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CHEMICAL ENGINEERING

Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is

CHEMICAL ENGINEERING, Mass Transfer /
Question: Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is
[A].

25%

[B].

50%

[C].

70%

[D].

90%

Answer: Option B

Explanation:

No answer description available for this question.

Compound A is extracted from a solution of A + B into a pure solvent S. A Co-current unit is used for the liquid-liquid extraction. The inlet rate of the solution containing A is 200 moles of B/hr.m2 and the solvent flow, rate is 400 moles of S/m2. hr. The equilibrium data is represented by Y = 3X2 , where Y is in moles of a A/moles of B and X is in moles A/moles of S. The maximum percentage extraction achieved in the unit is Read More »

CHEMICAL ENGINEERING, Mass Transfer

A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids))

CHEMICAL ENGINEERING, Mass Transfer /
Question: A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids))
[A].

M(Xo – XF)

[B].

M(Xo/XF)

[C].

M ln(Xo/XF)

[D].

MXo ln(Xo/XF)

Answer: Option D

Explanation:

No answer description available for this question.

A solid is being dried in the linear drying rate regime from moisture content Xo to XF. The drying rate is zero at X = 0 and the critical moisture content is the same as the initial moisture Xo. The drying time for M = (Ls/ARc) is (where, L = total mass of dry solid, A = total surface area for drying Rc = Constant maximum drying rate per unit area X = moisture content (in mass of water/mass of dry solids)) Read More »

CHEMICAL ENGINEERING, Mass Transfer

A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is

CHEMICAL ENGINEERING, Mass Transfer /
Question: A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is
[A].

0.54

[B].

1.04

[C].

1.24

[D].

1.94

Answer: Option C

Explanation:

No answer description available for this question.

A pure drug is administered as a sphere and as a cube. The amount of drug is the same in the two tablets. Assuming that the shape and size do not influence the mass transfer, the ratio of rate of dissolution in water at t = 0 for the cubic to spherical tablet is Read More »

CHEMICAL ENGINEERING, Mass Transfer

H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is

CHEMICAL ENGINEERING, Mass Transfer /
Question: H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is
[A].

2.0

[B].

1.56

[C].

1.0

[D].

0.56

Answer: Option A

Explanation:

No answer description available for this question.

H2S is being absorbed in a gas absorber unit. The height of the transfer unit based on the overall mass transfer coefficient on the gas side is 0.4 m. The equilibrium data is given by, y = 1.5 x. The bulk concentration of H2S has to be reduced from 0.05 to 0.001 mole fraction in the gas side. The height of the tower (in metres) corresponding to an operating line given by, y = 5x + 0.001 is Read More »

CHEMICAL ENGINEERING, Mass Transfer

Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is

CHEMICAL ENGINEERING, Mass Transfer /
Question: Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is
[A].

5.6 x 10-4

[B].

8.5 x 10-4

[C].

5.6 x 10-3

[D].

8.5 x 10-3

Answer: Option B

Explanation:

No answer description available for this question.

Experiments were conducted to determine the flux of a species A in a stagnant medium across a gas-liquid interface. The overall mass transfer co-efficient based on the liquid side for dilute systems for the above was estimated to be 4 x 10-3 kg mole/m2.s. The equilibrium data for the system is given as y = 2x. The flux across the interface (in kg mole/m2 .s) for bulk concentrations of A in gas phase and liquid phase as y = 0.4 and x = 0.01 respectively is Read More »

CHEMICAL ENGINEERING, Mass Transfer

Component A is diffusing in a medium B. The flux NA relative to a stationary point is equal to the flux due to molecular difusion, if

CHEMICAL ENGINEERING, Mass Transfer /
Question: Component A is diffusing in a medium B. The flux NA relative to a stationary point is equal to the flux due to molecular difusion, if
[A].

mass transfer is accompanied by reaction.

[B].

diffusion of A is in stagnant medium B.

[C].

molecular mean free path is high.

[D].

there is equimolar counter diffusion.

Answer: Option D

Explanation:

No answer description available for this question.

Component A is diffusing in a medium B. The flux NA relative to a stationary point is equal to the flux due to molecular difusion, if Read More »

CHEMICAL ENGINEERING, Mass Transfer

It takes 6 hours to dry a wet solid from 50% moisture content to the critical moisture content of 15%. How much longer it will take to dry the solid to 10% moisture content, under the same drying conditions? (The equilibrium moisture content of the solid is 5%).

CHEMICAL ENGINEERING, Mass Transfer /
Question: It takes 6 hours to dry a wet solid from 50% moisture content to the critical moisture content of 15%. How much longer it will take to dry the solid to 10% moisture content, under the same drying conditions? (The equilibrium moisture content of the solid is 5%).
[A].

15 min

[B].

51 min

[C].

71 min

[D].

94 min

Answer: Option C

Explanation:

No answer description available for this question.

It takes 6 hours to dry a wet solid from 50% moisture content to the critical moisture content of 15%. How much longer it will take to dry the solid to 10% moisture content, under the same drying conditions? (The equilibrium moisture content of the solid is 5%). Read More »

CHEMICAL ENGINEERING, Mass Transfer