Power point slide on Refridgeration second mod.pptx

Aie conditioning

1. Types of Refrigeration Ice refrigeration Throttling Refrigeration Air Refrigeration Vapour Compression refrigeration System (VCRS) Vapour Absorption Refrigeration System

2. Ice Refrigeration

3. Ice refrigeration Advantages Simplicity portability low or no energy consumption Cost- effectiveness Disadvantages L imited cooling capacity Frequent ice replacement Temperature fluctuations Space constraints

4. Throttling refrigeration The enthalpy of an ideal gas is a function of temperature only During an isenthalpic process, the temperature of the ideal gas remains constant. In case of real gases, whether the temperature decreases or increases during the isenthalpic throttling process This increase or decrease in temperatre depends on a property of the gas called Joule- Thomson coefficient, μ JT , given by

5. Joule- Kelvin Effect

6. Isenthalpic states of a gas

7. Isenthalpic Curve and the Inversion curve

8. Air Refrigeration It is a method of cooling or refrigeration that uses air as the working medium to transfer heat and achieve a cooling effect. It's a less common method compared refrigeration technologies like VCRS, which to other uses refrigerant gases.

9. Advantages of Air Refrigeration system No risk of fire Since air is non- flammable It is cheaper as air is easily available as compared to the other refrigerants. The weight of this system per ton of refrigeration is quite low as compared to other systems. Because of this reason, this system is employed in aircrafts.

10. Disadvantages of Air Refrigeration system The C.O.P of the system is very low in comparison to other systems. The weight of the air required to be circulated is comparatively more. High Volume Flow Limited Temperature Range(Not suitable for deep freezing)

11. Carnot Refrigeration Cycle

12. Four processes are Isentropic compression process ( 1 – 2 ) Isothermal compression process ( 2 – 3 ) Isentropic expansion process ( 3 – 4 ) Isothermal expansion process ( 4 – 1 )

13. Heat absorbed by the refrigerant Q L = T L ( S 1 - S 4 ) Heat rejected from the refrigerant Q H = T H ( S 2 - S 3 ) Net Work input = Heat Rejected – Heat absorbed W= Q H - Q L = T H ( S 2 - S 3 ) - T L ( S 1 - S 4 ) W = (T H - T L )( S 1 - S 4 ) RE = Q L = T L ( S 1 - S 4 )

14. T- Temperature in Kelvin

15. For reversible cycle

16. For Irreversible Cycle For Reversible Cycle

17. EX 1. A machine working on a carnot cycle operates between 305 and 260 K. Determine the COP when it is operated as A refrigerating machine A heat pump and A heat engine. Solution : T L = 260 K, T H = 305 K. When it is operated as a refrigerator , C O P = T L / (T H - T L ) = 260 / ( 305 – 260 ) = 5.78 When is working as a heat pump, C O P = T H / ( T H - T L ) = 305 / ( 305 – 260) = 6.78 When it is working as a heat engine, C O P = (T H - T L ) / T H = ( 305 – 260 ) / 305 = 0.147

18. EX 2. A cold storage is to be maintained at - 5 C while the C surroundings are at 35 . The heat leakage from the surroundings in to the cold storage is estimated to be 29 kW. The actual COP of the refrigeration plant is 1/3 rd of an ideal plant working between the same temperature. Find the power required to drive the plant. November 2022 Rev(15) – Mark 8 Solution: T L = - 5 C = - 5 + 273 = 268 K , T H = 35 C = 35 + 273 = 308 K . Heat leakage from the surroundings , Q L = 29 kW Theoretical COP = T L /(T H – T L ) = 268 / ( 308 – 268) = 6.7 Actual COP =(1/3 ) x Theoretical COP = (1/3) x 6.7 = 2.23 Work done = Refrigerating effect = 29/2.23 = 13 kW Actual COP