Effect of clearance and volumetric efficiency

Clearance volume $V_c$ is the volume left in the cylinder (including the valve ports) when the piston is at the end of the delivery stroke. It is provided so that the piston does not strike the cylinder head and to leave room for the valves. The clearance ratio is $C = V_c/V_s$, where $V_s$ is the swept volume.

Effect of clearance

• The trapped air re-expands during the suction stroke, so suction starts only after the piston has moved some distance.
• The volume of fresh air drawn in is less than the swept volume, so the capacity falls.
• The work per kg of air delivered is not changed (the compression and re-expansion work of the clearance air cancel for the same index), but the machine must be larger for a given output.
• A greater pressure ratio and a larger clearance reduce the volume drawn in; at a high enough ratio no air is delivered at all.

Derivation of volumetric efficiency

   p
   p2 |   3+--+2
      |    |   \
      |     \   \
      |      \   \
   p1 |      4+---\--+1
      +-----+-+----+--+---- V
            Vc V4      V1
              <---Va--->

The swept volume is $V_s = V_1 - V_c$, so $V_1 = V_c + V_s$. The effective suction volume (air drawn in) is $V_a = V_1 - V_4$, where $V_4$ is the volume of the re-expanded clearance air.

The clearance air expands polytropically from $p_2$ to $p_1$: $p_2V_c^{\,n} = p_1V_4^{\,n}$, so

$$V_4 = V_c\left(\frac{p_2}{p_1}\right)^{1/n}$$ $$\begin{aligned} V_a &= V_s + V_c - V_c\left(\frac{p_2}{p_1}\right)^{1/n}\\ \eta_v &= \frac{V_a}{V_s} = 1 + \frac{V_c}{V_s} - \frac{V_c}{V_s}\left(\frac{p_2}{p_1}\right)^{1/n} \end{aligned}$$ $$\boxed{\eta_v = 1 + C - C\left(\frac{p_2}{p_1}\right)^{1/n}}$$

Hence $\eta_v$ decreases when the clearance ratio $C$ or the pressure ratio $p_2/p_1$ increases. This is one reason for using multistage compression at high pressure ratios.