A piston-cylinder assembly contains 2 lb of water, initially at 100 lbf/in^2 and 350°F. The water undergoes two processes in series: a constant-pressure process followed by a const... A piston-cylinder assembly contains 2 lb of water, initially at 100 lbf/in^2 and 350°F. The water undergoes two processes in series: a constant-pressure process followed by a constant volume process. At the end of the constant-volume process, the temperature is 300°F and the water is a two-phase liquid-vapor mixture with a quality of 60%. Neglect kinetic and potential energy effects. Determine the work and heat transfer for each process, all in Btu. Read more

Steps to Solve

1. Define the initial state and find specific volume and enthalpy

The initial state (state 1) is defined by $P_1 = 100 \text{ lbf/in}^2$ and $T_1 = 350^\circ\text{F}$. From the steam tables (or a thermodynamics property calculator), we find that the specific volume $v_1 = 3.9784 \text{ ft}^3/\text{lb}$ and the specific enthalpy $h_1 = 1199.3 \text{ Btu/lb}$.

2. Define the intermediate state and find specific volume and enthalpy

State 2 is at the same pressure as state 1, so $P_2 = P_1 = 100 \text{ lbf/in}^2$. The specific volume $v_2$ will be the same as $v_3$ which we will find in the next step

3. Define the final state and determine specific enthalpy

State 3 is defined by $T_3 = 300^\circ\text{F}$ and a quality $x_3 = 0.6$. Since we have a saturated mixture at state 3, we can find the saturation pressure: $P_{sat}(300^\circ\text{F}) = 67.03 \text{ lbf/in}^2$. We also know that $v_3 = v_2$. Now we can find the specific volume at state 3 using: $v_3 = v_f + x_3(v_g - v_f)$, where $v_f$ and $v_g$ are the specific volumes of the saturated liquid and saturated vapor at $300^\circ\text{F}$, respectively. From the steam tables, $v_f = 0.01745 \text{ ft}^3/\text{lb}$ and $v_g = 6.3345 \text{ ft}^3/\text{lb}$. Thus, $v_3 = 0.01745 + 0.6(6.3345 - 0.01745) = 3.8077 \text{ ft}^3/\text{lb}$. Similarly, $h_3 = h_f + x_3(h_g - h_f)$, where $h_f = 269.73 \text{ Btu/lb}$ and $h_g = 1164.1 \text{ Btu/lb}$: $h_3 = 269.73 + 0.6(1164.1 - 269.73) = 806.35 \text{ Btu/lb}$.

4. Determine the state 2 temperature

Since $P_2 = 100 \text{ lbf/in}^2$ and $v_2 = v_3 = 3.8077 \text{ ft}^3/\text{lb}$, we can interpolate between $T=320^\circ\text{F}$ and $T=340^\circ\text{F}$ from the $P = 100 \text{ lbf/in}^2$ steam table. So $T_2 = 320 + (3.8077-3.4769)/(3.8475-3.4769)*(340-320) = 337.87^\circ\text{F}$. You can also use a thermodynamics property calculator using $P$ and $v$ as inputs. Using the calculator gives you $h_2 = 1197.1 \text{ Btu/lb}$.

5. Calculate the work for the constant-pressure process (1-2)

$W_{12} = mP_1(v_2 - v_1) = 2 \text{ lb} \times 100 \frac{\text{lbf}}{\text{in}^2} (3.8077 - 3.9784) \frac{\text{ft}^3}{\text{lb}} \times 144 \frac{\text{in}^2}{\text{ft}^2} \times \frac{1 \text{ Btu}}{778 \text{ ft lbf}} = -6.33 \text{ Btu}$.

6. Calculate the heat transfer for the constant-pressure process (1-2)

$Q_{12} = m(h_2 - h_1) = 2 \text{ lb} \times (1197.1 - 1199.3) \frac{\text{Btu}}{\text{lb}} = -4.4 \text{ Btu}$.

7. Calculate the work for the constant-volume process (2-3)

Since the volume is constant, $W_{23} = 0 \text{ Btu}$.

8. Calculate the heat transfer for the constant-volume process (2-3)

$Q_{23} = m(u_3 - u_2)$. We know that $u = h - Pv$, so $u_3=h_3-P_3v_3$ and $u_2=h_2-P_2v_2$. So, $Q_{23} = m[(h_3-P_3v_3) - (h_2-P_2v_2)]$. We need to convert $P$ in $\text{lbf/in}^2$ and $v$ in $\text{ft}^3/\text{lb}$ to units of Btu/lb. $P_3v_3 = 67.03 \frac{\text{lbf}}{\text{in}^2} \times 3.8077 \frac{\text{ft}^3}{\text{lb}} \times 144 \frac{\text{in}^2}{\text{ft}^2} \times \frac{1 \text{ Btu}}{778 \text{ ft lbf}} = 47.34 \text{ Btu/lb}$. $P_2v_2 = 100 \frac{\text{lbf}}{\text{in}^2} \times 3.8077 \frac{\text{ft}^3}{\text{lb}} \times 144 \frac{\text{in}^2}{\text{ft}^2} \times \frac{1 \text{ Btu}}{778 \text{ ft lbf}} = 70.59 \text{ Btu/lb}$. $Q_{23} = 2[(806.35 - 47.34)-(1197.1-70.59)] = 2[759.01 - 1126.51] = -735 \text{ Btu}$.