_3_Air_refrigeration

Air Refrigeration https://www.quizbroz.com/ MEL 405 Refrigeration & Air conditioning Amit Arora Assistant Professor Department of Mechanical Engineering ITM University, Gurgaon

Air Refrigeration cycle • As the name indicates, Air (gas) is used as the refrigerant in Air Refrigeration cycle • Initially, Air was a lucrative working fluid due to being abundantly available and at no cost • But heat carrying capacity of Air (per unit mass) is very small compared to phase change refrigerants – Bcoz only sensible heat exchange effect exists

Reversed Carnot cycle (when refrigerant is a Gas) • Carnot cycle with Air (gas) as a refrigerant lies in the superheated region on property diagram – outside the saturation dome

• Four processes of the cycle are already discussed • Process (1-2) = Isentropic compression Heat transfer, q12  0  p11 p22 1 R(T2  T1) Work done , w12     1 • Process (2-3) = Isothermal heat rejection to reservoir at high temp. (comp. / cond.) Work done , w 23  p22 In 2  RT2.In 2 3 3 For a perfect gas undergoing isothermal process, Heat transfer, q23  w23 ( du  0) • Process (3-4) = Isentropic expansion Heat transfer, q34  0 Work done , w 34  p33  p44  R(T3  T4 )  1  1 • Process (4-1) = Isothermal heat absorption from reservoir at low temp.(exp./ evap.) Work done , w 4 1  p44 In 1  RT4 .In 1 4 4 Heat transfer, q41  w41 ( du  0)

• Thus, Refrigerating effect produced by the cycle is Heat transfer, q41  w41 ( du  0) – Area under process (4-1) on T-s diagram where, w 4 1  RT4.In 1 4 • Work consumed by the cycle is wnet  w12  w23  w34  w41 – But work of isentropic compression (w1-2) & expansion (w3-4) is equal – Bcoz (T2-T1 )= (T3-T4) Work done , w12  R(T2  T1)  1 Work done , w 34  R(T3  T4 )  1 • Net work consumed by the cycle is wnet  w23  w41  RT2.In 2  RT4.In 1 3 4 • COP of the Carnot cycle with Gas as a refrigerant is COPR , Carnot  q ref  q41 w net w23  w41

COPR , Carnot  q ref  q 41 w net w23  w41 • Substitute numerator & denominator RT4.In 1 • Upon simplification 4 COPR, Carnot  2 1 RT2.In 3  RT4.In 4 COPR, Carnot  T4  For isentropic processes (1 2) & (3  4), T2  T4 Thot  r1 where 'r' comp. ratio for isen. process  Tcold Tcold Thot  Tcold  T2  1 1   T3  4   1   2   3  • Rewrite Bcoz  T1     T4   COPR, Carnot  1 Thus, 1  4 OR 2  1 Thot 1 Tcold 2 3 3 4  1 1 r  1

COPR , Carnot  1 r1 1 • Thus, COP of Carnot refrigerator is a function of compression ratio (r) of compressor/ expander • Note: • Recall that COP of a Carnot cycle depends only on operating temperatures (and not on working fluid i.e. refrigerant) COPR , Carnot  Tcold Thot  Tcold • But it can also be expressed in an equivalent form which is a function refrigerant property index ‘ϒ’ (as shown above)

Limitations of Reversed Carnot cycle (when refrigerant is a Gas) • Despite of highest COP of Reversed Carnot cycle, its practical application has serious hurdles • No refrigerator could be build to run on Reversed Carnot cycle • Why? 1. It requires Gas with infinite specific heat, which is impossible  When change in Gas temperature is zero, specific heat must be infinite for heat transfer to exist Q  m.c.(Tin  Tout )  Bcoz there is no phase change

2. Both isothermal processes are practically infeasible (when refrigerant is Gas)  Isothermal heat transfer process with Gas (no phase change) requires executing the process at infinitely slow speed, which is impractical 3. Extreme pressures are developed and large volume of fluid needs to be handled  This is due to the cycle operating in superheated region  When the cycle is inside the dome, process (2-3) is isobaric as well as isothermal, so pressure rise takes place only during process (1-2)  But for gaseous refrigerants, pressure rise takes place during both processes (1-2) and (2-3)  Cycle in the superheated region will obviously handle more volume of working fluid compared to cycle inside the dome  Consequently, heavy and volumous hardware is needed

4. Fourth, p-v diagram of Carnot cycle with Gas is very narrow and long (when drawn as per correct scale)  It results in long stroke and hence higher mechanical irreversibility  Consequently COP is very poor due to major part of the work getting wasted in overcoming the mechanical irreversibility 5. Isentropic process is also practically impossible • Bcoz perfect insulator does not exist • An engineering solution is to run it at a very high speed • But a cycle with very fast (isentropic) & very slow (isothermal) processes is mechanically unviable • Solution https://www.quizbroz.com/ • Isothermal processes of Reversed Carnot is replaced by Isobaric processes which are practically possible • Such a cycle is called  Reversed Brayton cycle OR Reversed Joule cycle  Famously known as Bell-Coleman cycle