i. Inside conditions 24°C DBT, 55% RH

ii. Outside conditions 37°C DBT, 25°C WBT

iii. Sensible heat gain 13.6kW

iv. Latent heat gain 8.25kW

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.

a)Supply air condition to the room

$t_d6=14℃, t_w6=11.8℃$

b)Amount of fresh air supplied

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

c)Cooling load

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