MSE 626: Transport Phenomena – End-Semester Examination: Model Solutions
Indian Institute of Technology Kanpur
Department of Materials Science and Engineering
MSE 626 Transport Phenomena
AY 2025-2026 Semester I
End Semester Examination Model Solutions
Problem 1
Consider flow of liquid film over an inclined plane. A shell force balance is applied to a control volume of length L, thickness Δx and width W along the z direction.
The balance gives the equation:
(P0 − PL) ΔxW + (τxz|x − τxz|x+Δx) WL + ρ(LWΔx) g cosβ = 0
Since both ends of the inclined plane are open, P0 = PL.
Dividing by the control volume and taking the limit as Δx → 0:
− dτxz/dx + ρ g cosβ = 0
Therefore:
dτxz/dx = ρ g cosβ
Integrating:
τxz = ρ g cosβ x + C1
Using Newton's law of viscosity:
τxz = − μ dvz/dx
Thus:
− μ dvz/dx = ρ g cosβ x + C1
Integrating again:
vz = − (ρ g cosβ / 2μ) x² − (C1/μ)x + C2
Boundary Conditions
1. No slip at the wall of inclined plane: vz = 0 at x = δ
2. At free surface x = 0 the shear stress is zero.
Using these conditions:
C1 = 0
C2 = ρ g cosβ δ² / 2μ
Velocity Distribution
vz = (ρ g cosβ / 2μ)(δ² − x²)
or
vz = (ρ g cosβ δ² / 2μ)(1 − x²/δ²)
Maximum Velocity
vz,max = ρ g cosβ δ² / 2μ
Average Velocity
v̄z = (2/3) vz,max
Volumetric Flow Rate
Q = v̄z (δW)
Q = ρ g cosβ δ³ W / 3μ
Mass Flow Rate
ṁ = ρ Q
ṁ = ρ² g cosβ δ³ W / 3μ
Wall Shear Force
τxz at x = δ multiplied by wall area gives the total wall shear force.
Problem 2
Consider a slab with internal heat generation under steady state conditions.
From shell energy balance:
d²T/dx² + S/k = 0
Integrating:
dT/dx = − Sx/k + C1
Integrating again:
T = − Sx²/(2k) + C1x + C2
Boundary Conditions
Right face at x = L: T = 25°C
Left face at x = −L is insulated: dT/dx = 0
Solving gives constants C1 and C2.
The temperature profile and heat flux profile can then be plotted across the slab.
Problem 3
Emissive power of a black body:
Eb = σ T⁴
For temperature T = 5800 K and σ = 5.673 × 10⁻⁸ W/m²K⁴:
Eb = 5.673 × 10⁻⁸ × (5800)⁴
Eb = 64.198 × 10⁶ W/m²
Eb = 64.198 MW/m²
Problem 4
Radiation exchange between diffuse gray surfaces in an enclosure can be written as:
qi = (Eb,i − Ji) / ((1 − εi) / (εi Ai))
The radiosity equations for each surface are solved simultaneously to determine J1, J2 and J3.
Once radiosities are obtained, net heat transfer from each surface can be calculated.
Problem 5
Hydrogen diffusion through steel vessel is analyzed using steady state shell mass balance in spherical coordinates.
The governing equation becomes:
(1/r²) d/dr (r² NH) = 0
Integrating:
NH = C1 / r²
Using Fick's law:
JH = − DH-Fe dCH/dr
Integrating with boundary conditions at inner and outer radii gives concentration distribution.
The hydrogen molar flow rate through the spherical wall is:
WH = 2.63 × 10⁻⁵ mole of H per second