The fallowing question is from a reeds book on applied heat i am studying. Although the answer is provided, im struggling to understand the whole answer and the formulas and the steps theyre using. Also where some ov the values such as Hg and Hf come from in part i for example. Please explain step per step in detail thanks In an NH, refrigerator, the ammonia leaves the evaporatorand enters the cornpressor as dry saturated vapour at 2.68 bar,it leaves the compressor and enters the condenser at 8.57 bar with50" of superheat. it is condensed at constant pressure and leavesthe condenser as saturated liquid. If the rate of flow of the refrigerantthrough the circuit is 0.45 kglmin calculate (i) the compressorpower, (ii) the heat rejected to the condenser cooling water in kJ/s,an (iii) the refrigerating effect in kJ/s. From tables page 12, NH,:2.68 bar, hg= 1430.58.57 bar, hf = 275.1 h supht 50" = 1597.2Mass flow of refrigerant--- - - 0.0075 kgls 60Enthalpy gain per kg of refrigerant in cornpressor= h, - h,= 1597.2 - 1430.5 = 166.7 W/kgEnthalpy gain per second= mass flow [kgls] x 166.7 [mg]= 0.0075 x 166.7 = 1.25 MISMIS = kW, therefore useful compressor power= 1.25 kW Ans. (i)Enthalpy drop per kg of refrigerant through condenser= h2 - h3= 1597.2 - 275.1 = 1322.1 W/kgHeat rejected = 0.0075 x 1322.1= 9.915 Ws Ans. (ii) Enthalpy gain per kg of refrigerant through evaporator= hl--h4= 1430.5 - 275.1 = 1155.4 kJ/kg(Note that the value of h4 is the same as h, because there is nochange of enthalpy in the throttling process through the expansionvalve between the condenser exit and the evaporator inlet).Refrigerating effect = 0.0075 x 1155.4= 8.666 M/s Ans, (iii)
The fallowing question is from a reeds book on applied heat i am studying. Although the answer is provided, im struggling to understand the whole answer and the formulas and the steps theyre using. Also where some ov the values such as Hg and Hf come from in part i for example. Please explain step per step in detail thanks
In an NH, refrigerator, the ammonia leaves the evaporator
and enters the cornpressor as dry saturated vapour at 2.68 bar,
it leaves the compressor and enters the condenser at 8.57 bar with
50" of superheat. it is condensed at constant pressure and leaves
the condenser as saturated liquid. If the rate of flow of the refrigerant
through the circuit is 0.45 kglmin calculate
(i) the compressorpower,
(ii) the heat rejected to the condenser cooling water in kJ/s,
an (iii) the refrigerating effect in kJ/s.
From tables page 12, NH,:
2.68 bar, hg= 1430.5
8.57 bar, hf = 275.1 h supht 50" = 1597.2
Mass flow of refrigerant
--- - - 0.0075 kgls 60
Enthalpy gain per kg of refrigerant in cornpressor
= h, - h,
= 1597.2 - 1430.5 = 166.7 W/kg
Enthalpy gain per second
= mass flow [kgls] x 166.7 [mg]
= 0.0075 x 166.7 = 1.25 MIS
MIS = kW, therefore useful compressor power
= 1.25 kW Ans. (i)
Enthalpy drop per kg of refrigerant through condenser
= h2 - h3
= 1597.2 - 275.1 = 1322.1 W/kg
Heat rejected = 0.0075 x 1322.1
= 9.915 Ws Ans. (ii)
Enthalpy gain per kg of refrigerant through evaporator
= hl--h4
= 1430.5 - 275.1 = 1155.4 kJ/kg
(Note that the value of h4 is the same as h, because there is no
change of enthalpy in the throttling process through the expansion
valve between the condenser exit and the evaporator inlet).
Refrigerating effect = 0.0075 x 1155.4
= 8.666 M/s Ans, (iii)
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