Chapter 9, Problem 9.32P

To determine

(a)

To compute:

The pressure in the tank.

Answer to Problem 9.32P

$p_0=141kPa$

Explanation of Solution

Given information:

The nozzle exit velocity is equal to $235m/s$

$h=30cm$

The stagnation temperature is defined as,

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

Where, $C_p=1005m^2/s^2.°K$

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 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:

Convert,

$T_0=30°C=303K$

Calculate the exit temperature,

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

Substitute for known values,

$303K=T+\frac{{\left(235m/s\right)}^2}{2\left(1005m^2/s^2.°K\right)}$

Therefore,

$T=275.52K$

Calculate the Mach number at exit,

$Ma=\frac{V}{a}=\frac{V}{\sqrt{kRT}}=\frac{235m/s}{\sqrt{1.4\left(287m^2/s^2.°K\right)\left(275.52K\right)}}=0.706$

Calculate the pressure ratio,

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

In above equation,

$p_0$ - Tank pressure

$p$ - Exit pressure

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.706\right)}^2\right]}^{1.4/1.4-1}=1.395$

Therefore,

$p_0=1.395p\to \left(1\right)$

Apply hydrostatic formula for above system,

$p_0-{\gamma }_{mercury}h=p$

Assume, the specific weight of mercury as,

${\gamma }_{mercury}=133100N/m^3$

$p_0-p=\left(133100N/m^3\right)\left(0.3m\right)=39930Pa$

Therefore,

$p_0-p=39930Pa\to \left(2\right)$

According to equation 1 and 2,

The pressure inside the tank is equal to,

$p_0=141kPa$

Conclusion:

The pressure inside the tank is equal to,

$p_0=141kPa$.

To determine

(b)

The atmospheric pressure.

Answer to Problem 9.32P

$p=101kPa$

Explanation of Solution

Given information:

The nozzle exit velocity is equal to $235m/s$

$h=30cm$

The stagnation temperature is defined as,

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

Where, $C_p=1005m^2/s^2.°K$

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 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:

Convert,

$T_0=30°C=303K$

Calculate the exit temperature,

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

Substitute for known values,

$303K=T+\frac{{\left(235m/s\right)}^2}{2\left(1005m^2/s^2.°K\right)}$

Therefore,

$T=275.52K$

Calculate the Mach number at exit,

$Ma=\frac{V}{a}=\frac{V}{\sqrt{kRT}}=\frac{235m/s}{\sqrt{1.4\left(287m^2/s^2.°K\right)\left(275.52K\right)}}=0.706$

Calculate the pressure ratio,

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

In the above equation,

$p_0$ - Tank pressure

$p$ - Exit pressure

$\frac{p_0}{p}={\left[1+\frac{1}{2}\left(1.4-1\right){\left(0.706\right)}^2\right]}^{1.4/1.4-1}=1.395$

Therefore,

$p_0=1.395p\to \left(1\right)$

Apply hydrostatic formula for the above system,

$p_0-{\gamma }_{mercury}h=p$

Assume, the specific weight of mercury as,

${\gamma }_{mercury}=133100N/m^3$

$p_0-p=\left(133100N/m^3\right)\left(0.3m\right)=39930Pa$

Therefore,

$p_0-p=39930Pa\to \left(2\right)$

According to equation 1 and 2,

The atmospheric pressure is equal to,

$p=101kPa$

Conclusion:

The atmospheric pressure is equal to,

$p=101kPa$.

To determine

(c)

To calculate:

The Mach number at exit.

Answer to Problem 9.32P

$Ma=0.706$

Explanation of Solution

Given information:

The nozzle exit velocity is equal to $235m/s$

$h=30cm$

The stagnation temperature is defined as,

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

Where, $C_p=1005m^2/s^2.°K$

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

Calculation:

Convert,

$T_0=30°C=303K$

Calculate the exit temperature,

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

Substitute for known values,

$303K=T+\frac{{\left(235m/s\right)}^2}{2\left(1005m^2/s^2.°K\right)}$

Therefore,

$T=275.52K$

Calculate the Mach number at exit,

$Ma=\frac{V}{a}=\frac{V}{\sqrt{kRT}}=\frac{235m/s}{\sqrt{1.4\left(287m^2/s^2.°K\right)\left(275.52K\right)}}=0.706$

Conclusion:

The Mach number at exit is equal to $0.706$.