Chapter 10, Problem 10.1P

The reservoir pressure and temperature for a convergent-divergent nozzle are 5 atm and $\text{52}0°\text{R}$, respectively. The flow is expanded isentropically to supersonic speed at the nozzle exit. If the exit-to-throat area ratio is 2.193, calculate the following properties at the exit: $M_e,\text{}{p}_e,T_e,{\rho }_e,u_e,p_{0,e}$.

To determine

The Mach Number $M_e$ .

The exit pressure $p_e$ .

The exit Temperature $T_e$ .

The exit density ${\rho }_e$ .

The velocity of flow at exit $u_e$

The stagnation pressure at exit $p_{0,e}$ .

The stagnation temperature at exit $T_{0,e}$ .

Answer to Problem 10.1P

The Mach Number $M_e$ is $2.3$ .

The exit pressure $p_e$ is $846.55\text{lb}/{\text{ft}}^{\text{3}}$

The exit Temperature $T_e$ is $252.72°\text{R}$ .

The exit density ${\rho }_e$ is $0.00195\text{slug}/{\text{ft}}^{\text{3}}$ .

The velocity of flow at exit $u_e$ is $1792.318\text{ft}/\text{s}$ .

The stagnation pressure at exit $p_{0,e}$ is $10581.875\text{lb}/{\text{ft}}^{\text{2}}$ .

The stagnation temperature at exit $T_{0,e}$ is $520.056°\text{R}$ .

Explanation of Solution

Given:

The Reservoir pressure is $p_0=5\text{atm}$ .

The Reservoir temperature is $T_0=520°\text{R}$ .

The ratio of exit area to throat area is $\frac{A_e}{A*}=2.193$ .

Formula used:

The expression for calculating pressure is given as,

  $p_e=\frac{p_e}{p_{0,e}}{p}_0$

The expression for calculating temperature is given as,

  $T_e=\frac{T_e}{T_{0,e}}{T}_0$

The expression for density is given as,

  ${\rho }_e=\frac{p_e}{R{T}_e}$

Here, $R$ is the gas constant.

The expression for velocity of sound is given as,

  $a_e=\sqrt{\gamma R{T}_e}$

Here $\gamma$ is the adiabatic constant.

The expression for speed of velocity is given as,

  $u_e=a_e{M}_e$

Calculation:

Refer to the “isentropic flow properties” for the Mach number at the ratio of exit area to throat area. The Mach number is obtained as,

  $M_e=2.3$

Refer to the “isentropic flow properties” for the pressure ratio at the ratio of exit area to throat area. The pressure ratio is obtained as,

  $\frac{p_{0,e}}{p_e}=12.5$

Refer to the “isentropic flow properties” for the temperature ratio at the ratio of exit area to throat area. The temperature ratio is obtained as,

  $\frac{T_{0,e}}{T_e}=2.058$

The pressure at the exit is calculated as,

  $\begin{array}{c}{p}_e=\frac{p_e}{p_{0,e}}{p}_0\\ =\frac{1}{12.5}\times 5\text{atm}\left(\frac{1.0133\times {10}^5\text{Pa}}{1\text{atm}}\right)\times \left(\frac{0.020886\text{lb}/{\text{ft}}^{\text{2}}}{1\text{Pa}}\right)\\ =\text{846}\text{.55}\text{lb}/{\text{ft}}^{\text{2}}\end{array}$

The stagnation pressure at the exit is calculated as,

  $\begin{array}{c}\frac{p_{0,e}}{p_e}=12.5\\ p_{0,e}=12.5\times 846.55\\ =10581.875\text{lb}/{\text{ft}}^{\text{2}}\end{array}$

The Temperature at the exit is calculated as,

  $\begin{array}{c}{T}_e=\frac{T_e}{T_{0,e}}{T}_0\\ =\frac{1}{2.058}\times 520°\text{R}\\ =\text{252}\text{.72}°\text{R}\end{array}$

The stagnation temperature at the exit is calculated as,

  $\begin{array}{c}\frac{T_{0,e}}{T_e}=2.058\\ p_{0,e}=2.058\times 252.7°\text{R}\\ =520.056°\text{R}\end{array}$

The Density at the exit is calculated as,

The value of gas constant is in English units is $R=1716.49\text{lbf}\cdot \text{ft}/\text{slug}\cdot \text{°R}$ .

  $\begin{array}{c}{\rho }_e=\frac{p_e}{R{T}_e}\\ =\frac{846.55\text{lb}/{\text{ft}}^{\text{2}}}{1716.49\text{lbf}\cdot \text{ft}/\text{slug}\cdot \text{°R}\times 252.7°\text{R}}\\ =1.91\times {10}^{-3}\text{slug}/{\text{ft}}^{\text{3}}\end{array}$

The velocity at exit is calculated as,

The value of adiabatic constant is $\gamma =1.4$ .

  $\begin{array}{c}{u}_e=a_e{M}_e\\ =\sqrt{\gamma R{T}_e}\times M_e\\ =\sqrt{1.4\times 1716.49\text{lbf}\cdot \text{ft}/\text{slug}\cdot °\text{R}\times 252.7°\text{R}}\times 2.3\\ =1792.318\text{ft}/\text{s}\end{array}$

Conclusion:

Therefore, The Mach Number $M_e$ is $2.3$ .

Therefore, The exit pressure $p_e$ is $846.55\text{lb}/{\text{ft}}^{\text{3}}$

Therefore, The exit Temperature $T_e$ is $252.72°\text{R}$ .

Therefore, The exit density ${\rho }_e$ is $0.00195\text{slug}/{\text{ft}}^{\text{3}}$ .

Therefore, The velocity of flow at exit $u_e$ is $1792.318\text{ft}/\text{s}$ .

Therefore, The stagnation pressure at exit $p_{0,e}$ is $10581.875\text{lb}/{\text{ft}}^{\text{2}}$ .

Therefore, The stagnation temperature at exit $T_{0,e}$ is $520.056°\text{R}$ .