Question

al 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 20 m/min of outdoor 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 Psychometric chart only Plot the processes on the ASHRAE Psychrometric chart and submit. Show all the states clearly on the chart. Determine: the required capacity of the steam heater. kW 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 i saturated wiporis supplied at 150 kPa and ects the heaters saturated liquid at the same pressure, kamin

Answer 1

by creating cloth. thermometer a) Wet Bulb Temperature (WBT). - It is the temperature measured When its bulb is covered will wet - It tells how tells how easy it is for water to evaporate if Which can be cooled evaptration upto 100% relative Can't add more watery in it) then the that ling effect OCwrs adiabatic evaporation OF Water is called wet bulb temperature bli gou consider qir by humidity cie? we Cooling effect by h hz 3 o [kgling dry of airt 27 45°C 1 Jo'c 30'c OBT [°C] State 4:- State 2:- State 1:- T2 - lo'c T4 = 45°c Ti = 30°C 0,= 0.5 V. = 200 mg /min 02=0:2 V2 = 40 m3/min CS Scanned with CamScanner

ii] from psychrometric chant at Ti = 30°c Wi - 13.5 of dA & 0-0.5 gm/ng v; • 0.877 merking of dA kJ/kg hi 65 KJ at T2. lo'c & $2 = 0.2 h2 = 14 W2 1.5 kJ/kg Vz 0-804 me /kg 1.5 gm /kg 228.05 kg/min Of da mi Vi 11 200 0.877 m2 V2 V2 0.804 40 - 49.75 Kg/min - 217.8 kg/min m3 = m + m2 228.05 + 49.75 V3 = Vi + V2 200+ 40 V3 mz . 0-8639 mi/kg mit me 211.8 of dA → Draw line State 1 & 2 then find a point joining at which V3 = 0.8639 m3 and mark it as point 3. 13 /kg and cs Scanned with CamScanner

At point 3:- ha · 55.86 kJ/kg T3 = 27.1°C W3 of dA $3: 0:48 → Dram a horizontal line from point 3 to a 4 point Where T: 45°C to point 4. 10.8 gm /kg = 10.8 gm /kg W4: W3 44= 0.18 h4-13 KJ/kg of heater ( QHeater) iij capacity QHeater : M3 (24-h3) 277-8 (73 - 55.86) 60 QHeater 79. 3582 kW iji] Relative humidity at room inlet (64) = 04- 18% iv) according to humidity is 18% , It is not acceptable to ASHRAE condition for comfort norms, Standard Relative is 30 to 50%. As relative humidity Supply cs Scanned with CamScanner

v] from ha Steam table Steam at entry has 3) bg @p @p - 150 kpa 2693. 1 KJ/kg Steam at exit heiq: hf @p: 150 kpa hliq = 467.13 KJ KJ/kg balance in heater, • Applying energy M3 Cha-h3) = ms Cha-htig) 277.8(73 - 55.86) = ms ( 2693.1 – 467-13) mis 2.139 Kg/n /min mis - 0-03565 kgls Mass flow rate of Steam. Ooo cs Scanned with CamScanner