Question

QUESTION 1 [15 marks] Due: 5 August 2020 An air conditioner operating at steady state takes in moist air at 28°C, 1 atm and 70% relative humidity. The moist air first passes over a cooling coil in the dehumidifier unit and some water vapor is condensed. The rate of heat transfer between the moist air and the cooling coil is 38.69 kW. Both saturated moist air and condensate streams exit the dehumidifier unit at the same temperature. The moist air then passes through a heating unit, exiting at 24°C, 1 atm and 40% relative humidity. Neglecting kinetic and potential energy effect. Sketch the process path of air on a Psychrometry Chart. (Please download the Psychrometry chart from Moodle Uniten and submit it together with your solutions), determine: (a) The specific humidity of the air before entering the dehumidifier unit. [1 marks] (6) The dew point temperature and the specific humidity of the air at the exit of the heating unit. [2 marks] (c) The dry bulb and wet bulb temperature of the saturated moist air leaving the dehumidifier unit. [2 marks] (d) The mass flow rate of the air entering the dehumidifier unit. [6 marks) (e) The rate of condensate leaving the dehumidifier unit, [2 marks] (1) The rate of heat transfer into the air at the heating unit. [2 marks] Schematic of component: 1 Cooling with dehumidification 2 Heating 3 T7=28 °C elell teee T3= 24°C RH = 40% RH = 70% = 36.89 kW P = 1 atm Condensed water Tcondensed water = TZ

Answer 1

Given: line 1 hg = 10.74 kj [kg dry air Ihz Enthalpy Humidi- 43.00 0-70% ty h₂ 128.31 Wi Ratio (kgrap 10.01669 kg dry saturated line p=100% p=404 v 12 Wz = | Wz -0.00742 24 9.6°C 28°c boy Bulb Temperature (°C) cooling 1 with dehumi difi- 2 3 cation Ti = 28°C Heating ell T = 24°C RH=701 T2 RU= 100 % (saturated) state _20__ beated) RH = 40% I condensate T=T water = 36.89 kW 2 from chrometric chart, nee find- Also can be found using formula W- air : 0.01669 ug uaplegday W=0.622 Pv where Pu=0Ps and we = W₂ { No moisture semoval P-Pv Ps = Saturation or addition as heating there? pressure at given dry bulb temperature : W₂ = W₂ = 0.00142 kg vap1 kg day air es

&= Pu b) ne see at exit of heating unit, - d = 40%. Tab= T3 = 28°c so from steam table saturation pressure at 28° c = 2.9856 kPa 0.40 Ps = Pu Ps Py = 0.40 (2.9856)=1-1945 kPa Dew point temp erature = satweation temperature at vapour pressure from steam table at Py = 1.1945 kPa. To = 9.58510 Dew point temperature at exit (Top) 59.5857 of heating unit unit- specific humidity at exit of heating Wag = 0,00742 kg vap1 kg dry air Dew point temper- ature c) for satwrated moist air - heet bulb temperature - Doy buls temperature - during heating there is no humidity along as moistwee addition and lemoral and thus Ratio Remains constant same horizontal line and thus 2 and 3 lie psychromenic chart and then. (Tap); Dew point temperature at 2 =9.5857°C (Tap)2 = 9.58516 à 9.6°f (Tadb = 9.58570 29.6°C (1) wb = 9.58576 à 9.6°C

air unit - the dehumi khumidifier d) Mass flow low rate of Applying mass balance to Jainit entering (cooling of day +Dehumidfyong ma, na = ma air Remains constant mass flow rate muz maa Balance in, - my mu m (mass flow rate of condensed water ) = nie ma co ~ W₂) Applying energy (mah,)-(mung + maha) somefic chart - = 90.74 kg 1kg dry 28.31 kg lieg dry from saturated water table- (hg) @ 9.58510 = 40.269 kg lg dry air ma (70.74)-ma (0.01669-0.00742) (40.269) - ma (28.31) from psycho hi air hr ais = 36.89 ma = 36.89 (70.74) -40.269 (0.01669-0.00742) - 28-31 of ma=0.8771 mass flow rate kg/sec air e) Rate of condensate leaving the Dehumidifier whit- m ma (w, W2) - 0.8771 (0.01669 -0.00742) in = 0.00813 kg/ see

f) Rate of heat transfer into air at heating unit (ha) ma (hz-h₂) c0.8711 (43 - 28.31) Q = 12.884 kW qus