a. What does 2650watts mean?
b. eff=(tempIN-TempOUT)/TemOUT in Kelvins
c. work=energy/time
2650=2330/time * efficience
solve for time of cycle
With each cycle, a 2650-W engine extracts 2330 J from a thermal reservoir at 85.1°C and expels 1840 J into a thermal reservoir at 20.9°C.
a) What is the work done for each cycle?
b) What is the engine's efficiency?
c) How much time does each cycle take?
2 answers
b. Bob Pursley's equation only applies for Carnot cycle engines. Real engines are not built that way. Stirling cycle engine are supposed to come close.
In your case, you should use the actual work out and heat out data.
Wout (per cycle) = Qin - Qout = 490 J
Efficiency = Wout/Qin = 490/2330 = 21.0%
The Carnot cycle formula predicts a maximum possible efficiency of 64.2/358.3 = 17.9%
I note that your heat reservoir temperature is quite low.
Since the stated efficiecy exceeds the Carnot efficiency, the numbers you have been given are not possible.
In your case, you should use the actual work out and heat out data.
Wout (per cycle) = Qin - Qout = 490 J
Efficiency = Wout/Qin = 490/2330 = 21.0%
The Carnot cycle formula predicts a maximum possible efficiency of 64.2/358.3 = 17.9%
I note that your heat reservoir temperature is quite low.
Since the stated efficiecy exceeds the Carnot efficiency, the numbers you have been given are not possible.
3 answers