-1}}\\ {\text{P}}_{\text{3}}\text{ =15}\text{.146 Psia}\end{array}$

Temperature at nozzle exit is given by

  $\begin{array}{l}{\text{T}}_3\text{ =}{\text{T}}_{\text{0}}\left(\frac{\text{2}}{\text{2+}\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{2}}{}^{\text{2}}}\right)\\ {\text{T}}_3\text{ =740 }\left(\frac{\text{2}}{\text{2+}\left(\text{1}\text{.667-1}\right)\text{2}\text{.40}{\text{5}}^{\text{2}}}\right)\\ {\text{T}}_3\text{ =252}\text{.6 R}\end{array}$

Area of exit at nozzle is given by

  ${\text{A}}_{\text{2}}\text{=}\frac{\stackrel{\text{.}}{\text{m}}}{{\text{ρ}}_{\text{2}}{\text{V}}_{\text{2}}}\mathrm{.......}(2)$

Density at exit of nozzle is given by

  $\begin{array}{l}{\text{ρ}}_2\text{=}\frac{P_2}{\text{R}{\text{T}}_2}\\ {\text{ρ}}_2\text{=}\frac{15}{\text{2}\text{.6809}\times 252.6}\\ {\text{ρ}}_2\text{=0}\text{.002215 lbm/f}{\text{t}}^{\text{3}}\end{array}$

Velocity at exit of nozzle is given by

  $\begin{array}{l}{\text{V}}_2\text{=M}{\text{a}}_2{c}_2\text{=M}{\text{a}}_2\sqrt{\text{kR}{\text{T}}_2}\\ {\text{V}}_2\text{=}\sqrt{\text{1}\text{.667×0}\text{.4961}\left(\text{252}\text{.6}\right)\left(\frac{\text{25,037 f}{\text{t}}^{\text{2}}\text{/}{\text{s}}^{\text{2}}}{\text{1Btu/lbm}}\right)}\\ {\text{V}}_2\text{=5500 ft/s}\end{array}$

Substitute, ${\text{ρ}}_2\text{=0}\text{.002215 lbm/f}{\text{t}}^{\text{3}}$, ${\text{V}}_2\text{=5500 ft/s}$ and $\stackrel{\text{.}}{\text{m}}\text{ = 0}\text{.2}$ in equation $\text{(2)}$ is given by

  $\begin{array}{l}{\text{A}}_{\text{2}}\text{=}\frac{\stackrel{\text{.}}{\text{m}}}{{\text{ρ}}_{\text{2}}{\text{V}}_{\text{2}}}\\ {\text{A}}_{\text{2}}\text{=}\frac{\text{0}\text{.2}}{\text{0}\text{.02215 ×5500 }}\\ {\text{A}}_{\text{2}}\text{=0}\text{.00164 f}{\text{t}}^{\text{2}}\end{array}$

Conclusion:

Thus, the throat area of the nozzle is $\text{A*=8}\text{.19×1}{\text{0}}^{\text{-4}}\text{f}{\text{t}}^{\text{2}}$

The exit area of the nozzle is ${\text{A}}_{\text{2}}\text{=0}\text{.00164 f}{\text{t}}^{\text{2}}$

The exit Mach number comes out to be greater than the 1, hence the nozzle must be of converging-diverging type.