Question Bank • QB_FP_050 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Power efficiencies and measurement, engine performance curves

Question Statement:

A Morse test was conducted on a 4-cylinder 4-stroke tractor diesel engine running at $1800\text{ rpm}$. The total brake power with all cylinders firing was $BP_{\text{all}} = 32.0\text{ kW}$. The brake powers measured when each cylinder was successively cut out were: Cyl 1 cut out = $23.2\text{ kW}$, Cyl 2 cut out = $23.5\text{ kW}$, Cyl 3 cut out = $23.0\text{ kW}$, and Cyl 4 cut out = $23.3\text{ kW}$. The mechanical efficiency of the engine in percentage is ________ (round off to 1 decimal place).

Answer Key:

91.4

Step-by-Step Solution & Explanation

1. Indicated power of each individual cylinder ($IP_i = BP_{\text{all}} - BP_{-i}$): $IP_1 = 32.0 - 23.2 = 8.8\text{ kW}$ $IP_2 = 32.0 - 23.5 = 8.5\text{ kW}$ $IP_3 = 32.0 - 23.0 = 9.0\text{ kW}$ $IP_4 = 32.0 - 23.3 = 8.7\text{ kW}$ 2. Total indicated power ($IP_{\text{total}}$): $IP_{\text{total}} = 8.8 + 8.5 + 9.0 + 8.7 = 35.0\text{ kW}$ 3. Mechanical efficiency ($\eta_m$): $\eta_m = \frac{BP_{\text{all}}}{IP_{\text{total}}} \times 100 = \frac{32.0}{35.0} \times 100 = 91.428\% \approx 91.4\%$
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