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
a) What is the work done for each cycle?
b) What is the engine's efficiency?
c) How much time does each cycle take?
b. eff=(tempIN-TempOUT)/TemOUT in Kelvins
c. work=energy/time
2650=2330/time * efficience
solve for time of cycle
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.
a) To determine the work done for each cycle, we need to calculate the net heat transfer during one cycle. The net heat transfer can be calculated using the equation:
Net Heat Transfer = Heat Input - Heat Output
Heat Input = Q1 = 2330 J (energy extracted from the hot reservoir)
Heat Output = Q2 = 1840 J (energy expelled into the cold reservoir)
Therefore, the net heat transfer is:
Net Heat Transfer = Q1 - Q2
= 2330 J - 1840 J
= 490 J
Since work done in one cycle is equal to the net heat transfer, the work done for each cycle is:
Work Done = Net Heat Transfer
= 490 J
b) The engine's efficiency is the ratio of the work done to the heat input:
Efficiency = (Work Done / Heat Input) * 100%
Heat Input = Q1 = 2330 J (energy extracted from the hot reservoir)
Therefore, the engine's efficiency is:
Efficiency = (Work Done / Heat Input) * 100%
= (490 J / 2330 J) * 100%
= 0.21 * 100%
= 21%
The engine's efficiency is 21%.
c) To calculate the time taken for each cycle, we need to use the power equation:
Power = Work Done / Time
Given that the power is 2650 Watts (W) and the work done is 490 J, we rearrange the equation to solve for time:
Time = Work Done / Power
= 490 J / 2650 W
= 0.1851 seconds
Therefore, each cycle takes approximately 0.1851 seconds.