GATE GUIDE

GATE ME Heat Transfer Formula Sheet 2027

By MD ANISH AHAMADUpdated 4 Oct 20267 min read
GATE ME Heat Transfer Formula Sheet 2027

Heat transfer in GATE ME is steady and calculation-heavy. Conduction, convection, radiation and heat exchangers each drew between 9 and 12 marks across the six papers from 2022 to 2026, so none of them can be dropped. This sheet gives the formulas each topic runs on, a worked example for each, and the trap behind it.

In this guide
  1. Key takeaways
  2. The terms this sheet uses
  3. Conduction and thermal resistance
  4. Fins
  5. Unsteady conduction: the lumped model
  6. Convection
  7. Boiling and condensation (new in 2027)
  8. Heat exchangers: LMTD and NTU
  9. Radiation
  10. Using this sheet in the exam
  11. Quick revision

Key takeaways

The terms this sheet uses

Thermal resistance RR is temperature difference divided by heat rate, so resistances in series add like electrical ones. hh is the convection coefficient and kk the conductivity. θb=Tb−T∞\theta_b = T_b - T_\infty is the excess temperature at a fin base.

Every temperature inside T4T^4, a ratio or an exponential is absolute, in kelvin. Celsius is safe only in a difference. σ=5.67×10−8 W/m2K4\sigma = 5.67 \times 10^{-8}\ \text{W/m}^2\text{K}^4.

Conduction and thermal resistance

Formula Watch out for
q=−kAdTdx\displaystyle q = -kA\frac{dT}{dx}; q=hA(Ts−T∞)q = hA(T_s - T_\infty); q=εσA(Ts4−Tsur4)q = \varepsilon\sigma A(T_s^4 - T_{sur}^4) Kelvin in the radiation term
Plane wall LkA\displaystyle \frac{L}{kA}; cylinder ln⁡(r2/r1)2πkL\displaystyle \frac{\ln(r_2/r_1)}{2\pi kL}; sphere r2−r14πkr1r2\displaystyle \frac{r_2 - r_1}{4\pi kr_1r_2}; convection 1hA\displaystyle \frac{1}{hA} The cylinder takes the ratio of radii inside the logarithm
1UA=∑R\displaystyle \frac{1}{UA} = \sum R Parallel paths add as conductances
Critical radius rc=kh\displaystyle r_c = \frac{k}{h} (cylinder), 2kh\displaystyle \frac{2k}{h} (sphere) Below rcr_c, insulation raises the loss
Generation, wall of half-thickness LL: Tmax⁡−Ts=q˙L22k\displaystyle T_{\max} - T_s = \frac{\dot{q}L^2}{2k}; cylinder q˙R24k\displaystyle \frac{\dot{q}R^2}{4k}; sphere q˙R26k\displaystyle \frac{\dot{q}R^2}{6k} Unequal face temperatures move the maximum off-centre

Example: a wall with L/k=0.2L/k = 0.2, inside h=10h = 10 and outside h=25 W/m2Kh = 25\ \text{W/m}^2\text{K} has ∑R=0.2+0.1+0.04=0.34 m2K/W\sum R = 0.2 + 0.1 + 0.04 = 0.34\ \text{m}^2\text{K/W} per square metre. A 34 K overall difference drives 100 W/m2100\ \text{W/m}^2.

Example: insulation with k=0.1 W/m Kk = 0.1\ \text{W/m K} and h=5h = 5 gives rc=20r_c = 20 mm. A 10 mm wire loses more heat as insulation is added, up to 20 mm.

Trap: a steady profile without heat generation cannot have an interior maximum. If a question shows one, there must be a source.

Fins

m=hPkAc,qlong=hPkAc θb,qinsulated tip=hPkAc θbtanh⁡(mL)m = \sqrt{\frac{hP}{kA_c}}, \qquad q_{\text{long}} = \sqrt{hPkA_c}\,\theta_b, \qquad q_{\text{insulated tip}} = \sqrt{hPkA_c}\,\theta_b\tanh(mL)

Fin efficiency is tanh⁡(mL)mL\displaystyle \frac{\tanh(mL)}{mL} for an insulated tip. For a convecting tip, use the corrected length L+t2\displaystyle L + \frac{t}{2} for a thin plate or L+D4\displaystyle L + \frac{D}{4} for a pin.

Example: a 10 mm pin with h=20h = 20, k=200k = 200 and θb=80\theta_b = 80 K has m=4hkD=6.32 m−1\displaystyle m = \sqrt{\frac{4h}{kD}} = 6.32\ \text{m}^{-1}. A very long fin passes 7.95 W; a 0.1 m fin with an insulated tip passes 7.95tanh⁡(0.632)=4.457.95\tanh(0.632) = 4.45 W, at efficiency 0.885.

Fins pay where hh is low, as with gases, and kk is high.

Unsteady conduction: the lumped model

Formula Watch out for
Bi=hLck<0.1\displaystyle Bi = \frac{hL_c}{k} < 0.1, Lc=VAs\displaystyle L_c = \frac{V}{A_s} Sphere r3\displaystyle \frac{r}{3}, cylinder r2\displaystyle \frac{r}{2}, plate half-thickness
T−T∞Ti−T∞=e−t/τ\displaystyle \frac{T - T_\infty}{T_i - T_\infty} = e^{-t/\tau}, τ=ρVchAs\displaystyle \tau = \frac{\rho Vc}{hA_s} Same as e−Bi⋅Foe^{-Bi \cdot Fo}
Fo=αtLc2\displaystyle Fo = \frac{\alpha t}{L_c^2}, α=kρc\displaystyle \alpha = \frac{k}{\rho c} kk of the solid

Example: a 10 mm steel ball with k=40k = 40, ρ=7800\rho = 7800, c=500c = 500 in air at h=100h = 100 has Lc=1.67L_c = 1.67 mm and Bi=0.0042Bi = 0.0042. Then τ=65\tau = 65 s, and the excess falls to a tenth in 65ln⁡10=15065\ln 10 = 150 s.

Both unsteady conduction questions in the six counted papers were lumped-model questions.

Convection

Formula Watch out for
Nu=hLk\displaystyle Nu = \frac{hL}{k} (kk of the fluid); Pr=να\displaystyle Pr = \frac{\nu}{\alpha}; Gr=gβΔTL3ν2\displaystyle Gr = \frac{g\beta\Delta TL^3}{\nu^2}; Ra=Gr PrRa = Gr\,Pr BiBi uses kk of the solid
GrRe2≫1\displaystyle \frac{Gr}{Re^2} \gg 1 free; ≪1\ll 1 forced Near 1 is mixed
Laminar plate: Nux=0.332 Rex1/2Pr1/3Nu_x = 0.332\,Re_x^{1/2}Pr^{1/3}; mean 0.664 ReL1/2Pr1/30.664\,Re_L^{1/2}Pr^{1/3} Mean hh is twice the local hh at LL
Turbulent plate: mean 0.037 ReL0.8Pr1/30.037\,Re_L^{0.8}Pr^{1/3} Transition near Rex=5×105Re_x = 5 \times 10^5
Laminar pipe Nu=3.66Nu = 3.66 (uniform wall temperature), 4.364.36 (uniform flux) Fully developed only
Dittus–Boelter Nu=0.023 Re0.8PrnNu = 0.023\,Re^{0.8}Pr^{n} n=0.4n = 0.4 heating, 0.30.3 cooling
Uniform flux: m˙cpΔTm=q′′πDL\dot{m}c_p\Delta T_m = q''\pi DL Needs no correlation

Example: doubling the free-stream velocity raises a laminar hh by 2=1.41\sqrt{2} = 1.41 times and a turbulent one by 20.8=1.742^{0.8} = 1.74 times. With Re=50 000Re = 50\,000 and Pr=0.7Pr = 0.7, heating, Dittus–Boelter gives Nu=0.023×5743×0.867=114.5Nu = 0.023 \times 5743 \times 0.867 = 114.5.

Remember: turbulent correlations go as Re0.8Re^{0.8} against Re0.5Re^{0.5} for laminar flow. That exponent is what the syllabus line on the effect of turbulence comes down to.

The book's last-minute sheet lists every Heat Transfer formula with its condition, and its chapter traces the 29 counted questions topic by topic. Both are in the GATE ME 2027 book, with 942 questions with worked solutions and 10 full mock tests.

Boiling and condensation (new in 2027)

The pool boiling curve plots heat flux against the excess temperature Ts−TsatT_s - T_{sat}. It runs through free convection, nucleate boiling (very high hh, the useful regime), transition boiling and film boiling. Past the critical heat flux, a heat-flux-controlled surface jumps to film boiling, which is burnout. The Leidenfrost point sits at the minimum heat flux.

Dropwise condensation gives a much higher hh than filmwise. For a laminar film on a vertical plate, Nusselt's result is:

h=0.943[ρl(ρl−ρv)g hfgkl3μlL(Tsat−Ts)]1/4h = 0.943\left[\frac{\rho_l(\rho_l - \rho_v)g\,h_{fg}k_l^3}{\mu_l L(T_{sat} - T_s)}\right]^{1/4}

The heat released is m˙hfg\dot{m}h_{fg}. A horizontal tube uses a smaller constant, with the diameter in place of LL.

Heat exchangers: LMTD and NTU

Formula Watch out for
ΔTlm=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\displaystyle \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}; Q=UAF ΔTlmQ = UAF\,\Delta T_{lm} End differences, not inlet minus outlet of one stream
ε=QCmin⁡(Th,in−Tc,in)\displaystyle \varepsilon = \frac{Q}{C_{\min}(T_{h,in} - T_{c,in})}; NTU=UACmin⁡\displaystyle NTU = \frac{UA}{C_{\min}}; Cr=Cmin⁡Cmax⁡\displaystyle C_r = \frac{C_{\min}}{C_{\max}} C=m˙cpC = \dot{m}c_p
Cr=0C_r = 0 (condensing or evaporating stream): ε=1−e−NTU\varepsilon = 1 - e^{-NTU} Holds for every flow arrangement
Counterflow, Cr=1C_r = 1: ε=NTU1+NTU\displaystyle \varepsilon = \frac{NTU}{1 + NTU} ΔT\Delta T constant along the length
Parallel flow: ε=1−e−NTU(1+Cr)1+Cr\displaystyle \varepsilon = \frac{1 - e^{-NTU(1 + C_r)}}{1 + C_r} Never above 11+Cr\displaystyle \frac{1}{1 + C_r}

Example: hot fluid 150 to 90 °C and cold 30 to 70 °C give end differences of 80 and 60 K in counterflow, so ΔTlm=20ln⁡(4/3)=69.5\displaystyle \Delta T_{lm} = \frac{20}{\ln(4/3)} = 69.5 K. In parallel flow the ends are 120 and 20 K, giving 55.8 K. Counterflow always gives the larger LMTD.

Example: a condenser with NTU=2NTU = 2 has ε=1−e−2=0.865\varepsilon = 1 - e^{-2} = 0.865; a balanced counterflow unit with NTU=2NTU = 2 has ε=0.667\varepsilon = 0.667.

Radiation

Formula Watch out for
Eb=σT4E_b = \sigma T^4; Wien λmax⁡T=2898 μm K\lambda_{\max}T = 2898\ \mu\text{m K} Kelvin
α+ρ+τ=1\alpha + \rho + \tau = 1; grey surface ε=α\varepsilon = \alpha Opaque: τ=0\tau = 0
A1F12=A2F21A_1F_{12} = A_2F_{21}; ∑jF1j=1\sum_j F_{1j} = 1 Flat or convex surface: F11=0F_{11} = 0
Surface resistance 1−εεA\displaystyle \frac{1 - \varepsilon}{\varepsilon A}; space resistance 1A1F12\displaystyle \frac{1}{A_1F_{12}} Re-radiating surface is a floating node
Parallel plates q=σ(T14−T24)1/ε1+1/ε2−1\displaystyle q = \frac{\sigma(T_1^4 - T_2^4)}{1/\varepsilon_1 + 1/\varepsilon_2 - 1} Per unit area
Small body in large enclosure q=ε1σA1(T14−T24)q = \varepsilon_1\sigma A_1(T_1^4 - T_2^4) Enclosure emissivity drops out
nn equal shields cut the flux to 1n+1\displaystyle \frac{1}{n + 1} All emissivities equal

Example: plates at 800 K and 400 K with ε=0.5\varepsilon = 0.5 each exchange 5.67×10−8×3.84×10113=7.26 kW/m2\displaystyle \frac{5.67 \times 10^{-8} \times 3.84 \times 10^{11}}{3} = 7.26\ \text{kW/m}^2. One shield of the same emissivity halves it to 3.63 kW/m23.63\ \text{kW/m}^2. Wien puts the peak of a 1000 K black body at 2.898 μm2.898\ \mu\text{m}.

Example: for long concentric cylinders of radii 0.1 and 0.4 m, F12=1F_{12} = 1, so reciprocity gives F21=0.25F_{21} = 0.25 and summation F22=0.75F_{22} = 0.75.

In one line: find every view factor by summation and reciprocity first, then build the network.

Using this sheet in the exam

Write the radiation network resistances and the fin formula on the rough sheet early; the exam-day rules guide explains what is provided. Logarithms, tanh⁡\tanh and fourth powers all run on the GATE virtual calculator, so keep full precision until the last step. Many questions here are NAT with no negative marking, as the marking scheme, common to every GATE paper, explains.

More formula sheets: all of GATE ME · Fluid Mechanics · Strength of Materials · Thermodynamics

Quick revision

  1. Resistances: LkA\displaystyle \frac{L}{kA}, ln⁡(r2/r1)2πkL\displaystyle \frac{\ln(r_2/r_1)}{2\pi kL}, 1hA\displaystyle \frac{1}{hA}; series add.
  2. Critical radius kh\displaystyle \frac{k}{h} for a cylinder, 2kh\displaystyle \frac{2k}{h} for a sphere.
  3. Fin: m=hPkAc\displaystyle m = \sqrt{\frac{hP}{kA_c}}; insulated tip multiplies by tanh⁡(mL)\tanh(mL).
  4. Lumped only if Bi<0.1Bi < 0.1 on Lc=V/AsL_c = V/A_s.
  5. Dittus–Boelter exponent 0.4 heating, 0.3 cooling; laminar pipe 3.66 or 4.36.
  6. ε=1−e−NTU\varepsilon = 1 - e^{-NTU} when one stream changes phase.
  7. Parallel plates divide by 1ε1+1ε2−1\displaystyle \frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1; Wien 2898 μm K\mu\text{m K}.
  8. Boiling: nucleate is useful, beyond the critical heat flux lies burnout.

Frequently asked questions

When can I use the lumped parameter model in GATE ME?

Only when the Biot number Bi=hLck\displaystyle Bi = \frac{hL_c}{k} is below 0.1, with the characteristic length Lc=V/AsL_c = V/A_s. That is r/3r/3 for a sphere, r/2r/2 for a long cylinder and the half-thickness for a plate. Using the radius itself is the standard error. Once the test passes, the temperature excess decays as e−t/τe^{-t/\tau} with τ=ρVchAs\displaystyle \tau = \frac{\rho Vc}{hA_s}.

What is the critical radius of insulation?

For a cylinder it is rc=k/hr_c = k/h, and for a sphere 2k/h2k/h, with kk the conductivity of the insulation and hh the outside convection coefficient. If the bare radius is below rcr_c, adding insulation first increases the heat loss, because the extra outer area lowers the convection resistance faster than the insulation adds conduction resistance.

Is boiling and condensation in the GATE ME 2027 syllabus?

Yes. Boiling and condensation is newly named in the 2027 Heat Transfer paragraph, and Heisler's charts are no longer named. Learn the pool boiling curve with its regimes, the critical heat flux and burnout, the Leidenfrost point, and why dropwise condensation gives a much higher coefficient than filmwise. Nusselt's laminar film result for a vertical plate is the main formula.

Which is better for heat exchanger questions, LMTD or NTU?

Use LMTD when all four terminal temperatures are known or easy to find, for sizing an exchanger with Q=UA ΔTlmQ = UA\,\Delta T_{lm}. Use the effectiveness-NTU method when outlet temperatures are unknown, for rating. Many questions need only ε=QCmin⁡(Th,in−Tc,in)\displaystyle \varepsilon = \frac{Q}{C_{\min}(T_{h,in} - T_{c,in})} and an energy balance, so start there.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

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