Question

A 4-cylinder, 4-stroke diesel engine operating at 3000 rpm has a compression ratio \(r\) of 12 and cut-off ratio \(r_c\) of 2.5. The temperature rise during the heat addition process is 2400 K. The efficiency of an air-standard diesel cycle is given by: \[ \eta = 1 - \frac{1}{r^{\gamma-1}} \left( \frac{r_c^\gamma - 1}{\gamma (r_c - 1)} \right). \] Assume the working fluid as air with a mass flow rate of 0.05 kg/s, \(\gamma = 1.4\), and \(C_p = 1.004 \, \text{kJ/kg-K}\). The power output of the engine is …… kW (rounded off to the nearest integer).
 

Hint:

For diesel cycle problems: 1. Use the given efficiency formula and substitute parameters carefully.
2. Calculate heat supplied using \( q_s = C_p \Delta T \).
3. Multiply work output per unit mass by the mass flow rate to get total power output.

Answer 1

Step 1: Calculate the efficiency of the diesel cycle.
The efficiency of the diesel cycle is given by: \[ \eta = 1 - \frac{1}{r^{\gamma-1}} \left( \frac{r_c^{\gamma} - 1}{\gamma (r_c - 1)} \right). \] Substitute \( r = 12 \), \( r_c = 2.5 \), and \( \gamma = 1.4 \): \[ \eta = 1 - \frac{1}{12^{1.4-1}} \left( \frac{2.5^{1.4} - 1}{1.4 (2.5 - 1)} \right). \] 1. Calculate \( 12^{1.4-1} = 12^{0.4} \approx 2.297 \). 2. Calculate \( 2.5^{1.4} \approx 3.302 \). 3. Substitute into the formula: \[ \eta = 1 - \frac{1}{2.297} \left( \frac{3.302 - 1}{1.4 (1.5)} \right). \] Simplify: \[ \eta = 1 - \frac{1}{2.297} \left( \frac{2.302}{2.1} \right) = 1 - \frac{1}{2.297} (1.096). \] Calculate: \[ \eta = 1 - \frac{1.096}{2.297} \approx 1 - 0.477 = 0.523. \] Thus, the efficiency of the cycle is: \[ \eta = 0.523 \, \text{or} \, 52.3\%. \] Step 2: Calculate the heat supplied.
The heat supplied per unit mass flow rate is: \[ q_s = C_p \cdot \Delta T, \] where \( C_p = 1.004 \, \text{kJ/kg.K} \) and \( \Delta T = 2400 \, \text{K} \). \[ q_s = 1.004 \cdot 2400 = 2409.6 \, \text{kJ/kg}. \] Step 3: Calculate the work output per unit mass.
The work output per unit mass is: \[ w = \eta \cdot q_s. \] Substitute \( \eta = 0.523 \) and \( q_s = 2409.6 \): \[ w = 0.523 \cdot 2409.6 \approx 1260.23 \, \text{kJ/kg}. \] Step 4: Calculate the power output.
The total power output is given by: \[ \text{Power} = w \cdot \dot{m}, \] where \( \dot{m} = 0.05 \, \text{kg/s} \). \[ \text{Power} = 1260.23 \cdot 0.05 = 63.0115 \, \text{kW}. \] Rounding off to the nearest integer: \[ \text{Power} = 64 \, \text{kW}. \] Conclusion: The power output of the engine is \( 64 \, \text{kW} \).