i. Outside design conditions $42°C DBT, 26°C WBT$

ii. Inside design conditions $25°C DBT, 50% RH$

iii. Solar heat gain through glass 5.6kW

iv. Solar heat gain through walls, roof and floor 5.87kW

v. Occupants 35

vi. Sensible heat gain per person 57W

vii. Latent heat gain per person 57W

viii. Internal lighting load

12 lamps of 100 W 7 lamps of 60 W

12 fluorescent tubes of 40 W

ix. Infiltrared air 16 CMM

If 25% fresh air and 75% recirculated air is mixed and passed through the conditioner coil find

a) The amount of total air required in m3/hr

b) The DPT of the coil

c) The condition of supply air to the room

d) The capacity of conditioning plant

Subject:- Refrigeration and Air Conditioning

Topic:- Design of air conditioning systems

Difficulty:- Low

$t_d1=42℃,t_w1=26℃, t_d2=25℃, 〖RH〗_2=50%,BPF=0.2$

Locate points 1 and 2 on psychrometric chart, and draw vertical from 1 and horizontal from 2 to obtain point A.

∴$Q ̇_si=0.2914×(68.5-51)=5.1 \frac{kJ}{s}$

Let $Q ̇_li$ be rate at which latent heat is added by infiltrated air=$ m ̇_i ×(h1-hA)$

∴$Q ̇_li=0.2914×(80.5-68.5)=3.4968 \frac{kJ}{s}$

Now RSH=$5.87+5.6+\frac{(35×57)}{1000}+\frac{(12×100+7×60+12×40)}{1000}+Q ̇_si=20.665 kW$

RLH=$\frac{(35×57)}{1000}+Q ̇{_li}$=5.4918 kW

$RSHF=\frac{RSH}{(RSH+RLH)}$=0.79

Now draw RSHF line. To locate point 3, we know it lies on line 1-2

$\frac{m ̇_1}{m ̇_3} =\frac{length(23)}{length(12)}$

$\frac{m ̇_1}{m ̇_3} =0.25$ (given)

length(23)=1.475 cm

Now mark point 3 and assume coil ADP as 11°C and mark point C on saturation curve. Let 4 be the point of intersection of C-3 line and RSHF line.

BPF=$\frac{length(C4)}{length(C3)} =\frac{1.4}{6.2}$

=0.225 which is close enough to 0.2,hence we proceed with this

Calculations: By energy balance of room

$m ̇_4×h4+RSH+RLH=m ̇_2×h2$

But from the layout, $m ̇_4=m ̇_2$

$m ̇_4=2.09$ kg of dry air/sec

Also,$v_4=0.83 \frac{m3}{kg}$

∴ $m ̇_4=\frac{V ̇_a4}{v_4}$

$V ̇_a4$=total air required in $\frac{m3}{hr}=6244.92 \frac{m3}{hr}$

DPT of coil= 11°C (by trial and error approach)

Supply air conditions i.e. point 4

$t_d4=15.8℃,t_w4=13.6℃, 〖RH〗_4=78% $

Capacity of conditioning plant,

$Q ̇_a=m ̇_3×(h3-h4)$

Also, $m ̇_3=m ̇_4$

3.5×capacity=$m ̇_4×(58-38.5)$

Capacity=11.644 TR