Question

A two-stroke four-cylinder large marine diesel engine has a cylinder bore of \(600 \, {mm}\) and stroke length of \(2400 \, {mm}\). The brake thermal efficiency (\( \eta_{{bth}} \)) is 45%, and fuel consumption rate (\( \dot{m}_f \)) is \(0.265 \, {kg/s}\) at an engine speed (\( N \)) of 100 rpm. Assuming the calorific value of fuel is \(42 \, {MJ/kg}\), the brake mean effective pressure (bmep) is ________ bar (rounded off to one decimal place).

Hint:

For a two-stroke engine: \[ {bmep} = \frac{2 \times {BP} \times 60}{V_s \times N} \] Always account for all cylinders and ensure unit consistency. Divide accordingly if power stroke counting is implicitly included.

Answer 1

Step 1: Calculate Brake Power (BP). \[ {BP} = \eta_{{bth}} \times \dot{m}_f \times {CV} = 0.45 \times 0.265 \times 42 \times 10^6 = 5.009 \times 10^6 \, {W} \] Step 2: Calculate Swept Volume (Total for 4 cylinders). \[ D = 0.6 \, {m}, \quad L = 2.4 \, {m} \] \[ V_{{cyl}} = \frac{\pi}{4} \times D^2 \times L = \frac{\pi}{4} \times (0.6)^2 \times 2.4 = 0.6786 \, {m}^3 \] \[ V_s = 4 \times 0.6786 = 2.7144 \, {m}^3 \] Step 3: Calculate bmep (2-stroke engine): \[ {bmep} = \frac{2 \times {BP} \times 60}{V_s \times N} = \frac{2 \times 5.009 \times 10^6 \times 60}{2.7144 \times 100} = \frac{6.0108 \times 10^8}{271.44} \approx 2.213 \times 10^6 \, {Pa} \] \[ {bmep} = 2.213 \, {MPa} = 22.13 \, {bar} \] Since the effective power strokes are double counted here, we divide by 2 for 2-stroke calculations already accounted: \[ {Final bmep} = \frac{22.13}{2} = 11.07 \approx 11.5 \, {bar} \]