Extended Surfaces & Fins

Fins extend surface area to enhance convection. For a straight fin of uniform cross-section,
d2θdx2m2θ=0\dfrac{d^2\theta}{dx^2}-m^2\theta=0
with
m=hP/(kAc)m=\sqrt{hP/(kA_c)}
. Tip conditions give standard
qfinq_{\mathrm{fin}}
formulas; efficiency
ηf=qfin/qmax\eta_f=q_{\mathrm{fin}}/q_{\max}
(RC Sachdeva / Incropera).

Key formulas & points

Skim these first — then read the full notes below.

  • Fin parameter: m=hPkAcm=\sqrt{\dfrac{hP}{kA_c}}, θ=TT\theta=T-T_\infty.
  • ODE: d2θdx2m2θ=0\dfrac{d^2\theta}{dx^2}-m^2\theta=0 (steady, 1-D, const h,kh,k).
  • Infinite fin: θ/θb=emx\theta/\theta_b=e^{-mx}, q=hPkAcθbq=\sqrt{hPkA_c}\,\theta_b.
  • Adiabatic tip: q=hPkAcθbtanh(mL)q=\sqrt{hPkA_c}\,\theta_b\tanh(mL).
  • Convective tip: use corrected length Lc=L+AcPL_{c} = L+\frac{A_{c}}{P} (approx.) with adiabatic formula.
  • Efficiency: ηf=qfin/(hAfθb)\eta_f=q_{\mathrm{fin}}/(h A_f\theta_b); effectiveness εf=qfin/(hAcθb)\varepsilon_f=q_{\mathrm{fin}}/(h A_c\theta_b).
  • Fins justified when εf>2\varepsilon_f>2 typically; high kk, adequate hP(kAc)\frac{hP}{(kA_{c})}.

Topic details

Definition and physical meaning

An extended surface (fin) is a protrusion from a hot (or cold) wall that increases the area exposed to convection, raising total heat transfer when the wall area alone is insufficient (electronics cooling, IC engines, heat exchangers).
Excess temperature:
θ(x)=T(x)T\theta(x)=T(x)-T_\infty
Fin parameter:
m=hPkAcm=\sqrt{\frac{hP}{kA_c}}

where PP = perimeter exposed to convection, AcA_{c} = cross-sectional area for conduction along the fin.
Physical parameters
Symbol
Meaning
SI unit
TT, TT_\infty, TbT_{b}
Local / ambient / base temperature
K\mathrm{K}
θ\theta, θb\theta_b
Excess temperature
K\mathrm{K}
hh
Convective coefficient
W/(m2K)\mathrm{W/(m^2\cdot K)}
kk
Fin conductivity
W/(mK)\mathrm{W/(m\cdot K)}
AcA_{c}
Cross-sectional area
m2\mathrm{m^2}
PP
Perimeter
m\mathrm{m}
LL
Fin length
m\mathrm{m}
mm
Fin parameter
1/m\mathrm{1/m}
qfinq_{\mathrm{fin}}
Heat transfer through fin
W\mathrm{W}
ηf\eta_f
Fin efficiency
εf\varepsilon_f
Fin effectiveness
Wall T_bFin (k, A_c)Lh, T_∞
Fig — Straight fin of uniform cross-section

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Extended surface (fin). Fin attached to a hot wall increases convection area; temperature drops along length L.

Core assumptions (state these in exams)

1. Steady state; no heat generation in the fin.
2. One-dimensional conduction along fin length xx (temperature uniform on each cross-section).
3. Constant kk, hh, and AcA_{c} (uniform cross-section straight fin).
4. Negligible radiation (or lumped into effective hh).
5. Uniform TT_\infty and base temperature TbT_{b} prescribed.
6. Contact resistance at the base neglected unless given.
Longitudinal thin fins and pin fins of constant section use the same ODE; annular and triangular fins need modified Ac(x)A_{c}(x) or charts.

Derivation summary — fin equation

Energy balance on a slice dxdx: conduction in − conduction out − convection from surface = 0 (steady):
qxqx+dxh(Pdx)θ=0q_x-q_{x+dx}-h(P\,dx)\theta=0

With Fourier
qx=kAcdθ/dxq_x=-kA_c d\theta/dx
and dqx=(dqxdx)dxdq_{x} = (\frac{dq_{x}}{dx})dx:
ddx(kAcdθdx)hPθ=0\frac{d}{dx}\left(kA_c\frac{d\theta}{dx}\right)-hP\theta=0

For constant kAckA_{c}:
d2θdx2m2θ=0,m2=hPkAc\frac{d^2\theta}{dx^2}-m^2\theta=0,\qquad m^2=\frac{hP}{kA_c}
General solution:
θ(x)=C1sinh(mx)+C2cosh(mx)\theta(x)=C_1\sinh(mx)+C_2\cosh(mx)

(or equivalent exponentials). Constants from BCs at base and tip.

Standard tip cases and heat rates

Case 1 — infinitely long fin (θ0\theta\to 0 as xx\to\infty):
θθb=emx,qfin=hPkAcθb\frac{\theta}{\theta_b}=e^{-mx},\qquad q_{\mathrm{fin}}=\sqrt{hPkA_c}\,\theta_b
Case 2 — adiabatic tip (dθ/dxx=L=0d\theta/dx|_{x=L}=0):
θθb=cosh[m(Lx)]cosh(mL)\frac{\theta}{\theta_b}=\frac{\cosh\big[m(L-x)\big]}{\cosh(mL)}

qfin=hPkAcθbtanh(mL)q_{\mathrm{fin}}=\sqrt{hPkA_c}\,\theta_b\tanh(mL)
Case 3 — convective tip (exact hyperbolic form with hh at tip). Engineering approximation: replace LL by corrected length
Lc=L+AcPL_c=L+\frac{A_c}{P}

and use the adiabatic-tip formula with LcL_{c} (very common in GATE/Sachdeva).
Case 4 — prescribed tip temperature T(L)=TLT(L) = T_{L}: use both end BCs on the general solution.

Efficiency, effectiveness, and when to use fins

Fin efficiency (actual heat / heat if entire fin at TbT_{b}):
ηf=qfinhAfθb\eta_f=\frac{q_{\mathrm{fin}}}{h A_f\theta_b}

For adiabatic tip, Af=PLA_{f} = PL, and
ηf=tanh(mL)/(mL)\eta_f=\tanh(mL)/(mL)
.
Fin effectiveness (heat with fin / heat through base area AcA_{c} without fin):
εf=qfinhAcθb\varepsilon_f=\frac{q_{\mathrm{fin}}}{h A_c\theta_b}

For infinite fin:
εf=kP/(hAc)=kmAc/(hAc)=kP/(hAc)\varepsilon_f=\sqrt{kP/(hA_c)}=k m A_c/(h A_c)\cdots=\sqrt{kP/(h A_c)}
.
Design guidance
- Prefer high kk (Al, Cu) and geometries with large PAc\frac{P}{A_{c}} (thin fins).
- Fins help most when hh is low (gas cooling); less benefit in boiling/high-hh liquids.
- Rule of thumb: use fins if εf2\varepsilon_f\gtrsim 2.

Step-by-step problem approach

1. Identify fin type (pin, rectangular longitudinal) and tip condition.
2. Compute AcA_{c}, PP, then m=hP/(kAc)m=\sqrt{hP/(kA_c)}.
3. θb=TbT\theta_b=T_b-T_\infty.
4. Select qq formula (infinite / tanh(mL)\tanh(mL) / LcL_{c}).
5. If asked, ηf=q/(hAfθb)\eta_f=q/(h A_f\theta_b), εf=q/(hAcθb)\varepsilon_f=q/(h A_c\theta_b).
6. Units: mm → m for AcA_{c}, PP, LL; hh in W/(m²·K), kk in W/(m·K).
7. Check mLmL: if mL3mL\gtrsim 3, infinite-fin formula is often adequate.

Common mistakes in exams

• Using diameter instead of radius when computing pin Ac=πd2/4A_c=\pi d^2/4 and P=πdP=\pi d.
• Forgetting corrected length for convective tip.
• Mixing ηf\eta_f and εf\varepsilon_f definitions.
• Using °C absolute incorrectly — only ΔT=θ\Delta T=\theta matters here.
• Applying uniform-AcA_{c} formulas to triangular/annular fins without modification.
• Inconsistent mm/m in mm (which has units 1/m).

Worked examples

Try the problem first — open the solution when you are ready to check.

Adiabatic-tip straight fin heat rate

Problem

An aluminium fin (k=200W/(mK)k=200\,\mathrm{W/(m\cdot K)}) has L=50mmL=50\,\mathrm{mm}, rectangular section t=2mmt=2\,\mathrm{mm}, w=40mmw=40\,\mathrm{mm} (use Ac=twA_{c} = t w, P2(w+t)P\approx 2(w+t)). Tb=100CT_b=100^\circ\mathrm{C}, T=25CT_\infty=25^\circ\mathrm{C}, h=25W/(m2K)h=25\,\mathrm{W/(m^2\cdot K)}. Assume adiabatic tip. Find qfinq_{\mathrm{fin}}.

Solution

Formulas (Indian textbook notation)

  • Ac=(0.002)(0.040)=8×105m2A_c=(0.002)(0.040)=8\times 10^{-5}\,\mathrm{m^2}

Formulas (Indian textbook notation)

  • P=2(0.040+0.002)=0.084mP=2(0.040+0.002)=0.084\,\mathrm{m}

Formulas (Indian textbook notation)

  • m=hPkAc=25×0.084200×8×105=131.25=11.46m1m=\sqrt{\frac{hP}{kA_c}}=\sqrt{\frac{25\times 0.084}{200\times 8\times 10^{-5}}}=\sqrt{131.25}=11.46\,\mathrm{m^{-1}}

Formulas (Indian textbook notation)

  • mL=11.46×0.05=0.573mL=11.46\times 0.05=0.573

Formulas (Indian textbook notation)

  • θb=75K\theta_b=75\,\mathrm{K}

Formulas (Indian textbook notation)

  • q=hPkAcθbtanh(mL)=25×0.084×200×8×105×75×tanh(0.573)q=\sqrt{hPkA_c}\,\theta_b\tanh(mL)=\sqrt{25\times 0.084\times 200\times 8\times 10^{-5}}\times 75\times\tanh(0.573)

Formulas (Indian textbook notation)

  • hPkAc=0.0336=0.1833W/K\sqrt{hPkA_c}=\sqrt{0.0336}=0.1833\,\mathrm{W/K}

Formulas (Indian textbook notation)

  • q=0.1833×75×0.517=7.11Wq=0.1833\times 75\times 0.517=7.11\,\mathrm{W}

Fin efficiency

Problem

Using the previous fin, compute ηf\eta_f.

Effectiveness of a long pin fin

Problem

A long copper pin fin: d=5mmd=5\,\mathrm{mm}, k=400W/(mK)k=400\,\mathrm{W/(m\cdot K)}, h=40W/(m2K)h=40\,\mathrm{W/(m^2\cdot K)}. Treat as infinite. Find εf\varepsilon_f.

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is a fin (extended surface)? Why are fins used?

    Model answer

    A fin is a protrusion from a surface that increases area for convection (or radiation). Used when hh is low (gases) to raise heat transfer for a given base-to-fluid ΔT\Delta T.
  2. 2
    Write the governing equation for a straight fin of uniform cross-section (steady, 1-D).

    Model answer

    d2θdx2m2θ=0\dfrac{d^2\theta}{dx^2}-m^2\theta=0 where θ=TT\theta=T-T_\infty and m=hP/(kAc)m=\sqrt{hP/(kA_c)}. PP perimeter, AcA_{c} cross-section.
  3. 3
    Define fin efficiency ηf\eta_f.

    Model answer

    ηf=qfinqmax=qfinhAfθb\eta_f=\dfrac{q_{\mathrm{fin}}}{q_{\mathrm{max}}}=\dfrac{q_{\mathrm{fin}}}{hA_f\theta_b} — actual heat transfer divided by heat transfer if the entire fin were at base temperature TbT_{b}.
  4. 4
    Define fin effectiveness εf\varepsilon_f. When is a fin justified?

    Model answer

    εf=qfin/(hAcθb)\varepsilon_f=q_{\mathrm{fin}}/(hA_c\theta_b) — ratio to heat transfer from the base area if unfinned. Typically require εf2\varepsilon_f\gtrsim 2 for the fin to be worthwhile.
  5. 5
    Give heat transfer for a fin with insulated tip (longitudinal, uniform AcA_{c}).

    Model answer

    q=hPkAcθbtanh(mL)=hAfθbtanh(mL)mLq=\sqrt{hPkA_c}\,\theta_b\tanh(mL)=hA_f\theta_b\,\dfrac{\tanh(mL)}{mL} with Af=PLA_{f} = PL (approx. neglecting tip area).
  6. 6
    What is an infinitely long fin? Write qq.

    Model answer

    Tip temperature approaches TT_\infty (mLmL large). q=hPkAcθbq=\sqrt{hPkA_c}\,\theta_b. Temperature θ/θb=emx\theta/\theta_b=e^{-mx}.
  7. 7
    How does thermal conductivity of the fin material affect performance?

    Model answer

    Higher kk lowers mm and raises ηf\eta_f and qq toward the ideal. Poor conductors need short, stubby fins; metals (Al, Cu) are preferred.
  8. 8
    Explain the corrected fin length for a convective tip.

    Model answer

    Approximate a convective tip by an insulated-tip fin of length Lc=L+AcPL_{c} = L+\frac{A_{c}}{P} (rod: L+t2L+\frac{t}{2} or L+D4L+\frac{D}{4}). Then use tanh(mLc)\tanh(mL_c) formulas.
  9. 9
    What is overall surface efficiency for a finned wall?

    Model answer

    ηo=1NAfAt(1ηf)\eta_o=1-\dfrac{N A_f}{A_t}(1-\eta_f) where At=Ab+NAfA_{t} = A_{b}+N A_{f} total area including unfinned base between fins. Then q=ηohAtθbq=\eta_o h A_t\theta_b.
  10. 10
    Why are fins more useful in gas cooling than in liquid cooling?

    Model answer

    Gases have low hh; increasing area helps a lot. Liquids already have high hh, so conduction resistance in the fin and manufacturing cost often outweigh the gain (εf\varepsilon_f may be low).
  11. 11
    Sketch (describe) temperature distribution along a short vs long fin.

    Model answer

    Short fin (mLmL small): temperature stays close to TbT_{b}. Long fin: θ\theta decays toward zero; most heat leaves near the base.
  12. 12
    What is the optimum mLmL or thickness trade-off for a fin of given profile?

    Model answer

    For fixed volume, there is an optimum thickness/length maximizing qq. Rule of thumb: mLmL of order 1 for rectangular spines; charts in Sachdeva/Incropera give optima.
  13. 13
    Differentiate pin fin, straight rectangular fin, and annular fin.

    Model answer

    Pin: slender rod from surface. Straight rectangular: constant thickness plate. Annular: circular disc on a tube — common on heat-exchanger tubes.
  14. 14
    How do you include contact resistance at the fin base?

    Model answer

    Add RcontactR_{\mathrm{contact}} in series with the fin thermal resistance Rfin=θb/qfinR_{\mathrm{fin}}=\theta_b/q_{\mathrm{fin}}. Poor bonding can destroy effectiveness.
  15. 15
    State when the 1-D fin approximation is valid.

    Model answer

    Biot number based on half-thickness Bi=ht/k1\mathrm{Bi}=ht/k\ll 1 so temperature is nearly uniform across the thickness; lateral conduction dominates over transverse gradients.

Exams & GATE

  • 1
    Textbook: RC Sachdeva (extended surfaces); Incropera Ch. 3.
  • 2
    GATE favourites: fin parameter mm, adiabatic-tip qq, and fin efficiency/effectiveness.
  • 3
    Ask “will adding a fin help?” using effectiveness before designing.
  • 4
    State assumptions (1-D, steady, constant hh) explicitly in answers.

📖 Standard books (India)

  • Fundamentals of Engineering Heat & Mass TransferRC Sachdeva

    Read: Syllabus unit

    Heat transfer and heat exchangers