Let's go through the calculations step-by-step:

Given Data:

• Indicator Power (IP) = 30 kW
• Brake Power (BP) = 26 kW
• Engine Speed = 1000 rpm (not needed for these calculations)
• Fuel per brake power hour = 0.35 kg
• Calorific Value (C.V.) of fuel = 43,900 kJ/kg

We'll address each part of the question individually.

Step 1: Indicated Thermal Efficiency

Indicated thermal efficiency ($\eta_{th,ind}$) is defined as the ratio of the indicator power (IP) to the fuel energy input rate. The fuel energy input rate is given by: $\text{Fuel Energy Input Rate} = \text{Fuel Consumption Rate} \times \text{Calorific Value}$

Calculating Fuel Consumption Rate: Since the fuel consumption per brake power hour is 0.35 kg, the fuel consumption per hour is: $\text{Fuel Consumption} = 0.35 \, \text{kg} \times \text{Brake Power (BP)} = 0.35 \times 26 = 9.1 \, \text{kg/hr}$

Converting this rate to seconds: $\text{Fuel Consumption Rate} = \frac{9.1 \, \text{kg}}{3600 \, \text{s}} \approx 0.00253 \, \text{kg/s}$

Calculating Fuel Energy Input Rate: $\text{Fuel Energy Input Rate} = 0.00253 \times 43900 \approx 1110.67 \, \text{kJ/s} = 1110.67 \, \text{kW}$

Indicated Thermal Efficiency: $\eta_{th,ind} = \frac{\text{Indicator Power (IP)}}{\text{Fuel Energy Input Rate}} = \frac{30}{1110.67} \approx 0.027 = 2.7\%$

Step 2: Brake Thermal Efficiency

Brake thermal efficiency ($\eta_{th,brake}$) is the ratio of the brake power (BP) to the fuel energy input rate.

$\eta_{th,brake} = \frac{\text{Brake Power (BP)}}{\text{Fuel Energy Input Rate}} = \frac{26}{1110.67} \approx 0.0234 = 2.34\%$

Step 3: Mechanical Efficiency

Mechanical efficiency ($\eta_{mech}$) is the ratio of brake power (BP) to indicator power (IP).

$\eta_{mech} = \frac{\text{Brake Power (BP)}}{\text{Indicator Power (IP)}} = \frac{26}{30} \approx 0.867 = 86.7\%$

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Final Answers:

1. Indicated Thermal Efficiency: $2.7\%$
2. Brake Thermal Efficiency: $2.34\%$
3. Mechanical Efficiency: $86.7\%$

Would you like more details on any part of this solution?

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Further Questions for Practice:

1. How would an increase in engine speed affect the efficiencies calculated above?
2. What is the role of the calorific value in determining thermal efficiency?
3. How would a decrease in fuel per brake power hour impact brake thermal efficiency?
4. Why is mechanical efficiency typically less than 100%?
5. If the brake power increased to 28 kW, what would be the new mechanical efficiency?

Tip: Always check if the units are consistent, especially when calculating energy and power efficiencies.