EXERCISE 5.10

EXERCISE 5.10
EDWARD ANDRES CASTAÑEDA GIRLADO
INGENERIA ELECTROMECANICA
THERMAL MACHINES
CARLOS ALBERTO ACEVEDO ALVAREZ
INSTITUTO TECNOLOGICO METROPOLITANO
ITM
28/09/2021
shows two power cycles, denoted 1 and 2, operating in series, together
with three thermal reservoirs. The energy transfer by heat into cycle 2 is
equal in magnitude to the energy transfer by heat from cycle 1. All
energy transfers are positive in the directions of the arrows.(a)
Determine an expression for the thermal efficiency of an overall cycle
consisting of cycles 1 and 2 expressed in terms of their individual
thermal efficiencies.(b) If cycles 1 and 2 are each reversible, use the
result of part (a) to obtain an expression for the thermal efficiency of
the overall cycle in terms of the temperatures of the three reservoirs, TH, T,
and TC, as required. Comment.(c) If cycles 1 and 2 are each reversible
and have the same thermal efficiency, obtain an expression for the
intermediate temperature T in terms of TH and TC.
SOLUTION
a) problem data:
two power cycles are working in series
for the overall cycle (W1+W2) =QH-QC
+,
n=!"#!$
' )*
%&
+cycle 1
+
++
QH=)*.)
n1= 1-
we clear QH it and it looks like this.
Cycle 2
+/
n2=1- + we clear QC it and it looks like this.
QC=Q(1-n2)
of the formula of cycle 1 and clicks 2 a comparison is made
𝑄𝐶
= (1 − 𝑛1)(1 − 𝑛2)
𝑄𝐻
n=1-(1-n1) (1-n2) = 1-(1-n1-n2+n1n2) = n1+n2-n1n2
b) with cycles 1 and 2 reversible
:
:/
we have to n1= 1-:- and n2=1- :
we replace in the result of the values that we have in n1 and in n2
:
n= (1- :- ) + (1-
:,
:
) – (1-
:
:-
:,
)(1- 1- : )
We carry out the sum and multiplication in the previous operation and it
looks like this:
:
n=1- :- + 1-
:,
:
:
- [ 1- :- -
:,
:
+
: :
]
:- :-
:,
= 1 - :;
c) if cycles 1 and 2 each reversible
:
n1= 1- :- and n2=1-
:,
:
we have to:
n1=n2
1-
:
:-
= 1-
:,
:
𝑇 = = 𝑇𝑐𝑇𝐻 = T=√𝑇𝑐𝑇𝐻
With this formula we can already find the intermediate temperature if we
have the values of Tc and Th, as long as they are reversible.