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0.004 kg of water vapour per kg of atmospheric air is removed and temperature of air after removing the water vapours becomes

$20^{\circ}$Determine (without using psychrometric chart)

i. Relative humidity

ii. Dew point temperature

Assume that condition of atmospheric air is 30°C and 55% R.H. and pressure is 1.0132 bar.

Subject:- Refrigeration and Air Conditioning

Topic:- Psychrometry

Difficulty:- Low

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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 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℃

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0.004 kg of water vapour per kg of atmospheric air is removed and temperature of air after removing the water vapour becomes 20 "C. Determine: 1- Relative humidity 2- Dew point temperature. Assume that conditions of atmospher pressure is 1.0132 bar. air is 30 °C and 55 % RH and