Inside conditions $24^°C$ DBT, 55% RH, Outside conditions 37$°C$ DBT, 25$°C$ WBT, Sensible heat gain 13.6kW,.Latent heat gain 8.25k Return air is mixed with outdoor air before entering coil in ratio 3.5:1 and return from room is also mixed after cooling coil in ratio 1:3.5 Coil BPF is 0.15 Air may be reheated if necessary before supplying to conditioned room. Assuming ADP 7°C. Determine: a) Supply air condition to the room b) Amount of fresh air supplied c) Cooling load

Subject:- Refrigeration and Air Conditioning

Topic:- Design of air conditioning systems

Difficulty:- Low

$t_d2=24℃$,∅2=55%, $t_d1=37℃$, $ t_w1=25℃,ADP=7℃,$

RSH=13.6 kW,RLH=8.25 kW, BPF=0.15

Locate points 1 and 2 on psychrometric chart. Since return air from room and outside air are mixed before entering cooling coil in ratio 3.5:1, therefore mark this point 3 on line 1-2 as

length(23)=$\frac{length(21)}{4.5}$

From psychrometric chart,$t_d3$=27℃ and h3=56 $\frac{kJ}{kg}$ of dry air

We know, BPF=$\frac{(t_d4-ADP)}{(t_d3-ADP)}$

∴$t_d4=10℃$

mark point 4 on the line joining point 3 and ADP=7℃ on saturation curve such that its DBT is 10℃ From chart, h4=28 $\frac{kJ}{kg}$ of dry air

Since return air from room is also mixed with air leaving the cooling coil in ratio 1:3.5, therefore mark this point 5 on line 4-2 as length(45)=$\frac{length(42)}{4.5}$

From psychrometric chart, $t_d5=13.3℃$ and h5=33 $\frac{kJ}{kg}$ of dry air

RSHF=$\frac{RSH}{(RSH+RLH)}$=0.6224 Draw horizontal from point 5 to intersect RSHF line at point 6 which represents condition of supply air to room.

Supply air condition to the room

$t_d6=14℃$, $t_w6=11.8℃$

Amount of fresh air supplied

From chart, h6=33.5 $\frac{kJ}{kg}$ of dry air and h2=51$\frac{kJ}{kg}$ of dry air

Total amount of supply air =$\frac{(Room total heat)}{(Total heat removed per kg)}$=$\frac{(RSH+RLH)}{(h2-h6)}$=$\frac{1.248kg}{sec}$=74.91 kg/min

Amount of dehumidified air= $m_D=74.91\frac{3.5}{4.5}$=58.27 kg/min

Quantity of fresh air =$\frac{m_D}{4.5}$=12.94 kg/min

*Cooling load

Cooling load=$m_D (h3-h4)$=1631.56 $\frac{kJ}{min}$= $\frac{1631.56}{210}$=7.77 TR