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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

20Determine (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

2 Answers
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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℃

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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

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