written 6.3 years ago by
teamques10
★ 69k
|
•
modified 6.3 years ago
|
Given:
td1=30℃,td2=20℃,RH1=55
ω_1-ω_2=0.004 \frac{(kg \ of \ water \ vapour)}{(kg \ of \ dry \ air)}
To find:
RH_2
DPT_2 or t_dp2
We know,
Humidity ratio =
ω=0.622×\frac{p_v}{(p_b-p_v )}
∴ω_1-ω_2=0.622[\frac{p_v1}{(p_b-p_v1 )}-\frac{p_v2}{(p_b-p_v2 )]}]
∴0.004=0.622[\frac{p_v1}{(1.0132×100-p_v1 )}-\frac{p_v2}{(1.0132×100-p_v2 )]}]……..(1)
Also,
RH_1= \frac{p_v1}{p_vs1} ……(2)
Now,p_{vs1}=p_(sat_{t_d1})=0.04242 bar (from steam table)
Substituting in (2) we get,
p_v1=0.023331 bar=0.023331×100 kPa
Substituting in (1) we get,
p_v2=1.70725 kPa
Now, RH_2= \frac{p_v2}{p_vs2}
Substituting,
p_v2 and p_vs2=p_{sat_{t_{d2}}}=2.337 kPa (from steam table)
We get, RH_2=73.05%
We know, DPT_2 or t_dp2=T_{sat{p_v2}}
From steam table using interpolation we find T_sat at pressure of 0.0170725 bar,
\frac{0.02-0.0170725}{0.02-0.015}=\frac{17.51-T_sat}{17.51-13.04}
∴DPT_2 or t_dp2=T_{sat_{p_v2 }}=14.893℃