slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
CP302 Separation Process Principles PowerPoint Presentation
Download Presentation
CP302 Separation Process Principles

Loading in 2 Seconds...

play fullscreen
1 / 25

CP302 Separation Process Principles - PowerPoint PPT Presentation


  • 77 Views
  • Uploaded on

CP302 Separation Process Principles. Mass Transfer - Set 8. Treated gas G out , y out. Inlet solvent L in , x in. We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class. G y. L x. Control volume.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'CP302 Separation Process Principles' - rafael-knight


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

CP302 Separation Process Principles

Mass Transfer - Set 8

Prof. R. Shanthini

slide2

Treated gas

Gout, yout

Inlet solvent

Lin, xin

We have learned to determine the height of packing in packed columns with dilute solutions in the last lecture class.

G

y

L

x

Control

volume

Today, we will learn to determine the height of packing in packed columns with concentrated solutions.

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide3

Notations

Gs - inert gas molar flow rate (constant)

Ls - solvent molar flow rate (constant)

G - total gas molar flow rate (varies as it looses the solute)

L - total liquid molar flow rate (varies as it absorbs the solute)

Y - mole ratio of solute A in gas

= moles of A / moles of inert gas

y - mole fraction of solute A in gas

= moles of A / (moles of A + moles of inert gas)

X - mole ratio of solute A in liquid

= moles of A / moles of solvent

x - mole fraction of solute A in liquid

= moles of A / (moles of A + moles of solvent)

Solute in the gas phase = Gs Y = G y

Solute in the liquid phase = Ls X = L x

Prof. R. Shanthini

slide4

Equations for Packed Columns

Treated gas

Gout, yout

Inlet solvent

Lin, xin

Mass of solute lost from the gas over the differential height of packing dz

= G y - G (y + dy) = - G dy

L

x

G

y

was used for packed column with dilute solution assuming G can be taken as a constant for dilute solutions.

dz

L

x+dx

G

y+dy

Z

z

For concentrated solution we ought to use Gs (inert gas molar flow rate) which is constant across the column along with Y (mole ratio of solute A in gas).

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide5

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

Mass of solute lost from the gas over the differential height of packing dz

= Gs Y - Gs (Y + dY) = - Gs dY

(85)

Gs

Y

Ls

X

Relate Gs to G:

dz

Gs = G (1 – y)

(86)

Ls

X+dX

Z

Gs

Y+dY

z

Relate Y to y:

From y = Y / (Y+1), we get

Y = y / (1 – y)

(87)

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide6

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

Using (86) and (87), mass of solute lost from the gas over the differential height of packing dz, given by (85), can be written as follows:

Gs

Y

Ls

X

  • Gs dY = - G (1 – y) d[y / (1 – y)]

dz

Ls

X+dX

dy

Z

= - G (1 – y)

Gs

Y+dY

(1 – y)2

z

dy

= - G

(88)

(1 – y)

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide7

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

Mass of solute transferred from the gas to the liquid

= Kya (y – y*) S dz

where S is the inside cross-sectional area of the tower.

Gs

Y

Ls

X

dz

Relating (88) to the above at steady state, we get

Ls

X+dX

Z

Gs

Y+dY

z

dy

-G = Kya (y – y*) S dz

(89)

(1 – y)

Compare (89) with (79) used for packed columns with dilute solutions.What are the differences?

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide8

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

Rearranging and integrating (89) gives the following:

yin

dy

G

(90)

Z =

Gs

Y

Ls

X

KyaS

(1 – y)(y – y*)

yout

dz

Ls

X+dX

Z

Gs

Y+dY

z

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide9

Equations for Packed Columns with concentrated solutions

Multiply the numerator and denominator of (90) by (1 – y)LM:

yin

(1 – y)LM dy

G

Z =

(91)

Kya(1 – y)LMS

(1 – y)(y – y*)

yout

where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows:

y* – y

(1 – y)LM

=

(92)

ln[(1 – y)/(1 – y*)]

Prof. R. Shanthini

slide10

Equations for Packed Columns with concentrated solutions

Even though G and (1 – y)LM are not constant across the column, we can consider the ratio of the two to be a constant and take G / [Kya(1 – y)LMS] out of the integral sign in (91) without incurring errors larger than those inherent in experimental measurements of Kya. (Usually average values of G and (1 – y)LM are used.)

Therefore (91) becomes the following:

yin

(1 – y)LM dy

G

Z =

(93)

Kya(1 – y)LMS

(1 – y)(y – y*)

yout

NOG

HOG

Prof. R. Shanthini

slide11

Equations for Packed Columns with concentrated solutions

y* in (93) can be related to the bulk concentration using the equilibrium relationship as follows:

y* = K x

(94)

x in (94) can be related to y in (93) using the operating line equation.

We will determine the operating line equation next

Prof. R. Shanthini

slide12

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

The operating equation for the packed column is obtained by writing a mass balance for solute over the control volume:

Lin xin + G y = L x + Gout yout

(74)

G

y

L

x

If dilute solution is assumed, then Lin = L = Lout and Gin = G = Gout.

Control

volume

We somehow have to relate L to Lin and G to Gout in (74).

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide13

Equations for Packed Columns with concentrated solutions

Treated gas

Gout, yout

Inlet solvent

Lin, xin

To relate L to Lin, write a mass balance for solvent over the control volume:

Lin (1 – xin) = L (1 – x)

L = Lin (1 – xin)/ (1 – x)

(95)

G

y

L

x

To relate G to Gout, write the overall mass balance over the entire column:

Control

volume

G + Lin = Gout + L

(96)

Inlet gas

Gin, yin

Spent solvent

Lout, xout

Prof. R. Shanthini

slide14

Equations for Packed Columns with concentrated solutions

Use (95) to eliminate L from (96):

G + Lin = Gout + Lin (1 – xin)/ (1 – x)

G = Gout + Lin (x – xin) / (1 – x)

(97)

Combining (74), (95) and (97), we get the following:

Lin xin + [Gout+Lin(x – xin)/ (1 – x)] y = Lin(1 – xin)x/(1 – x) + Gout yout

Gout yout + [Lin (1 – xin) x / (1 – x)] - Lin xin

y =

(98)

Gout + [Lin(x – xin) / (1 – x)]

Equation (98) is the operating line equation for packed columns with concentrated solutions. Compare it with equation (76) used for packed columns with dilute solutions.What are the differences?

Prof. R. Shanthini

slide15

Equations for Packed Columns with concentrated solutions

If the solvent fed to the column is pure then xin = 0.

Therefore, (98) becomes

Gout yout + [ Lin x / (1 – x) ]

y =

(99)

Gout + [ Lin x / (1 – x) ]

Prof. R. Shanthini

slide16

Example 1:

Draw the operating curve for a system where 95% of the ammonia from an air stream containing 40% ammonia by volume is removed in a packed column. Solvent used in 488 lbmol/h per 100 lbmol/h of entering gas.

Solution:

Lin = 488 lbmol/h; Gin = 100 lbmol/h; yin = 0.4; xin = 0

In the inlet air stream: ammonia = 40 lbmol/h; air = 60 lbmol/h

Ammonia removed from the air stream = 0.95 x 40 = 38 lbmol/h

In the outlet air stream: ammonia = 02 lbmol/h; air = 60 lbmol/h

Therefore, Gout = (60+2) = 62 lbmol/h, and

yout = 2 / 62 = 0.0323

Prof. R. Shanthini

slide17

Example 1:

Using Lin = 488 lbmol/h, Gout = 62 lbmol/h, yout = 0.0323 and xin = 0 in (98), we get the operating curve as follows:

62 x 0.0323+ [488 x / (1 – x)]

y =

62 + [488x / (1 – x)]

yin (bottom of the tower)

Operating curve

yout (top of the tower)

Prof. R. Shanthini

slide18

Example 2:

Draw the equilibrium curve, which is approximately described by K = 44.223x + 0.4771, on the same plot as in Example 1.

Solution:

Operating curve

Equilibrium curve

y = K x

= 44.223x2 + 0.4771x

Prof. R. Shanthini

slide19

Example 3:

Determine the NOG using the data given in Examples 1 and 2.

Solution:

yin

(1 – y)LM dy

NOG =

(1 – y)(y – y*)

yout

y* – y

where (1 – y)LM

=

ln[(1 – y)/(1 – y*)]

Given x, calculate y from the operating curve and y* from the equilibrium curve. Using those values, determine the integral above that gives NOG.

Prof. R. Shanthini

slide20

Example 3:

The shaded area gives NOG as 3.44

(Numerical integration is better suited to get the answer.)

yin = 0.4

yout = 0.0323

Prof. R. Shanthini

slide21

Example 4:

Determine the height of packing Z using the data given in Examples 1 and 2.

Solution:

G

Need more data to work it out.

HOG =

Kya(1 – y)LMS

Z = NOG HOG

Can be calculated once HOG is known.

Prof. R. Shanthini

slide22

Summary with overall gas-phase transfer coefficients for packed column with concentrated solutions

yin

(1 – y)LM dy

G

Z =

(93)

Kya(1 – y)LMS

(1 – y)(y – y*)

yout

NOG

HOG

where (1 – y)LM is the log mean of (1 – y) and (1 – y*) given as follows:

y* – y

(1 – y)LM

=

(92)

ln[(1 – y)/(1 – y*)]

Prof. R. Shanthini

slide23

Summary with overall liquid-phase transfer coefficients for packed column with concentrated solutions

xout

(1 – x)LM dy

L

Z =

(93)

Kxa(1 – x)LMS

(1 – x)(x* – x)

xin

NOL

HOL

where (1 – x)LM is the log mean of (1 – x) and (1 – x*) given as follows:

x* – x

(1 – x)LM

=

(92)

ln[(1 – x)/(1 – x*)]

Prof. R. Shanthini

slide24

Summary: Equations for Packed Columns for dilute solutions

Distributed already:

Photocopy of Table 16.4 Alternative mass transfer coefficient groupings for gas absorption

from

Henley EJ and Seader JD, 1981, Equilibrium-Stage Separation Operations in Chemical Engineering, John Wiley & Sons.

Prof. R. Shanthini

slide25

Gas absorption, Stripping and Extraction

Gas absorption: NOG and HOG are used

Stripping: NOL and HOL are used

Extraction: NOL and HOL are used

Humidification: NG and HG are used.

Prof. R. Shanthini