Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 3 (Q.No. 40)
40.
A cyclic engine exchanges heat with two reservoirs maintained at 100 and 300°C respectively. The maximum work (in J) that can be obtained from 1000 J of heat extracted from the hot reservoir is
Discussion:
3 comments Page 1 of 1.
Ajinkya said:
10 years ago
It will go like.
Q1/Q2 = T1/T2.
1000/Q2 = 573/373.
So Q2 = 651 J.
And W = Q1-Q2.
= 1000-651 = 349 J.
Q1/Q2 = T1/T2.
1000/Q2 = 573/373.
So Q2 = 651 J.
And W = Q1-Q2.
= 1000-651 = 349 J.
(6)
Jasher said:
9 years ago
N = 1-Tc/Th where, Tc = 100+273.15 = 373.15 and Th = 300+273.15 = 573.15.
W = Q*n.
W = 1000 (1 - 373.15 / 573.15) = 349 J.
W = Q*n.
W = 1000 (1 - 373.15 / 573.15) = 349 J.
(4)
Melese Ayalew said:
6 years ago
Qh=1000J
Qc=?
Th=300oc,
Tc=100oc
W=Qh-Qc.
Qh/Qc = Th/Tc.
Qc=Qh * Tc/Th = 1000 * 100/300=333.3.
Then,
W = Qh-Qc = 1000-333.3 = 666.7 = 666J.
Qc=?
Th=300oc,
Tc=100oc
W=Qh-Qc.
Qh/Qc = Th/Tc.
Qc=Qh * Tc/Th = 1000 * 100/300=333.3.
Then,
W = Qh-Qc = 1000-333.3 = 666.7 = 666J.
(1)
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