An absorber is used to remove acetone from a nitrogen carrier gas. the feed (str

Question

An absorber is used to remove acetone from a nitrogen carrier gas. the feed (stream 1). with acetone weight fraction of 0.213, enters at a rate of 2 m^3/h. the absorbing liquid is water and enters the absorber (stream 2) with a mass flow rate five times more than the feed mass flow rate. the exit gas (stream 3) is 0.8 wt% acetone and 2.9 wt% water vapor. the mass flow rate of stream 3 decreases by 18% compared to feed stream. Assume no N_2 goes with liquid. Draw the process flow diagram and completely label all streams accordingly. Determine the acetone weight fraction in the bottom stream removed by water (stream 4)

Explanation / Answer

Stream-1 :   Basis : 1 gm of acetone, it contains 0.213 gm of acetone and 1-0.213= 0.787 gm of nitrogen.

Moles : Mass/molar mass, Molar masses : acetone =58 and nitrogen =28

Moles : acetone =0.213/58 =0.0037, moles of Nitrogen = 0.787/28= 0.028

Total moles = 0.0037+0.028= 0.032

Assuming the flow rate mentioned is at standard conditions, 1 mole of any gas at STP occupies 22.4 Liters. 2m3/hr= 2000L/hr correspond to 2000/22.4 moles = 89.3 moles/hr

0.032 moles of mixture is present in 1 gm

89.3 moles/hr is present in 89.3/0.032 =2791 gm of mixture.

Moles of acetone entering in stream-1= total moles* mole fraction acetone = 89.3* 0.0037/(0.0037+0.028) =10.29 moles of acetone/hr

Stream-2 flow rate = 5*2791 gm = 13955 gm/hr, moles = 13955/18=775 moles/hr

Stream-3 : taking a basis of 1 gm of stream-3, moles of acetone =0.8/(100*58)=0.000138

Moles of water = 2.9/(18*100)= 0.001611, nitrogen = .968/(28*100)= 0.000346

Total mole in 1 gm = 0.000138+0.001611+0.000346= 0.002095

Moles in Stream-3 = 0.82*m1= 0.82*89.3 =73 moles/hr

   moles in stream-1 : m1, moles in stream-2 = m2, moles in stream-3 =m3 and moles in stream-4= m4

Writing overall moles balance, m1+m2= m3+m4

89.3+775= 73+ m4, m4= 791.3 moles/hr

Writing acetone balance

Moles of acetone in stream-1 + moles of acetone in stream-2 = moles of acetone in stream-3+ moles of acetone in stream-4 =

10.29+ 0 = 73*0.000138/(0.000138+0.001061+0.000346) + 791.3*x, x= mole   fraction acetone

Hence x= 0.004764

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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