Question

# # V=320 P=latin Air flowing at 320 m/min is to be conditioned in a constant pressure process at 1 atm from 43°C& and 40 percent relative humidity to 25°C dry bulb and 18.5°C wet bulb. The air is first passed ove T= 43°C coils to remove all the moisture necessary to achieve the final moisture content and then is heating coils to achieve the final state. Fill in the properties table [8%) and determine the following: =40% passed T₁=25°C Twbs=18.5°C 1 2 3 Name: Problem 3: [30%] Tab (°C) 43 -Y = Pv 8.71 25 Pu 117057 P- sat _=W3 Pu = Pesat Heating coils teleeeeeeeeeeeee W₂ =W₂ from then ermine \$ (%) 40 too 2 chart (0 2 state 2 \$/2=100%. 0.0714 1.635 1.635 a) [3%] Sketch the process on the provided psychrometric chart. b) [2%] The mass flow rate of air in "kg/s" →Pv₁ = 3.484 kpa Cooling coils P=1₁7057 15°C h (kJ/kga) llllllle 1 atm Condensate Condensate removal (m³/kga) T=43 Pat= 8.71 kpap W 0.622 Pu P-PU = 0.622 (3.484) 8.71 (3,484) = 0.0714 T₁=43°C ₁-40% State Twb (°C) اسپنسه 18.5 |Tz=15°C1 Sot = 1.7057 Ⓒ 0,622(1.7057) 1.50 1176037-(1.1257) التي 1/2 = 1.639 8=. 12 2 1.55 T=25°C =316= 0.60 065 03/nName: c) [5%] The rate of condensation in cooling section in "kg/s" d) [7%] The rate of heat removal in the cooling section in "kW" e) [5%] The rate of heat input in the heating section in "kW"  Fig: 1  Fig: 2