Chapter 9, Problem 9.84P

To determine

(a)

The mass flow rate of air through a duct with $A_1=24$ cm² , $A_2=18$ cm², and $A_3=32$ cm² if a normal shock stands at section 2.

Answer to Problem 9.84P

The mass flow rate is $0.963$ kg/s.

Explanation of Solution

Given:

$A_1=24$ cm² $=24\times {10}^{-4}$ m² $A_2=18$ cm² $=18\times {10}^{-4}$ m² $A_3=32$ cm² $=32\times {10}^{-4}$ m² $T_0=500$ K.

$p_1=40$ kPa $=40000$ Pa.

$T_1={30}^{°}$ C $=303$ K.

$M{a}_1=2.5$

Concept Used:

Mass flow rate, $\dot{m}=\rho Av$

Where, $\dot{m}$ is the mass flow rate in kg/s.

$A$ is the area of the exit in m².

The temperature at the exit, $T_e=\frac{T_0}{1+0.2\text{x}M{a}^2}$

The density of air ${\rho }_1=\frac{p_1}{R{T}_1}$

Local sound speed, $a_1=\sqrt{kR{T}_1}$

$R=287$ kJ/kgK for air.

Velocity at the exit, $v_1=M{a}_1\text{x}{a}_1$

Where, $Ma$ is the Mach number

$k$ is the specific heat ratio of the perfect gas.

Calculation:

Local sound speed, $a_1=\sqrt{kR{T}_1}$

$=\sqrt{1.4\times 287\times 303}$

$=348.92\approx 349$ m/s

Velocity of air, $v_1=M{a}_1\text{x}{a}_1$

$=0.25\text{x}349\approx 872$ m/s.

The density of air ${\rho }_1=\frac{p_1}{R{T}_1}$

$=\frac{40\times {10}^3}{287\text{x}303}\approx 0.4599\approx 0.46$ kg/m³.

Mass flow rate, $\dot{m}={\rho }_1{A}_1{v}_1$

Substituting the values of density, area and the velocity in the above relation, we get

$\dot{m}=0.46\times 24\times {10}^{-4}\times 872\approx 0.9627$

$\dot{m}\approx 0.963$ kg/s

Conclusion:

The mass flow rate is $0.963$ kg/s.

To determine

(b)

The Mach number at the section 2.

Answer to Problem 9.84P

The Mach number is $M{a}_3\approx 0.27$.

Explanation of Solution

Given:

$A_1=24$ cm² $=24\times {10}^{-4}$ m² $A_2=18$ cm² $=18\times {10}^{-4}$ m² $A_3=32$ cm² $=32\times {10}^{-4}$ m² $T_0=500$ K.

$p_1=40$ kPa $=40000$ Pa.

$T_1={30}^{°}$ C $=303$ K.

$M{a}_1=2.5$

Concept Used:

Mass flow rate, $\dot{m}=\rho Av$

Where, $\dot{m}$ is the mass flow rate in kg/s.

$A$ is the area of the exit in m².

The temperature at the exit, $T_e=\frac{T_0}{1+0.2\text{x}M{a}^2}$

The density of air ${\rho }_1=\frac{p_1}{R{T}_1}$

Local sound speed, $a_1=\sqrt{kR{T}_1}$

$R=287$ kJ/kgK for air.

Velocity at the exit, $v_1=M{a}_1\text{x}{a}_1$

Where, $Ma$ is the Mach number

$k$ is the specific heat ratio of the perfect gas.

Calculation:

From section 1 to 2 the flow is isentropic.

For the Mach number $M{a}_1=2.5$, the area ratio $\frac{A_1}{A_1^{*}}=2.6367$.

$A_1^{*}=\frac{24}{2.64}=9.0909\approx 9.1$ cm².

The area ratio $\frac{A_2}{A_1^{*}}=\frac{18}{9.1}=1.978\approx 1.98$ cm²

Corresponding to this area ratio, the Mach number $M{a}_2=2.19$.

The pressure across the shock is given by

$p_{01}=p_{02}=40{\left[1+0.2{\left(2.5\right)}^2\right]}^{3.5}\approx 683.43\approx 683$ kPa

The area ratio is given by

$\frac{A_3^{*}}{A_2^{*}}=1.57$

$A_3^{*}=9.1\times 1.57\approx 14.28\approx 14.3$ cm².

The area ratio is given by

$\frac{A_3}{A_3^{*}}=\frac{32}{14.3}=2.2377\approx 2.24$

Corresponding to this area ratio, the Mach number $M{a}_3\approx 0.2697$.

Thus, the Mach number $M{a}_3\approx 0.27$.

Conclusion:

The Mach number is $M{a}_3\approx 0.27$.

To determine

(c)

The stagnation pressure at the section 3.

Answer to Problem 9.84P

The stagnation pressure at the section 3 $p_{03}\approx 435$ kPa

Explanation of Solution

Given:

$A_1=24$ cm² $=24\times {10}^{-4}$ m² $A_2=18$ cm² $=18\times {10}^{-4}$ m² $A_3=32$ cm² $=32\times {10}^{-4}$ m² $T_0=500$ K.

$p_1=40$ kPa $=40000$ Pa.

$T_1={30}^{°}$ C $=303$ K.

$M{a}_1=2.5$

Concept Used:

Mass flow rate, $\dot{m}=\rho Av$

Where, $\dot{m}$ is the mass flow rate in kg/s.

$A$ is the area of the exit in m².

The temperature at the exit, $T_e=\frac{T_0}{1+0.2\text{x}M{a}^2}$

The density of air ${\rho }_1=\frac{p_1}{R{T}_1}$

Local sound speed, $a_1=\sqrt{kR{T}_1}$

$R=287$ kJ/kgK for air.

Velocity at the exit, $v_1=M{a}_1\text{x}{a}_1$

Where, $Ma$ is the Mach number

$k$ is the specific heat ratio of the perfect gas.

Calculation:

The stagnation pressure ratio across the shock can be determined using the Mach number at the section 2 $M{a}_2=2.18$.

For the Mach number $M{a}_2=2.18$, the pressure ratio is $\frac{p_{03}}{p_{02}}\approx 0.6367\approx 0.637$.

$p_{03}=0.637p_{02}=0.637\text{x}683\approx 435.071\approx 435$ kPa.

Thus, the stagnation pressure at the section 3 $p_{03}\approx 435$ kPa

Conclusion:

The stagnation pressure at the section 3 $p_{03}\approx 435$ kPa