Chapter 9, Problem 9.33P

To determine

(a)

To calculate:

The Mach number at point 1.

Answer to Problem 9.33P

$M{a}_1=1.467$

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

$\begin{array}{l}{p}_0=300kPa\\ T_0=500K\end{array}$

At section 1,

$\begin{array}{l}{A}_1=0.2m^2\\ V_1=550m/s\end{array}$

Speed of sound is defined as,

$a=\sqrt{kRT}$

Where,

$R$ - Gas constant

$R=287m^2/s^2.K$

$k$ - Specific heat capacity

The Mach number is defined as,

$Ma=\frac{V}{a}$

Where,

$V$ - Air velocity

The stagnation temperature is defined as,

$T_0=T+\frac{V^2}{2C_p}$

Where,

$C_p=1005m^2/s^2.K$

Calculation:

Calculate the temperature at point 1,

$T_0=T_1+\frac{V_1{}^2}{2C_p}$

Substitute for known values,

$500K=T_1+\frac{{\left(550m/s\right)}^2}{2\left(1005m^2/s^2.K\right)}$

Solve for temperature $T_1$,

$T_1=349.5K$

Calculate the speed of sound,

$a=\sqrt{kR{T}_1}=\sqrt{1.4\left(287m^2/s^2.K\right)\left(349.5K\right)}=374.74m/s$

Calculate the Mach number at section 1,

$M{a}_1=\frac{V_1}{a}=\frac{550m/s}{374.74m/s}=1.467$

Conclusion:

The Mach number at section 1 is equal to $M{a}_1=1.467$.

To determine

(b)

To calculate:

The Temperature at point 1.

Answer to Problem 9.33P

$T_1=349.5K$

Explanation of Solution

Given information:

At stagnation point inside the reservoir,

$\begin{array}{l}{p}_0=300kPa\\ T_0=500K\end{array}$

At section 1,

$\begin{array}{l}{A}_1=0.2m^2\\ V_1=550m/s\end{array}$

The stagnation temperature is defined as,

$T_0=T+\frac{V^2}{2C_p}$

Where,

$C_p=1005m^2/s^2.K$

Calculation:

Calculate the temperature at point 1,

$T_0=T_1+\frac{V_1{}^2}{2C_p}$

Substitute for known values,

$500K=T_1+\frac{{\left(550m/s\right)}^2}{2\left(1005m^2/s^2.K\right)}$

Solve for temperature $T_1$,

$T_1=349.5K$

Conclusion:

The temperature at section 1 is equal to $T_1=349.5K$.

To determine

(c)

To calculate:

The pressure at point 1.

Answer to Problem 9.33P

$p_1=85.7kPa$

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

$\begin{array}{l}{p}_0=300kPa\\ T_0=500K\end{array}$

At section 1,

$\begin{array}{l}{A}_1=0.2m^2\\ V_1=550m/s\end{array}$

The pressure ratio is defined as,

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]}^{k/k-1}$

Where,

$k=1.4$

Calculation:

Calculate the pressure at point 1,

$\frac{p_0}{p_1}={\left[1+\frac{1}{2}\left(k-1\right)M{a}_1{}^2\right]}^{k/k-1}$

According to sub-part a,

$M{a}_1=1.467$

Therefore,

$\frac{300kPa}{p_1}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(1.467\right)}^2\right]}^{1.4/1.4-1}$

Solve for pressure at point 1,

$p_1=85.7kPa$

Conclusion:

The pressure at section 1 is equal to $p_1=85.7kPa$.

To determine

(d)

To calculate:

The mass flow.

Answer to Problem 9.33P

$m=93.982kg/s$

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

$\begin{array}{l}{p}_0=300kPa\\ T_0=500K\end{array}$

At section 1,

$\begin{array}{l}{A}_1=0.2m^2\\ V_1=550m/s\end{array}$

The density at section 1 is defined as,

${\rho }_1=\frac{p_1}{R{T}_1}$

The pressure ratio is defined as,

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(k-1\right)M{a}^2\right]}^{k/k-1}$

The mass flow is defined as,

$m={\rho }_1{A}_1{V}_1$

Where,

$A_1$ - Area at section 1,

For ideal gas,

$\begin{array}{l}R=287m^2/s^2.K\\ k=1.4\end{array}$

Calculation:

Calculate the density at section1,

${\rho }_1=\frac{p_1}{R{T}_1}=\frac{85700Pa}{\left(287m^2/s^2.K\right)\left(349.5K\right)}=0.854kg/m^3$

Calculate the mass flow,

$m={\rho }_1{A}_1{V}_1=\left(0.854kg/m^3\right)\left(0.2m^2\right)\left(550m/s\right)=93.982kg/s$

Conclusion:

The mass flow is equal to $m=93.982kg/s$.

To determine

(e)

Calculate $A^{\ast }$.

Answer to Problem 9.33P

$A^{\ast }=0.173m^2$

The flow is choked

Explanation of Solution

Given information:

At stagnation point, inside the reservoir,

$\begin{array}{l}{p}_0=300kPa\\ T_0=500K\end{array}$

At section 1,

$\begin{array}{l}{A}_1=0.2m^2\\ V_1=550m/s\end{array}$

For perfect gas, where $k=1.4$

Area change is defined as,

$\frac{A}{A^{\ast }}=\frac{1}{Ma}\frac{{\left(1+0.2M{a}^2\right)}^3}{1.728}$

Calculation:

Calculate $A^{\ast }$

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

According to subpart a,

$M{a}_1=1.467$

Therefore,

$\begin{array}{l}\frac{0.2m^2}{A^{\ast }}=\frac{1}{1.467}\frac{{\left(1+0.2{\left(1.467\right)}^2\right)}^3}{1.728}\\ A^{\ast }=0.173m^2\end{array}$

Conclusion:

The flow is choked, because to produce supersonic flow in the duct

$A^{\ast }=0.173m^2$.