A piston-cylinder assembly contains 0.7 lb of propane. The propane expands from an initial state where p1 = 60 lbf/in² and T1 = 90°F to a final state where p2 = 10 lbf/in². During... A piston-cylinder assembly contains 0.7 lb of propane. The propane expands from an initial state where p1 = 60 lbf/in² and T1 = 90°F to a final state where p2 = 10 lbf/in². During the process, the pressure and specific volume are related by pv² = constant. Determine the energy transfer by work, in Btu. Read more

Steps to Solve

1. Write down the given information

$m = 0.7 \text{ lb}$ $p_1 = 60 \text{ lbf/in}^2$ $T_1 = 90^\circ \text{F}$ $p_2 = 10 \text{ lbf/in}^2$ $pv^2 = \text{constant}$

2. Write the equation for work for a process where $pv^2 = \text{constant}$

The work done during a process where $pv^n =$ constant is expressed as:

$W = \int_{V_1}^{V_2} pdV = \frac{p_2V_2 - p_1V_1}{1-n}$

In our situation $n=2$, so the equation simplifies to:

$W = \frac{p_2V_2 - p_1V_1}{1-2} = p_1V_1 - p_2V_2$

Since we're working with specific volume, we multiply both sides by mass $m$:

$W = m(p_1v_1 - p_2v_2)$

3. Use the relation $pv^2 = \text{constant}$ to express $v_2$ in terms of $v_1$

Since $p_1v_1^2 = p_2v_2^2$, then $v_2 = v_1\sqrt{\frac{p_1}{p_2}}$

4. Substitute $v_2$ into the work equation

$W = m(p_1v_1 - p_2v_1\sqrt{\frac{p_1}{p_2}}) = m p_1v_1(1 - \sqrt{\frac{p_2}{p_1}})$

5. Find $v_1$ using the ideal gas law: $p_1v_1 = R T_1$

From Table A-1E, we have the gas constant for propane $R = \frac{1544}{44.09} = 35.02 \frac{\text{ft} \cdot \text{lbf}}{\text{lb} \cdot ^\circ\text{R}}$

Convert the temperature from Fahrenheit to Rankine: $T_1 = 90 + 460 = 550 \text{ }^\circ\text{R}$

$v_1 = \frac{RT_1}{p_1} = \frac{35.02 \frac{\text{ft} \cdot \text{lbf}}{\text{lb} \cdot ^\circ\text{R}} \cdot 550 \text{ }^\circ\text{R}}{60 \frac{\text{lbf}}{\text{in}^2} \cdot 144 \frac{\text{in}^2}{\text{ft}^2}} = 2.224 \text{ ft}^3/\text{lb}$

6. Calculate the work

$W = m p_1v_1(1 - \sqrt{\frac{p_2}{p_1}}) = 0.7 \text{ lb} \cdot 60 \frac{\text{lbf}}{\text{in}^2} \cdot 144 \frac{\text{in}^2}{\text{ft}^2} \cdot 2.224 \frac{\text{ft}^3}{\text{lb}} (1 - \sqrt{\frac{10}{60}}) = 9747.8 \text{ ft} \cdot \text{lbf}$

7. Convert the work from ft$\cdot$lbf to Btu

$1 \text{ Btu} = 778 \text{ ft} \cdot \text{lbf}$

$W = 9747.8 \text{ ft} \cdot \text{lbf} \cdot \frac{1 \text{ Btu}}{778 \text{ ft} \cdot \text{lbf}} = 12.53 \text{ Btu}$