LECTURER : Dr. Nabilah Binti Zaini

GROUP : 4

MEMBERS:

1. MOHAMAD FAEZ AMIR BIN ROSLI2. WAN NUR SYAHIDA BINTI WAN MOHAMED

ZURI 3. NURUL NAJIHAH BINTI MOHD RAZALI4. MAISARAH BINTI MOHD ZAKI5. FATIN NAJWA BINTI MOHAMMAD6. NURHAMIZAH BINTI MOHD NORDIN

QUESTION 1One thousand kilograms per hour of a mixture containing equal parts by mass of methanol and water is distilled. Product streams leave the top and the bottom of the distillation column. The flow rate of the bottom stream is measured and found to be 673 kg/h, and the overhead stream is analyzed and found to contain 96.0 wt% methanol.

a) Draw and label a flowchart of the process and do the degree-of-freedom analysis.

b) Calculate the mass and mole fractions of the methanol and the molar flow rates of methanol and water in the bottom product stream.

c) Suppose the bottom product stream is analyzes and the mole fraction of methanol is found to be significantly higher than the value calculated in part (b). List as many possible reasons for the discrepancy as you can think of. Include in your list possible violations of assumptions made in part (b).

a) Draw and label a flowchart of the process and do the degree-of-freedom analysis.

? ṁ₂ (kg/h)

0.96 kg CH3OH / kg

? kg H2O/kg (1 - 0.96 = 0.04kg)

1000 kg/h

500 kg/h CH3OH

500 kg/h H2O

673 kg/h

X METHANOL (kg CH3OH/kg)

1ーX METHANOL (kg H2O/kg)

DISTILLATIONCOLUMN

DOF : ndf = nunknown - nindependent

Unknown : ṁ₂, X METHANOL

Independent species balances :

2 material balances : methanol, water

DOF : 2 -2 = 0 (can be solved)

b) Calculate the mass and mole fractions of the methanol and the molar flow rates of methanol and water in the bottom product stream.

ṁ₂(kg/h)

0.96 kg CH3OH 0.04 kg H2O

100 kg/h

500 kg/h CH3OH 500 kg/h H2O

673 kg/h

X METHANOL (kg CH3OH/kg) X WATER (kg H2O/kg)

DISTILLATIONCOLUMN

0 0 0

Input + generation - output - consumption = accumulation

Input = Output

Overall balances: ṁ1 = ṁ2+ ṁ3

1000 kg/h = ṁ2 + 673 kg/h

ṁ2 = 327 kg/h

Material balance for CH3OH :

ṁ1x1 = ṁ2x2 + ṁ3x3

1000 kg/h (0. 5) = 327 kg/h (0.96) + 673 kg/h (X METHANOL)

X METHANOL = 0.276 kg CH3OH/kg

Material balance for H2O (bottom):

X WATER = 1 - X METHANOL

= 1 - 0.276

= 0.724 kg H2O/kg

Top streamṁMETAHANOL = 227 kg/h (0.96)

= 313.92 kg/h

ṁWATER = 327 kg/h - 313.92 kg/h

= 13.08 kg/h

Bottom stream (input = output)ṁMETAHANOL = 673 kg/h (0.276)

= 185.75 kg/h

ṁWATER = 673 kg/h (0.724)

= 487.25kg/h

Molar flow rates:

1. Methanol, CH3OH185.75 kg/h = 5.80 x 103 mol/h

32.0 kg/kmol

2. Water 487.25 kg/h = 2.71 x 104 mol/h 18.0 kg/kmol

Mole fraction of methanol:

Mole of CH3OH = 5.80 x 103

Total of mole 5.80 x 103 + 2.71 x 104

= 0.176 mol CH3OH/mol

c) Suppose the bottom product stream is analyzes and the mole fraction of methanol is found to be significantly higher than the value calculated in part (b). List as many possible reasons for the discrepancy as you can think of. Include in your list possible violations of assumptions made in part (b).

ANSWER:

I. Methanol is being consumed as a reactant or generated as a product within the unit.

II. Methanol is accumulating in the unit - possibly adsorbing on the walls or other surfaces in the vessel.

III. Methanol is leaking from the unit.IV. The measurement are wrong.

QUESTION 2

Wet air containing 4.0 mole% water vapor is passed through a column of calcium chloride pellets.The pellets adsorb 97%of the water and none of the other constituents of the air. The column packing was initially dry and mass of 3.40kg. Following 5.0 hours of operation , the pellets are reweighed and found to have a mass of 3.45kg.

a) Calculate the molar flow rate (mol/h) of the feed gas and the mole fraction of water vapor in the product gas

b) The mole fraction of water in the product gas is monitored and found to have the value calculated in part (a) for the first 10 hours operation , but then begins to increase. What is the most likely cause of the increase? If the process continues to run, what will the mole fraction of water in the product gas eventually be?

Flow chart..

ṅ1 = (mole/h)0.04 mole H2O/mole0.96 mole air/mole

ṅ2 =? mole H2O/h ( 97% of water adsorbed)

ṅ3=?mole H2O/hx = ? mole water/mole (1-x) = ? mole air/mole

Calcium chloridepellets

ṁ =0.028kg/h

General balance equation:

Input+ generation - output - consumption = accumulation Input = output

a) Mole flow rate at feed

96% of H2O from feed is adsorb

(ṅ1)(0.04)(0.96)=(ṅ2)(1)

ṅ2 = 0.028 kg/h ✕ 1 kmol ✕ 1000 mol 18.01kg 1 kmol

=1.554mol/h

ṅ1= 1.554 = 40.486 mol/h (0.04)(0.96)

ṅ1(0.04)(0.03) = ( ṅ3) (x) ( ṅ3) (x) = (40.485) (0.04) (0.03)

= 0.0485 H2O mol/h

Overall balances

ṅ1=ṅ2 + ṅ3

ṅ3= 40.486 -1.554 = 38.932 mol/h

X ( mol fraction of water vapor ) = 0.0485 mol/h H2O/air 38.932 mol/h H2O/air

=0.00124

b) As the time increase , the rate of absorption become slow

Time = 1/ absorption

This is because the pellet will eventually become fully with H2O , so no absorption occurs .Thus 4% of H2O will remain in the product stream.