Answer to Question #166341 in Chemical Engineering for Piyush Mishra

Question #166341

It is required to absorb 95 % of the acetone from a mixture with nitrogen containing 1.5% mol% of the compound in a counter current tray tower. The total gas input is 30 kmol/h and water enter the tower at a rate of 90 kmol/h. The tower operates at 300K and 1 atm. The equilibrium relation is y =2.53x. Determine the number of ideal trays necessary for this separation. Use the graphical method as well as the Kremser analytical method.


1
Expert's answer
2021-02-25T04:16:24-0500

Basis: 1 hour

GN+1 = 30 kmol

yN+1= 0.015

L0 = 90 kmol

Moles acetone in = 30×0.015 moles = 0.45 moles

Moles nitrogen in = (30-0.45) moles = 29.55 moles Moles acetone leaving (95% absorbed) = 0.45×(1-0.95) moles = 0.0225 moles

Gs = 29.55 moles

Ls = 90 moles

α = 2.53 [as, Y = 2.53X] 

Y1 = 0.0225/29.55 =7.61×10-4

YN+1 = 0.015

Gs(YN+1 - Y1) = Ls(XN - X0)

29.55 × (0.015 − 7.61×10−4) = 90(Xn −0)

XN = 4.68×10-3

a) Solution by graphical method Construction of operating line PQ:

P(X0, Y1) = P(0, 7.61×10-4)

Q(XN, YN+1) = Q(4.68×10-3, 0.015) 

Construction of equilibrium line (Y = 2.53X): X 0 0.001 0.002 0.003. 0.004 0.005

Y 0 0.00253 0.00506 0.00759 0.01012 0.01265

From graphical construction, the number of triangles obtained is more than 7. Hence number of ideal stages is 8. 



b) Solution by Kremser analysis method As Ā≠1, according to Kremser analysis method: 

N = ln {[(YN+1 - X0)/(Y1 - X0)][1 - 1/Ā] + 1/Ā} / lnĀ

N = ln {[(0.015 - 2.53x0)/(7.61×10-4 - 2.53x0)][1 - 1/1.204] + 1/1.204} / ln(1.204)

N = ln[(20)(0.204/1.204) + 1/1.204] / 0.1856

N = 1.43966 / 0.1856 = 7.75

Number of ideal stages is 8


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