# PROBLEM problem Basis of Calculation, = 1 min operation

PROBLEM
– 1

Fig. 1. Schematic diagram of the problem

We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!

order now

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