Question

A spark ignition engine burns a fuel of calorific value 45MJkg. It compresses the air-ful mixture in accordance with PV^1.3=constant. The pressure developed in the cylinder during 30% and 70% of the compression stroke is 1.5 bar and 2.75 bar, respectively. The relative efficiency of the engine with respect to the air standard efficiency is 50%. Determine the (i) compression ratio (ii) fuel consumption per hour per kW of power generated

Answer 1

i). Compression ratio = 3.678

ii). fuel consumption = 0.4947 kg/hr

Step-by-step explanation:

Given :

`PV^(1.3)=C`

Fuel calorific value = 45 MJ/kg

We know, engine efficiency is given by,

`\eta = 1-\left ( (1)/(r_(c)) \right )^(1.3-1)`

where
`r_(c)` is compression ratio =
`(v_(c)+v_(s))/(v_(c))`

`r_(c)` =
`1+(v_(s))/(v_(c))`

where
`v_(c)` is compression volume

`v_(s)` is swept volume

Now it is given that swept volume at 30% of compression, 70% of the swept volume remains.

Then,
`v_(30)=v_(c)+0.7v_(s)`

and at 70% compression, 30% of the swept volume remains

∴
`v_(70)=v_(c)+0.3v_(s)`

We know,

`(P_(2))/(P_(1))=\left ( (V_(1))/(V_(2)) \right )^(n)`

`(2.75)/(1.5)=\left ( (v_(c)+0.7* v_(s))/(v_(c)+0.3* v_(s)) \right )^(1.3)`

`\left ( 1.833 \right )^{(1)/(1.3)}=(v_(c)+0.7v_(s))/(v_(c)+0.3v_(s))\\`

`1.594=(v_(c)+0.7v_(s))/(v_(c)+0.3v_(s))`

`v_(c)+0.7v_(s)=1.594v_(c)+0.4782v_(s)`

`0.7v_(s)-0.4782v_(s)=1.594v_(c)-v_(c)`

`0.2218v_(s) = 0.594v_(c)`

`v_(c)=0.3734 v_(s)`

∴
`r_(c)= 1+(v_(s))/(0.3734v_(s))`

Therefore, compression ratio is
`r_(c)` = 3.678

Now efficiency,
`\eta =\left ( 1-(1)/(r_(c)) \right )^(0.3)`

`\eta =\left ( 1-(1)/(3.678) \right )^(0.3)`

`\eta =0.32342` , this is the ideal efficiency

Therefore actual efficiency,
`\eta_(act) =0.5* \eta _(ideal)`

`\eta_(act) =0.5* 0.32342`

`\eta_(act) =0.1617`

Therefore total power required = 1 kW x 3600 J

= 3600 kJ

∴ we know efficiency,
`\eta=(W_(net))/(Q_(supply))`

`Q_(supply)=(W_(net))/(\eta _(act))`

`Q_(supply)=(3600)/(0.1617)`

`Q_(supply)=22261.78 kJ`

Therefore fuel required =
`(22261.78)/(45000)`

= 0.4947 kg/hr