The following results refer to a test on a petrol engine.



Indicator power - 30Kw, Brake power - 26Kw, Engine speed-1000rpm, Fuel per brake power hr. - 0.35kg, C.V. of fuel used - 43, 900 KJ/kg,

Calculate,

i)

Indicated thermal efficiency

ii) Brake thermal efficiency

iii) Mechanical efficiency

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).

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

---

### 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?

---

**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.

The following results refer to a test on a petrol engine.

Indicator power - 30Kw, Brake power - 26Kw, Engine speed-1000rpm, Fuel per brake power hr. - 0.35kg, C.V. of fuel used - 43, 900 KJ/kg,

Calculate,

i)

Indicated thermal efficiency

ii) Brake thermal efficiency

iii) Mechanical efficiency

Calculate indicated thermal efficiency, brake thermal efficiency, and mechanical efficiency for a petrol engine with given power and fuel parameters. This solution includes efficiency formulas and step-by-step calculation details, suitable for engineering students.

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