Chapter 22, Problem 31P

(a)

To determine

The temperature at exit.

Answer to Problem 31P

The temperature at exit is $564\text{K}$.

Explanation of Solution

Write the pressure volume relation in adiabatic expansion.

  $P_i{V}_i^{\gamma }=P_f{V}_f^{\gamma }$                                                                                                              (I)

Here, $P_i$ is the initial pressure, $P_f$ is the final pressure, $V_i$ is the initial volume, $V_f$ is the final volume and $\gamma$ is the adiabatic constant.

Write pressure volume temperature relation in adiabatic process.

  ${\left(\frac{P_i{V}_i}{T_i}\right)}^{\gamma }={\left(\frac{P_f{V}_f}{T_f}\right)}^{\gamma }$                                                                                                  (II)

Here, $T_i$ is the initial temperature and $T_f$ is the final temperature.

Divide equation (I) by (II) to get $T_f$.

  $\begin{array}{c}\frac{P_i{V}_i^{\gamma }}{{\left(\frac{P_i{V}_i}{T_i}\right)}^{\gamma }}=\frac{P_f{V}_f^{\gamma }}{{\left(\frac{P_f{V}_f}{T_f}\right)}^{\gamma }}\\ \frac{T_i^{\gamma }}{P_i^{\gamma -1}}=\frac{T_f^{\gamma }}{P_f^{\gamma -1}}\\ T_f^{\gamma }=T_i^{\gamma }\frac{P_f^{\gamma -1}}{P_i^{\gamma -1}}\\ T_f=T_i{\left(\frac{P_f}{P_i}\right)}^{\frac{\gamma -1}{\gamma }}\end{array}$                                                                                              (III)

Write the expression to convert temperature in degree Celsius into Kelvin scale.

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$                                                                                              (IV)

Here, $T\left(\text{K}\right)$ is the temperature in Kelvin scale and $T\left(°\text{C}\right)$ is the temperature in degree Celsius.

Conclusion:

For argon, value for adiabatic constant is $\frac{5}{3}$.

Substitute $800°\text{C}$ for $T\left(°\text{C}\right)$ in equation (IV) to get $T\left(\text{K}\right)$.

  $\begin{array}{c}T\left(\text{K}\right)=800°\text{C}+273\\ =1073\text{K}\end{array}$

Substitute $1073\text{K}$ for $T_i$, $\frac{5}{3}$ for $\gamma$, $300\text{kPa}$ for $P_f$ and $1.50\text{MPa}$ for $P_i$ in equation (III) to get $T_f$.

  $\begin{array}{c}{T}_f=\left(1073\text{K}\right){\left(\frac{300\text{kPa}\frac{1000\text{ Pa}}{1\text{ kPa}}}{1.50\text{MPa}\frac{{10}^6\text{ Pa}}{1\text{ MPa}}}\right)}^{\frac{\frac{5}{3}-1}{\frac{5}{3}}}\\ =\left(1073\text{K}\right){\left(\frac{300\times {10}^3\text{Pa}}{1.50\times {10}^6\text{Pa}}\right)}^{\frac{2}{5}}\\ =564\text{K}\end{array}$

Therefore, the temperature at exit is $564\text{K}$.

(b)

To determine

The maximum power output of the engine.

Answer to Problem 31P

The maximum power output of the engine is $212\text{kW}$ .

Explanation of Solution

Write the expression for the change in internal energy.

  $\Delta E_{\text{int}}=n{C}_V\Delta T$                                                                                                         (V)

Here, $\Delta E_{\text{int}}$ is the change in internal energy, $n$ is the number of moles, $C_V$ is the specific heat capacity at constant volume and $\Delta T$ is the change in temperature.

Write the expression for the change in internal energy using first law of thermodynamics.

  $\Delta E_{\text{int}}=Q-W_{\text{eng}}$                                                                                                      (VI)

Here, $Q$ is the heat absorbed by argon gas and $W_{\text{eng}}$ is the work done by the engine.

For an adiabatic process total heat change will be zero.

Substitute $0$ for $Q$ in equation (VI) to get $W_{\text{eng}}$.

  $\begin{array}{l}\Delta E_{\text{int}}=0-W_{\text{eng}}\\ W_{\text{eng}}=-\Delta E_{\text{int}}\end{array}$                                                                                                       (VII)

Substitute $n{C}_V\Delta T$ for $\Delta E_{\text{int}}$ in equation (VII) to get $W_{\text{eng}}$.

  $W_{\text{eng}}=-n{C}_V\Delta T$                                                                                                    (VIII)

Write the expression for the power output of the turbine.

  $P=\frac{W_{\text{eng}}}{\Delta t}$                                                                                                                (IX)

Here, $\Delta t$ is the time.

Substitute $-n{C}_V\Delta T$ for $W_{\text{eng}}$ in above equation to get $P$.

  $P=\frac{-n{C}_V\Delta T}{\Delta t}$                                                                                                            (X)

Write the expression for $n$.

  $n=m\left(\text{kg}\right)\times \frac{1\text{mol}}{0.0399\text{kg}}$

Here, $m\left(\text{kg}\right)$ is the mass in kilogram.

Write the expression for $\Delta T$.

  $\Delta T=T_f-T_i$

Here, $T_i$ is the initial temperature and $T_f$ is the final temperature.

Substitute $T_f-T_i$ for $\Delta T$ and $m\left(\text{kg}\right)\times \frac{1\text{mol}}{0.0399\text{kg}}$ for $n$ in equation (X) to get $P$.

  $P=\frac{-\left(m\left(\text{kg}\right)\times \frac{1\text{mol}}{0.0399\text{kg}}\right)C_V\left(T_f-T_i\right)}{\Delta t}$                                                                 (XI)

Conclusion:

Substitute $80.0\text{kg}$ for $m\left(\text{kg}\right)$, $\frac{3}{2}$ for $C_V$, $564\text{K}$ for $T_f$ and $1073\text{K}$ for $T_i$ and $1\text{min}$ for $\Delta t$ in equation (XI) to get $P$.

  $\begin{array}{c}P=\frac{-\left(80.0\text{kg}\times \frac{1\text{mol}}{0.0399\text{kg}}\right)\frac{3}{2}\left(564\text{K}-1073\text{K}\right)}{1\text{min}\times \frac{60\text{s}}{1\text{min}}}\\ =2.12\times {10}^5\text{W}\times \frac{1\text{kW}}{1000\text{W}}\\ =212\text{kW}\end{array}$

Therefore, the maximum power output of the engine is $212\text{kW}$ .

(c)

To determine

The maximum efficiency of the engine.

Answer to Problem 31P

The maximum efficiency of the engine is $0.475$.

Explanation of Solution

Write the expression for the efficiency of Carnot engine.

  $e_c=1-\frac{T_c}{T_h}$

Conclusion:

Substitute $564\text{K}$ for $T_c$ and $1073\text{K}$ for $T_h$ in above equation to get maximum efficiency.

  $\begin{array}{c}{e}_c=1-\frac{564\text{K}}{1073\text{K}}\\ =0.475\text{K}\end{array}$

Therefore, the maximum efficiency of the engine is $0.475$.