Chapter 9. Refrigeration and Liquefaction-student

Chapter 9. Refrigeration and Liquefaction-student - Chapter...

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Application Air conditioning of buildings, transportation, and preservation of foods and beverages Manufacture of ice Dehydration of gases Low-temperature reactions Separation of volatile hydrocarbons Continuous absorption of heat at a low temperature level. Carnot Refrigerator The ideal refrigerator, operating on a Carnot cycle: Two isothermal steps: heat |Q C | is absorbed at low temperature T C , rejected | Q H | at the higher temperature T H Two adiabatic steps : expansion and compression Chapter 9 Refrigeration and Liquefaction
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For ideal ( Carnot) cycle, For a refrigeration at H C = 5 o C and T H = 30 o C Coefficient of performance: It gives maximum possible coefficient for any refrigerator operating between T H and T C The coefficient of performance:
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The vapor-compression cycle (Rankine refrigeration cycle) 1→2: liquid (absorb heat) evaporating at constant pressure (so constant T) 2→3: isentropic (reversible and adiabatic) compression to a higher pressure 3→4: cooled and condensed with rejection of heat at a higher temperature level 4→1: expansion throttling process or boiler
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On the basis of a unit mass of fluid The heat absorbed in the evaporator (constant P): The heat rejected in the condenser (constant P): The work of compression (Eq. (2.32)): Define the rate of circulation of refrigerant: Rankine refrigeration cycle or boiler
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ln P H Constant S 1 2 3 4 liquid gas C P-H diagram is more commonly used than T-S diagram because they show directly the required enthalpy.
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Example 9.1 A refrigerated space is maintained at 10°F and cooling water is available at 70°F. Refrigeration capacity is 120000 Btu/hr. The evaporator and condenser are of sufficient size that a 10°F minimum-temperature difference for heat transfer can be realized in each. The refrigerant is tetrafluoreothane (HFC – 134a), for which data are given in Table 9.1 and Fig G.2 (App. G). (1) what is the value of ω for a Carnot refrigerator? (2) Calculate ω and for the vapor-compression cycle of Fig 9.1 if the compressor efficiency is 0.80. m (1) For a Carnot refrigerator: 75 . 5 ) 67 . 459 0 ( ) 67 . 459 80 ( 67 . 459 0 = + - + + = - = C H C T T T ϖ Note: K = ( o F + 459.67)x5/9 T H and T C should be zero for maximum performance Temperature of the cooling water, T cw Temperature of the refrigerated region, T ref Condensation T (80 o F) is 10 o F higher than cooling water T Evaporation T (0 o F) is 10 o F lower than refrigerator T T S 1 2 3 3’ 70 o F 10 o F
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C H C T T T - = ϖ (b) For irreversible refrigerator, Carnot engine cannot be used: We cannot calculate coefficient of performance using We need to use 2 3 1 2 H H H H - - = 2 3 4 1 3
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At 0°F, HFC – 134a vaporizes at 21.162 (psia) (using H v in Table 9.1): m lb Btu H 015 . 103 2 = R lb Btu S m 22525 . 0 2 = At 80°F, HFC – 134a condenses at 101.37 (psia) (using H l ) m lb Btu H 978 . 37 4 = Data from Table 9.1:
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The choice of refrigerant Dependence?
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Chapter 9. Refrigeration and Liquefaction-student - Chapter...

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