Question: revolution of crankshaft. Comment on the type of engine.
5. A tractor engine delivering 75 kW uses fuel at the rate of 23 L/hr.
The fuel density is 0.84 kg/L and its energy value is 36000 kJ/L.
Calculate (a) specific fuel consumption in kg/kWh (b) percent thermal efficiency.
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1. **Problem Statement:**
Calculate the (a) specific fuel consumption in kg/kWh and (b) percent thermal efficiency of a tractor engine delivering $75$ kW power, using fuel at $23$ L/hr, with fuel density $0.84$ kg/L and energy value $36000$ kJ/L.
2. **Given Data:**
- Power output, $P = 75$ kW
- Fuel consumption rate, $V = 23$ L/hr
- Fuel density, $\rho = 0.84$ kg/L
- Energy value of fuel, $E = 36000$ kJ/L
3. **Formulas:**
- Specific fuel consumption (SFC) in kg/kWh:
$$\text{SFC} = \frac{\text{mass of fuel consumed per hour}}{\text{power output}} = \frac{m}{P}$$
- Thermal efficiency ($\eta$) in percentage:
$$\eta = \frac{\text{power output (kW)} \times 3600}{\text{energy input per hour (kJ)}} \times 100$$
4. **Calculations:**
- Mass of fuel consumed per hour:
$$m = V \times \rho = 23 \times 0.84 = 19.32 \text{ kg/hr}$$
- Specific fuel consumption:
$$\text{SFC} = \frac{19.32}{75} = 0.2576 \text{ kg/kWh}$$
- Energy input per hour:
$$Q_{in} = V \times E = 23 \times 36000 = 828000 \text{ kJ/hr}$$
- Power output in kJ/hr:
$$P_{out} = 75 \times 3600 = 270000 \text{ kJ/hr}$$
- Thermal efficiency:
$$\eta = \frac{270000}{828000} \times 100 = 32.61\%$$
5. **Final Answers:**
- (a) Specific fuel consumption = $0.2576$ kg/kWh
- (b) Thermal efficiency = $32.61\%$
This means the engine consumes about $0.2576$ kg of fuel to produce one kWh of power and converts about $32.61\%$ of the fuel's energy into useful work.