Chapter 7.13, Problem 142P

a)

To determine

The rate of heat transfer to the air.

Answer to Problem 142P

The rate of heat transfer to the air is $205.7\text{kW}$.

Explanation of Solution

Write the expression for the energy balance equation for closed system.

${\dot{E}}_{in}-{\dot{E}}_{out}=\Delta {\dot{E}}_{system}$ (I).

Here, rate of net energy transfer into the control volume is ${\dot{E}}_{in}$, rate of net energy transfer exit from the control volume is ${\dot{E}}_{out}$ and rate of change in internal energy of system is $\Delta {\dot{E}}_{system}$.

Conclusion:

The rate of change in internal energy of the system is zero at the steady flow system.

Substitute 0 for $\Delta {\dot{E}}_{system}$ in Equation (I).

$\begin{array}{l}{\dot{E}}_{in}-{\dot{E}}_{out}=0\\ \dot{m}{h}_1={\dot{Q}}_{out}+\dot{m}{h}_2\\ {\dot{Q}}_{out}=\dot{m}\left(h_1-h_2\right)\\ {\dot{Q}}_{out}={\left[\dot{m}{c}_p\left(T_1-T_2\right)\right]}_{gas}\end{array}$ (II).

Here, mass flow rate is $\dot{m}$, enthalpy at initial state is $h_1$, enthalpy at final state is $h_2$, rate of heat transfer to the air is ${\dot{Q}}_{out}$, specific heat at constant pressure is $c_p$, inlet temperature is $T_1$ and exit temperature is $T_2$.

The specific heat at constant pressure $\left(c_p\right)$ for combustion gas is $1.10\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute $2.2\text{kg}/\text{s}$ for $\dot{m}$, $1.10\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $180°\text{C}$ for $T_2$ and $95°\text{C}$ for $T_1$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{out}=\left(2.2\text{kg}/\text{s}\right)\left(1.10\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(180°\text{C}-95°\text{C}\right)\\ =\left(2.2\text{kg}/\text{s}\right)\left(1.10\text{kJ}/\text{kg}\cdot °\text{C}\right)\left[90°\text{C}\right]\\ =205.7\text{kJ}/\text{s}\left(\frac{1\text{kW}}{\text{1}\text{kJ}/\text{s}}\right)\\ =205.7\text{kW}\end{array}$

Thus, the rate of heat transfer to the air is $205.7\text{kW}$.

b)

To determine

The outlet temperature of the air.

Answer to Problem 142P

The outlet temperature of the air is $133.2°\text{C}$.

Explanation of Solution

Write the expression to calculate the mass flow rate of air.

${\dot{m}}_{air}=\frac{P_1\dot{V}}{R{T}_3}$ (III)

Here, volume flow rate of air at inlet is $\dot{V}$, the temperature of air at inlet is $T_3$, gas constant is $R$ and the pressure of air inlet is $P_1$.

Write the expression for the rate of heat transfer output.

${\dot{Q}}_{in}={\dot{m}}_{\text{air}}{c}_p\left(T_4-T_3\right)$ (IV).

Here, heat gained by the air is ${\dot{Q}}_{in}$ , outlet temperature of air is $T_4$ and inlet temperature of air is $T_3$.

Conclusion:

Substitute $95\text{kPa}$ for $P_1$, $0.6{\text{m}}^3/\text{s}$ for $\dot{V}$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$ and $20°\text{C}$ for $T_3$ in Equation (I).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\frac{95\text{kPa}\left(0.6{\text{m}}^3/\text{s}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(20°\text{C}\right)}\\ =\frac{95\text{kPa}\left(0.6{\text{m}}^3/\text{s}\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(20+273\right)\text{K}}\\ =1.808\text{kg}/\text{s}\end{array}$

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

${\dot{Q}}_{in}={\dot{Q}}_{out}$

Substitute $20°\text{C}$ for $T_3$, $205.7\text{kW}$ for ${\dot{Q}}_{out}$, $1.808\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, and $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$ in Equation (III).

$\begin{array}{l}205.7\text{kW}=1.808\text{kg}/\text{s}\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(T_4-20°\text{C}\right)\\ T_4=20°\text{C+}\frac{205.7\text{kW}}{\left(1.808\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)}\\ T_4=133.2°\text{C}\end{array}$

Thus, the outlet temperature of the air is $133.2°\text{C}$.

c)

To determine

The rate of entropy generation

Answer to Problem 142P

The rate of entropy generation is $0.091\text{kW}/\text{K}$.

Explanation of Solution

Write the expression for the entropy balance in the heat exchanger.

${\dot{S}}_{in}-{\dot{S}}_{out}+{\dot{S}}_{gen}=\Delta {\dot{S}}_{system}$ (V).

Here, rate of net input entropy is ${\dot{S}}_{in}$, rate of net output entropy is ${\dot{S}}_{out}$, rate of entropy generation is ${\dot{S}}_{gen}$, and rate of change of entropy of the system is $\Delta {\dot{S}}_{system}$.

Conclusion:

Substitute ${\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3$ for ${\dot{S}}_{in}$, ${\dot{m}}_2{s}_2+{\dot{m}}_3{s}_4$ for ${\dot{S}}_{out}$ and 0 for $\Delta {\dot{S}}_{system}$ in Equation (V).

$\begin{array}{l}{\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-\left({\dot{m}}_2{s}_2+{\dot{m}}_3{s}_4\right)+{\dot{S}}_{gen}=0\\ {\dot{m}}_1{s}_1+{\dot{m}}_3{s}_3-{\dot{m}}_2{s}_2-{\dot{m}}_3{s}_4+{\dot{S}}_{gen}=0\end{array}$

$\begin{array}{l}{\dot{m}}_{\text{exhaust}}{s}_1+{\dot{m}}_{\text{air}}{s}_3-{\dot{m}}_{\text{exhaust}}{s}_2-{\dot{m}}_{\text{air}}{s}_4+{\dot{S}}_{gen}=0\\ {\dot{S}}_{gen}={\dot{m}}_{\text{exhaust}}\left(s_2-s_1\right)+{\dot{m}}_{\text{air}}\left(s_4-s_3\right)\\ {\dot{S}}_{gen}={\dot{m}}_{\text{exhaust}}{c}_p\ln \left(\frac{T_2}{T_1}\right)+{\dot{m}}_{\text{air}}{c}_p\ln \left(\frac{T_4}{T_3}\right)\end{array}$ (VI)

Here, mass flow rate at entry 1 is ${\dot{m}}_1$, mass flow rate at entry 2 is ${\dot{m}}_2$, mass flow rate at exit is ${\dot{m}}_3$, entropy at state 3 is $s_3$ , entropy at state 4 is $s_4$, entropy at state 2 is $s_2$ and entropy at state 1 is $s_1$.

Substitute $1.808\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,air}$, $95°\text{C}$ for $T_2$, $180°\text{C}$ for $T_1$.$2.2\text{kg}/\text{s}$ for ${\dot{m}}_{\text{exhaust}}$, $1.1\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{p,exhaust}$, $20°\text{C}$ for $T_3$, and $133.2°\text{C}$ for $T_4$ in Equation (VI).

$\begin{array}{c}{\dot{S}}_{gen}=\left[\begin{array}{l}\left(2.2\text{kg}/\text{s}\right)\left(1.1\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(95°\text{C}\right)}{\left(180°\text{C}\right)}\\ +\left(1.808\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(133.2°\text{C}\right)}{\left(20°\text{C}\right)}\end{array}\right]\\ =\left[\begin{array}{l}\left(2.2\text{kg}/\text{s}\right)\left(1.1\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(95+273\right)\text{K}}{\left(180+273\right)\text{K}}\\ +\left(1.808\text{kg}/\text{s}\right)\left(1.005\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{\left(133.2+273\right)\text{K}}{\left(20+273\right)\text{K}}\end{array}\right]\\ =0.091\text{kJ}/\text{s}\cdot \text{K}\left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\\ =0.091\text{kW}/\text{K}\end{array}$

Thus, the rate of entropy generation is $0.091\text{kW}/\text{K}$.