MECH3610 · Advanced Thermofluids
Introduction to Heat Transfer & Conduction I
Week 1 of UNSW MECH3610 sets up the three modes of heat transfer and their rate laws — Fourier's law of conduction, Newton's law of cooling and Stefan-Boltzmann radiation — plus the general heat-diffusion equation and the four boundary-condition types. These are the foundations the whole heat-transfer block (the 20% lab and the 30% open-book mid-term) is built on, and combined-mode problems that add a conduction, convection and radiation term in one balance recur throughout the exam.
What this chapter covers
- 01The three modes and when each needs a medium: conduction and convection require matter, radiation does not
- 02Fourier's law q'' = -k dT/dx, plane-wall form q'' = k(T1 - T2)/L, and heat flows down the gradient
- 03Newton's law of cooling q'' = h(T_s - T-inf); h is a flow/geometry property, not a material constant
- 04Stefan-Boltzmann radiation q'' = epsilon*sigma*(T_s^4 - T_surr^4) with temperatures in kelvin; sigma = 5.67e-8 W/m2-K4
- 05The radiation coefficient h_r = epsilon*sigma*(T_s + T_surr)(T_s^2 + T_surr^2) linearising radiation to Newton form
- 06The heat-diffusion equation d2T/dx2 + g_dot/k = (1/alpha) dT/dt and thermal diffusivity alpha = k/(rho*c_p)
- 07The 1-D steady, no-generation solution: linear T(x) = T_s,1 + (T_s,2 - T_s,1)(x/L)
- 08The four boundary conditions: prescribed temperature, prescribed flux, adiabatic/symmetry, and convective
Conduction rate through a furnace wall, then the effect of doubling k
- +1Assume one-dimensional, steady conduction with constant k. From Fourier's law for a plane wall q'' = k*(T1 - T2)/L = 1.0 * (1200 - 200) / 0.20 = 1000 / 0.20 = 5000 W/m2.
- +1The total rate is the flux times the face area A = 3 * 2 = 6 m2: Q = q'' * A = 5000 * 6 = 30000 W = 30 kW.
- +1The flux is directly proportional to k at fixed geometry and fixed face temperatures, so doubling the conductivity doubles the flux to 10000 W/m2 (and Q to 60 kW). A more conductive wall leaks more heat, not less — insulation wants a low k.
Key terms
- Heat flux (q'')
- Heat transfer rate per unit area, q'' = Q/A [W/m2]. The primed quantity used when no area is given; multiply by area for the total rate Q [W].
- Thermal conductivity (k)
- A material property [W/m-K] measuring how readily a material conducts heat; high for metals, low for insulators. It sets the constant in Fourier's law.
- Thermal diffusivity (alpha)
- alpha = k/(rho*c_p) [m2/s], the ratio of a material's ability to conduct heat to its ability to store it. It governs how fast a temperature disturbance spreads and appears in the transient heat-diffusion equation.
- Emissive power (E)
- The radiation emitted per unit area of a surface, E = epsilon*sigma*T^4 [W/m2] (T in kelvin). For a small surface in large surroundings the net exchange is q'' = epsilon*sigma*(T_s^4 - T_surr^4).
- Radiation heat-transfer coefficient (h_r)
- A linearised radiation coefficient h_r = epsilon*sigma*(T_s + T_surr)(T_s^2 + T_surr^2) [W/m2-K] that lets radiation be written in Newton-cooling form q'' = h_r(T_s - T_surr) and combined in parallel with convection.
- Boundary condition (thermal)
- One of the four conditions that close the heat equation at a surface: prescribed temperature, prescribed flux (-k dT/dx = q''_s), adiabatic/symmetry (dT/dx = 0), or convective (-k dT/dx = h(T-inf - T_surface)).
Introduction to Heat Transfer & Conduction I FAQ
Do I use Celsius or Kelvin in these equations?
For a temperature difference in conduction or convection either scale works, because a difference of 1 C equals a difference of 1 K. But any radiation term (epsilon*sigma*T^4) and any absolute-temperature group must use kelvin, so convert with T[K] = T[C] + 273.15. A common exam slip is leaving Celsius inside a T^4 term.
Why is there a minus sign in Fourier's law?
Heat flows from high to low temperature, i.e. down the temperature gradient, but dT/dx points up the gradient. The minus sign makes the flux positive in the direction of decreasing temperature. In the plane-wall form you simply write q'' = k(T_hot - T_cold)/L and the sign is already handled.
Is h a material property like k?
No. The conductivity k is a property of the material, but the convection coefficient h depends on the fluid, the geometry and the flow regime (forced vs free, laminar vs turbulent). That is why the convection chapters spend so long building empirical correlations to predict h, whereas k is simply read from a table.
How does this show up in the MECH3610 exam?
As the opening moves of almost every heat-transfer question and as standalone combined-mode problems: a surface losing heat by convection and radiation at once, or a wall whose flux you set equal to a convective loss. Because the mid-term is open-book, the marks are for choosing the right rate law and setting up the balance, not for recalling the formula. Confirm the exam format on Moodle.
Exam move
Memorise the three rate laws as a set — conduction q'' = k*deltaT/L, convection q'' = h*deltaT, radiation q'' = epsilon*sigma*(T_s^4 - T_surr^4) — and practise deciding which modes act at a given surface. Drill the surface energy balance where conduction into a face equals the convection plus radiation leaving it, since that combined-mode setup is the course's signature exam move. Get fluent converting to kelvin for radiation, and rehearse the four boundary conditions so you can write the correct one by inspection. Because Week 1 feeds every later chapter, over-learn it now; confirm the mid-term's coverage and format on Moodle.
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