CH2004 - Heat and Mass Transfer

Quiz

Total Marks: 25

Time: 1.5 hours

1. At 25oC, the haemoglobin molecule has a diffusivity of 0.069× 10−9m2/sin water. Using the Stokes-Einstein relation, estimate the diameter of thismolecule. Assume water has a viscosity of 10−3kg/m/s. (2 points)

Using Stokes-Einstein relation

R =KB T

6 π η D=

1.38 × 10−23 × 298

6 × 3.14 × 10−3 × 0.069 × 10−9≈ 3.16nm (1)

so diameter is 6.32nm.

2. In a two component mixture consisting of component A and B, usingFick’s law write down the expression for the velocity difference vA − vB

in terms of mass fractions. (2 Point)

vA = v − D∇∇∇ log ωA

vB = v − D∇∇∇ log ωB

(2)

or,

vA − vB = −D∇∇∇ logωA

ωB

(3)

3. Calculate the mean free path of carbon mono-oxide CO molecule. Assumethe CO molecules are traveling in air at T=300K, P=1.00 atm, and thatthe diameters of CO and air molecules are both 3.75×10−10 m. (2 Point)

Assuming ideal gas law, we can compute number density of CO moleculeas

n =P

kB T=

1.01 × 105

1.38 × 10−23 × 300= 2.415× 1025 (4)

λ =1√

2n π d2=

1√2 × 2.415 × 1025 × 3.14 × (3.75 × 10−10)2

= 65nm

(5)

4. Use the value of mean-free path of CO molecule found in previous problemto estimate the viscosity and the thermal conductivity of the CO molecule.(3 Point)

Average speed is

v̄ =

√

8 kB T

π m=

√

8 R T

π M=

√

8 × 8.314× 300

3.14 × 28 × 10−3= 476.39m/s (6)

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mass of one molecule of CO is

m =28 × 10−3

6.023 × 1023= 4.648× 10−26 kg (7)

η =m n

3λ v̄ =

4.648× 10−26 × 2.415× 1025

3× 65 × 10−9 × 476.39 (8)

which gives η = 11.18 × 10−5 N-s/m2

and thermal conductivity is

κ =3

2 mkB η =

3

2 × 4.648 × 10−26× 1.38 × 10−23 × 11.18 × 10−5 (9)

5. Provide typical values of Prandtl number number for ideal gases as pre-dicted by kinetic theory of gases(1 Point)

Pr = 2/3

6. The fact that Pyrex glass is almost impermeable to every gas except heliumcan be used to separate Helium from natural gas. Suppose a natural gasmixture is contained in a Pyrex tube of radius 2cm and length 50cm.It is given that the thickness of tube is 0.1 cm. Obtain the rate (inunit of mole/s) at which helium will leak out of the tube in terms ofthe interfacial concentrations of the helium in the pyrex and diffusioncoefficient of helium-Pyrex system. (4 Point)

The solution is dilute and external force is not present. So, we can useFick’s second law

∂tC = D∇2C (10)

which at steady state and one-dimensional diffusion along radial directionbecomes

∂

∂r

(

r∂

∂rC

)

= 0 (11)

which can be solved to obtain

C = K1 log r + K2 (12)

But, at R = R0, C = C10. so,

C10 = K1 log R0 + K2 (13)

which meansC = C10 + K1 log

r

R0

(14)

But, at R = Ri, C = C1i. so,

C = C10 + (C1i − C10)log r

R0

log Ri

R0

(15)

2

So, flux is

N

∣

∣

∣

∣

∣

r=R0

= −D∂

∂rC

∣

∣

∣

∣

∣

r=R0

= D (C1i − C10)1

R0 log R0

Ri

(16)

so the molar rate is

dn

dt= 2π R0 L N

∣

∣

∣

∣

∣

r=R0

= 2π L D (C1i − C10)1

log R0

Ri

(17)

7. Jasmone (C11H16O) is a valuable aroma from Jasmine flowers that is usedin soaps and cosmetics. We are recovering this from its water solution(Jasmine flowers in water) with benzene drops. We can model the systemas combination of two thick films in contact with each other (one in theBenzene drop and one in water) . The mass transfer coefficient of Jasmonein Benzene is kB = 3.0 × 10−4 cm/ s while in water it is kW = 2.4 ×10−3cm/s . Find out the mass flux of Jasomene in terms of its bulkconcentration in Benzene and Water. (5 Point)

Hint: At steady state the flux of Jasmone from Benzene should be samein magnitude as flux of Jasmone from water.

Assuming steady state

N1 = kW (C10W − C1iW ) = kB (C1iB − C10B) (18)

also,C1iB = H C1iW (19)

so,kW (C10W − C1iW ) = kB (H C1iW − C10B) (20)

or,

C1iW =kW C10W + kB C10B

(kB H + kW )(21)

Substituting it in flux equation, we have

N1 = kW

(

C10W − kW C10W + kB C10B

(kB H + kW )

)

=1

1

KW

+ 1

KB

(H C10W − C10B)

(22)

8. Two immisible solvents I and II are in contact at the plane z = 0. Attime t = 0 the concentration of solute A is cAI = C0

Iin phase I and

cAII = C0

IIin phase II. For t > 0 diffusion takes place across the liquid-

liquid interface.

(a) Write down the species conservation equations in both phases. (2Point)

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As there is no mechanism for bulk flow, it is reasonable to assumethat bulk velocity v = 0. Then we have,

∂t ωAI = DI ∂2

zωAI for −∞ ≤ z ≤ 0 (23)

and∂t ωAII = DII ∂2

zωAII for 0 ≤ z ≤ ∞ (24)

(b) Provide the initial and Boundary condition to solve the problem. (Donot try to solve the equations !) (4 Point) Initial Condition:

at t = 0, ωAI = ω0

AI for −∞ ≤ z ≤ 0 (25)

andat t = 0, ωAII = ω0

AII for0 ≤ z ≤ ∞ (26)

Boundary Conditions:

i.at z = ∞, ωAII = ω0

AII (27)

ii.at z = −∞, ωAI = ω0

AI (28)

iii.at z = 0, ωAII = H ωAI (29)

iv.at z = 0, DI ∂zωAII = DII ∂zωAII (30)

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