Chapter 5.5, Problem 55P

Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.15-m² opening with a velocity of 30 m/s. It exits through a 0.08-m² opening. Calculate the mass flow rate of air and the required power input.

To determine

The mass flow rate of air and required power input.

Answer to Problem 55P

The mass flow rate of air is $5.3513\text{kg/s}$.

The required power input is $2084.0946\text{kW}$.

Explanation of Solution

Consider the air flows at steady state. Hence, the inlet and exit mass flow rates are equal.

${\dot{m}}_1={\dot{m}}_2=\dot{m}$

Write the formula for mass flow rate with inlet condition.

$\dot{m}=\frac{A_1{V}_1{P}_1}{R{T}_1}$ (I)

Here, the cross sectional area is $A$, the velocity is $V$, the pressure is $P$, the gas constant of air is $R$ and the temperature is $T$; the suffix 1 indicates the inlet condition.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant of air is $0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$.

Write the formula for exit velocity of the air.

$V_2=\frac{\dot{m}R{T}_2}{A_2{P}_2}$ (II)

Write the energy rate balance equation for one inlet and one outlet system.

$\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$ (III)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The air flows at steady state through the compressor. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Since, the compressor is adiabatic compressor, hence neglect the heat transfer rates. And also neglect the potential energy changes. The work done is on the system (compressor) and the work done by the system is zero i.e. ${\dot{W}}_2=0$.

The Equations (III) reduced as follows to obtain the work input.

$\begin{array}{l}\left[0+{\dot{W}}_1+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[0+0+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ {\dot{W}}_1+\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)=0\\ {\dot{W}}_1+\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\\ {\dot{W}}_1=\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)-\dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)\end{array}$

${\dot{W}}_1=\dot{m}\left(h_2-h_1+\frac{V_2{}^2-V_1{}^2}{2}\right)$ (IV)

Write the formula for change in enthalpy $\left(h_2-h_1\right)$.

$h_2-h_1=c_p\left(T_2-T_1\right)$

Substitute $c_p\left(T_2-T_1\right)$ for $\left(h_2-h_1\right)$ in Equation (IV).

${\dot{W}}_1=\dot{m}\left[c_p\left(T_2-T_1\right)+\frac{V_2{}^2-V_1{}^2}{2}\right]$ (V)

The average temperature of air is calculated as follows.

$\begin{array}{c}{T}_{avg}=\frac{T_1+T_2}{2}\\ =\frac{20°\text{C}+400°\text{C}}{2}\\ =210°\text{C}\\ =\left(\text{210+273}\right)\text{K}\end{array}$

$=483\text{K}$

Refer Table A-2 (b), “Ideal-gas specific heats of various common gases”.

Obtain the specific heat of air at constant pressure $\left(c_p\right)$ corresponding to the average temperature of $483\text{K}$ - using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (VI)

Show the temperature and enthalpy values from the Table A-2 (b) as in below table.

S.No.	x	y
	Temperature $\left(T\right)$,in $\text{K}$	Specific heat $\left(c_p\right)$, in $\text{kJ/kg}\cdot \text{K}$
1	450	1.020
2	483	?
3	500	1.029

Substitute $450$ for $x_1$, $483$ for $x_2$, $500$ for $x_3$, $1.020$ for $y_1$, and $1.029$ for $y_3$ in Equation (VI).

$\begin{array}{c}{y}_2=\frac{\left(483-450\right)\left(1.029-1.020\right)}{\left(500-450\right)}+1.020\\ =1.026\text{kJ/kg}\cdot \text{K}\end{array}$

Thus, the specific heat of air at constant pressure $\left(c_p\right)$ corresponding to the average temperature of $483\text{K}$ is $1.026\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $0.15{\text{m}}^2$ for $A_1$, $30\text{m/s}$ for $V_1$, $100\text{kPa}$ for $P_1$, $0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$ and $20\text{°C}$ for $T_1$ in Equation (I).

$\begin{array}{c}\dot{m}=\frac{\left(0.15{\text{m}}^2\right)\left(30\text{m/s}\right)\left(100\text{kPa}\right)}{\left(0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(20\text{°C}\right)}\\ =\frac{450\text{kPa}\cdot {\text{m}}^3\text{/s}}{\left(0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(20+273\right)\text{K}}\\ =\frac{450\text{kPa}\cdot {\text{m}}^3\text{/s}}{84.091\text{kPa}\cdot {\text{m}}^3\text{/kg}}\\ =5.3513\text{kg/s}\end{array}$

Thus, the mass flow rate of air is $5.3513\text{kg/s}$.

Substitute $5.3513\text{kg/s}$ for $\dot{m}$, $0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}$ for $R$, $400°\text{C}$ for $T_2$, $0.08{\text{m}}^2$ for $A_2$ and $1.8\text{MPa}$ for $P_2$ in Equation (II).

$\begin{array}{c}{V}_2=\frac{\left(5.3513\text{kg/s}\right)\left(0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(400°\text{C}\right)}{\left(0.08{\text{m}}^2\right)\left(1.8\text{MPa}\right)}\\ =\frac{\left(5.3513\text{kg/s}\right)\left(0.2870\text{kPa}\cdot {\text{m}}^3\text{/kg}\cdot \text{K}\right)\left(400+273\right)\text{K}}{\left(0.08{\text{m}}^2\right)\left(1.8\text{MPa}\times \frac{{10}^3\text{}\text{kPa}}{1\text{MPa}}\right)}\\ =\frac{1033.6177\text{kPa}\cdot {\text{m}}^3\text{/s}}{144\text{kPa}\cdot {\text{m}}^2}\\ =7.1779\text{m/s}\end{array}$

Substitute $5.3513\text{kg/s}$ for $\dot{m}$, $1.026\text{kJ/kg}\cdot \text{K}$ for $c_p$, $400°\text{C}$ for $T_2$, $20°\text{C}$ for $T_1$, $7.1779\text{m/s}$ for $V_2$ and $30\text{m/s}$ for $V_1$ in Equation (V).

$\begin{array}{c}{\dot{W}}_1=\left(5.3513\text{kg/s}\right)\left[\begin{array}{l}\left(1.026\text{kJ/kg}\cdot \text{K}\right)\left(400°\text{C}-20°\text{C}\right)\\ +\frac{{\left(7.1779\text{m/s}\right)}^2-{\left(30\text{m/s}\right)}^2}{2}\end{array}\right]\\ =\left(5.3513\text{kg/s}\right)\left[\begin{array}{l}\left(1.026\text{kJ/kg}\cdot \text{K}\right)\left[\left(400+273\right)\text{K}-\left(20+273\right)\text{K}\right]\\ -\left(424.2389{\text{m}}^2{\text{/s}}^2\times \frac{1\text{}\text{kJ/kg}}{1000{\text{m}}^2{\text{/s}}^2}\right)\end{array}\right]\\ =\left(5.3513\text{kg/s}\right)\left(389.88\text{kJ/kg}-0.42424\text{kJ/kg}\right)\\ =\left(5.3513\text{kg/s}\right)\left(389.4558\right)\end{array}$

$\begin{array}{c}=2084.0946\text{kJ/s}\times \frac{1\text{W}}{\text{J/s}}\\ =2084.0946\text{kW}\end{array}$

Thus, the required power input is $2084.0946\text{kW}$.