Chapter 6, Problem 163RQ

(a)

To determine

The exit temperature of the air.

Explanation of Solution

Given:

The inlet pressure $\left(P_1\right)$ is $1\text{ MPa}$.

The inlet temperature $\left(T_1\right)$ is $550\text{ K}$.

The mass flow rate $\left(\dot{m}\right)$ is $800\text{kg/min}$.

The heat transfer rate $\left({\dot{Q}}_1\right)$ is $2700\text{kJ/s}$.

The inlet temperature of regenerator ${\left(T_{in}\right)}_{reg}$ is $800\text{ K}$.

The exit temperature of regenerator ${\left(T_{out}\right)}_{reg}$ is $600\text{ K}$.

The inlet pressure of regenerator ${\left(P_{in}\right)}_{reg}$ is $140\text{ kPa}$.

The exit pressure of regenerator ${\left(P_{out}\right)}_{reg}$ is $130\text{ kPa}$.

Consider the system is in steady state. Hence, the inlet and exit mass flow rates are equal.

The mass flow rate of air $\left({\dot{m}}_{\text{air}}\right)$ is as follows.

  ${\dot{m}}_1={\dot{m}}_2={\dot{m}}_{\text{air}}$

The mass flow rate of exhaust gas $\left({\dot{m}}_{\text{eg}}\right)$ is as follows.

  ${\dot{m}}_3={\dot{m}}_4={\dot{m}}_{\text{eg}}$

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

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

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 system is at steady state. Hence, the rate of change in net energy of the system becomes zero.

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

Neglect the work transfer, potential and kinetic energies. The heat transfer occurs from the exhaust gas to the air.

Consider the air inlet and outlet alone.

The Equations (I) reduced as follows for air.

  $\begin{array}{l}\left[{\dot{Q}}_1+0+{\dot{m}}_{\text{air}}\left(h_1+0+0\right)\right]-\left[0+0+{\dot{m}}_{\text{air}}\left(h_2+0+0\right)\right]=0\\ {\dot{Q}}_1+{\dot{m}}_{\text{air}}{h}_1-{\dot{m}}_{\text{air}}{h}_2=0\\ {\dot{Q}}_1={\dot{m}}_{\text{air}}\left(h_2-h_1\right)\end{array}$        (II)

Rearrange the Equation (II) to obtain exit enthalpy $\left(h_2\right)$.

  $h_2=\frac{{\dot{Q}}_1}{{\dot{m}}_{\text{air}}}+h_1$        (III)

For air:

At inlet:

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the inlet enthalpy $\left(h_1\right)$ corresponding to the temperature of $550\text{K}$.

  $h_1=555.74\text{kJ/kg}$

Calculation:

Substitute  ${\dot{Q}}_{\text{1}}=2700\text{kJ/s}$, ${\dot{m}}_{\text{air}}=800\text{kg/min}$ and $h_1=555.74\text{kJ/kg}$ in

Equation (III)

  $\begin{array}{c}{h}_2=\frac{2700\text{kJ/s}}{800\text{kg/min}}+555.74\text{kJ/kg}\\ =\frac{2700\text{kJ/s}}{800\text{kg/min}\times \frac{1\text{min}}{60\text{s}}}+555.74\text{kJ/kg}\\ =202.5\text{kJ/kg}+555.74\text{kJ/kg}\\ =758.24\text{kJ/kg}\end{array}$

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the exit temperature of air $\left(T_2\right)$ corresponding to the enthalpy of $758.24\text{kJ/kg}$-using interpolation method.

Write the formula of interpolation method of two variables.

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

Show the temperature and enthalpy values from the Table A-17 as in below table.

S.No.	x	y
	Enthalpy $\left(h\right)$, in $\text{kJ/kg}$	Temperature $\left(T\right)$,in $\text{K}$
1	756.44	740
2	758.24	?
3	767.29	750

Substitute $x_1=756.44$, $x_2=758.24$, $x_3=767.29$, $y_1=740$, and $y_3=750$ in Equation (IV).

  $\begin{array}{c}{y}_2=\frac{\left(758.24-756.44\right)\left(750-740\right)}{\left(767.29-756.44\right)}+740\\ =741.6589\text{K}\\ \simeq 741\text{K}\end{array}$

Thus, the exit temperature of the air is $741\text{K}$.

(b)

To determine

The mass flow rate of exhaust gas.

Explanation of Solution

Consider the exhaust gas inlet and outlet alone.

The heat transfer occurs from the exhaust gas to the air. The heat losses from the exhaust gas.

The Equations (I) reduced as follows for exhaust gas.

  $\begin{array}{l}\left[0+0+{\dot{m}}_{\text{eg}}\left(h_3+0+0\right)\right]-\left[{\dot{Q}}_2+0+{\dot{m}}_{\text{eg}}\left(h_4+0+0\right)\right]=0\\ {\dot{m}}_{\text{eg}}{h}_3-\left({\dot{Q}}_2+{\dot{m}}_{\text{eg}}{h}_4\right)=0\\ {\dot{m}}_{\text{eg}}{h}_3-{\dot{Q}}_2-{\dot{m}}_{\text{eg}}{h}_4=0\\ {\dot{m}}_{\text{eg}}\left(h_3-h_4\right)={\dot{Q}}_2\end{array}$

  ${\dot{m}}_{\text{eg}}=\frac{{\dot{Q}}_2}{h_3-h_4}$        (V)

Heat loss by the exhaust gas is equal to the heat gained by the air.

  ${\dot{Q}}_{\text{eg}}={\dot{Q}}_{\text{air}}={\dot{Q}}_2=2700\text{kJ/s}$

For exhaust gas:

Consider the exhaust gas as air and refer the property tables of air.

At inlet:

Refer Table A-17, “Ideal-gas properties of air”.

Obtain the inlet enthalpy $\left(h_3\right)$ corresponding to the temperature of $800\text{K}$.

  $h_3=821.95\text{kJ/kg}$

Obtain the exit enthalpy $\left(h_4\right)$ corresponding to the temperature of $600\text{K}$.

  $h_4=607.02\text{kJ/kg}$

Calculation:

Substitute ${\dot{Q}}_{\text{2}}=2700\text{kJ/s}$, $h_3=821.95\text{kJ/kg}$ and $h_4=607.02\text{kJ/kg}$ in

Equation (V).

  $\begin{array}{c}{\dot{m}}_{\text{eg}}=\frac{2700\text{kJ/s}}{821.95\text{kJ/kg}-607.02\text{kJ/kg}}\\ =\frac{2700\text{kJ/s}}{214.93\text{kJ/kg}}\\ =12.5622\text{kg/s}\\ \simeq 12.6\text{kg/s}\end{array}$

Thus, the mass flow rate of exhaust gas is $12.6\text{kg/s}$.