12.50 An air conditioner operating at steady state takes in moist air at 28°C, 1 bar, and 70% relative humidity. The moist air f
irst passes over a cooling coil in the dehumidifier unit and some water vapor is condensed. The rate of heat transfer between the moist air and the cooling coil is 11 tons. Saturated moist air and condensate streams exit the dehumidifier unit at the same temperature. The moist air then passes through a heating unit, exiting at 24°C, 1 bar, and 40% relative humidity. Neglecting kinetic and potential energy effects, determine a. the temperature of the moist air exiting the dehumidifier unit, in °C. b. the volumetric flow rate of the air entering the air conditioner, in m3/min. c. the rate water is condensed, in kg/min. d. the rate of heat transfer to the air passing through the heating unit, in kW.
1 answer:
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Answer:
The mass flow rate of cooling water required to cool the refrigerant is .
Explanation:
A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:
The mass flow rate of the cooling water is now cleared:
Given that ,
,
and
, the mass flow of the cooling water is:
The mass flow rate of cooling water required to cool the refrigerant is .
Answer:
A) ΔS_refrigerant = 0.70754 Kj/K
B) ΔS_space = -0.68441 Kj/K
C) ΔS_total = 0.02313 Kj/K
Explanation:
A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C
Converting to degree kelvin yields;
T = -18.77 + 273 = 255.23 K
Formula for entropy change of refrigerant is given as;
ΔS_refrigerant = Q_in/T_refrigerant
We are given Q = 180 KJ
Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K
B) Formula for entropy change of cooled space is given as;
ΔS_space = Q_out/T_s pace
T_space = -10°C = 273 - 10 = 263K
Thus, ΔS_space = -180/263 = -0.68441 Kj/K
C) the total entropy change would be;
ΔS_total = ΔS_refrigerant + ΔS_space
Thus,
ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K
