Question

The stroke and cylinder diameter of a compression ignition engine are 250 mm and 150 mm respectively. If the clearance volume is 0.0004 m3 and fuel injection takes place at constant pressure for 5 per cent of the stroke determine the efficiency of the engine. Assume the engine working on the diesel cycle

Answer 1

Hey mate don't know

Answer 2

The efficiency of the engine is $0.593$ or $59.3$ %

Given :

Length of the stroke ⇒ $L=250mm=0.25m$

Cylinder Diameter    ⇒ $D=150mm=0.15m$

Clearance volume    ⇒ $V_{2}= 0.0004m^{3}$

Swept volume =  $V_{S}=\frac{\pi}{4} D^{2} L$

After substitution,

Swept volume ⇒ $0.15^{2} X 0.25 = 0.004418 m^{3}$

Total cylinder volume ⇔ Swept volume + Clearance volume  

                                      ⇒ $0.004418 + 0.0004= 0.004818 m^{3}$

Volume at point of cut-off,

$V_{3} = V_{2} +\frac{5}{100} V_{S}$

$V_{3}=0.0004 +\frac{5}{100} X 0.004418 = 0.000621 m^{3}$

Cut-off ratio,

ρ ⇒  $\frac{V_{3}}{V_{2}} =\frac{0.000621}{0.0004}=1.55$

Compression ratio,

$r= \frac{V_1}{V_2} =\frac{V_2+V_s}{V_2} =\frac{0.004418 + 0.0004}{0.0004} = 12.04$

η$_{diesel}$ ⇒ $1-0.264 X 1.54=0.593$ or $59.3$ %