a certain machine changes a large input force into a smaller output force. how will the machine affect the distance over which t
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Answer:
The net work produced is 30.37 KJ
The efficiency of cycle is 72.3%
Explanation:
For the net work produced, we have the formula:
Work = (Th - Tc)(Sh - Sc)(M)
Where,
Th = higher temperature = 800° C + 273 = 1073 k
Tc = lower temperature = 25° C + 273 = 298 k
Sh = specific entropy at higher temperature
Sc = specific entropy at lower temperature
M = molar mass of air
Using, ideal gas table to find entropy. The table is attached.
therefore,
Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)
Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)
<u>Work = 30.37 KJ</u>
Now, for the efficiency (n), we have a formula:
n = 1 - Tc/Th
n = 1 - (298 k)/(1073 k)
<u>n = 0.723 = 72.3 %</u>
