## Related questions with answers

Air is compressed in an axial-flow compressor operating at steady state from $27^{\circ} \mathrm{C}, 1\ \text{bar}$ to a pressure of $2.1\ \text{bar}$. The work required is $94.6 \mathrm{~kJ}$ per $\mathrm{kg}$ of air flowing. Heat transfer from the compressor occurs at an average surface temperature of $40^{\circ} \mathrm{C}$ at the rate of $14 \mathrm{~kJ}$ per kg of air flowing. The effects of motion and gravity can be ignored. Let $T_0=20^{\circ} \mathrm{C}, p_0=1\ \text{bar}$. Assuming ideal gas behavior, determine (a) the temperature of the air at the exit, in ${ }^{\circ} \mathrm{C}$, and (b) the rate of exergy destruction within the compressor, in $\mathrm{kJ}$ per $\mathrm{kg}$ of air flowing,

Solution

Verified$\text{\color{#4257b2} a)}$ In this part of the problem we need to calculate the final temperature $T_2$. We will use the energy balance equation with the given assumptions that we can neglect motion and gravity effects and the ideal gas model for air. We will also need the given initial temperature $T_1=300\,\text{K}$, work done $w=94.6\,\text{kJ/kg}$ and heat transfer $q=14\,\text{kJ/kg}$.

$\begin{align*} 0&=q - w + (h_2 - h_1) \\ 0&=q - w + c_p \cdot (T_2 - T_1) \\ 0&=14\,\frac{\text{kJ}}{\text{kg}} - 94.6\,\frac{\text{kJ}}{\text{kg}} + 1.01\,\frac{\text{kJ}}{\text{kg K}} \cdot (T_2 - 300\,\text{K}) \\ T_2&=\boxed{\color{#c34632}380\,\text{K}} \\ \end{align*}$

$\text{\color{#4257b2} b)}$ In this part of the problem we will calculate the exergy destruction $\dot{e}_d$ using the exergy rate balance equation. For the calculation we will also need the given surface temperature $T_b=313\,\text{K}$ and ambient temperature $T_0=293\,\text{K}$ and pressures $p_1=1\,\text{bar}$ and $p_2=2.1\,\text{bar}$.

$\begin{align*} \dot{e}_d&=\left(1-\frac{T_0}{T_b}\right) \cdot q - w + (h_2 - h_1) - T_0 \cdot \left(s_2 - s_1\right) \\ \dot{e}_d&=\left(1-\frac{T_0}{T_b}\right) \cdot q - w + c_p \cdot (T_2 - T_1) - T_0 \cdot \left(c_p \ln \frac{T_2}{T_1} - R \ln \frac{p_2}{p_1}\right) \\ \dot{e}_d&=\left(1-\frac{293\,\text{K}}{313\,\text{K}}\right) \cdot 14\,\frac{\text{kJ}}{\text{kg}} - 94.6\,\frac{\text{kJ}}{\text{kg}} + 1.01\,\frac{\text{kJ}}{\text{kg K}} \cdot (380\,\text{K}- 300\,\text{K}) - \\ -& 293\,\text{K}\cdot \left(1.01\,\frac{\text{kJ}}{\text{kg K}} \ln \frac{380\,\text{K}}{300\,\text{K}} - 0.287\,\frac{\text{kJ}}{\text{kg K}} \ln \frac{2.1\,\text{bar}}{1\,\text{bar}}\right) \\ \dot{e}_d&=\boxed{\color{#c34632}-20.47\,\frac{\text{kJ}}{\text{kg}}} \\ \end{align*}$

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