Question

A two-stroke CI. engine delivers 5000 kWwhile using 1000 kW to overcome friction losses. It consumes 2300 kg of fuel per hour at an air-fuel ratio of 20 to 1. The heating value of fuel is 42000 kJkg Find the (a) indicated power (b) mechanical efïiciency, (c) air consumption per hr, (d) indicated themal efficiency, and (e) brake thermal efficiency.

Answer 1

(a) Indicating power(IP)=6000 KW

(b)
`\eta_(mech)`=0.833

(c) Consumption of air per hour =46000 kg/hr

(d)
`\eta_(BPth)`=0.1865

Step-by-step explanation:

Break power(BP) =5000 KW

Friction power(FP)=1000 KW

Consumption of fuel per hour=2300 kg/hr

CV=42000 KJ/kg

We know that

Indicating power(IP)=Break power(BP)+Friction power(FP)

⇒IP=5000+1000 KW

IP=6000 KW

(a)

Indicating power(IP)=6000 KW

(b)

Mechanical efficiency
`\eta_(mech)=(BP)/(IP)`

`\eta_(mech)=(5000)/(6000)`

`\eta_(mech)`=0.833

(c)

Air fuel ratio=
`(mass \ of \ air)/(mass \ of \ fuel)`

consumption of air per hour=20
`*`2300 kg/hr

So consumption of air per hour =46000 kg/hr

(d)

Break thermal efficiency
`\eta_(BPth)=\frac{IP}{\dot{m_f}* CV}`

`\dot{m_f}=(2300)/(3600)`

=0.638 kg/s

`\eta_(BPth)=\frac{5000}{{0.638}* 42000}`

`\eta_(BPth)`=0.1865