Subtopic: Power efficiencies and measurement, engine performance curves
Question Statement:
A Morse test is conducted on a four-cylinder, four-stroke agricultural diesel engine running at a constant speed of $2000\text{ rpm}$. The total brake power with all four cylinders firing is $BP = 36.0\text{ kW}$. When each cylinder is cut off one by one, the measured brake powers are: $BP_1 = 25.5\text{ kW}$, $BP_2 = 25.8\text{ kW}$, $BP_3 = 25.2\text{ kW}$, and $BP_4 = 25.5\text{ kW}$. The mechanical efficiency of the engine expressed as a percentage is ________ $\%$ (round off to 2 decimal places).
Answer Key:
85.71
Step-by-Step Solution & Explanation
```
Morse Test Calculations:
Total BP = 36.0 kW
IP_i = BP_total - BP_cut_i
```
**Step-by-step Calculation:**
1. Indicated power of each individual cylinder:
$IP_1 = BP - BP_1 = 36.0 - 25.5 = 10.5\text{ kW}$
$IP_2 = BP - BP_2 = 36.0 - 25.8 = 10.2\text{ kW}$
$IP_3 = BP - BP_3 = 36.0 - 25.2 = 10.8\text{ kW}$
$IP_4 = BP - BP_4 = 36.0 - 25.5 = 10.5\text{ kW}$
2. Total indicated power ($IP$):
$IP = IP_1 + IP_2 + IP_3 + IP_4 = 10.5 + 10.2 + 10.8 + 10.5 = 42.0\text{ kW}$
3. Engine friction power ($FP$):
$FP = IP - BP = 42.0 - 36.0 = 6.0\text{ kW}$
4. Mechanical efficiency ($\eta_m$):
$\eta_m = \frac{BP}{IP} \times 100 = \frac{36.0}{42.0} \times 100 = \frac{6}{7} \times 100 = 85.714\% \approx 85.71\%$