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 specific volume and enthalpy at the initial state (State 1)

Given: $P_1 = 100 , \text{lbf/in}^2$, $T_1 = 350 , ^\circ\text{F}$. We look up the properties of water at this state in the steam tables. Since $T_1 > T_{sat}$ at $P_1$, the water is in a superheated vapor state. From superheated steam tables, we find $v_1 = 4.255 , \text{ft}^3/\text{lb}$ and $h_1 = 1205.3 , \text{Btu/lb}$.

2. Determine the specific volume at State 2

Since process 1-2 is a constant pressure process, $P_2 = P_1 = 100 , \text{lbf/in}^2$. We need to find the specific volume $v_2$. To find $v_2$, we need another property at state 2. We know that process 2-3 is constant volume, so $v_2 = v_3$.

3. Determine the specific volume at the final state (State 3)

Given: $T_3 = 300 , ^\circ\text{F}$, $x_3 = 0.6$. The specific volume for a two-phase mixture is given by $v_3 = v_{f3} + x_3(v_{g3} - v_{f3})$. From saturated water tables at $300 , ^\circ\text{F}$, we find $v_{f3} = 0.01745 , \text{ft}^3/\text{lb}$ and $v_{g3} = 6.344 , \text{ft}^3/\text{lb}$. Therefore, $v_3 = 0.01745 + 0.6(6.344 - 0.01745) = 3.813 , \text{ft}^3/\text{lb}$. Since $v_2 = v_3$, we have $v_2 = 3.813 , \text{ft}^3/\text{lb}$.

4. Find the enthalpy at State 2

We know $P_2 = 100 , \text{lbf/in}^2$ and $v_2 = 3.813 , \text{ft}^3/\text{lb}$. Comparing $v_2$ with $v_g$ at 100 lbf/in$^2$, we have $v_2 < v_g$. Thus, state 2 must be in the two-phase region. Looking up saturated water tables at $P_2 = 100 , \text{lbf/in}^2$ we have $v_f = 0.01774 , \text{ft}^3/\text{lb}$ and $v_g = 4.434 , \text{ft}^3/\text{lb}$. We can calculate the quality $x_2 = \frac{v_2 - v_f}{v_g - v_f} = \frac{3.813 - 0.01774}{4.434 - 0.01774} = 0.859$. Then $h_2 = h_{f2} + x_2(h_{g2} - h_{f2})$. From saturated water tables at $100 , \text{lbf/in}^2$, we have $h_{f2} = 298.33 , \text{Btu/lb}$ and $h_{g2} = 1150.5 , \text{Btu/lb}$. Thus, $h_2 = 298.33 + 0.859(1150.5 - 298.33) = 1029.8 , \text{Btu/lb}$.

5. Find the enthalpy at State 3

Given: $T_3 = 300 , ^\circ\text{F}$, $x_3 = 0.6$. The enthalpy for a two-phase mixture is given by $h_3 = h_{f3} + x_3(h_{g3} - h_{f3})$. From saturated water tables at $300 , ^\circ\text{F}$, we find $h_{f3} = 269.62 , \text{Btu/lb}$ and $h_{g3} = 1164.1 , \text{Btu/lb}$. Therefore, $h_3 = 269.62 + 0.6(1164.1 - 269.62) = 806.3 , \text{Btu/lb}$.

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

The work done during a constant pressure process is $W_{12} = mP_1(v_2 - v_1)$. Substituting our values: $W_{12} = 2 , \text{lb} \cdot 100 , \text{lbf/in}^2 \cdot (3.813 - 4.255) , \text{ft}^3/\text{lb} = -88.4 , \text{lbf} \cdot \text{ft}^3/\text{in}^2$. We need to convert the units to Btu: $W_{12} = -88.4 \cdot \frac{144 , \text{in}^2}{1 , \text{ft}^2} \cdot \text{lbf} \cdot \text{ft} \cdot \frac{1 , \text{Btu}}{778 , \text{lbf} \cdot \text{ft}} = -16.4 , \text{Btu}$. Applying the first law: $Q_{12} = m(h_2 - h_1) + W_{12} = 2 , \text{lb} \cdot (1029.8 - 1205.3) , \text{Btu/lb} + (-16.4) , \text{Btu} = -367.4 , \text{Btu}$.

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

Since the volume is constant, $W_{23} = 0$. Applying the first law: $Q_{23} = m(h_3 - h_2) + W_{23} = 2 , \text{lb} \cdot (806.3 - 1029.8) , \text{Btu/lb} + 0 = -447.0 , \text{Btu}$.