Analyzing Polytropic Processes of Air as an Ideal Gas
Air undergoes a polytropic compression in a piston–cylinder assembly from p_{1}=1 atm , T_{1}=70^{\circ} F \text { to } p_{2}=5 atm. Employing the ideal gas model with constant specific heat ratio k, determine the work and heat transfer per unit mass, in Btu/lb, if (a) n = 1.3, (b) n = k. Evaluate k at T_{1}.
Known Air undergoes a polytropic compression process from a given initial state to a specified final pressure.
Find Determine the work and heat transfer, each in Btu/lb.
Schematic and Given Data:
Engineering Model
1. The air is a closed system.
2. The air behaves as an ideal gas with constant specific heat ratio k evaluated at the initial temperature.
3. The compression is polytropic and the piston is the only work mode.
4. There is no change in kinetic or potential energy.
Analysis The work can be evaluated in this case from the expression
W=\int_{1}^{2} p d V
With Eq. 3.57
\int_{1}^{2} p d V=\frac{m R\left(T_{2}-T_{1}\right)}{1-n} \quad \text { (ideal gas, } n \neq 1) (3.57)
\frac{W}{m}=\frac{R\left(T_{2}-T_{1}\right)}{1-n} (a)
The heat transfer can be evaluated from an energy balance. Thus,
\frac{Q}{m}=\frac{W}{m}+\left(u_{2}-u_{1}\right)
Inspection of Eq. 3.47b shows that when the specific heat ratio k is constant, c_{v} is constant. Thus,
c_{v}(T)=\frac{R}{k-1} (3.47b)
\frac{Q}{m}=\frac{W}{m}+c_{v}\left(T_{2}-T_{1}\right) (b)
a. For n = 1.3, the temperature at the final state, T_{2}, can be evaluated from Eq. 3.56 as follows
\frac{T_{2}}{T_{1}}=\left(\frac{p_{2}}{p_{1}}\right)^{(n-1) / n}=\left(\frac{V_{1}}{V_{2}}\right)^{n-1} \quad \text { (ideal gas) } (3.56)
T_{2}=T_{1}\left(\frac{p_{2}}{p_{1}}\right)^{(n-1) / n}=530^{\circ} R \left(\frac{5}{1}\right)^{(1.3-1) / 1.3}=768^{\circ} R \left(308^{\circ} F \right)
Using Eq. (a), the work is then
\frac{W}{m}=\frac{R\left(T_{2}-T_{1}\right)}{1-n}=\left(\frac{1.986 Btu }{28.97 lb \cdot{ }^{\circ} R }\right)\left(\frac{768^{\circ} R -530^{\circ} R }{1-1.3}\right)
=-54.39 Btu / lb
At 70°F, Table A-20E gives k=1.401 \text { and } c_{v}=0.171 Btu / lb \cdot{ }^{\circ} R. Alternatively, c_{v} can be found using Eq. 3.47b, as follows:
c_{v}=\frac{R}{k-1}
=\frac{(1.986 / 28.97) Btu / lb \cdot{ }^{\circ} R }{(1.401-1)}=0.171 \frac{ Btu }{ lb \cdot{ }^{\circ} R } (c)
Substituting values into Eq. (b), we get
\frac{Q}{m}=-54.39 \frac{ Btu }{ lb }+\left(0.171 \frac{ Btu }{ lb \cdot{ }^{\circ} R }\right)\left(768^{\circ} R -530^{\circ} R \right)
=-13.69 \frac{ Btu }{ lb }
b. For n = k, substituting Eqs. (a) and (c) into Eq. (b) gives
\frac{Q}{m}=\frac{R\left(T_{2}-T_{1}\right)}{1-k}+\frac{R\left(T_{2}-T_{1}\right)}{k-1}=0
That is, no heat transfer occurs in the polytropic process of an ideal gas for which n = k.
1 The states visited in a polytropic compression process are shown by the curve on the accompanying p–v diagram. The magnitude of the work per unit of mass is represented by the shaded area below the curve.
Skills Developed
Ability to…
• evaluate work using Eq. 2.17.
W=\int_{V_{1}}^{V_{2}} p d V (2.17)
• apply the energy balance using the ideal gas model.
• apply the polytropic process concept.
Quick Quiz
For n = k, evaluate the temperature at the final state, in °R and °F. Ans. 840°R (380°F)
