Question

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 m/min of room return air at 30°C, 50% relative humidity is first mixed with 40 m/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 ill) 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 at vap 12 MPa 200 30C, -50% Moned air 45°C Room 40 min 100, 20% Steam heater Sa. 1501

Answer 1

Que a) Physical interpretation of WBT. - 0 WBT signifies that if we will cool a fluid or a medium with the help of atmosphoric air them the fluid will be cooled down to minimum temperature which is Wet bulb temp, not less than that. o Whan air will be saturated then DBT = WBT = Dew point temp. states (e) Crinen - State 1 V,= 200 m/min T=30°c d=0.5 state 2-40 m/min T2 = 10°C O2 = 0.2 Ta = 45°C Solyn: (i) Procus on Psychromitrie Chart- ina 4 3 quch ( 49/ vg of da.) (95c 10°c T (°C) (i) From Psychrometric chart - at T1=30°C : 0,=0.5 W, = 13.5 gr/kg of da. V, = 0.877 m²/long of dia. h = 65 kulkey of dia. at Tz=10°c : 82=0.2 h2=14 Kalkg of da. U2 = 0.804 m²/kg of da. W2 = 1.5 grm/1s of da.

entay ha= hg@p- 1 Soleta - 26931 1 steam es't he hf @ P= Isoppa = 467.13 K3 200 228.05 k/min 0.877 miz 40 0.804 49.75 kg/min miz=mi + m2 = 277. Sol k/min 200+40 0.8639 m3/rs of dha mitma 277.81 3. (86 From Chart draw a line jorking the state 0 16 then find the point which meets &z= 0. 8639 m²/leg of dier to get the point from Chart will get a h₂=55.86 Kalkg of d.a. T3 = 27. 1° c Wz=10.8gm/sof d.o. $3=0.48 Nom draw a horizontal line from point 3 to point and met where To=45°C. to get pt. 4. WqWz = 10.89m [oss [as steam do not kg min -0.18 h = 73 kJ/kg of d.a. Now From steam table table- Steam = (ii) Capacity of heater- tg Q Heater min (ha-hs) Qreater 277.801 ( 73 - 55.86) Kals 60 Qleath= 79.35 kW

(iii) R.M. at room inlet: - 04 = 0.18 We have calculated Parley. 04 =184 Mostly from ASHRAF guidelines relative humidity at inlet should be (40N 55% RH) so for comfortable of human bringe. So This value $4=184. is not acceptable. It will Create discomfort and minor disase. Steam Apply energy balance of Heater! - kha 4 air wiz (ha - hz) = mg (har hle) 277.80) (73 - 55.86) mis (2693.1- 467.13) mis = 2139 /min mass flow rate of steam mis = 0.0356 kole # PLS RATE!