Question

The atmospheric air 32 °C dry bulb temperature and 70% relative humidity supplied to the cooling coil at a rate of 45m3/min. The air cooling to the

saturated state and leaving at a temperature of 16 °C. Determine: a- Specific humidity at each state.

b- Wet bulb and dew point temperatures at the final state

c- Final relative humidity.

d- Mass of water condensed.

e- Rate of heat removed from the air in kW. f- Show the process on Psychrometric chart.

Answer 1

Sol alr atmospheric air → 32 °C = DBT & Relative humidity flow rate = 45 m/min $- 70% cooling coil I 2 terre? atm. air our 25.5°C Ws= 0.018 79 Rg water Rg of dry air cooled aie our leaving after cooled at the saturated zlate & temp lóc =16c TDPT at saturation state TDBT - TWBT & 100% teoperties at stale 1 (D67= 32°C 6.8=70%) So from peychromobic chart WB TE specific humidity 98.22 $5) agaf doyait mg of day air specific valume.. properties at state 2 ( D87=160 & $ = 100%) from peycheometric chart [OPT = W87 -087=180 160] W2 = 0.01136 kg water ha = 64.88 kJ/kg of dry air kg of dry air W = specific humidity 0.8 267 ming of dey air. specific valume. w,-W2 = (0.01879 -0.01136) Rg water kg of dry cair I W, -wa) → drop of water vapour quantity ny U = 0. 8867 =) air 닐 nlow 743 100 as it poss from cooling coll

743 now that deep in water vapow will condense. Rg of water 100 kg of dryain flow eate 45m/min li specfic volume at stale 1 now au -0-88667 m/bg afdeyair = flow eate of air mas mass 45 60 x 0.8867 Polsace = ( 2 ) liño = 0.0650 kg of dry au s 743x0.0650 106 ti mass of water condensed = 0.4g4 gm/sec -> Rate of heat removed from air = (h-ha) mina OR (98.22 – 64.88) 0.6650 de 22.12 kw

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Hope you can see process on psychrometric chart, it is shown in red colored line.