Chapter 12, Problem 127P

To determine

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

Answer to Problem 127P

The throat area of the nozzle is $\text{A*=5}\text{.96 c}{\text{m}}^{\text{2}}$

The exit area of the nozzle is ${\text{A}}_{\text{2}}\text{=8}\text{.97c}{\text{m}}^{\text{2}}$

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:

Ambient temperature at inlet ${\text{T}}_1\text{= 500 K}$

Stagnation Pressure ${\text{P}}_{01}\text{=}{\text{P}}_1\text{=220 psia}$

Exit pressure ${\text{P}}_{\text{2}}\text{ = 0}\text{.1 MPa}$

Mass flow rate $\stackrel{\text{.}}{\text{m}}\text{ = 0}\text{.34 kg/s}$

Calculation:

Properties of helium,

  $\text{R=2}\text{.0769}\times \text{1}{\text{0}}^{\text{3}}\text{J/kg}\text{.K}$

  $c_p\text{=5}\text{.1926 KJ/kg}\text{.K}$

  $\text{k=1}\text{.667}$

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

Stagnation temperature of helium at inlet is given by

  $\begin{array}{l}{\text{T}}_{\text{01}}\text{=}{\text{T}}_{\text{1}}\text{+}\frac{{\text{V}}_{\text{1}}{}^{\text{2}}}{\text{2}{\text{c}}_{\text{p}}}\\ \text{The ambient temperature at inlet is }{\text{T}}_{\text{1}},\text{inlet velocity of fluid is }{\text{V}}_{\text{1}}.\\ {\text{T}}_{\text{01}}\text{=500+}\frac{{\left(\text{0}\right)}^{\text{2}}}{\text{2}\left(\text{5}\text{.1926}\right)}\times \left(\frac{\text{1}}{\text{1000 }}\right)\\ {\text{T}}_{\text{01}}\text{=500}\text{.0K}\end{array}$

Stagnation pressure of helium at inlet is given by

  $\begin{array}{l}{\text{P}}_{\text{0}}\text{=}{\text{P}}_{\text{1}}{\left(\frac{{\text{T}}_{\text{01}}}{{\text{T}}_{\text{1}}}\right)}^{\text{k/k-1}}\\ \text{Pressure of the fluid }{\text{P}}_{\text{1}}.\\ {\text{P}}_{\text{01}}\text{=0}\text{.8}{\left(\frac{\text{500}}{\text{500}}\right)}^{\text{1}\text{.667/1}\text{.667-1}}\\ {\text{P}}_{\text{01}}\text{=0}\text{.8MPa}\end{array}$

The flow is considered to be isentropic, so the stagnation temperature and stagnation pressure remain same throughout the nozzle.

  ${\text{T}}_{02}\text{=}{\text{T}}_{01}\text{= 500 K}$

  ${\text{P}}_{02}\text{=}{\text{P}}_{01}\text{=0}\text{.8 Mpa}$

Now the critical temperature is given by

  $\begin{array}{l}\text{T*=}{\text{T}}_{\text{0}}\left(\frac{\text{2}}{\text{k+1}}\right)\\ \text{T*=500}\left(\frac{\text{2}}{\text{1}\text{.667+1}}\right)\\ \text{T*=374}\text{.95 K}\end{array}$

And critical pressure is given by

  $\begin{array}{l}\text{P*=}{\text{P}}_{\text{0}}{\left(\frac{\text{2}}{\text{k+1}}\right)}^{k/k-1}\\ \text{P*=0}\text{.8}{\left(\frac{\text{2}}{\text{1}\text{.667+1}}\right)}^{1.667/1.667-1}\\ \text{P*0}\text{.3896 MPa}\end{array}$

The critical density is given by

  $\begin{array}{l}\text{ρ*=}\frac{\text{P*}}{\text{RT*}}\\ \text{ρ*=}\frac{0.3896\times \text{1}{\text{0}}^6}{\text{2}\text{.0769×1}{\text{0}}^3\times \text{374}\text{.9531}}\\ \text{ρ*=0}\text{.50039 kg/}{\text{m}}^{\text{3}}\end{array}$

Inlet velocity is given by

  $\begin{array}{l}\text{V*=c*=}\sqrt{\text{kRT*}}\\ \text{V*=}\sqrt{\text{1}\text{.667×2}\text{.0769}\left(\text{374}\text{.95}\right)\left(\frac{\text{1000}}{\text{1}}\right)}\\ \text{V*=1139}\text{.36 m/s}\end{array}$

Exit area of throat is given by

  $\begin{array}{l}\text{A*=}\frac{\stackrel{\text{.}}{\text{m}}}{\text{ρ*V*}}\\ \stackrel{\text{.}}{\text{m}}\text{ is mass flow rate,ρ*is critical density,V\&*\#x00A0;is inlet velocity}\text{.}\\ \text{A*=}\frac{\text{0}\text{.34}}{\text{1139}\text{.36×0}\text{.50039}}\\ \text{A*=5}\text{.96×1}{\text{0}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{A*=5}\text{.96 c}{\text{m}}^{\text{2}}\end{array}$

Pressure at the exit of nozzle is given as ${\text{P}}_{\text{2}}\text{ = 0}\text{.1 MPa}\text{.}$

  $\begin{array}{l}\frac{P_0}{P_2}\text{ = }{\left(1+\frac{k-1}{2}M{a}_2{}^2\right)}^{k/k-1}\\ \text{exit pressure is }{P}_2.\\ \frac{0.8}{15}\text{ = }{\left(1+\frac{1.667-1}{2}M{a}_2{}^2\right)}^{1.667/1.667-1}\\ M{a}_2=1.972806>1\text{ nozzle is converging-diverging}\text{.}\end{array}$

Temperature at nozzle exit is given by

  $\begin{array}{l}{\text{T}}_{\text{2}}\text{ =}{\text{T}}_{\text{0}}\left(\frac{\text{2}}{\text{2+}\left(\text{k-1}\right)\text{M}{\text{a}}_{\text{2}}{}^{\text{2}}}\right)\\ {\text{T}}_{\text{2}}\text{ =500 }\left(\frac{\text{2}}{\text{2+}\left(\text{1}\text{.667-1}\right)\text{1}\text{.97280}{\text{6}}^{\text{2}}}\right)\\ {\text{T}}_{\text{2}}\text{ =217}\text{.58 K}\end{array}$

Density at exit of nozzle is given by

  $\begin{array}{l}{\text{ρ}}_2\text{=}\frac{P_2}{\text{R}{\text{T}}_2}\\ {\text{ρ}}_2\text{=}\frac{0.1\times \text{1}{\text{0}}^6}{\text{2}\text{.0769}\times \text{1}{\text{0}}^3\times 217.58}\\ {\text{ρ}}_2\text{=0}\text{.2212 kg/}{\text{m}}^{\text{3}}\end{array}$

Velocity at exit of nozzle is given by

  $\begin{array}{l}{c}_2\text{=}\sqrt{\text{kR}{\text{T}}_2}\\ \text{R is gas constant,k specific heat ratio,}{\text{T}}_2\text{ exit temperature}\text{.}\\ c_2\text{=}\sqrt{\text{1}\text{.667×0}\text{.4961}\left(\text{217}\text{.58}\right)\left(\frac{\text{1000}}{\text{1}}\right)}\\ c_2\text{=867}\text{.93 m/s}\\ {\text{V}}_2\text{=M}{\text{a}}_2{c}_2\\ {\text{V}}_2\text{=}\left(1.972806\right)\text{867}\text{.93}\\ {\text{V}}_2\text{=}1712.272\text{ m/s}\end{array}$

Area of exit at nozzle is given by

  $\begin{array}{l}{\text{A}}_{\text{2}}\text{=}\frac{\stackrel{\text{.}}{\text{m}}}{{\text{ρ}}_{\text{2}}{\text{V}}_{\text{2}}}\\ \stackrel{\text{.}}{\text{m}}\text{ is mass flow rate,}{\text{ρ}}_{\text{2}}\text{is exit density,}{\text{V}}_{\text{2}}\text{ is exit velocity}.\\ {\text{A}}_{\text{2}}\text{=}\frac{\text{0}\text{.34}}{\text{1712}\text{.272 ×0}\text{.2212 }}\\ {\text{A}}_{\text{2}}\text{=8}\text{.97}\times {10}^{-4}{\text{m}}^2\\ {\text{A}}_{\text{2}}\text{=8}\text{.97 c}{\text{m}}^2\end{array}$

Conclusion:

Therefore, the throat area of the nozzle is $\text{A*=5}\text{.96 c}{\text{m}}^{\text{2}}.$

The exit area of the nozzle is ${\text{A}}_{\text{2}}\text{=8}\text{.97c}{\text{m}}^{\text{2}}\text{.}$

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