UNIT 2: Advanced Conduction, Extended Surfaces, and Transient Analysis
1. Conduction in Cylindrical and Spherical Systems
1.1 One-Dimensional Steady-State Conduction in a Cylinder
Assumptions:
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Radial symmetry, no angular or axial variation.
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Constant thermal conductivity \(k\).
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No internal heat generation.
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Steady-state conditions.
Governing Differential Equation (in radial coordinates \(r\)):
For a cylinder of length \(L\), the heat conduction equation simplifies to:
$$ \frac{d}{dr} \left( r \frac{dT}{dr} \right) = 0 $$
Integrating twice yields the general solution:
$$ T(r) = C_1 \ln r + C_2 $$
Boundary Conditions:
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At inner radius \(r = r_i\), \(T = T_i\).
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At outer radius \(r = r_o\), \(T = T_o\).
Temperature Distribution:
Applying boundary conditions:
$$ T(r) = \frac{T_i - T_o}{\ln(r_i/r_o)} \ln\left(\frac{r}{r_o}\right) + T_o $$
Heat Transfer Rate:
Fourier's law for radial conduction:
$$ Q = -k A_r \frac{dT}{dr} = 2\pi k L \frac{T_i - T_o}{\ln(r_o/r_i)} $$
Note: Heat flow is independent of \(r\) in steady state.
Thermal Resistance:
$$ R_{\text{cond, cyl}} = \frac{\ln(r_o/r_i)}{2\pi k L} $$
1.2 One-Dimensional Steady-State Conduction in a Sphere
Assumptions:
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Spherical symmetry, no angular variation.
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Constant \(k\), no heat generation, steady-state.
Governing Differential Equation:
For a sphere of radius \(r\):
$$ \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dT}{dr} \right) = 0 $$
Integrating:
$$ r^2 \frac{dT}{dr} = C_1 \quad \Rightarrow \quad \frac{dT}{dr} = \frac{C_1}{r^2} $$
Integrating again:
$$ T(r) = -\frac{C_1}{r} + C_2 $$
Boundary Conditions:
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At \(r = r_i\), \(T = T_i\).
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At \(r = r_o\), \(T = T_o\).
Temperature Distribution:
$$ T(r) = \frac{r_o - r}{r_o - r_i} \frac{r_i r_o}{r} T_i + \frac{r - r_i}{r_o - r_i} T_o $$
Heat Transfer Rate:
$$ Q = -k A_r \frac{dT}{dr} = 4\pi k \frac{r_i r_o}{r_o - r_i} (T_i - T_o) $$
Thermal Resistance:
$$ R_{\text{cond, sph}} = \frac{r_o - r_i}{4\pi k r_i r_o} $$
1.3 Composite Cylindrical Walls
For multiple cylindrical layers in series (e.g., tube with insulation), total thermal resistance is sum of individual resistances:
$$ R_{\text{total}} = \sum_{j} \frac{\ln(r_{j,\text{out}}/r_{j,\text{in}})}{2\pi k_j L} $$
Overall heat transfer rate per unit length:
$$ Q = \frac{T_i - T_\infty}{R_{\text{total}}} $$
Interface temperatures found by applying \(Q\) across each layer sequentially.
Example (May 2023 Q3): Stainless steel tube (\(k_s=19\ \mathrm{W/m\,K}\)) with ID=2 cm, OD=5 cm, insulated with 3 cm asbestos (\(k_a=0.2\ \mathrm{W/m\,K}\)). Given \(\Delta T = 600\,^\circ\mathrm{C}\) across wall, find \(Q\) per unit length.
Solution: Use series resistance with \(r_1=0.01\ \mathrm{m}\), \(r_2=0.025\ \mathrm{m}\), \(r_3=0.055\ \mathrm{m}\).
2. Critical Radius of Insulation
2.1 Concept and Physical Significance
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Paradox: For cylinders/spheres, adding insulation can increase heat loss if outer radius is below critical radius.
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Reason: Insulation adds conductive resistance but also increases outer surface area for convection. Below \(r_{\text{cr}}\), area effect dominates.
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Design Implication: In refrigeration/cables, if goal is to minimize heat loss, avoid operating below \(r_{\text{cr}}\). If goal is to maintain temperature (e.g., freeze protection), operating below \(r_{\text{cr}}\) may be acceptable.
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Plane Wall: No critical radius—insulation always reduces heat loss (area constant).
2.2 Derivation for Cylindrical Insulation
Total thermal resistance per unit length:
$$ R_{\text{total}} = \underbrace{\frac{\ln(r_o/r_i)}{2\pi k}}_{\text{conduction}} + \underbrace{\frac{1}{2\pi r_o h}}_{\text{convection}} $$
Heat loss per unit length:
$$ Q = \frac{T_i - T_\infty}{R_{\text{total}}} $$
Condition for critical radius: \(\frac{dQ}{dr_o} = 0\) or \(\frac{dR_{\text{total}}}{dr_o} = 0\).
Differentiate \(R_{\text{total}}\) w.r.t. \(r_o\):
$$ \frac{dR_{\text{total}}}{dr_o} = \frac{1}{2\pi k r_o} - \frac{1}{2\pi h r_o^2} = 0 $$
$$ \Rightarrow \frac{1}{r_o} = \frac{k}{h r_o^2} \quad \Rightarrow \quad r_{\text{cr}} = \frac{k}{h} $$
Critical thickness: \(t_{\text{cr}} = r_{\text{cr}} - r_i = \frac{k}{h} - r_i\).
2.3 Derivation for Spherical Insulation
For a sphere with insulation from \(r_i\) to \(r_o\):
$$ R_{\text{total}} = \frac{1}{4\pi k} \left( \frac{1}{r_i} - \frac{1}{r_o} \right) + \frac{1}{4\pi r_o^2 h} $$
Set \(\frac{dR_{\text{total}}}{dr_o} = 0\):
$$ \frac{1}{4\pi k r_o^2} - \frac{1}{2\pi h r_o^3} = 0 \quad \Rightarrow \quad r_{\text{cr}} = \frac{2k}{h} $$
Comparison:
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Cylinder: \(r_{\text{cr}} = k/h\)
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Sphere: \(r_{\text{cr}} = 2k/h\)
Sphere has twice the critical radius of cylinder for same \(k, h\).
2.4 Graphical Representation
Plot of \(Q\) vs. \(r_o\) for a cylinder:
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For \(r_o < r_{\text{cr}}\): \(Q\) increases with \(r_o\) (area effect dominates).
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At \(r_o = r_{\text{cr}}\): \(Q\) maximum.
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For \(r_o > r_{\text{cr}}\): \(Q\) decreases with \(r_o\) (conduction resistance dominates).
Design Tip: To reduce heat loss, ensure \(r_o > r_{\text{cr}}\). For pipes with small \(r_i\), often \(r_i < r_{\text{cr}}\), so initial insulation layer increases loss—use carefully.
3. Extended Surfaces (Fins)
3.1 Introduction and Applications
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Definition: Surfaces engineered to extend from a base to enhance heat transfer by increasing surface area.
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Applications: Heat sinks (electronics), radiators (engines), cooling fins (airfoils), refrigerator coils.
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Purpose: Reduce required \(\Delta T\) or increase \(Q\) for given \(\Delta T\).
3.2 Classification of Fins
| Basis | Types |
|---|---|
| Geometry | Straight (rectangular, tapered), annular (circular disk), pin (circular, elliptical) |
| Cross-section | Uniform (UCS) vs. non-uniform (varying \(A_c\)) |
| Tip Condition | Infinite, convective, insulated, prescribed temperature |
3.3 Governing Differential Equation
Assumptions:
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1D conduction along fin (\(x\)-direction).
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Constant \(k\), \(h\), no heat generation.
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Negligible radiation.
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Uniform cross-sectional area \(A_c\) and perimeter \(P\) (for UCS).
Energy Balance on differential element \(dx\):
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Conduction in: \(-k A_c \frac{dT}{dx}\)
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Conduction out: \(-k A_c \frac{dT}{dx} - k A_c \frac{d^2T}{dx^2} dx\)
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Convection loss: \(h P (T - T_\infty) dx\)
Net energy accumulation = 0 at steady state:
$$ k A_c \frac{d^2T}{dx^2} dx - h P (T - T_\infty) dx = 0 $$
$$ \frac{d^2T}{dx^2} - \frac{hP}{k A_c} (T - T_\infty) = 0 $$
Let \(\theta = T - T_\infty\) and \(m^2 = \frac{hP}{k A_c}\):
$$ \boxed{\frac{d^2\theta}{dx^2} - m^2 \theta = 0} $$
Physical significance of \(m\): Fin parameter (m⁻¹) representing ratio of convective to conductive conductance.
3.4 Boundary Conditions
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Base (\(x=0\)): \(\theta = \theta_b = T_b - T_\infty\)
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Tip (\(x=L\)):
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Infinite fin: \(\theta \to 0\) as \(x \to \infty\)
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Convective tip: \(-k A_c \frac{d\theta}{dx} = h A_t \theta_L\) (where \(A_t\) is tip area)
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Insulated tip: \(\frac{d\theta}{dx} = 0\) at \(x=L\)
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Prescribed tip temperature: \(\theta = \theta_L\) at \(x=L\)
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3.5 Solution for Straight Rectangular Fin (UCS)
General solution: \(\theta = C_1 \cosh(mx) + C_2 \sinh(mx)\)
| Tip Condition | Temperature Distribution \(\theta(x)\) | Heat Transfer Rate \(Q_{\text{fin}}\) |
|---|---|---|
| Infinite | \(\theta_b e^{-mx}\) | \(\sqrt{hPkA_c} \ \theta_b\) |
| Insulated | \(\theta_b \frac{\cosh[m(L-x)]}{\cosh(mL)}\) | \(\sqrt{hPkA_c} \ \theta_b \tanh(mL)\) |
| Convective | \(\theta_b \frac{\cosh[m(L-x)] + \frac{h}{mk} \sinh[m(L-x)]}{\cosh(mL) + \frac{h}{mk} \sinh(mL)}\) | \(\sqrt{hPkA_c} \ \theta_b \frac{\sinh(mL) + \frac{h}{mk} \cosh(mL)}{\cosh(mL) + \frac{h}{mk} \sinh(mL)}\) |
Infinite fin approximation valid when \(mL > 3\) (error < 0.7%).
3.6 Fin Performance Parameters
Fin efficiency:
$$ \eta_f = \frac{Q_{\text{fin}}}{h A_f \theta_b} $$
where \(A_f = A_c + A_t \approx PL\) (for thin fins).
Fin effectiveness:
$$ \varepsilon_f = \frac{Q_{\text{fin}}}{h A_b \theta_b} $$
where \(A_b\) is base area.
Relationship:
$$ \varepsilon_f = \eta_f \frac{A_f}{A_b} $$
Design Insight: High \(\eta_f\) requires small \(mL\) (short/thick fins or high \(k\)). But small \(A_f/A_b\) limits \(\varepsilon_f\). Trade-off exists.
3.7 Design Considerations
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Material: High \(k\) (Cu, Al) to minimize temperature drop.
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Geometry: Optimize thickness/height/spacing to balance area vs. efficiency.
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Fin arrays: Adjacent fins interfere with airflow; spacing critical.
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Manufacturing: Cost vs. performance (e.g., extruded vs. bonded fins).
4. Transient Heat Conduction
4.1 Introduction
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Definition: Temperature varies with time and position.
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Governing equation: \(\frac{\partial T}{\partial t} = \alpha \nabla^2 T\), where \(\alpha = k/(\rho c_p)\).
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Complexity: Requires solving PDE; often use approximate methods.
4.2 Lumped Capacitance Method
Assumptions:
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Uniform temperature throughout solid at any instant (no internal gradients).
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Negligible internal conduction resistance vs. convective resistance.
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Constant \(k, c_p, \rho, h\).
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Biot number criterion: \(\text{Bi} = \frac{h L_c}{k} < 0.1\), where \(L_c = V/A_s\) (characteristic length).
Energy Balance:
Rate of energy decrease = net convective loss:
$$ -\rho V c_p \frac{dT}{dt} = h A_s (T - T_\infty) $$
Separate variables:
$$ \frac{dT}{T - T_\infty} = -\frac{h A_s}{\rho V c_p} dt $$
Integrate from \(t=0\) (\(T=T_i\)) to \(t\) (\(T=T\)):
$$ \ln \frac{T - T_\infty}{T_i - T_\infty} = -\frac{h A_s}{\rho V c_p} t $$
Temperature-time relation:
$$ \boxed{\frac{T - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h A_s}{\rho V c_p} t\right)} $$
Time to reach temperature \(T\):
$$ \boxed{t = \frac{\rho V c_p}{h A_s} \ln \frac{T_i - T_\infty}{T - T_\infty}} $$
Plot: Exponential decay of \((T - T_\infty)/(T_i - T_\infty)\) vs. \(t\).
Exam Tip: Always check \(\text{Bi} < 0.1\) first. For sphere, \(L_c = r/3\); for cylinder, \(L_c = r/2\); for slab, \(L_c = \text{thickness}/2\).
4.3 Infinite Thermal Conductivity Method
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Explanation: If \(k \to \infty\), internal resistance → 0, temperature uniform → same as lumped capacitance.
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Derivation: Same as Section 4.2.
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Use: Approximate when Bi is small but not necessarily <0.1; gives upper bound on cooling/heating rate.
4.4 Examples and Applications
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Cooling of spheres/cylinders/slabs: Use appropriate \(L_c\) for Bi check.
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Time calculations: Given \(T_i, T_\infty, T\), find \(t\).
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Limitations: Invalid for high Bi (needs Heisler charts or numerical methods).
Common Pitfall: Forgetting to use correct \(L_c\) for geometry.
Example (May 2024 Q13): Brass rod (\(k=120\ \mathrm{W/m\,K}\)), \(d=2\ \mathrm{cm}\), \(L=15\ \mathrm{cm}\), \(h=100\ \mathrm{W/m^2K}\), \(\rho=8500\ \mathrm{kg/m^3}\), \(c_p=380\ \mathrm{J/kgK}\). Check Bi: \(L_c = V/A_s\). For cylinder, \(L_c = r/2 = 0.01/2 = 0.005\ \mathrm{m}\), \(\text{Bi} = 100 \times 0.005 / 120 \approx 0.0042 < 0.1\) → valid. Then compute time to cool to \(100\,^\circ\mathrm{C}\) from \(500\,^\circ\mathrm{C}\) with \(T_\infty=30\,^\circ\mathrm{C}\).
END OF UNIT 2 NOTES