PROBLEM

– 1

Fig. 1. Schematic diagram of the problem

Basis of Calculation, = 1 min operation

Now, In the feed,

SO2 content = 1.38×5.5×10-4 = 7.59 x 10-4

lbmol.

Non diffusing component in gas phase, Air in feed =

Gs = 1.38 – 7.54X10-4 lbmol = 1.379241 lb mol.

As per the problem, mole fraction of the SO2 in feed

= y1 = 5.5 x 10-4

So mole ratio, Y1 = 5.5X10-4 /1.379241 =

0.55X10-4 lbmol

According to the material balance of the absorber,

Y1 – Y = (Ls/Gs)(X1 – X)

Now, according to the problem, when the liquid flow

rate is minimum, the operating line of the absorber touches the equilibrium

curve,

So, (Y1 – Y2)/(X1 – 0) = Lmin/Gs = slope of the

operating line,

Now, X1, Y1 is lie on the equilibrium curve,

So, Y1 = 33. X1

So, (4×10-4 -0.55×10-4 )/( 4×10-4

/33) = Lmin/1.379241

Or, Lmin = 38.99 lb.mol

b) Now, the water flow rate is the 1.5 times of the

minimum flow rate,

Ls = 38.99 x 1.5 = 58.49 lb mol.

Now, according to the Kremser equation,

Where, A is the absorption factor, = Ls/m.Gs =

48.49/(33×1.379241) = 1.285

m is the slope of the equilibrium line = 33.

And x0 = 0 for pure water,

Or, N = 3.47 = 3.5

Mccabe thiele method,

The equation of the operating line, y1 – y = (Ls/Gs)(x1 – x)

So, the slope of the operating line, 58.49/1.379241

Fig. 2 Mccabe thiele method graphical stage estimation

Problem

– 2

Fig. 3 Schematic diagram of problem 2

Assume a solute A is diffusing through a still and stagnant gas

and then into a still or stagnant liquid, the overall material balance of

component A is,

Ls(x2/1-x2) + Vs(y1/1-y1) = Ls(x1/1-x1) + Vs(y2/1-y2)

Where Ls is kg mol of inert liquid/s or kg mol of non-diffusing

liquid/s.m2. Vs is kg mol of non diffusing gas /s and y and x are mole fraction

of A in gas and liquid.

Now, a is the interfacial area in m2/m3 volume of packed

section, the volume of packed height dz m is S.dz and,

dA = a.s.dz

where s is cross sectional area of the tower then, film mass

transfer coefficient and overall mass transfer coefficient is defined as, kx. a

= kg mol/s.m3. packing.mole fraction. and,

Kx. a = kg mol/s.m3.

packing.mole fraction,

For the absorption of component A from B the operating line

equation holds. For the height of the tower dz shown in the figure, material

balance gives,

D(Vy) = d(Lx)

V and L is the kg mol of total gas and liquid /s and D(Vy) =

d(Lx) is the total moles of A transferred across dz.

Now from the mass transfer total mole transferred,

NA = where, (1-xA)*M =

So, NA dA=

Where, NA. dA = kg mol A

transferred/ s in dz height,

So, d(LxAL) =

Since, L = Ls/(1-xAL)

So, Ldx/(1-xAL) =

Now, dropping the subscript, we get,

Now for dilute solution, (1-xA)*M = (1-x)

So the equation reduces to,

Or,

PROBLEM

– 3

Fig. 4 Equilibrium line and operating line for counter current

absorber.

It has been found from the graph that if the operating line and

equilibrium line are both straight and parallel then for any values of x the

different between y and y* is constant.

So, NOy = = =

a) In

the gas mixture inlet gas mole fraction

= yb = 0.001

Removal of gas = 90% so outlet gas mole fraction = ya =

0.001×0.1 = 0.0001

And (y – y*) = 0.0001

So, Noy = (0.001-0.0001)/0.0001 = 9

b) As

per the equilibrium relationship the Mccabe thiele graphical stage is

calculated and the graph is given below. In the gas mixture inlet gas mole

fraction = yb = 0.001

Removal of gas = 90% so outlet gas mole fraction = ya =

0.001×0.1 = 0.0001

And (y – y*) = 0.0001

So, the operating line and equilibrium line are parallel and the

distance is 0.0001

Fig. 5. Mccabe thiele method graphical stage estimation

So number of stage = 28.6 = 29