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What is the ideal efficiency of an automobile engine where fuel is heated to 2700 K and the outdoor air is at 270 K?

how do i do this..i dont need the answer just the way to do it

2007-06-04 04:13:30 · 4 answers · asked by Lou 1 in Science & Mathematics Physics

4 answers

In an ideal cyclical engine, efficiency is defined as e = w / Qh (work / heat flow from hot reserviour). Based on the 2nd law of thermodynamics, in a closed system with each reserviour at constant temperature, Qh / Th + Qc / Tc = 0. Work is also the difference between Qh and Qc. Qh - Qc = w.
So:

e = 1 - Qc/Qh = 1 - Tc/Th

where Tc is 270 K and Th is 2700 K.

2007-06-04 04:29:20 · answer #1 · answered by Wraith 2 · 0 0

Not at all clear what you are asking here. Ideally we'd like 100% efficiency. That is, we'd like to get out as useful work all the energy put into a system.

But if you're asking the theoretical maximum efficiency from a Carnot cycle; then that's [(TH - TC)/TH] X 100% = ((2700-270)/2700] X 100% efficiency...you can do the arithmetic. [See source.]

2007-06-04 04:46:57 · answer #2 · answered by oldprof 7 · 0 0

The Carnot cycle is, through definition, probably the most effective cycle among any 2 warmth reservoirs. So, even though you've a truly gasoline, the potency might pass down however it's going to nonetheless be one hundred% of the Carnot potency. However, I query whether or not a truly gasoline may also be real reversible and keep the features of a Carnot cycle.

2016-09-05 21:31:08 · answer #3 · answered by bradberry 3 · 0 0

Combustion engine is not quite equivalent to Carnot engine.
The formula (T-To)/T does not apply here.

For example if you consider AA battery, operating
at temperature, equal to ambient temperature, then
its efficiency must be what? zero?

2007-06-04 04:48:16 · answer #4 · answered by Alexander 6 · 0 0

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