written 5.5 years ago by
teamques10
★ 66k

•
modified 5.5 years ago

Given:
$t_d1=30℃,t_d2=20℃,RH_1=55%,p_b=1.0132 bar=1.0132×100 kPa$
$ω_1ω_2=0.004 \frac{(kg \ of \ water \ vapour)}{(kg \ of \ dry \ air)}$
To find:
$RH_2$
$DPT_2$ or $t_dp2$
We know,
Humidity ratio =
ω=0.622×$\frac{p_v}{(p_bp_v )}$
∴$ω_1ω_2=0.622[\frac{p_v1}{(p_bp_v1 )}\frac{p_v2}{(p_bp_v2 )]}]$
∴0.004=0.622$[\frac{p_v1}{(1.0132×100p_v1 )}\frac{p_v2}{(1.0132×100p_v2 )]}]$……..(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,
$\frac{0.020.0170725}{0.020.015}=\frac{17.51T_sat}{17.5113.04}$
∴$DPT_2$ or $t_dp2$=$T_{sat_{p_v2 }}$=14.893℃