Midterm 1 Solutions
MCEN 3022 : Heat Transfer
P1
1. Heat Transfer Modes (Examples)
Conduction
• Heat conducts along a metal spoon left in hot soup (energy moves through a solid).
• Heat loss through a wall/window pane in winter (energy conducts through solid material).
Convection
• Forced convection: A fan blowing air over a laptop heat sink increases heat removal.
• Natural convection: Warm air rises above a radiator due to buoyancy-driven flow.
Radiation
• Sunlight warming the Earth (radiation travels through space; no medium required).
• Feeling heat from a furnace/fire without touching it (thermal radiation).
2. Radiative Heat Loss from Earth to Space
Assuming blackbody exchange with deep space:
4
Q̇ = εσA TE4 − Tspace
,
2
A = 4πRE
Given:
ε = 1,
σ = 5.67 × 10−8 W/m2 K4 ,
RE = 6.0 × 106 m,
TE = 288.15 K,
Tspace = 3 K.
Area:
A = 4π(6.0 × 106 )2 ≈ 4.52 × 1014 m2
Radiative flux:
1
4
σ(TE4 − Tspace
) ≈ 391 W/m2
Thus,
Q̇ ≈ (391)(4.52 × 1014 ) ≈ 1.77 × 1017 W
Q̇ ≈ 1.8 × 1017 W
3. Solar Radiation Absorbed by Earth
Using solar flux G = 1000 W/m2 and projected area:
2
)
Q̇ = G(4πRE
Q̇ = 4000π(6.0 × 106 )2 ≈ 4.52 × 1017 W
Q̇ ≈ 4.52 × 1017 W
4. True/False
Statement: Steady-state 1D temperature distribution is independent of material.
Answer: False
For steady 1D conduction:
dT
d
kA
+ q ′′′ A = 0
dx
dx
If internal heat generation exists (q ′′′ ̸= 0), temperature depends on k. It may be independent when q ′′′ = 0.
5. True/False
Statement: Metals always have higher thermal conductivity than insulators.
Answer: False.
Some insulators (e.g., diamond) have thermal conductivity exceeding many metals.
2
P2
Composite plane wall:
• Glass: 0 ≤ x ≤ 6 m,
k = 1 W/mK
• Wood: 6 ≤ x ≤ 14 m,
k = 5 W/mK
1. Heat Flux Profile
At steady state without generation, heat flux is constant:
q ′′ (x) = −k
dT
= constant
dx
2. Heat Source in Glass (0–4 m)
Answer: False.
Straight-line T (x) implies:
d2 T
= 0 ⇒ q ′′′ = 0
dx2
Thus no volumetric heat generation.
3. Heat Source in Wood (10–12 m)
Answer: Yes.
Curved T (x) implies:
d2 T
> 0 ⇒ q ′′′ < 0
dx2
Indicating volumetric heat removal.
3
4
P3
Two materials in series:
• Material A (blue): heat generation present.
• Material B (purple): no heat generation.
• Left end insulated.
• Right end convective cooling:
T∞ = 20◦ C,
h = 100 W/m2 K.
1. Governing Equation (Material A)
ρ A cA
∂ 2 TA
∂TA
= kA
+ q̇A
∂t
∂x2
2. Steady-State Form
kA
d2 TA
+ q̇A = 0
dx2
3. Material B (No Generation)
kB
d2 TB
=0
dx2
4. Left Boundary Condition (Insulated)
dTA
=0
dx x=0
5. Right Boundary Condition
Energy balance:
−kB
dTB
+ qr = h(Ts − T∞ )
dx x=L
where:
Ts = TB (L)
5
6. Interface Conditions
Temperature continuity:
TA (xi ) = TB (xi )
Heat flux continuity:
−kA
dTB
dTA
= −kB
dx x−
dx x+
i
i
7. Maximum Temperature Location
Maximum temperature occurs at the insulated left end:
x=0
Reason:
• Adiabatic boundary (dT /dx = 0) gives an extremum.
• Heat removal occurs only toward the right via convection.
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