written 7.0 years ago by
teamques10
★ 70k
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modified 7.0 years ago
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Given:
$t_d1=30℃,t_d2=20℃,RH_1=55%,p_b=1.0132 bar=1.0132×100 kPa$
$ω_1-ω_2=0.004 \frac{(kg \ of \ water \ vapour)}{(kg \ of \ dry \ air)}$
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 …
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