AEE Mechanical Engineering Core · Chapter 10

Heat Transfer

What to remember

Heat moves by conduction (Fourier's law, Q = -kA dT/dx), convection (Newton's law, Q = hAΔT) and radiation (Stefan-Boltzmann law, E = σT⁴). Use thermal resistances in series and parallel just like electrical resistances.
Dimensionless numbers decide convection: Nu = hL/k, Re = ρVL/μ, Pr = μcp/k, Gr = gβΔT L³/ν². Flow in a pipe is laminar when Re < 2300; on a flat plate the transition is near Re = 5 × 10⁵.
Heat exchanger design uses LMTD (Q = U A F LMTD) or the effectiveness-NTU method. Counterflow gives the largest LMTD and the best performance.

1. Conduction

Fourier's law: Q = -k A (dT/dx). Thermal conductivity k has unit W/m·K. Metals have high k, gases have very low k, insulators low k. Thermal diffusivity α = k / (ρ c) (m²/s) tells how fast temperature spreads.

Plane wall: Q = k A (T1 - T2) / L. Thermal resistance R = L / (k A) (K/W).

Composite wall in series: Q = ΔT / (R1 + R2 + ...). Add convection resistances 1/(hA) at the surfaces.

Overall heat transfer coefficient: 1/U = 1/h1 + L1/k1 + L2/k2 + 1/h2 (per unit area).

Cylinder (radii r1 < r2, length L): Q = 2π k L (T1 - T2) / ln(r2/r1). Resistance = ln(r2/r1) / (2π k L).

Sphere: Q = 4π k (T1 - T2) / (1/r1 - 1/r2). Resistance = (1/r1 - 1/r2) / (4π k).

Worked example. Wall 0.2 m thick, k = 0.5 W/m·K, faces at 100 °C and 20 °C. Heat flux q = 0.5 × 80 / 0.2 = 200 W/m².

Critical radius of insulation: for a cylinder r_c = k / h; for a sphere r_c = 2k / h. Adding insulation on a pipe of radius smaller than r_c actually increases heat loss until r_c is reached. Insulation must be thicker than r_c to help.

Heat generation: for a plane wall of thickness 2L with both faces at Ts, centre temperature rise = qg L² / (2k). For a solid cylinder radius R: qg R² / (4k). For a solid sphere: qg R² / (6k). The maximum temperature is at the centre.

Fins: extended surfaces that increase heat transfer area. With m = √(hP / (kA)):

Infinitely long fin: Q = √(hPkA) × θ0.
Fin with insulated tip: Q = √(hPkA) × θ0 × tanh(mL).
Fin efficiency = actual heat transfer / heat if the entire fin were at base temperature.
Fin effectiveness = heat with fin / heat without fin. Fins are useful when h is small (gas side) and k is large.

Transient conduction: Biot number Bi = h Lc / k (Lc = V/A). If Bi < 0.1, the lumped system analysis is valid: (T - T∞) / (Ti - T∞) = exp(-t/τ), with time constant τ = ρ V c / (h A). After one time constant, 63.2% of the temperature difference is gone. Fourier number Fo = α t / L².

2. Convection

Newton's law of cooling: Q = h A (Ts - T∞). Forced convection: fluid moved by pump or fan. Natural (free) convection: fluid moves due to density difference caused by temperature.

Number	Formula	Meaning
Nusselt Nu	hL/k	Convection to conduction ratio
Reynolds Re	ρVL/μ	Inertia to viscous force
Prandtl Pr	μ cp / k = ν / α	Momentum to thermal diffusivity
Grashof Gr	g β ΔT L³ / ν²	Buoyancy to viscous force
Stanton St	Nu / (Re Pr)	Heat transferred to heat capacity of flow
Rayleigh Ra	Gr × Pr	Natural convection regime

Pr is about 0.7 for gases, large for oils, very small for liquid metals.
Flat plate, laminar: hydrodynamic boundary layer thickness δ = 5x / √Re_x. Local Nu_x = 0.332 Re_x^0.5 Pr^(1/3). Average Nu over length L is twice the local value at L. Ratio of thermal to velocity boundary layer thickness = Pr^(-1/3). Transition at Re about 5 × 10⁵.
Pipe flow: laminar if Re < 2300, turbulent if Re > 4000. Fully developed laminar flow: Nu = 3.66 (constant wall temperature) and Nu = 4.36 (constant wall heat flux). Turbulent: Dittus-Boelter equation Nu = 0.023 Re^0.8 Pr^n, with n = 0.4 for heating and 0.3 for cooling.
Hydraulic diameter Dh = 4 A / P.
Reynolds analogy links friction and heat transfer: St = Cf / 2.
Boiling: pool boiling curve has natural convection, nucleate boiling, transition boiling and film boiling. The peak of nucleate boiling is the critical heat flux (burnout point). The Leidenfrost point is the minimum film boiling heat flux.
Condensation: dropwise condensation gives a much higher h (5 to 10 times) than filmwise condensation. Condensers in thermal plants generally have filmwise condensation.

3. Radiation

Thermal radiation travels as electromagnetic waves and needs no medium.

Stefan-Boltzmann law: black body emissive power Eb = σ T⁴, with σ = 5.67 × 10⁻⁸ W/m²·K⁴ and T in kelvin. Real surface: E = ε σ T⁴.
Wien's displacement law: λmax × T = 2898 μm·K. A body at 1000 K has peak emission near 2.9 μm.
Planck's law gives spectral distribution of black body radiation.
Kirchhoff's law: at thermal equilibrium, emissivity equals absorptivity (ε = α). For a gray body this holds at all conditions. A black body has α = ε = 1.
Lambert's cosine law: radiation intensity from a diffuse surface is the same in all directions; emissive power E = π I.
Absorptivity + reflectivity + transmissivity = 1. An opaque body has zero transmissivity.

View factor (shape factor): F12 is the fraction of radiation leaving surface 1 that reaches surface 2. Reciprocity: A1 F12 = A2 F21. Summation: sum of F1j over all surfaces = 1. For a convex surface, F11 = 0.

Exchange between two surfaces:

Two black surfaces: Q12 = A1 F12 σ (T1⁴ - T2⁴).
Two large parallel gray plates: Q/A = σ (T1⁴ - T2⁴) / (1/ε1 + 1/ε2 - 1).
Small body (area A1, emissivity ε1) inside a large enclosure: Q = ε1 A1 σ (T1⁴ - T2⁴).
Radiation shields: n thin shields of same emissivity as the plates reduce the heat transfer to 1/(n+1) of the unshielded value. Low emissivity shields (polished) are best.

Worked example. A black surface of 1 m² at 1000 K emits σ T⁴ = 5.67 × 10⁻⁸ × 10¹² = 56,700 W.

4. Heat exchangers

Types: double pipe, shell and tube, cross flow, plate, compact. Flow arrangements: parallel flow (both fluids enter at the same end), counterflow (opposite directions), cross flow.

Heat balance: Q = mh cph (Th,in - Th,out) = mc cpc (Tc,out - Tc,in). Capacity rate C = m cp.
LMTD = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2), where ΔT1 and ΔT2 are the temperature differences at the two ends. Q = U A LMTD. For shell and tube or cross flow, multiply by correction factor F (F ≤ 1).
Counterflow has a higher LMTD than parallel flow for the same terminal temperatures, so it needs less area. In parallel flow the cold fluid outlet can never be hotter than the hot fluid outlet.
When one fluid changes phase (condenser, boiler), its temperature is constant, and the capacity rate ratio Cr = 0.
Effectiveness-NTU: ε = actual heat transfer / maximum possible = Q / (Cmin (Th,in - Tc,in)). NTU = U A / Cmin. Cr = Cmin / Cmax.
Counterflow, Cr = 1: ε = NTU / (1 + NTU).
Parallel flow, maximum possible: ε = 1 / (1 + Cr) as NTU becomes large.
Cr = 0 (condenser or evaporator): ε = 1 - exp(-NTU), for all flow arrangements.
Fouling factor Rf is the added resistance due to scale and deposits; it reduces U. The NTU method is preferred when outlet temperatures are unknown.
In APGENCO power plants, important heat exchangers include the surface condenser, feed water heaters, air preheater and economiser, and the lubricating oil coolers of turbines. These follow the same principles.

5. Comparison table

Mode	Law	Medium needed	Main parameter
Conduction	Fourier	Yes (solids mainly)	k
Convection	Newton	Yes (fluid motion)	h
Radiation	Stefan-Boltzmann	No	ε, T⁴

Exam traps

Thermal conductivity k is a property; heat transfer coefficient h is not a property (it depends on flow).
Critical radius of cylinder is k/h, of sphere is 2k/h; do not interchange them.
Biot number uses solid conductivity; Nusselt number uses fluid conductivity.
Prandtl number is a fluid property; Reynolds number is a flow property.
Wien's law gives wavelength of maximum emission, not the total emissive power.
Kirchhoff's law: emissivity equals absorptivity, not reflectivity.
Pipe flow laminar limit is Re = 2300; flat plate transition is Re = 5 × 10⁵.
NTU method works with unknown outlet temperatures; LMTD needs all four temperatures.

One-liners

1. Unit of thermal conductivity is W/m·K.
2. Thermal diffusivity α = k / (ρ c).
3. The lumped system analysis needs Bi < 0.1.
4. Time constant τ = ρ V c / (h A).
5. Fin efficiency is highest for short, thick fins of high conductivity.
6. Fully developed laminar flow in a tube at constant wall temperature has Nu = 3.66.
7. Dropwise condensation has a higher h than filmwise condensation.
8. Stefan-Boltzmann constant is 5.67 × 10⁻⁸ W/m²·K⁴.
9. Wien's constant is 2898 μm·K.
10. LMTD for counterflow is higher than for parallel flow.
11. Radiation shields reduce heat exchange by the factor 1/(n+1).
12. Effectiveness of a condenser with NTU is 1 - e^(-NTU).

Practice questions

The SI unit of thermal conductivity is
1. J/kg·K
2. W/m·K
3. W/m²·K
4. W·m/K²
Answer
B. W/m·K
k = Q L / (A ΔT), so its unit is W/m·K.
The critical radius of insulation for a long cylinder is
1. k / h
2. k / (2h)
3. h / k
4. 2k / h
Answer
A. k / h
Heat loss is maximum at r = k/h for a cylinder.
The critical radius of insulation for a sphere is
1. h / 2k
2. k / h
3. 2k / h
4. k / 2h
Answer
C. 2k / h
For a sphere r_c = 2k/h.
The Nusselt number is defined as
1. ρ V L / μ
2. h L / k (solid)
3. h L / k (fluid)
4. μ cp / k
Answer
C. h L / k (fluid)
Nu is the ratio of convective to conductive heat transfer in the fluid layer.
Flow in a circular pipe is laminar when the Reynolds number is below about
1. 5 × 10⁵
2. 500
3. 4000
4. 2300
Answer
D. 2300
Pipe flow transition begins at Re about 2300; 5 × 10⁵ is for a flat plate.
A plane wall of area 2 m², thickness 0.3 m and k = 1.5 W/m·K has a temperature difference of 60 K across it. The heat flow is
1. 600 W
2. 300 W
3. 1200 W
4. 900 W
Answer
A. 600 W
Q = kAΔT/L = 1.5 × 2 × 60 / 0.3 = 600 W.
Two layers in series: layer 1 has L = 0.1 m, k = 0.5 W/m·K; layer 2 has L = 0.2 m, k = 0.4 W/m·K. The overall temperature difference is 70 K. The heat flux is
1. 70 W/m²
2. 100 W/m²
3. 140 W/m²
4. 35 W/m²
Answer
B. 100 W/m²
R = 0.1/0.5 + 0.2/0.4 = 0.2 + 0.5 = 0.7 m²K/W; q = 70/0.7 = 100.
For a hollow cylinder of fixed k and length, the conduction resistance when r2/r1 increases from 2 to 4 becomes
1. Four times
2. Unchanged
3. Half
4. Twice
Answer
D. Twice
R is proportional to ln(r2/r1); ln 4 = 2 ln 2.
Lumped system analysis is valid when the Biot number is
1. Equal to 1 exactly
2. Greater than 100
3. Less than 0.1
4. Greater than 10
Answer
C. Less than 0.1
Small Bi means the internal conduction resistance is negligible, so the body is nearly uniform in temperature.
A body has h = 100 W/m²·K, characteristic length V/A = 0.01 m and k = 50 W/m·K. Its Biot number is
1. 0.2
2. 0.02
3. 2
4. 50
Answer
B. 0.02
Bi = h Lc / k = 100 × 0.01 / 50 = 0.02.
A metal body: ρ = 8000 kg/m³, c = 500 J/kg·K, V/A = 0.005 m, h = 100 W/m²·K. The time constant of cooling is
1. 200 s
2. 20 s
3. 40 s
4. 2000 s
Answer
A. 200 s
τ = ρ c (V/A) / h = 8000 × 500 × 0.005 / 100 = 200 s.
In lumped cooling, after one time constant the temperature difference from the surroundings falls by about
1. 63.2%
2. 36.8%
3. 90%
4. 50%
Answer
A. 63.2%
The ratio is e⁻¹ = 0.368, so 63.2% of the difference is lost.
For a long fin: h = 10 W/m²·K, perimeter 0.1 m, k = 100 W/m·K, cross-section 10⁻⁴ m², base excess temperature 100 K. The heat flow is
1. 31.6 W
2. 10 W
3. 1 W
4. 100 W
Answer
B. 10 W
Q = √(hPkA) θ0 = √(10 × 0.1 × 100 × 10⁻⁴) × 100 = 0.1 × 100 = 10 W.
Fin effectiveness is the ratio of
1. Tip to base temperature
2. Fin length to thickness
3. Heat transfer with the fin to that without the fin
4. Actual fin heat to ideal fin heat
Answer
C. Heat transfer with the fin to that without the fin
Fin efficiency compares actual with ideal; effectiveness compares with the bare surface.
The Prandtl number is given by
1. h L / k
2. ρ V L / μ
3. g β ΔT L³ / ν²
4. μ cp / k
Answer
D. μ cp / k
Pr = ν / α = μ cp / k.
For laminar flow over a flat plate, δ = 5x/√Re_x. At x = 1 m with Re_x = 10⁴, the boundary layer thickness is
1. 0.05 m
2. 0.1 m
3. 0.5 m
4. 0.005 m
Answer
A. 0.05 m
δ = 5 × 1 / √10⁴ = 5/100 = 0.05 m.
For a fluid with Pr greater than 1, the thermal boundary layer compared with the velocity boundary layer is
1. Equal
2. Absent
3. Thinner
4. Thicker
Answer
C. Thinner
δt / δ is proportional to Pr^(-1/3), which is below 1 when Pr > 1.
In the Dittus-Boelter equation Nu = 0.023 Re^0.8 Pr^n, n for heating of the fluid is
1. 0.3
2. 0.8
3. 0.5
4. 0.4
Answer
D. 0.4
n = 0.4 for heating and 0.3 for cooling.
For fully developed laminar flow in a circular tube at constant wall temperature, the Nusselt number is
1. 4.36
2. 3.66
3. 2.0
4. 0.023
Answer
B. 3.66
4.36 is for constant wall heat flux.
In a pipe flow, Nu = 50, fluid k = 0.6 W/m·K and D = 0.05 m. The heat transfer coefficient is
1. 60 W/m²·K
2. 6 W/m²·K
3. 1500 W/m²·K
4. 600 W/m²·K
Answer
D. 600 W/m²·K
h = Nu k / D = 50 × 0.6 / 0.05 = 600.
Compared with filmwise condensation, dropwise condensation gives
1. A lower heat transfer coefficient
2. No heat transfer
3. A higher heat transfer coefficient
4. The same coefficient
Answer
C. A higher heat transfer coefficient
Drops leave bare surface so the liquid film resistance is absent.
In the pool boiling curve, the critical heat flux is the
1. Peak of nucleate boiling
2. Film boiling minimum
3. Start of natural convection
4. Condensation point
Answer
A. Peak of nucleate boiling
Beyond it, a vapour film forms and heat transfer drops sharply (burnout).
A black surface at 500 K emits energy (σ = 5.67 × 10⁻⁸ W/m²K⁴) of about
1. 7088 W/m²
2. 3544 W/m²
3. 1417 W/m²
4. 35,440 W/m²
Answer
B. 3544 W/m²
σT⁴ = 5.67 × 10⁻⁸ × 6.25 × 10¹⁰ = 3543.75 W/m².
A black body has maximum emission at 2 μm. Using Wien's constant 2898 μm·K, its temperature is
1. 724 K
2. 5796 K
3. 2898 K
4. 1449 K
Answer
D. 1449 K
T = 2898/2 = 1449 K.
Kirchhoff's law states that, in thermal equilibrium, for a surface
1. Emissivity equals reflectivity
2. Emissivity equals absorptivity
3. Absorptivity equals transmissivity
4. Reflectivity equals one
Answer
B. Emissivity equals absorptivity
ε = α at equilibrium (gray surface).
Surface 1 has area 2 m², surface 2 has area 4 m², and F12 = 0.5. Then F21 is
1. 2.0
2. 0.5
3. 0.25
4. 1.0
Answer
C. 0.25
A1 F12 = A2 F21, so F21 = 2 × 0.5 / 4 = 0.25.
Two large parallel plates each of emissivity 0.5 exchange radiation. The effective emissivity factor 1/(1/ε1 + 1/ε2 - 1) is
1. 1/3
2. 1/2
3. 1/4
4. 1
Answer
A. 1/3
1/(2 + 2 - 1) = 1/3.
Three thin radiation shields of the same emissivity are placed between two parallel plates. The radiation exchange becomes what fraction of the original?
1. 1/2
2. 1/6
3. 1/4
4. 1/3
Answer
C. 1/4
With n shields the exchange is 1/(n+1) = 1/4.
Consider: 1. For the same terminal temperatures, counterflow has a larger LMTD than parallel flow. 2. In parallel flow, the cold fluid outlet can be hotter than the hot fluid outlet. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
In parallel flow the outlet temperatures cannot cross, so 2 is wrong.
In a counterflow heat exchanger the temperature difference is 30 K at both ends. The LMTD is
1. 15 K
2. 30 K
3. 60 K
4. Zero
Answer
B. 30 K
When both end differences are equal, the LMTD equals that difference.
A counterflow heat exchanger with Cr = 1 has NTU = 1. Its effectiveness is
1. 1.0
2. 0.63
3. 0.37
4. 0.5
Answer
D. 0.5
ε = NTU / (1 + NTU) = 1/2.
A hot fluid with capacity rate 1000 W/K cools by 20 K. The cold fluid has capacity rate 2000 W/K. The cold fluid temperature rise is
1. 20 K
2. 10 K
3. 5 K
4. 40 K
Answer
B. 10 K
Q = 1000 × 20 = 20,000 W; ΔT = 20,000 / 2000 = 10 K.
Two fluids with h1 = 100 and h2 = 100 W/m²·K are separated by a very thin wall of high conductivity. The overall coefficient U is
1. 200 W/m²·K
2. 100 W/m²·K
3. 25 W/m²·K
4. 50 W/m²·K
Answer
D. 50 W/m²·K
1/U = 1/100 + 1/100 = 0.02, so U = 50.
Fouling of heat exchanger tubes generally
1. Reduces the overall heat transfer coefficient
2. Doubles the LMTD
3. Has no effect
4. Increases it
Answer
A. Reduces the overall heat transfer coefficient
Scale adds a thermal resistance Rf.
Consider: 1. Biot number uses the conductivity of the solid. 2. Nusselt number uses the conductivity of the fluid. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Bi = hLc/ks (solid) and Nu = hL/kf (fluid).
A black surface of area 2 m² is at 1000 K. Its total emission is about (σ = 5.67 × 10⁻⁸ W/m²K⁴)
1. 113 kW
2. 56.7 kW
3. 227 kW
4. 11.3 kW
Answer
A. 113 kW
E = σT⁴ A = 5.67 × 10⁴ × 2 = 113,400 W.
A wall of thickness 0.2 m (faces at the same temperature) generates heat at 10⁶ W/m³; k = 50 W/m·K. The centre-to-surface temperature rise qg L²/(2k) with L = 0.1 m is
1. 200 K
2. 100 K
3. 10 K
4. 50 K
Answer
B. 100 K
ΔT = 10⁶ × 0.01 / (2 × 50) = 100 K.
For a solid sphere with uniform heat generation, the centre-to-surface rise is
1. qg R² / (4k)
2. qg R² / k
3. qg R² / (2k)
4. qg R² / (6k)
Answer
D. qg R² / (6k)
Wall: L²/2k, cylinder: R²/4k, sphere: R²/6k.
Natural convection is mainly governed by the product
1. Nusselt number × Stanton number
2. Reynolds number × Prandtl number
3. Grashof number × Prandtl number
4. Biot number × Fourier number
Answer
C. Grashof number × Prandtl number
Ra = Gr Pr.
Consider: 1. Fins are most useful when the heat transfer coefficient is low. 2. Thermal radiation needs a material medium to travel. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Radiation travels through vacuum, so 2 is wrong.
A rectangular duct 0.1 m × 0.3 m has a hydraulic diameter of
1. 0.3 m
2. 0.15 m
3. 0.2 m
4. 0.1 m
Answer
B. 0.15 m
Dh = 4A/P = 4 × 0.03 / 0.8 = 0.15 m.
For laminar flow over a plate of length L, the average Nusselt number is
1. Equal to the local value at L
2. Four times the local value
3. Half of the local value at L
4. Twice the local value at x = L
Answer
D. Twice the local value at x = L
Nu local ∝ x^(1/2), so the average over L is 2 × local at L.
A small body of area A1 and emissivity ε1 sits inside a large enclosure. Net radiation depends on
1. ε1 of the small body only
2. ε of the enclosure only
3. Product of both emissivities
4. Neither emissivity
Answer
A. ε1 of the small body only
The enclosure acts almost as a black body, so Q = ε1 A1 σ (T1⁴ - T2⁴).
In a thermal power plant condenser, the hot-side fluid is condensing steam. The capacity rate ratio Cr is
1. Infinite
2. One
3. Zero
4. Equal to NTU
Answer
C. Zero
Condensing fluid has effectively infinite capacity rate, so Cmin/Cmax tends to zero.
Reynolds analogy relates
1. Biot and Fourier numbers
2. Emissivity and absorptivity
3. Friction coefficient and heat transfer (St = Cf / 2)
4. Radiation and conduction
Answer
C. Friction coefficient and heat transfer (St = Cf / 2)
For Pr near 1, Stanton number is half the skin friction coefficient.

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