written 6.2 years ago by
teamques10
★ 69k
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modified 6.2 years ago
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$t_d1=40℃,t_w1=27℃, t_d2=25℃, RH_2=50%,BPF=0$
Locate points 1 and 2 on psychrometric chart and join them. Mark RSHF point and draw RSHF line passing through point 2. Wherever the RSHF line intersects the saturation curve is point 4 since BPF =0.
To locate point 3, we know it lies on line 1-2
$\frac{m ̇_1}{m ̇_3} =\frac{length(23)}{length(12)} $
$\frac{m ̇_1}{m ̇_3}$ =0.5 (given)
length(23)=2.95 cm
Now mark point 3 and join 3 and 4. Draw vertical from 1 and horizontal from 2 to obtain point A.

Let $Q ̇_sf$ be sensible heat load added by fresh air=$ m ̇_1 ×(hA-h2)$
$V ̇_a1=100\frac{m3}{min}=1.667 m3/s $
$m ̇_1=\frac{V ̇_a1}{v_1} =\frac{1.667}{0.915}=1.82 kg/s$
∴$Q ̇_sf=1.82×(67-51)=29.14 \frac{kJ}{s}$
Let $Q ̇_lf$ be latent heat load added by fresh air= $m ̇_1 ×(h1-hA)$
∴$Q ̇_lf=1.82×(85-67)=32.76 \frac{kJ}{s}$
ADP=$t_d4=11.4℃ $
Since $\frac{m ̇_1}{m ̇_3} =0.5 ∴m ̇_3=m ̇_4=m ̇_2=3.64 kg/s$
By energy balance of room
$m ̇_4×h4+RSH+RLH=m ̇_2×h2$
RSH+RLH=65.52…(1)
Also,
RSHF=$\frac{RSH}{(RSH+RLH)}$=0.8…(2)
From (1) and (2),
RSH= 52.416 kW
RLH=13.104 kW
Humidity ratioof entering air at air conditioning room=$ω_4$=$\frac{0.0086 (kg \ of \ moisture)}{(kg \ of \ dry \ air)}$
DBT of entering air at air conditioning room =$t_d4=11.4℃$