Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 5 (Q.No. 33)
33.
The following heat engine produces power of 100000 kW. The heat engine operates between 800 K and 300 K. It has a thermal efficiency equal to 50% of that of the Carnot engine for the same temperature. The rate at which heat is absorbed from the hot reservoir is
Discussion:
5 comments Page 1 of 1.
Ice Wizard said:
8 months ago
n = 1-300/800 = 0.625.
n = 0.5 given,
n(total) = 0.625 * 0.5 = 0.3125.
Wmax = n * Q.
100000/0.3125 = Q.
Q = 320,000.
n = 0.5 given,
n(total) = 0.625 * 0.5 = 0.3125.
Wmax = n * Q.
100000/0.3125 = Q.
Q = 320,000.
Sunil yadav said:
2 years ago
n = W/Qh.
Qh = W/n.
= 100000/.3125,
= 320000Kw.
Qh = W/n.
= 100000/.3125,
= 320000Kw.
Muhammad Moazzam said:
5 years ago
Thanks @Kamran.
Kamran Khan said:
6 years ago
As for carnot engine, thermal efficiency= nc=1-tc/th=1-800/300=0.625.
As given n=0.50nc.
Because, Thermal efficiency = 0.50(Carnot thermal efficiency),
So n=0.50(0.625)=0.3125.
As efficiency=n=W/Q then Q=nW=0.3125(100000)=319693.094 KW Approximately 320,000 kW.
As given n=0.50nc.
Because, Thermal efficiency = 0.50(Carnot thermal efficiency),
So n=0.50(0.625)=0.3125.
As efficiency=n=W/Q then Q=nW=0.3125(100000)=319693.094 KW Approximately 320,000 kW.
(2)
Aamir said:
7 years ago
Please explain the answer.
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