An air-conditioning system tasked with taking in air at 1 atmosphere pressure, 34°C, and with a relative humidity of 70%. At the outlet, it delivers air at 22°C with a reduced relative humidity of 50%. This process involves two main stages: first, the air passes over cooling coils where it's cooled and dehumidified, and then it moves over resistance heating wires to reach the desired temperature. Condensate is extracted from the cooling section at 10°C. Calculate:
(a) The temperature of the air before it enters the heating section.
(b) The amount of heat removed in the cooling section by also considering the enthalpy of the
removed water.
(c) The amount of heat transferred in the heating section, all expressed in kJ per kilogram of dry
air
Guide On Rating System
Vote
To solve this problem, we can use psychrometric charts to determine the properties of the air at different stages of the process.
(a) To find the temperature of the air before it enters the heating section, we first need to determine the properties of the air at the outlet of the cooling section. Using the psychrometric chart, we can find that the relative humidity of the air at the outlet is 50% and the specific enthalpy is h1. We can also determine that the dewpoint temperature of the air at the outlet is 10°C, as the condensate is extracted at that temperature.
Using the relative humidity and the dewpoint temperature, we can find the specific volume of the air at the outlet. From the chart, the specific volume at the dewpoint temperature of 10°C is approximately 0.856 m^3/kg.
Now, we can use the specific volume and the specific enthalpy at the outlet to determine the temperature of the air before it enters the heating section. We know that the specific volume at the outlet is equal to the specific volume at the inlet (assuming no leakage or mixing with the external environment). From the psychrometric chart, we can find that the specific volume at 34°C and 1 atm is approximately 0.987 m^3/kg.
Therefore, the temperature of the air before it enters the heating section is:
T2 = 34°C * (0.987 m^3/kg / 0.856 m^3/kg) = 39.6°C
So, the temperature of the air before it enters the heating section is approximately 39.6°C.
(b) The amount of heat removed in the cooling section can be calculated using the specific enthalpies of the air at the inlet and outlet of the cooling section. Using the psychrometric chart, we can find that the specific enthalpy of the air at the inlet is h2.
Therefore, the amount of heat removed in the cooling section is:
Q = (h2 - h1) * m
Where m is the mass flow rate of the air. Since we are given the properties of the air per kilogram of dry air, we can assume that the mass flow rate of the dry air is 1 kg.
(c) The amount of heat transferred in the heating section can be calculated using the specific enthalpies of the air at the outlet of the cooling section (h1) and the temperature at the outlet of the heating section (22°C).
Therefore, the amount of heat transferred in the heating section is:
Q = (h1 - h3) * m
Where h3 is the specific enthalpy of the air at 22°C.
Note: To get the specific enthalpies and other properties from the psychrometric chart, you may need to interpolate between the given values on the chart.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99