Chapter 7.13, Problem 111P

Nitrogen gas is compressed from 80 kPa and 27°C to 480 kPa by a 10-kW compressor. Determine the mass flow rate of nitrogen through the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n = 1.3, (c) isothermal, and (d) ideal two-stage polytropic with n = 1.3.

To determine

The mass flow rate of nitrogen through the compressor for isentropic compression process

Answer to Problem 111P

The mass flow rate of nitrogen through the compressor for isentropic compression process is $0.048\text{kg}/\text{s}$.

Explanation of Solution

Write the expression for the power input of the compressor for isentropic compression process.

${\dot{W}}_{comp,in}=\dot{m}\frac{kR{T}_1}{k-1}\left[{\left(\frac{P_2}{P_1}\right)}^{\left(k-1\right)/k}-1\right]$ (I)

Here, power input of the compressor is ${\dot{W}}_{comp,in}$, mass flow rate of nitrogen is $\dot{m}$, specific heat ratio is k, gas constant is R, initial temperature is $T_1$, initial pressure is $P_1$ and final pressure is $P_2$.

Conclusion:

From Table A-1, “Ideal-gas specific heats of various common gases”, the value of gas constant $\left(R\right)$ is $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for nitrogen gas.

From Table A-2, “Ideal-gas specific heats of various common gases”, the value of specific heat ratio $\left(k\right)$ is $1.4$ for the temperature of $300\text{K}$.

Substitute 10 kW for ${\dot{W}}_{comp,in}$, 1.4 for k, $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for R, $27°\text{C}$ for $T_1$, $\text{80 kPa}$ for $P_1$ and $\text{480 kPa}$ for $P_2$ in Equation (I).

$\begin{array}{l}10\text{kW}=\dot{m}\frac{\left(1.4\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left(27°\text{C}\right)}{1.4-1}\left[{\left(\frac{480\text{kPa}}{80\text{kPa}}\right)}^{\left(1.4-1\right)/1.4}-1\right]\\ 10\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)=\dot{m}\frac{\left(1.4\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(300+273\right)\text{K}\right]}{0.4}\left[{\left(\frac{480\text{kPa}}{80\text{kPa}}\right)}^{\left(0.4\right)/1.4}-1\right]\\ 10\text{kJ}/\text{s}\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)=\dot{m}\left(208.3346\right)\\ \dot{m}=0.048\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of nitrogen through the compressor for isentropic compression process is $0.048\text{kg}/\text{s}$.

To determine

The mass flow rate of nitrogen through the compressor for polytropic compression process.

Answer to Problem 111P

The mass flow rate of nitrogen through the compressor for polytropic compression process is $0.051\text{kg}/\text{s}$.

Explanation of Solution

Write the expression for the the power input of the compressor for polytropic compression process.

${\dot{W}}_{comp,in}=\dot{m}\frac{nR{T}_1}{n-1}\left[{\left(\frac{P_2}{P_1}\right)}^{\left(n-1\right)/n}-1\right]$ (II)

Here, power input of the compressor is ${\dot{W}}_{comp,in}$, mass flow rate of nitrogen is $\dot{m}$, polytropic index is n, gas constant is R, initial temperature is $T_1$, initial pressure is $P_1$ and final pressure is $P_2$.

Conclusion:

Substitute 10 kW for ${\dot{W}}_{comp,in}$, 1.3 for n, $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for R, $27°\text{C}$ for $T_1$, $\text{80 kPa}$ for $P_1$ and $\text{480 kPa}$ for $P_2$ in Equation (II).

$\begin{array}{l}10\text{kW}=\dot{m}\frac{\left(1.3\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left(27°\text{C}\right)}{1.3-1}\left[{\left(\frac{480\text{kPa}}{80\text{kPa}}\right)}^{\left(1.3-1\right)/1.3}-1\right]\\ 10\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)=\dot{m}\frac{\left(1.3\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(300+273\right)\text{K}\right]}{0.3}\left[{\left(\frac{480\text{kPa}}{80\text{kPa}}\right)}^{\left(0.3\right)/1.3}-1\right]\\ 10\text{kJ}/\text{s}\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)=\dot{m}\left(197.579\right)\\ \dot{m}=0.051\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of nitrogen through the compressor for polytropic compression process is $0.051\text{kg}/\text{s}$.

To determine

The mass flow rate of nitrogen through the compressor for isothermal compression process.

Answer to Problem 111P

The mass flow rate of nitrogen through the compressor for isothermal compression process is $0.063\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to calculate the power input of the compressor for isothermal compression process.

${\dot{W}}_{comp,in}=\dot{m}R{T}_1\ln \left(\frac{P_2}{P_1}\right)$ (III)

Here, power input of the compressor is ${\dot{W}}_{comp,in}$, mass flow rate of nitrogen is $\dot{m}$, gas constant is R, initial temperature is $T_1$, initial pressure is $P_1$ and final pressure is $P_2$.

Conclusion:

Substitute 10 kW for ${\dot{W}}_{comp,in}$, $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for R, $27°\text{C}$ for $T_1$, $\text{80 kPa}$ for $P_1$ and $\text{480 kPa}$ for $P_2$ in Equation (III).

$\begin{array}{l}10\text{kW}=\dot{m}\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left(27°\text{C}\right)\ln \left(\frac{480\text{kPa}}{80\text{kPa}}\right)\\ 10\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)=\dot{m}\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(300+273\right)\text{K}\right]\ln \left(\frac{480\text{kPa}}{80\text{kPa}}\right)\\ 10\text{kJ}/\text{s}\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)=\dot{m}\left(159.538\right)\\ \dot{m}=0.063\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of nitrogen through the compressor for isothermal compression process is $0.063\text{kg}/\text{s}$.

To determine

The mass flow rate of nitrogen for two stage compression process.

Answer to Problem 111P

The mass flow rate of nitrogen for two stage compression process is $0.056\text{kg}/\text{s}$.

Explanation of Solution

Write the expression to calculate the even pressure or pressure ratio $\left(P_x\right)$.

$P_x=\sqrt{P_1{P}_2}$ (IV)

Write the expression to calculate the power input of the compressor for two stage compression process.

${\dot{W}}_{comp,in}=2\dot{m}\frac{nR{T}_1}{n-1}\left[{\left(\frac{P_x}{P_1}\right)}^{\left(n-1\right)/n}-1\right]$ (V)

Here, power input of the compressor is ${\dot{W}}_{comp,in}$, mass flow rate of nitrogen is $\dot{m}$,  index is n, gas constant is R, initial temperature is $T_1$, even pressure is $P_x$ and final pressure is $P_2$.

Conclusion:

Substitute $\text{80 kPa}$ for $P_1$ and $\text{480 kPa}$ for $P_2$ in Equation (IV).

$\begin{array}{c}{P}_x=\sqrt{\left(80\text{kPa}\right)\left(480\text{kPa}\right)}\\ =196\text{kPa}\end{array}$

Substitute 10 kW for ${\dot{W}}_{comp,in}$, 1.3 for n, $0.2968\text{kJ}/\text{kg}\cdot \text{K}$ for R, $27°\text{C}$ for $T_1$, $\text{80 kPa}$ for $P_1$ and $\text{196 kPa}$ for $P_x$ in Equation (V).

$\begin{array}{l}10\text{kW}=2\dot{m}\frac{\left(1.3\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left(27°\text{C}\right)}{1.3-1}\left[{\left(\frac{196\text{kPa}}{80\text{kPa}}\right)}^{\left(1.3-1\right)/1.3}-1\right]\\ 10\text{kW}\left(\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right)=2\dot{m}\frac{\left(1.3\right)\left(0.2968\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(300+273\right)\text{K}\right]}{0.3}\left[{\left(\frac{196\text{kPa}}{80\text{kPa}}\right)}^{\left(0.3\right)/1.3}-1\right]\\ 10\text{kJ}/\text{s}\left(\frac{1\text{kPa}\cdot {\text{m}}^{\text{3}}}{1\text{kJ}}\right)=\dot{m}\left(177.273\right)\\ \dot{m}=0.056\text{kg}/\text{s}\end{array}$

Thus, the mass flow rate of nitrogen for two stage compression process is $0.056\text{kg}/\text{s}$.