Air Refrigeration
Air cycle refrigeration systems belong to the general class of gas cycle refrigeration systems, in which a gas/air is used as the working fluid. The gas does not undergo any phase change during the cycle, consequently, all the internal heat transfer processes are sensible heat transfer processes. Gas cycle refrigeration systems find applications in air craft cabin cooling and also in the liquefaction of various gases.
Ideal reverse Brayton cycle
Process 1-2: Reversible, adiabatic compression in a compressor Process 2-3: Reversible, isobaric heat rejection in a heat exchanger Process 3-4: Reversible, adiabatic expansion in a turbine Process 4-1: Reversible, isobaric heat absorption in a heat exchanger
Process 1-2: Gas at low pressure is compressed isentropically from state 1 to state 2. Applying steady flow energy equation and neglecting changes in kinetic and potential energy, we can write:
w 1 − 2 = m( h 2 − h 1 ) = mc p ( T 2 − T 1 )
S 2 = S1
where rp = (P 2 /P 1 ) = pressure ratio
Process 2-3: Hot and high pressure gas flows through a heat exchanger and rejects heat sensibly and isobarically to a heat sink. The enthalpy and temperature of the gas drop during the process due to heat exchange, no work transfer takes place and the entropy of the gas decreases. Again applying steady flow energy equation and second T ds equation:
q 2 − 3 = m ( h 2 − h 3 ) = m c p ( T 2 − T 3 )
s 2 − s 3 = cp lnT 2/T 3
P 2 = P 3
w 3 − 4 = m ( h 3 − h 4 ) = m c p ( T 3 − T 4 )
s 3 = s 4
where r p = (P 3 /P 4 ) = pressure ratio
Process 4-1: Cold and low pressure gas from turbine flows through the low temperature heat exchanger and extracts heat sensibly and isobarically from a heat source, providing a useful refrigeration effect. The enthalpy and temperature of the gas rise during the process due to heat exchange, no work transfer takes place and the entropy of the gas increases. Again applying steady flow energy equation and second T ds equation:
q 4 − 1 = m ( h 1 − h 4 ) = m c p ( T 1 − T 4 )
P 4= P 1
From the above equations, it can be easily shown that:
( T2/T1 ) = ( T3/T4 )
Applying 1 st law of thermodynamics to the entire cycle:
( q4 − 1 −q2 − 3 ) = ( w 3 − 4 −w 1 − 2 ) = −wnet
The COP of the reverse Brayton cycle is given by:
COP = q4-1/ Wnet = ( T l − T 4 ) / ( T 2 − T 1 ) − ( T 3 − T 4)
COP = ( T4 / T3 -T4 )
Dry Air Rated Temperature (DART)
A comparison between different aircraft refrigeration systems based on DART at different Mach numbers shows that:
- DART increases monotonically with Mach number for all the systems except the reduced ambient system
- The simple system is adequate at low Mach numbers
- At high Mach numbers either bootstrap system or regenerative system should be used
- Reduced ambient temperature system is best suited for very high Mach number, supersonic aircrafts