Advantages and Disadvantages of Air-refrigeration System

Advantages

• Air is a cheaper refrigerant and available easily compared to other refrigerants.
• There is no danger of fire or toxic effects due to leakage.
• The total weight of the system per ton of refrigerating capacity is less.

Disadvantages

• The quantity of air required per ton refrigerating capacity is far greater than other systems.
• The COP is low and hence maintenance cost is high.
• The danger of frosting at the expander valves is more as the air taken into the system always contains moisture.

Example 8.1: Carnot refrigeration cycle absorbs heat at 250 K and rejects heat at 300 K.

1. Calculate the coefficient of performance of this refrigeration cycle.
2. If the cycle is absorbing 1,050 kJ/min at 250 K, how many kJ of work is required per second.
3. If the Carnot heat pump operates between the same temperatures as the above refrigeration cycle, what is the coefficient of performance?
4. How many kJ/min will the heat pump deliver at 300 K if it absorbs 1,050 kJ/min at 250 K.

Solution:

Given: T₁ = 250 K; T₂ = 300 K

1. Coefficient of performance of Carnot refrigeration cycle 
2. Work required per secondLet W_R = Work required per secondHeat absorbed at 250 K (i.e., T₁), Q₁ = 1,050 kJ/min= 17.5 kJ/s
3. Coefficient of performance of Carnot heat pump 
4. Heat delivered by heat pump at 300 KLet Q₂ = Heat delivered by heat pump at 300 K.Heat absorbed at 250 K (i.e., T₁),Q₁ = 1,050 kJ/min (given)

Example 8.2: The capacity of a refrigerator is 150 TR when working between − 5 and 20°C. Determine the mass of ice produced per day from water at 20°C. Also find the power required to drive the unit. Assume that the cycle operates on reversed Carnot cycle and latent heat of ice is 336 kJ/kg.

Solution:

Given: Q = 150 TR; T₁ = −5°C = −5 + 273 = 268 K; T₂ = 20°C = 20 + 273 = 293 K

Mass of ice produced per day

Heat extraction capacity of the refrigerator = 150 × 210 = 31,500 kJ/min

Heat removed from 1 kg of water at 20°C

to form ice at 0°C (ITR = 210 kJ/min) = Mass × Sq. heat × (Change in temperature) + h_fg (ice)

8.4.2  Vapour Compression Refrigeration System

A simple vapour compression refrigeration system consists of the following equipments (Figure 8.4):

Figure 8.4 Schematic Diagram of Refrigeration Systems

• Compressor
• Condenser
• Expansion valve
• Evaporator

The low temperature and low pressure vapour is compressed by a compressor to high temperature and high pressure vapour. This vapour is then condensed into condenser at constant pressure and then passed through the expansion valve. Here, the vapour is throttled down to a low pressure liquid and passed through an evaporator, where it absorbs heat from the surroundings and vaporizes into low pressure vapour. The cycle then repeats again and again. The heat and work interaction in vapour compression process takes place in following way:

• Compressor requires work (W). The work is supplied to the system.
• During condensation, heat Q_H the equivalent of latent heat of condensation, etc. is removed from refrigerant.
• During evaporation, heat Q_L equivalent to latent heat of vaporization is absorbed by the refrigerant.
• There is no exchange of heat during throttling process through the expansion valve as this process occurs at constant enthalpy.

Figure 8.5 shows a simple vapour compression refrigeration cycle on T−S diagram for different compression processes. The cycle works between temperatures T₁ and T₂ representing the condenser and evaporator temperatures, respectively. The thermodynamic process of the cycle A−B−C−D is given as follows:

Figure 8.5 T−S Diagram of Vapour Compression Refrigeration Cycle

Process B–C: Isentropic compression of the vapour from state B to C.

Process C–D: Heat rejection in condenser is at constant pressure.

Process D–A: An irreversible adiabatic expansion of vapour through the expansion value. The pressure and temperature of the liquid are reduced. The process is accompanied by partial evaporation of some liquid. The process is shown by dotted line.

Process A–B: Heat absorption in evaporator at constant pressure. The final state depends on the quantity of heat absorbed and same may be wet, dry or superheated.