Air is compressed in a piston-cylinder assembly from p1 = 20 lbf/in², T1 = 500°R, V1 = 9 ft³ to a final volume of V2 = 1 ft³ in a process described by \(pv^{1.20} = constant\). Ass... Air is compressed in a piston-cylinder assembly from p1 = 20 lbf/in², T1 = 500°R, V1 = 9 ft³ to a final volume of V2 = 1 ft³ in a process described by \(pv^{1.20} = constant\). Assume ideal gas behavior and neglect kinetic and potential energy effects. Using constant specific heats evaluated at T1, determine the work and the heat transfer, in Btu. Read more

Steps to Solve

1. Calculate the final pressure $p_2$

Using the given relation $pv^{1.20} = constant$, we have: $p_1V_1^{1.20} = p_2V_2^{1.20}$ $p_2 = p_1 \left(\frac{V_1}{V_2}\right)^{1.20}$ $p_2 = 20 \left(\frac{9}{1}\right)^{1.20} = 20 \times 13.676 \approx 273.52 \text{ lbf/in}^2$

2. Calculate the work done, W

The work done for a polytropic process is given by: $W = \int_{V_1}^{V_2} pdV = \frac{p_2V_2 - p_1V_1}{1-n}$, where $n = 1.20$. $W = \frac{p_2V_2 - p_1V_1}{1-1.20} = \frac{(273.52 \text{ lbf/in}^2 \times 1 \text{ ft}^3) - (20 \text{ lbf/in}^2 \times 9 \text{ ft}^3)}{1-1.20}$

We need to convert $p_2V_2$ and $p_1V_1$ from $\text{lbf} \cdot \text{ft}^3/\text{in}^2$ to Btu. Recall that $1 \text{ Btu} = 778 \text{ lbf} \cdot \text{ft}$. Also, $1 \text{ ft}^2 = 144 \text{ in}^2$, so $1 \text{ ft}^3 = 144 \text{ in}^2 \cdot \text{ft}$.

$p_2V_2 = 273.52 \frac{\text{lbf}}{\text{in}^2} \times 1 \text{ ft}^3 = 273.52 \frac{\text{lbf}}{\text{in}^2} \times 144 \frac{\text{in}^2}{\text{ft}^2} \cdot \text{ft}^3 = 39386.88 \text{ lbf} \cdot \text{ft}$ $p_1V_1 = 20 \frac{\text{lbf}}{\text{in}^2} \times 9 \text{ ft}^3 = 180 \frac{\text{lbf}}{\text{in}^2} \times 144 \frac{\text{in}^2}{\text{ft}^2} \cdot \text{ft}^3 = 25920 \text{ lbf} \cdot \text{ft}$

$W = \frac{39386.88 - 25920}{1-1.20} = \frac{13466.88}{-0.20} = -67334.4 \text{ lbf} \cdot \text{ft}$ Converting to Btu: $W = \frac{-67334.4 \text{ lbf} \cdot \text{ft}}{778 \frac{\text{lbf} \cdot \text{ft}}{\text{Btu}}} = -86.55 \text{ Btu}$

3. Calculate the final temperature $T_2$

Using the ideal gas law and the polytropic process relation: $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{n-1}$ $T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{n-1} = 500 \left(\frac{9}{1}\right)^{1.20-1} = 500 \times 9^{0.20} = 500 \times 1.5518 \approx 775.9 \text{ °R}$

4. Calculate the change in internal energy $\Delta U$

$\Delta U = m c_v (T_2 - T_1)$

We need to find the mass $m$. Using the ideal gas law $p_1V_1 = mRT_1$ $R = 0.3704 \frac{\text{psia} \cdot \text{ft}^3}{\text{lbm} \cdot \text{°R}}$

$p_1 = 20 \frac{\text{lbf}}{\text{in}^2} = 20 \text{ psia}$ $m = \frac{p_1V_1}{RT_1} = \frac{20 \text{ psia} \times 9 \text{ ft}^3}{0.3704 \frac{\text{psia} \cdot \text{ft}^3}{\text{lbm} \cdot \text{°R}} \times 500 \text{ °R}} = \frac{180}{185.2} \approx 0.9719 \text{ lbm}$

$c_v$ at $T_1 = 500 \text{ °R}$ is $0.171 \frac{\text{Btu}}{\text{lbm} \cdot \text{°R}}$

$\Delta U = m c_v (T_2 - T_1) = 0.9719 \text{ lbm} \times 0.171 \frac{\text{Btu}}{\text{lbm} \cdot \text{°R}} \times (775.9 - 500) \text{ °R} = 0.9719 \times 0.171 \times 275.9 \approx 45.86 \text{ Btu}$

5. Calculate the heat transfer, Q

Using the first law of thermodynamics: $\Delta U = Q - W$, so $Q = \Delta U + W$ $Q = 45.86 \text{ Btu} + (-86.55 \text{ Btu}) = -40.69 \text{ Btu}$