Engine Efficiency: 1.4 kg Gasoline/hr = 3.5 kW Power

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Homework Help Overview

The discussion revolves around calculating the efficiency of an engine that burns 1.4 kg of gasoline per hour and produces 3.5 kW of power, using the heating value of gasoline as 43 MJ/kg.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formula for engine efficiency, questioning the relationship between power generated and heat energy. There are attempts to clarify the units involved and the correct interpretation of efficiency.

Discussion Status

Several participants are engaged in clarifying the concept of efficiency, with some suggesting different interpretations of the formula. There is an ongoing exploration of the calculations involved, but no consensus has been reached on the final value or its representation.

Contextual Notes

Participants are discussing the implications of unit conversions and the definition of efficiency, with some confusion about the appropriate units and whether to express efficiency as a percentage.

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Homework Statement


An engine burns 1.4 kg of gasoline per hour and yields 3.5 kW of power.

Homework Equations


Calculate the efficiency of the engine when the heating value of gasoline is 43 MJ/kg.

The Attempt at a Solution


Plz i just need the rule for engine efficiency.
 
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chawki said:

Homework Statement


An engine burns 1.4 kg of gasoline per hour and yields 3.5 kW of power.

Homework Equations


Calculate the efficiency of the engine when the heating value of gasoline is 43 MJ/kg.

The Attempt at a Solution


Plz i just need the rule for engine efficiency.

Coal burnt per second (A) = \frac{1.4}{60 * 60}
Power through coal burnt per second (P) = (43 * A)
P x Efficiency of engine (e) = 3.5kW

Find e. Be careful with the units and the conversion.
 
ok so:

efficiency = power generated/heat energy ?
and then:
power generated= 3.5kw
heat energy=mfuel*heating value

consumption per hour =mfuel*3600
mfuel=1.4/3600=3.88*10e-4 Kg/s

heat energy: Q= 3.88*10e-4*43=16.684kw

Efficiency= 3.5/16.684 = 0.209
 
chawki said:
ok so:

efficiency = power generated/heat energy ?
Not really.

Power of Engine = Efficiency of engine x Energy density of coal x Coal burnt per second
 
we would find same results, no?
 
I'm not sure, I didnt read beyond the first line because I assumed the rest of your post built upon that formula, which was wrong.

Because Efficiency = Power (W) /Heat Generated (J)
So it the unit for efficiency would be s^-1.

Efficiency has no unit.
 
in my work, efficiency has no unit.
it's kw/kw
 
chawki said:
in my work, efficiency has no unit.
it's kw/kw
You are correct. Efficiency is the ratio of energy output to energy input. This is the same as power output/power input.

AM
 
Andrew Mason said:
You are correct. Efficiency is the ratio of energy output to energy input. This is the same as power output/power input.

AM

Should we leave 0.209 as it is, or should we write 0.209*100% = 20.9% ?
 

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