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. Determine the initial specific volume and enthalpy

Given $P_1 = 100 \text{ lbf/in}^2$ and $T_1 = 350^\circ \text{F}$, we can find the specific volume $v_1$ and enthalpy $h_1$ from the saturated water tables or superheated water tables. Since $T_1 > T_{sat}$ at $P_1$, the water is in a superheated state. From the superheated water tables, we find $v_1 = 3.9724 \text{ ft}^3/\text{lb}$ and $h_1 = 1205.3 \text{ Btu/lb}$.

2. Determine the state after the constant-pressure process

Since the first process is constant pressure, $P_2 = P_1 = 100 \text{ lbf/in}^2$. The second process is constant volume, so $v_3 = v_2$. We can determine the specific volume and enthalpy at state 3. Given that at state 3 we have $T_3 = 300^\circ \text{F}$ and $x_3 = 0.6$, we can find $v_3$ and $h_3$:

$v_3 = v_f + x_3v_{fg}$

$h_3 = h_f + x_3h_{fg}$

From the saturated water tables at $300^\circ \text{F}$, we find $v_f = 0.01745 \text{ ft}^3/\text{lb}$, $v_{fg} = 6.354 \text{ ft}^3/\text{lb}$, $h_f = 269.73 \text{ Btu/lb}$, and $h_{fg} = 900.7 \text{ Btu/lb}$.

$v_3 = 0.01745 + 0.6(6.354) = 3.83 \text{ ft}^3/\text{lb}$

$h_3 = 269.73 + 0.6(900.7) = 810.15 \text{ Btu/lb}$

3. Determine the state after the constant pressure process

Since $v_2 = v_3 = 3.83 \text{ ft}^3/\text{lb}$ and $P_2 = 100 \text{ lbf/in}^2$, we can evaluate state 2 to determine the enthalpy $h_2$. Since $v_2 < v_1$, the water cools during the constant pressure process. From the superheated water tables at $100 \text{ lbf/in}^2$ we see that at $350^\circ \text{F}$, $v_1 = 3.9724 \text{ ft}^3/\text{lb}$. Also, from the saturated water tables at $100 \text{ lbf/in}^2$, $v_g = 4.434 \text{ ft}^3/\text{lb}$. Since $v_2 < v_g$ at $100 \text{ lbf/in}^2$, the water at state 2 is wet (saturated mixture). Thus at $100 \text{ lbf/in}^2$ we interpolate between $T = 300^\circ \text{F}$ and $T = 350^\circ \text{F}$.

We can calculate the quality $x_2$ as follows, using $v_f = 0.01774 \text{ ft}^3/\text{lb}$ and $v_{fg} = 4.416 \text{ ft}^3/\text{lb}$ at $P_2 = 100 \text{ lbf/in}^2$. $v_2 = v_f + x_2 v_{fg}$ $3.83 = 0.01774 + x_2 (4.416)$ $x_2 = (3.83 - 0.01774) / 4.416 = 0.8634$

The enthalpy at state 2 is therefore

$h_2 = h_f + x_2 h_{fg} = 298.35 + (0.8634)(889.1) = 1066.5 \text{ Btu/lb}$

4. Calculate the work for the constant-pressure process

The work during the constant-pressure process (1 to 2) is given by:

$W_{12} = \int_{V_1}^{V_2} P , dV = P(V_2 - V_1) = mP(v_2 - v_1)$

$W_{12} = 2 \text{ lb} \cdot 100 \frac{\text{lbf}}{\text{in}^2} \cdot (3.83 - 3.9724)\frac{\text{ft}^3}{\text{lb}} \cdot \frac{144 \text{ in}^2}{1 \text{ ft}^2} \cdot \frac{1 \text{ Btu}}{778 \text{ ft}\cdot\text{lbf}} = -5.27 \text{ Btu}$

5. Calculate the heat transfer for the constant-pressure process

The heat transfer during the constant-pressure process (1 to 2) is given by:

$Q_{12} = m(h_2 - h_1) = 2 \text{ lb}(1066.5 - 1205.3) \text{ Btu/lb} = -277.6 \text{ Btu}$

6. Calculate the work for the constant-volume process

Since the volume is constant during process 2 to 3, the work done is zero:

$W_{23} = 0 \text{ Btu}$

7. Calculate the heat transfer for the constant-volume process

The heat transfer during the constant-volume process (2 to 3) is given by:

$Q_{23} = m(u_3 - u_2) = m[(h_3 - P_3v_3) - (h_2 - P_2v_2)]$

Since $v_3=v_2$,

$Q_{23} = m(h_3 - h_2) - (P_3 - P_2)V = m(h_3 - h_2)$ since $W_{23} = 0$

From saturated water tables, at $T_3 = 300^\circ \text{F}$, $P_{sat} = 67.013 \text{ lbf/in}^2 = P_3$

$Q_{23} = 2 \text{ lb} \times (810.15 - 1066.5) \text{ Btu/lb} = -512.7 \text{ Btu}$