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 properties at the initial state (State 1)

Given: $P_1 = 100 \text{ lbf/in}^2$, $T_1 = 350^\circ \text{F}$, $m = 2 \text{ lb}$.

From the steam tables, we find that at $P_1 = 100 \text{ lbf/in}^2$ and $T_1 = 350^\circ \text{F}$, the specific volume $v_1 = 3.9775 \text{ ft}^3/\text{lb}$ and the specific enthalpy $h_1 = 1195.6 \text{ Btu/lb}$.

2. Determine the properties at the intermediate state (State 2)

The first process is a constant-pressure process, so $P_2 = P_1 = 100 \text{ lbf/in}^2$. State 2 is at the end of the constant pressure process and beginning of constant volume process. In the 3rd state, the water is a two-phase mixture, so we know the constant volume process has brought the water into a saturated mixture. Thus, state 2 must be a superheated vapor, as a first step towards becoming a saturated vapor.

3. Determine the properties at the final state (State 3)

Given: $T_3 = 300^\circ \text{F}$, $x_3 = 0.60$. Also, this is after the constant-volume process, so $v_3 = v_2$.

From the steam tables, we find the saturated liquid and vapor specific volumes at $T_3 = 300^\circ \text{F}$: $v_f = 0.01745 \text{ ft}^3/\text{lb}$ and $v_g = 6.3348 \text{ ft}^3/\text{lb}$.

Calculate $v_3$ using the quality $x_3$:

$v_3 = v_f + x_3(v_g - v_f)= 0.01745 + 0.60(6.3348 - 0.01745) = 3.8079 \text{ ft}^3/\text{lb}$

4. Determine the properties at state 2 continued

Since $v_2 = v_3 = 3.8079 \text{ ft}^3/\text{lb}$ and we know $P_2 = 100 \text{ lbf/in}^2$, we can find $h_2$ by interpolating from the superheated steam tables since the water changes into two-phase mixture in state 3.

At $100 \text{ lbf/in}^2$, $v_g =4.4324 \text{ ft}^3/\text{lb}$ at the saturated vapor state. And since the water changes from vapor to two-phase mixture in state 3, we can conclude that the second state is superheated. Thus, we can use the superheated tables. Find the two closest specific volumes: $v = 4.9372$ at $T = 400^\circ \text{F}$ with $h = 1231.5 \text{ Btu/lb}$ $v = 3.7256$ at $T = 300^\circ \text{F}$ with $h = 1192.4 \text{ Btu/lb}$

Now interpolate to find $h_2$:

$h_2 = 1192.4 + \frac{3.8079-3.7256}{4.9372-3.7256}(1231.5-1192.4) = 1195.02 \text{ Btu/lb}$

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

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

$W_{12} = (2 \text{ lb})(100 \text{ lbf/in}^2)(3.8079 \text{ ft}^3/\text{lb} - 3.9775 \text{ ft}^3/\text{lb})$ $W_{12} = -33.92 \text{ lbf} \cdot \text{ft}^3/\text{in}^2$

Convert to Btu:

$W_{12} = -33.92 \text{ lbf} \cdot \text{ft}^3/\text{in}^2 \times \frac{144 \text{ in}^2}{1 \text{ ft}^2} \times \frac{1 \text{ Btu}}{778 \text{ lbf} \cdot \text{ft}} = -6.27 \text{ Btu}$

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

$\Delta U = Q - W$, $Q = \Delta U + W$ Since $H = U + PV$, at constant pressure, $\Delta H = \Delta U + P\Delta V = \Delta U + W$. Also, $\Delta H = Q$.

$Q_{12} = m(h_2 - h_1)= 2 \text{ lb}(1195.02 \text{ Btu/lb} - 1195.6 \text{ Btu/lb}) = -1.16 \text{ Btu}$

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

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

8. Determine the specific enthalpy at state 3

Determine the saturated liquid and vapor enthalpies at $T_3 = 300^\circ \text{F}$ from the steam tables: $h_f = 269.62 \text{ Btu/lb}$ and $h_g = 1164.1 \text{ Btu/lb}$.

Calculate $h_3$ using the quality: $h_3 = h_f + x_3(h_g - h_f) = 269.62 + 0.60(1164.1 - 269.62) = 806.31 \text{ Btu/lb}$

9. Calculate the heat transfer the constant-volume process (Process 2-3)

$Q_{23} = \Delta U_{23} + W_{23}$ Since $W_{23} = 0$, $Q_{23} = \Delta U_{23}$ $Q_{23} = m(u_3 - u_2)$ $u = h - pv$, so $Q_{23} = m[(h_3 - p_3v_3) - (h_2 - p_2v_2)]$ Since $v_2 = v_3$, $Q_{23} = m[(h_3 - h_2) - v_2(p_3 - p_2)]$

To find $p_3$, since state 3 is a saturated mixture at $300^\circ \text{F}$, we can look up the saturation pressure $p_{sat}$ at this temperature: $p_3 = 67.03 \text{ lbf/in}^2$. Then,

$Q_{23} = 2[(806.31-1195.02) - 3.8079(67.03-100)\frac{144}{778}]$ $Q_{23} = 2[-388.71 - 3.8079(-32.97)(\frac{144}{778})] = -777.42 + 44.09= -733.3 \text{ Btu}$