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. Assum... 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 polytropic process equation

The polytropic process is given by $pv^n = \text{constant}$, where $n = 1.20$. Therefore, we can write

$p_1V_1^n = p_2V_2^n$

Solving for $p_2$:

$p_2 = p_1 \left(\frac{V_1}{V_2}\right)^n = 20 \text{ lbf/in}^2 \left(\frac{9 \text{ ft}^3}{1 \text{ ft}^3}\right)^{1.20} = 20 \cdot 9^{1.2} \text{ lbf/in}^2 \approx 347.75 \text{ lbf/in}^2$

2. Calculate the work done $W$ during the polytropic process

The work done during a polytropic process is given by:

$W = \frac{p_2V_2 - p_1V_1}{1-n}$

First, convert the pressures to lbf/ft²:

$p_1 = 20 \frac{\text{lbf}}{\text{in}^2} \cdot \frac{144 \text{ in}^2}{1 \text{ ft}^2} = 2880 \text{ lbf/ft}^2$

$p_2 = 347.75 \frac{\text{lbf}}{\text{in}^2} \cdot \frac{144 \text{ in}^2}{1 \text{ ft}^2} = 50076 \text{ lbf/ft}^2$

Next, calculate the work:

$W = \frac{(50076 \text{ lbf/ft}^2)(1 \text{ ft}^3) - (2880 \text{ lbf/ft}^2)(9 \text{ ft}^3)}{1 - 1.20} = \frac{50076 - 25920}{-0.20} \text{ ft-lbf} = \frac{24156}{-0.20} \text{ ft-lbf} = -120780 \text{ ft-lbf}$

Convert the work to Btu:

$W = -120780 \text{ ft-lbf} \cdot \frac{1 \text{ Btu}}{778 \text{ ft-lbf}} \approx -155.24 \text{ Btu}$

3. Calculate the final temperature $T_2$ using the ideal gas law and the polytropic equation

The polytropic relation can be written as:

$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{n-1}$

$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{n-1} = 500^{\circ}\text{R} \left(\frac{9}{1}\right)^{1.20-1} = 500 \cdot 9^{0.2} \approx 500 \cdot 1.5518 \approx 775.9^{\circ}\text{R}$

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

Assuming constant specific heats, the change in internal energy is given by:

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

To find $m$, we need to use the ideal gas law: $p_1V_1 = mRT_1$. We can look up $R$ and $c_v$ for air at $T_1 = 500^{\circ} R$.

$R = 0.3704 \frac{\text{psia ft}^3}{\text{lbm} ^\circ\text{R}} = 53.35 \frac{\text{ft lbf}}{\text{lbm} ^\circ\text{R}}$ $c_v = 0.172 \frac{\text{Btu}}{\text{lbm} ^\circ\text{R}}$

$m = \frac{p_1V_1}{RT_1} = \frac{(2880 \text{ lbf/ft}^2)(9 \text{ ft}^3)}{(53.35 \text{ ft lbf/lbm}^\circ\text{R})(500^\circ\text{R})} = \frac{25920}{26675} \text{ lbm} \approx 0.9717 \text{ lbm}$

$\Delta U = (0.9717 \text{ lbm}) \left(0.172 \frac{\text{Btu}}{\text{lbm}^\circ\text{R}}\right) (775.9^\circ\text{R} - 500^\circ\text{R}) \approx (0.9717)(0.172)(275.9) \text{ Btu} \approx 46.02 \text{ Btu}$

5. Calculate the heat transfer $Q$ using the first law of thermodynamics

The first law of thermodynamics states:

$Q = \Delta U + W$

Therefore,

$Q = 46.02 \text{ Btu} + (-155.24 \text{ Btu}) = -109.22 \text{ Btu}$