Solution Calculations with Entropy
1
2
3
4
Evaporator
Compres sor
Condenser
Throttling
Valve
The ammonia refrigeration cycle shown at the
left has the following properties for both parts
(a) and (b) below: P1 = P4 = 100 kPa; P2 = P3 = 2
MPa; point 1 is a saturated vapor and Point 3 is
a saturated liquid; for the throttling valve, h3 =
h4; the compressor has no heat transfer, and,
the low temperature heat input to the cycle,
which is in the evaporator, is 1 MW.
Determine the power input to the compressor if
(a) T2 = 250oC, (b) the compressor work input is
a minimum.
We can apply the first law to both the evaporator and the condenser as steady-flow devices with
one inlet and one outlet. If we neglect kinetic and potential energies we have the following
equation for the first law:
inoutuhhmWQ
. We are given that there is no heat transfer in
the compressor and we see that there is no useful work in the evaporator. Writing the first law for
each device gives the following pair of equations.
2141 hhmWhhmQ compevap
For a steady flow cycle, the mass flow rate must be the same in both devices. Eliminating the
mass flow rate between these two equations gives the following equation for the compressor
power:
41
21
21
41 hh
hh
QhhmW
hh
Q
mevapcomp
evap
From the problem data we find h1 = hg(P1 = 100 kPa) = 1544.14 kJ/kg; h4 = h3 = hf(P3 = 2 MPa) =
563.87 kJ/kg. These data will be the same for parts (a) and (b). For part a, h2 = h2(P2 = 2 MPa,
T2 = 250oC) = 2170.59 kJ/kg and the required compressor work is found from the combined first-
law equation derived above.
MW
kg
kJ
kg
kJ
kg
kJ
kg
kJ
MW
hh
hh
QW evapcomp 6391.0
874.56314.1544
59.217014.1544
1
41
21
MWWincomp 6391.0
,
The minimum work input for this adiabatic compressor work input occurs when s2 = s1 = sg(100
kPa) = 6.15296 kJ/kg·K. The value of h2 for the minimum compressor work input is found at the
given value of P2 = 2 MPa and s2 = 6.15296 kJ/kg·K; this pair of values is seen to be in the
superheat region so we can find h2 by interpolation at P2 = 2 MPa.
kg
kJ
Kkg
kJ
Kkg
kJ
Kkg
kJ
Kkg
kJ
kg
kJ
kg
kJ
kg
kJ
h91.203238592.651296.6
38592.652661.6
41.197420.2039
41.1974
2
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