Chapter 7.13, Problem 110P

Helium gas is compressed from 16 psia and 85°F to 120 psia at a rate of 10 ft³/s. Determine the power input to the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n = 1.2, (c) isothermal, and (d) ideal two-stage polytropic with n = 1.2.

To determine

The power input of the compressor for isentropic compression process

Answer to Problem 110P

The power input of the compressor for isentropic compression process is $129.8\text{hp}$.

Explanation of Solution

Write the expression to calculate the mass flow rate $\dot{m}$.

$\dot{m}=\frac{P_1{\dot{\nu }}_1}{R{T}_1}$ (I)

Here, initial pressure is $P_1$, rate of initial volume is ${\dot{\nu }}_1$, gas constant is $R$ , and initial temperature is $T_1$.

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

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

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-2E, “Ideal-gas specific heats of various common gases”, the value of gas constant $\left(R\right)$ is $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for and the value of specific heat ratio $\left(k\right)$ is $1.4$ for helium gas.

Substitute $16\text{psia}$ for $P_2$, $10{\text{ft}}^3/\text{s}$ for $V_2$, $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$ and $545\text{R}$ for $T_1$ in Equation (I).

$\begin{array}{c}\dot{m}=\frac{16\text{psia}\times 10{\text{ft}}^3/\text{s}}{0.496\text{Btu}/\text{lbm}\cdot \text{R}\times 545\text{R}}\\ =\frac{16\text{psia}\times 10{\text{ft}}^3/\text{s}}{0.496\text{Btu}/\text{lbm}\cdot \text{R}\left(\frac{5.40395\text{psia}\cdot {\text{ft}}^3}{1\text{Btu}}\right)\times 545\text{R}}\\ =0.1095\text{Ibm}/\text{s}\end{array}$

Substitute $0.1095\text{Ibm}/\text{s}$ for $\dot{m}$, $1.667$ for $k$, $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $545\text{R}$ for $T_1$, $120\text{psia}$ for $P_2$ and $16\text{psia}$ for $P_1$ in Equation (II).

$\begin{array}{c}{\dot{W}}_{comp,in}=\left\{\begin{array}{l}\left(0.1095\text{Ibm}/\text{s}\right)\frac{\left(1.667\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)}{1.667-1}\\ \times \left[{\left(\frac{120\text{psia}}{16\text{psia}}\right)}^{\left(1.667-1\right)/1.667}-1\right]\end{array}\right\}\\ {\dot{W}}_{comp,in}=\left\{\begin{array}{l}\left(0.1095\text{Ibm}/\text{s}\right)\frac{\left(1.667\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)}{0.667}\\ \times \left[{\left(\frac{120\text{psia}}{16\text{psia}}\right)}^{\left(0.667\right)/1.667}-1\right]\end{array}\right\}\end{array}$

$\begin{array}{c}{\dot{W}}_{comp,in}=91.74\text{Btu}/\text{s}\left(\frac{1\text{hp}}{\text{0}\text{.7068}\text{Btu}/\text{s}}\right)\\ =129.796\text{hp}\\ \text{=129}\text{.8}\text{hp}\end{array}$

Thus, the power input of the compressor for isentropic compression process is $129.8\text{hp}$.

To determine

The power input of the compressor for polytropic compression process.

Answer to Problem 110P

The power input of the compressor for polytropic compression process is $100.3\text{hp}$.

Explanation of Solution

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

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

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 $1.2$ for $n$ , $0.1095\text{Ibm}/\text{s}$ for $\dot{m}$, $1.667$ for $k$, $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $545\text{R}$ for $T_1$, $120\text{psia}$ for $P_2$ and $16\text{psia}$ for $P_1$ in Equation (III).

$\begin{array}{c}{\dot{W}}_{comp,in}=\left\{\begin{array}{l}\left(0.1095\text{Ibm}/\text{s}\right)\frac{\left(1.2\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)}{1.2-1}\\ \times \left[{\left(\frac{120\text{psia}}{16\text{psia}}\right)}^{\left(1.2-1\right)/1.2}-1\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left(0.1095\text{Ibm}/\text{s}\right)\frac{\left(1.2\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)}{0.2}\\ \times \left[{\left(\frac{120\text{psia}}{16\text{psia}}\right)}^{\left(0.2\right)/1.2}-1\right]\end{array}\right\}\\ =70.89\text{Btu}/\text{s}\left(\frac{1\text{hp}}{\text{0}\text{.7068}\text{Btu}/\text{s}}\right)\\ =100.2971\text{hp}\\ =100.3\text{hp}\end{array}$

Thus, the power input of the compressor for polytropic compression process is $100.3\text{hp}$.

To determine

The power input of the compressor for isothermal compression process.

Answer to Problem 110P

The power input of the compressor for isothermal compression process is $84.42\text{hp}$.

Explanation of Solution

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

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

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 $0.1095\text{Ibm}/\text{s}$ for $\dot{m}$, $1.667$ for $k$, $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $545\text{R}$ for $T_1$, $120\text{psia}$ for $P_2$ and $16\text{psia}$ for $P_1$ in Equation (IV).

$\begin{array}{c}{\dot{W}}_{comp,in}=\left(0.1095\text{Ibm}/\text{s}\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)\ln \left(\frac{120\text{psia}}{16\text{psia}}\right)\\ =59.67\text{Btu}/\text{s}\left(\frac{1\text{hp}}{\text{0}\text{.7068}\text{Btu}/\text{s}}\right)\\ =84.42\text{hp}\end{array}$

Thus, the power input of the compressor for isothermal compression process is $84.42\text{hp}$.

To determine

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

Answer to Problem 110P

The expression to calculate the power input of the compressor for two stage compression process is $105.28\text{hp}$.

Explanation of Solution

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

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

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

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

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 $120\text{psia}$ for $P_2$ and $16\text{psia}$ for $P_1$ in Equation (V).

$\begin{array}{c}{P}_x=\sqrt{\left(16\text{psia}\right)\left(120\text{psia}\right)}\\ =43.82\text{psia}\end{array}$

Substitute $43.82\text{psia}$ for $P_x$ , $1.2$ for $n$ , $0.1095\text{Ibm}/\text{s}$ for $\dot{m}$, $0.496\text{Btu}/\text{lbm}\cdot \text{R}$ for R, $545\text{R}$ for $T_1$, and $16\text{psia}$ for $P_1$ in Equation (VI).

$\begin{array}{c}{\dot{W}}_{comp,in}=\left\{\begin{array}{l}2\left(0.1095\text{Ibm}/\text{s}\right)\frac{\left(1.2\right)\left(0.496\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(545\text{R}\right)}{1.2-1}\\ \times \left[{\left(\frac{43.82\text{psia}}{16\text{psia}}\right)}^{\left(1.2-1\right)/1.2}-1\right]\end{array}\right\}\\ =74.41\text{Btu}/\text{s}\left(\frac{1\text{hp}}{\text{0}\text{.7068}\text{Btu}/\text{s}}\right)\\ =105.28\text{hp}\end{array}$

Thus, the expression to calculate the power input of the compressor for two stage compression process is $105.28\text{hp}$.