Question Bank • QB_FP_ADV_036 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses

Question Statement:

A tractor diesel engine operating at full load develops a brake power of $BP = 40\text{ kW}$ with a brake specific fuel consumption of $\text{BSFC} = 240\text{ g/kW}\cdot\text{h}$. The lower heating value of the diesel fuel is $CV = 42.5\text{ MJ/kg}$. The brake thermal efficiency of the engine expressed as a percentage is ________ (round off to 2 decimal places).

Answer Key:

35.29

Step-by-Step Solution & Explanation

``` Brake Thermal Efficiency Formula: eta_bth = (3600 / (BSFC * CV)) * 100 ``` **Step-by-step Calculation:** 1. Brake Specific Fuel Consumption in $\text{kg/kW}\cdot\text{h}$: $\text{BSFC} = 240\text{ g/kW}\cdot\text{h} = 0.240\text{ kg/kW}\cdot\text{h}$ 2. Heat energy supplied per $\text{kW}\cdot\text{h}$ of brake work: $Q_{\text{in}} = \text{BSFC} \times CV = 0.240\text{ kg/kW}\cdot\text{h} \times 42.5\text{ MJ/kg} = 10.20\text{ MJ/kW}\cdot\text{h}$ 3. Since $1\text{ kW}\cdot\text{h} = 3.60\text{ MJ}$: $\eta_{\text{bth}} = \frac{3.60\text{ MJ}}{10.20\text{ MJ}} \times 100 = \frac{3.60}{10.20} \times 100 = 35.294\% \approx 35.29\%$
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