Modern Engineering Thermodynamics

558 CHAPTER 14: Vapor and Gas Refrigeration Cycles
flash chamber functioning as the regenerator. As with the Rankine power cycle, the efficiency of a multistage
refrigeration cycle can be optimized through the proper choice of the regenerator (flash chamber) pressure. This
pressure then dictates the quality of the vapor entering the direct contact heat exchanger as
x flash = h f ðat the condenser pressureÞ − h f ðat the flash chamber pressureÞ
h fg ðat the flash chamber pressureÞ
(14.13)
The rate of cooling produced by a dual-stage refrigeration system is given by
and the total power input is
ð _Q L Þ dual
stage
= _m A ð1 − x flash Þðh 1B − h 4B Þ = _m B ðh 1B − h 4B Þ (14.14)
∑ _W compressors = _W A + _W B = _m A ðh 2A − h 1A Þ + _m B ðh 2B − h 1B Þ
= _m A ½ðh 2A − h 1A Þ + ð1 − x flash Þðh 2B − h 1B ÞŠ
= _m A ½ðh 2sA − h 1A Þ/ðη s Þ c−A + ð1 − x flash Þðh 2B − h 1B Þ/ðη s Þ c−B Š
The system coefficient of performance can then be computed from Eq. (14.10) as
(14.15)
COP dual
=
stage
_Q L
∑ _W compressors
=
_Q L
_W c − A + _W c−B
(14.16)
_m ref ð1 − x flash Þðh 1B − h 4B Þ
=
_m ref ½ðh 2sA − h 1A Þ/ðη s Þ c − A
+ ð1 − x flash Þðh 2B − h 1B Þ/ðη s Þ c−B
Š
An energy balance on the mixing heat exchanger in Figure 14.18 gives the value of the specific enthalpy at the
inlet of the compression stage in loop A as
h 1A = x flash h 5 + ð1 − x flash Þh 2B = x flash h g ðat p flash Þ + ð1 − x flash Þh 2B (14.17)
where we set h 5 = h g (at p flash ), and we compute h 2B from
h 2B = ðh 2sB − h 1B Þ/ðη s Þ c−B + h 1B (14.18)
The following example illustrates the effect of flash chamber pressure on the system’s overall coefficient of
performance.
EXAMPLE 14.7
A large food-processing plant needs a 14.0 ton refrigeration unit with an evaporator pressure of 100. kPa and a condenser
pressure of 1600. kPa. We are designing a two-stage, vapor-compression unit using refrigerant R-134a. The flash chamber is
to operate at 500. kPa, and the isentropic efficiency of both compressors is 80.0%. The following design specifications have
been established for the refrigerant loops shown in Figure 14.18:
Loop A
Station 1A Station 2sA Station 3A Station 4hA
Compressor A inlet Compressor A outlet Condenser A outlet Expansion valve A outlet
p 1A = 500: kPa p 2sA = 1600: kPa x 3A = 0:00 h 4hA = h 3A
s 2sA = s 1A
p 3A = 1600: kPa
Loop B
Station 1B Station 2sB Station 3B Station 4hB
Compressor B inlet Compressor B outlet Condenser B outlet Expansion valve B outlet
We now need to determine
x 1B = 1:00 p 2sB = 500: kPa x 3B = 0:00 h 4hB = h 3B
p 1B = 100: kPa s 2sA = s 1B p 3B = 500: kPa
a. The mass flow rate of the two refrigerants.
b. The system’s coefficient of performance.
c. The total power required by the compressors.