Stirring Water at Constant Volume
A well-insulated rigid tank having a volume of 10 ft ^{3} contains saturated water vapor at 212°F. The water is rapidly stirred until the pressure is 20 lbf / in ^{2} Determine the temperature at the final state, in °F, and the work during the process, in Btu.
Known By rapid stirring, water vapor in a well-insulated rigid tank is brought from the saturated vapor state at 212°F to a pressure of 20 lbf / in ^{2}
Find Determine the temperature at the final state and the work.
Schematic and Given Data:
Engineering Model
1. The water is a closed system.
2. The initial and final states are at equilibrium. There is no net change in kinetic or potential energy.
3. There is no heat transfer with the surroundings.
4. The tank volume remains constant.
Analysis To determine the final equilibrium state, the values of two independent intensive properties are required. One of these is pressure, p_{2}=20 lbf / in .^{2}, and the other is the specific volume: v_{2}=v_{1}. The initial and final specific volumes are equal because the total mass and total volume are unchanged in the process. The initial and final states are located on the accompanying T–υ and p–υ diagrams.
From Table A-2E, v_{1}=v_{ g }\left(212^{\circ} F \right)=26.80 ft ^{3} / lb , u_{1}=u_{ g }\left(212^{\circ} F \right)=1077.6 Btu / lb \text { . By using } v_{2}=v_{1} and interpolating in Table A-4E at p_{2}=20 lbf / in .^{2}
T_{2}=445^{\circ} F , \quad u_{2}=1161.6 Btu / lb
Next, with assumptions 2 and 3 an energy balance for the system reduces to
\Delta U+\Delta K E^{\nearrow0}+\Delta PE^{\nearrow0}={Q}^{\nearrow0}-W
On rearrangement
W=-\left(U_{2}-U_{1}\right)=-m\left(u_{2}-u_{1}\right)
To evaluate W requires the system mass. This can be determined from the volume and specific volume
m=\frac{V}{v_{1}}=\left(\frac{10 ft ^{3}}{26.8 ft ^{3} / lb }\right)=0.373 lb
Finally, by inserting values into the expression for W
W=-(0.373 lb )(1161.6-1077.6) Btu / lb =-31.3 Btu
where the minus sign signifies that the energy transfer by work is to the system.
1 Although the initial and final states are equilibrium states, the intervening states are not at equilibrium. To emphasize this, the process has been indicated on the T–υ and p–υ diagrams by a dashed line. Solid lines on property diagrams are reserved for processes that pass through equilibrium states only (quasiequilibrium processes). The analysis illustrates the importance of carefully sketched property diagrams as an adjunct to problem solving.
Skills Developed
Ability to…
• define a closed system and identify interactions on its boundary.
• apply the energy balance with steam table data.
• sketch T–υ and p–υ diagrams and locate states on them.
Quick Quiz
If insulation were removed from the tank and the water cooled at constant volume from T_{2}=445^{\circ} F \text { to } T_{3}=300^{\circ} F while no stirring occurs, determine the heat transfer, in Btu. Ans. −19.5 Btu
