Question

Nine hundred forty-seven cubic feet of wet air at 70 F and 29.2 in. Hg are dehydrated. If 0.94 lbs of water are removed, what was the relative humidity of the wet air?

Answer 1

Case Study :-Volume of Wet Air (V)=947 ft³,Temperature(T) =70 ⁰F,Pressure(Ps) =29.2 in Hg

Air id dehydrated and remove 0.94 lbs of water .Find out Relative Humidity of the wet air ?

Answer:-

(a)Volume of Wet Air (V)=947ft³=947/(3.281x3.281x3.281) m³=26.81 m3 thus V=26.81 m³

(b)Temperature (T)=70 ⁰F=0.555(70-32)⁰C=21.09⁰C=(21.09+273.16)⁰K=294.25 ⁰K Thus T=294.25 ⁰K

(c) Pressure (P)=29.2inHg=29.2x25.40 mmHg=741.68 mmHg=(741.68/760)atm=0.9759atm=(0.9759x101.325)KPa=98.88KPa Thus Ps=98.88 Kpa

Now According to Ideal Gas Law PV=nRT n=PV/RT=(98.88X26.81)/(8.314X294.25) here we are consider value of universal gas constant (R)=8.314 m³Kpa/Kmol^oK

n=1.084 Kmol ,Here n=Kmol of wet air present in system=1.084 Kmol

we are also known as 21% of oxygen (O₂) and 79 % of Nitrogen (N₂) present in Air

Oxygen (O₂)=1.084x0.21 Kmol=0.23 Kmol

As per chemical reaction between oxygen (O₂) and Hydrogen (H₂)

H₂+(1/2)O₂H₂O so that (1/2) Kmol of O₂ produce 1 Kmol of H₂O

Kmol of H₂O present =0.23x2=0.46 Kmol Water present in system=0.46x18=8.28 Kg

Now Water are removed 0.94 lbs=0.94/2.20 kg=0.427 kg

Water Balance in systen after dehydration process=8.28-0.427=7.857 kg

Kmol of water present after dehydration process=7.857/18=0.437 Kmol

O₂ present=0.438/2=0.219 kmol So moles of air present =0.219/0.21=1.04 kmol

n₂=Kmol of wet air present after dehydration process in system=1.04 Kmol

According to Idea gas law PV=n₂ RT   P=n₂ RT / V=(1.04X8.314X294.25)/(26.81) =94.90 KPa

Partial Pressure of Vapour (p)=94.90 KPa & Vapour pressure(Ps)=98.88 Kpa

Now Relative humidity(Rh) is ratio between partial pressure and vapour pressure

Rh=(p/Ps)x100=(94.90/98.88)x100=95.97 %

Relative Humidity after dehydration process or water removal is 95.97 %