Chapter 9, Problem 9.73P

To determine

(a)

Determine whether the nozzle is choked.

Answer to Problem 9.73P

The nozzle is choked since the flow is supersonic.

Explanation of Solution

Given Information:

Section pressure, $P_1=101kPa$

Section temperature, $T_1=300k$

Velocity, $V_1=868m/s$

Area of throat, $A_t=3c{m}^2$

Section 1 Mach number is given by,

$M{a}_1=\frac{V_1}{C_1}$

$=\frac{V_1}{\sqrt{kR{T}_1}}$

$=\frac{868}{\sqrt{1.4\times 287\times 300}}$

$M{a}_1=2.5$

The nozzle is choked since the flow is supersonic

Conclusion:

The nozzle is choked since the flow is supersonic.

To determine

(b)

Determine the area of section one.

Answer to Problem 9.73P

The area of section one is $A_1=7.92c{m}^2$.

Explanation of Solution

Given Information:

Section pressure, $P_1=101kPa$

Section temperature, $T_1=300k$

Velocity, $V_1=868m/s$

Area of throat, $A_t=3c{m}^2$

Section 1 area be A₁,

We know,

$\frac{A_1}{A_t}=\frac{1}{M{a}_1}\left(\frac{\begin{array}{l}\\ {\left(1+0.2M{a}_1{}^2\right)}^3\end{array}}{1.728}\right)$

$\frac{A_1}{3}=\frac{1}{2.5}\left(\frac{{\left(1+0.2{\left(2.5\right)}^2\right)}^3}{1.728}\right)$

$\frac{A_1}{3}=2.64$

$A_1=7.92c{m}^2$

Conclusion:

The area of section one is $A_1=7.92c{m}^2$.

To determine

(c)

Determine the mass flow. Also, determine the new $P_1$, $T_1$ and $V_1$ if there is change in gas properties at the section one

Answer to Problem 9.73P

The mass flow is $m=0.805kg/s$ . The new $V_1=926m/s$, $T_1=248.44K$ and $P_1=52.2kPa$.

Explanation of Solution

Given Information:

Section pressure, $P_1=101kPa$

Section temperature, $T_1=300k$

Velocity, $V_1=868m/s$

Area of throat, $A_t=3c{m}^2$

Let m = mass flow

If the process is isentropic then,

$\frac{P_0}{P_1}={\left(1+\frac{k-1}{2}M{a}_1{}^2\right)}^{\frac{k}{k-1}}$

$\frac{P_0}{101}={\left(1+\frac{1.4-1}{2}{\left(2.5\right)}^2\right)}^{\frac{1.4}{1.4-1}}$

$P_0=101{\left(1+0.2{\left(2.5\right)}^2\right)}^{3.5}$

$P_0=1725.68kPa$

Similarly, we know,

$\frac{T_0}{T_1}=\left(1+\frac{k-1}{2}M{a}_1{}^2\right)$

$\frac{T_0}{300}=\left(1+\frac{1.4-1}{2}{\left(2.5\right)}^2\right)$

$T_0=300\left(1+0.2{\left(2.5\right)}^2\right)$

$T_0=675k$

We know,

$m=m_{\max }$

$m=\frac{0.6847P_0{A}_t}{\sqrt{R{T}_0}}$

$m=\frac{0.6874\times 1725.68\times {10}^3\times 3\ast {10}^{-4}}{\sqrt{287\times 675}}$

$m=0.805kg/s$

Therefore, the mass flow is $m=0.805kg/s$

Throat area reduced to 2 cm²

We know,

$\frac{A_1}{A_t}=\frac{1}{M{a}_1}\left(\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\right)$

$\frac{7.92}{2}=\frac{1}{M{a}_1}\left(\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\right)$

$3.96=\frac{1}{M{a}_1}\left(\frac{{\left(1+0.2M{a}_1{}^2\right)}^3}{1.728}\right)$

${\left(1+0.2M{a}_1{}^2\right)}^3=7.057\times M{a}_1$

$\left(1+0.2M{a}_1{}^2\right)={\left(7.057\times M{a}_1\right)}^{1/3}$

By trial and error method,

$M{a}_1=2.93$

We know,

$\frac{P_0}{P_1}={\left(1+\frac{k-1}{2}M{a}_1{}^2\right)}^{\frac{k}{k-1}}$

$\frac{1725.68}{P_1}={\left(1+\frac{1.4-1}{2}{\left(2.93\right)}^2\right)}^{\frac{1.4}{1.4-1}}$

$\frac{1725.68}{P_1}={\left(1+0.2{\left(2.93\right)}^2\right)}^{3.5}$

$P_1=\frac{1725.68}{{\left(1+0.2{\left(2.93\right)}^2\right)}^{3.5}}$

$P_1=52.19kPa$

$P_1=52.2kPa$

We also know,

$\frac{T_0}{T_1}=\left(1+\frac{k-1}{2}M{a}_1{}^2\right)$

$\frac{675}{T_1}=\left(1+0.2{\left(2.93\right)}^2\right)$

$T_1=\frac{675}{\left(1+0.2{\left(2.93\right)}^2\right)}$

$T_1=248.44K$

$V_1=Ma\times \sqrt{kRT}$

$V_1=2.93\times \sqrt{1.4\times 287\times 248.44}$

$V_1=926m/s$

Conclusion:

The mass flow is $m=0.805kg/s$ . The new $V_1=926m/s$, $T_1=248.44K$ and $P_1=52.2kPa$.