A stream that consists of 2.00 mol% H_2O vapor and the balance, dry air (oxygen

Question

A stream that consists of 2.00 mol% H_2O vapor and the balance, dry air (oxygen and nitrogen), is mixed in a humidifier with a stream of liquid water to create a humidified stream of air that is 10.0 mol% water vapor. The flow rate of the liquid water stream entering the humidifier is 270 g/h. What is the total molar flow rate of the humidified air (mol/h) leaving the humidifier? Assume that the liquid water stream completely evaporates in the humidifier. Problem 4: Textbook problem 4.18 parts a & b Read Pages 4-1 to 4-6 of the student workbook, (posted on D2L in the content section under the heading F&R; workbook)

Explanation / Answer

let x= moles of vapor and air entering the humidifier/hr

moles of water vapor intering =x*2/100 moles/hr.

since flow rate of water added is given in g/hr, mass of water needs to be calculated.

Mass of water vapor entering= x*18*2/100= 0.36x x/hr ( Molecular weight= 18)

Molesof water vapor added= 270 g/hr

Total mass of water entering= 0.36x+270    (1)

Moles of air entering =98% is =0.98*x moles/hr (2)

Let y= moles/hr of humidifed air leaving

It contais 10 mole % water and rest 90% air, moles ofair at out let =0.9y

Moles of water vapor= 0.1y   mass of water entering= 18*0.1y= 1.8y

Sicne air is undergoing any change moles entering= moles leaving

0.98x= 0.9 y (3)   x= (0.9/0.98)y =0.92 y(4)

Mass of water entering= mass of water leaving

1.8y= 0.36x+270 ( from (1)

1.8 y= 0.36*0.92y +270

1.8y= 0.3312y+270

Therfore y= 188.8235 moles/hr

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.