Work of single-stage compressor without clearance

With negligible clearance, air is taken in at $p_1, V_1$, compressed to $p_2$ and delivered completely.

   p
   p2 |3 +--+ 2
      |  |   \
      |  |    \
      |  |     \_
   p1 |4 +--------- + 1
      +-------------------- V
        0  V2         V1

• 4-1: suction at constant $p_1$ (work done on the piston by the incoming air $= p_1V_1$).
• 1-2: compression ($pV^n = C$ or $pV = C$).
• 2-3: delivery at constant $p_2$ (work done by the piston on the delivered air $= p_2V_2$).
• 3-4: pressure drop at zero volume (valve change-over).

Work done per cycle

The work per cycle is the enclosed area 4-1-2-3: compression work plus delivery work, minus the suction work $p_1V_1$ done on the piston by the incoming air:

$$W = -\int_1^2 p\,dV + p_2V_2 - p_1V_1 = \int_{p_1}^{p_2} V\,dp$$

(a) Isothermal ($pV = p_1V_1 = C$)

$$\begin{aligned} W_{iso} &= \int_{p_1}^{p_2} \frac{p_1V_1}{p}\,dp\\ &= p_1V_1 \ln\frac{p_2}{p_1} = mRT_1 \ln\frac{p_2}{p_1} \end{aligned}$$

(b) Polytropic ($pV^n = C$)

Compression work done on the gas (area under 1-2):

$$-\int_1^2 p\,dV = \frac{p_2V_2 - p_1V_1}{n-1}$$

Adding the delivery work and subtracting the suction work:

$$\begin{aligned} W_{poly} &= \frac{p_2V_2 - p_1V_1}{n-1} + p_2V_2 - p_1V_1\\ &= \frac{n}{n-1}\left(p_2V_2 - p_1V_1\right) \end{aligned}$$

Since $p_2V_2 = p_1V_1 (p_2/p_1)^{(n-1)/n}$,

$$W_{poly} = \frac{n}{n-1}\,p_1V_1\left[\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}} - 1\right] = \frac{n}{n-1}\,mR\,(T_2 - T_1)$$

Isothermal efficiency

$$\eta_{iso} = \frac{\text{isothermal work}}{\text{indicated (polytropic) work}} = \frac{W_{iso}}{W_{poly}}$$

Isothermal compression needs the least work, so $\eta_{iso}<1$. Better cooling lowers $n$ towards 1 and raises $\eta_{iso}$. Typical values of $n$ for air are 1.25 to 1.35.