Known Two pounds of air are compressed between two specified states while there is heat transfer from the air of a known amount.
Find Determine the work, in Btu.
Schematic and Given Data:
Engineering Model
1. The air is a closed system.
2. The initial and final states are equilibrium states. There is no change in kinetic or potential energy.
2 3. The air is modeled as an ideal gas.
4. The piston is the only work mode.
Analysis An energy balance for the closed system is
\Delta KE ^{\nearrow0}+\Delta PE ^{\nearrow0}+\Delta U=Q-W
where the kinetic and potential energy terms vanish by assumption 2. Solving for W
3 W=Q-\Delta U=Q-m\left(u_{2}-u_{1}\right)
From the problem statement, Q=-20 Btu. Also, from Table A-22E at T_{1}=540^{\circ} R , u_{1}=92.04 Btu / lb , \text { and at } T_{2}=840^{\circ} R , u_{2}=143.98 Btu/lb. Accordingly
W=-20 Btu -(2 lb )(143.98-92.04) Btu / lb =-123.9 Btu
The minus sign indicates that work is done on the system in the process.
1 Although the initial and final states are assumed to be equilibrium states, the intervening states are not necessarily equilibrium states, so the process has been indicated on the accompanying p–v diagram by a dashed line. This dashed line does not define a “path” for the process.
2 Table A-1E gives p_{ c }=37.2 atm , T_{ c }=239^{\circ} R for air. Therefore, at state 1, p_{ R 1}=0.03, T_{ R 1}=2.26, \text { and at state } 2, p_{ R 2}=0.16, T_{ R 2}=3.51 Referring to Fig. A-1, we conclude that at these states Z \approx 1, as assumed in the solution.
3 In principle, the work could be evaluated through ∫ p dV, but because the variation of pressure at the piston face with volume is not known, the integration cannot be performed without more information.
Skills Developed
Ability to…
• define a closed system and identify interactions on its boundary.
• apply the energy balance using the ideal gas model.
Quick Quiz
Replacing air by carbon dioxide, but keeping all other problem statement details the same, evaluate work, in Btu. Ans. −125.1 Btu