In-Class Exercise 4-01: heat reservoirs T

In-Class Exercise 4-01:
A perfect heat engine with = 0.4 is used as a refrigerator (the
heat reservoirs Th and Tc remain the same). How much heat Qc
can be transferred in one cycle from the cold reservoir to the hot
one if the supplied work in one cycle is W = 10 kJ?
1
For one cycle:
=
W
Q − Qc
= h
Qh
Qh
Qc
=1−
Qh
Qc
Qc
Qc
Qh
COP =
=
=
Qc
W
Qh − Qc
1−Q
h
(1)
(2)
(3)
1−
1−
=
(4)
1 − (1 − )
1 − 0.4
COP =
= 1.5
(5)
0.4
So, if the supplied in one cycle work is W = 10 kJ, the heat that
be transferred from the cold reservoir to the hot one is
Qc = COP × W = 1.5 × 10 = 15 kJ.
COP =
In-Class Exercise 4-02 (textbook):
Prove that if one can had a heat engine whose efficiency was
better than the ideal value 1 − TTc , you could hook it up to an
h
ordinary Carnot refrigerator to make a refrigerator that requires
no work input.
2
Hot reservoir, Th Qh,r Qh,e Magic engine Carnot refrige W Qc,e Qc,r Cold reservoir, Th Step-1: Ideal engine, the heat input and output are in the same
ratio as the reservoir temperatures
Qh,e
Qc,e
=
Th,e
Tc,e
(6)
Step-2: The work to be generated W ,
Qh,e − Qc,e = W
(7)
Step-3: Ideal Carnot refrigerator, the heat input and output are
in the same ration as the reservoir temperatures
Qh,r
Qc,r
=
Th,r
Tc,r
(8)
Step-4: For a given work W ,
Qh,r − Qc,r = W
(9)
T
T
h,e
Step-5: Using equations above, and Th,r
=
Tc,e , we can get
c,r
Qh,e
=
Qh,r
=α
Qc,e
Qc,r
Qh,e − Qc,e = Qh,r − Qc,r
(10)
(11)
From this two equations, we can get:
Qh,r = Qh,e
(12)
Qc,r = Qc,e
(13)
Step-6: If the engine is better than ideal, Q0h,e < Qh,e = Qh,r .
For the same W , we also have Q0c,e < Qc,e = Qc,r
Step-7: The net heat flow from high T (to low T ) is: Q0h,e −Qh,r <
0. So, heat flow from low T to high T .
This is against the Second Law of Thermodynamics.