Morse test for friction power

Morse test

It is used for multi-cylinder engines. The engine is run at a fixed speed and load, and the brake power $B$ is measured with all cylinders firing. Then one cylinder is cut out (by shorting the spark plug in an SI engine, or cutting the fuel in a CI engine) and the load is reduced so that the speed returns to the original value. The new brake power $B_1$ is measured. The other cylinders keep running, and the friction and pumping power of the dead cylinder is still the same, so

$$IP_1 = B - B_1$$

is the indicated power of the cut-out cylinder. The process is repeated for each cylinder and $IP = \sum IP_i$. Then $FP = IP - B$ and $\eta_m = B/IP$. It gives a result without any indicator, but it assumes that friction does not change when a cylinder is cut out.

Calculation

$$\begin{aligned} IP &= 17.4 + 18.0 + 16.9 + 17.7 = 70.0\ \text{kW} \\ FP &= 70.0 - 62.5 = 7.5\ \text{kW} \\ \eta_m &= \frac{62.5}{70.0} = 0.893 \end{aligned}$$

Answer: cylinder indicated powers 17.4, 18.0, 16.9 and 17.7 kW; IP = 70.0 kW; FP = 7.5 kW; $\eta_m$ = 89.3%.