Bell Coleman Cycle
Bell Coleman Cycle also know as a Reversed Brayton Cycle or the Joule cycle. The fluid is always in a gaseous state which is compressed and expanded. It was one of the most punctual sorts of coolers utilized in boats conveying solidified meat. The cycle uses air as a refrigerant, which is effectively accessible and economical. Used both in cooling and heating effects.
Similarly, refrigeration working on this Cycle, has a less COP. The running cost is very high.
In this video, I have explained the working of the cycle.
Working of Bell Coleman Cycle
Bell Coleman Cycle process is defined below
Let p1, v1, and T1 be the pressure, volume, and temperature of the air.
- Isentropic Compression: The Cold air from the fridge is brought into the compressor and compressed isentropically. During this procedure, the pressure increments from p1 to p2. The specific volume diminishes from v1 to v2 and the temperature increments from T1 to T2. During this procedure Entropy ‘s’ stays steady (s1=s2). No heat is absorbed or rejected by the air.
- Steady Pressure Cooling Process: The warm air is from the compressor is presently passed into the cooler where it is cooled at consistent pressure, lessening the temperature from T3 to T2. Explicit Volume additionally diminishes from v2 to v3. Warmth is dismissed by the air during this procedure. Heat dismissed by the air: qr = Cp(T2 – T3 )
- Isentropic Expansion: Air from the cooler is presently brought into the expander and is extended isentropically. The pressure of the air stays steady during this procedure. Specific volume changes from v3 to v4 and the temperature diminishes from T3 to T4. No heat is dismissed or consumed by the air.
- Steady Pressure expansion process: The cold air from the expander is currently passed into the refrigerator and extended at a consistent pressure. The temperature of the air increments from T4 to T1. The specific volume of the air changes from v4 to v1.
Heat consumed by the air: qa = Cp( T1 – T4)
Work Done Of Bell Coleman Cycle
Work done during the cycle per kg of air = Heat rejected – Heat absorbed
= qr – qa
= Cp(T2 – T3 ) – Cp( T1 – T4)
Coefficient Of Performance Of Bell Coleman Cycle
C.O.P during the cycle per kg of air = Heat absorbed / work done
= qa/qr – qa
= Cp( T1 – T4 ) / Cp(T2 – T3 ) – Cp( T1 – T4)
Bell Coleman Cycle Problems
Let us take an example related to it.
A refrigeration plant working on the Bell Coleman Cycle, air is compressed to 6 bar from 1 bar. Its starting temperature is 15 C . After compression air is cooled to up to 25 C in a cooler before expanding back to 1 bar. Determine the C.O.P of the plant and net refrigerating effect.
Cp = 1.005 kJ/kg K and Cv= 0.718 kJ/kg K.
Given P2 = P3 = 6 bar, P1 = P4 = 1 bar
T1 = 15 + 273 = 288 K
T3 = 25 + 273 = 298 K
γ = Cp / Cv = 1.005/0.718 = 1.4
γ – 1 = 0.4
so γ – 1/ γ = 0.286
T2/T1 = (P2 / P1)γ – 1/γ
so T2/T1 = 1.669
similarly for process 3-4
T4 = T3 / 1.669
T4 = 178.55
C.O.P of the cycle is given by
C.O.P = T4 / T3 – T4
C.O.P = 178.55 / 298 – 178.55
C.O.P = 1.494
Net refrigerating effect = Cp ( T1 – T4)
= 1.005( 288 – 178.55)
= 109.99 kJ/kg