$t_d2=24℃,∅2=55%, t_d1=37℃, t_w1=25℃,ADP=7℃,$
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
![enter image description here](https://i.imgur.com/xkCaqox.png)