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Question z a) What is the physical interpretation of the wet-bulb temperature? b) During winter time...

Question:

Question z a) What is the physical interpretation of the wet-bulb temperature? b) During winter time a room is maintained atm heater, kW air supplied to the room, % dity of the supply air acceptable in common engineering practi steam through the hea

Question z a) What is the physical interpretation of the wet-bulb temperature? b) During winter time a room is maintained at a steady temperature and humidity by an air-conditioning system shown. 200 m3/min of room return air at 30°C, 50% relative humidity is first mixed with 40 m3/min of outdoor at 10°C, 20% relative humidity. The mixed air is then introduced to a heater where the temperature of the air is raised to 45°C and supplied to the room. [Note that air and the steam in the heater do not mix. Steam simply supplies the heat.] Do the following using the ASHRAE Psychrometric chart only: i) Plot the processes on the ASHRAE Psychrometric chart and submit. Show all the states clearly on the chart. Determine: ii) the required capacity of the steam heater, kW iii) the relative humidity, 6, of the air supplied to the room, % iv) Is the calculated relative humidity of the supply air acceptable in common engineering practices? Explain why. v) the required mass flow rate of steam through the heater if saturated vapor is supplied at 150 kPa and exits the heater as saturated liquid at the same pressure, kg/min Sat, vap, 40 150 K 200 m/min 30°C, 50% Mixed air 45°C Room 40 min 10°C, 20% Steam heater Sat. Na 150 kPa
m heater, kW air supplied to the room, % dity of the supply air acceptable in common engineering practi steam through the heater if saturated vapor is supplied at 150 kr ssure, kg/min Sat, vap. @ 150 kPa 1 200 m/min 30°C, 3 = 50% Mixed air 45°C Room 3 4 2 40 m/min 10°C, 0 = 20% Steam heater Sat. liq. @ 150 kPa

Answers

Que a) Physical interpretation of WBT. - 0 WBT signifies that if we will cool a fluid or a medium with the help of atmosphorientay ha= hg@p- 1 Soleta - 26931 1 steam est he hf @ P= Isoppa = 467.13 K3 200 228.05 k/min 0.877 miz 40 0.804 49.75 kg/min(iii) R.M. at room inlet: - 04 = 0.18 We have calculated Parley. 04 =184 Mostly from ASHRAF guidelines relative humidity at i

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