MECH3610 · Advanced Thermofluids
Conduction II: Thermal Resistance & Heat Generation
This chapter turns Fourier's and Newton's laws into the course's workhorse tool — the electrical-analogy thermal-resistance network — and then adds internal (volumetric) heat generation. Composite-wall and insulated-pipe resistance problems and generation-with-surface-temperature problems are canonical Weeks 1-2 tutorial types and recur directly on the 30% open-book mid-term, usually worked 'per unit area' or 'per unit length'.
What this chapter covers
- 01Thermal resistance R = deltaT/Q and the analogy Q<->current, deltaT<->voltage, R<->resistance
- 02Conduction resistances: plane wall L/kA, radial cylinder ln(r2/r1)/(2*pi*Lk), sphere (r2-r1)/(4*pi*k*r1*r2)
- 03Convection resistance 1/hA and radiation resistance 1/(h_r A); series add, parallel add as reciprocals
- 04Composite/multilayer walls and pipes; overall coefficient UA = 1/R_tot and per-layer drop deltaT = Q*R_layer
- 05Thermal contact resistance R''_t,c at imperfect interfaces adds a temperature jump q'' = deltaT/R''_t,c
- 06Volumetric generation g_dot: surface temperature from g_dot*V = h*A_s*(T_s - T-inf)
- 07Centre-to-surface maximum rise: wall g_dot*L^2/(2k), cylinder g_dot*r0^2/(4k), sphere g_dot*r0^2/(6k)
- 08Generation gives a parabolic T(x) or T(r); the biggest resistor owns the biggest temperature drop
Heat loss and outer-surface temperature of an insulated pipe (per unit length)
- +1Three resistances in series, per unit length (units m-K/W). Inside film R'_conv,i = 1/(h_i*2*pi*r1) = 1/(500*2*pi*0.025) = 0.0127. Insulation R'_ins = ln(r2/r1)/(2*pi*k) = ln(55/25)/(2*pi*0.05) = 0.7885/0.3142 = 2.510. Outside film R'_conv,o = 1/(h_o*2*pi*r2) = 1/(15*2*pi*0.055) = 0.193. Total R'_tot = 0.0127 + 2.510 + 0.193 = 2.716 m-K/W.
- +1The driving difference is 200 - 25 = 175 K, so the loss per metre is q' = deltaT/R'_tot = 175/2.716 = 64.4 W/m.
- +1The outer surface sits between the insulation and the outside air, so its temperature comes from the outside film alone: T_s,o = T-inf,o + q'*R'_conv,o = 25 + 64.4*0.193 = 37.4 C.
- +1The insulation resistance (2.510) is 92% of the 2.716 total, so it carries almost the whole 175 K drop; the two films are minor. Adding insulation thickness is the effective way to cut the loss, and the outer surface stays safe to touch at ~37 C.
Key terms
- Thermal resistance (R)
- R = (T1 - T2)/Q [K/W], the heat-transfer analogue of electrical resistance. Conduction plane R = L/kA, cylinder R = ln(r2/r1)/(2*pi*Lk); convection R = 1/hA.
- Series resistances
- Resistors a single heat flow passes through in turn (film, wall, film) add directly: R_tot = sum R_i, and Q = deltaT/R_tot. The temperature drop across any one layer is deltaT_i = Q*R_i.
- Parallel resistances
- Side-by-side heat paths (e.g. convection alongside radiation at one surface) combine as 1/R_tot = sum(1/R_i); two paths give R_tot = R1*R2/(R1 + R2).
- Overall heat-transfer coefficient (U)
- Defined by UA = 1/R_tot, so Q = UA*deltaT. It rolls a whole series network into a single coefficient and is the bridge to heat-exchanger analysis.
- Thermal contact resistance (R''_t,c)
- An interfacial resistance [m2-K/W] at an imperfect (bolted or pressed) joint that causes a temperature jump q'' = deltaT/R''_t,c. It adds another series resistor between two solids.
- Volumetric heat generation (g_dot)
- Internal heat release per unit volume [W/m3], e.g. resistive heating or nuclear/chemical sources. Balanced by surface convection it sets T_s = T-inf + g_dot*V/(h*A_s), and it makes the internal temperature profile parabolic.
Conduction II: Thermal Resistance & Heat Generation FAQ
When do I use ln(r2/r1) instead of L/kA?
Use the logarithmic cylinder resistance ln(r2/r1)/(2*pi*Lk) whenever conduction is radial (through a pipe or lagging), and the plane-wall L/kA only for flat slabs. Getting these mixed up is the most common radial-problem error. Spheres use a third form, (r2-r1)/(4*pi*k*r1*r2).
Why does the biggest resistance carry the biggest temperature drop?
In a series network the same heat rate Q flows through every resistor, and the drop across each is deltaT_i = Q*R_i. Since Q is common, the drops are in proportion to the resistances, so the largest resistor (usually the insulation) sees the largest deltaT. This is why lagging works and why a thin metal pipe wall barely matters.
Where is the hottest point in a body with heat generation?
At the point furthest from the cooled surface — the centreline of a wall, cylinder or sphere (or an adiabatic face). The rise above the surface is g_dot*L^2/(2k) for a wall, g_dot*r0^2/(4k) for a cylinder and g_dot*r0^2/(6k) for a sphere, giving a parabolic profile that is flat at the centre.
How is this assessed in MECH3610?
Composite-wall and insulated-pipe resistance networks and generation-with-surface-temperature problems are standard Weeks 1-2 tutorial questions and reappear on the 30% open-book mid-term, often with a qualitative closing part ('which layer to thicken', 'is the surface safe to touch'). Confirm the mid-term coverage on Moodle.
Exam move
Draw the thermal circuit before doing any algebra: label each resistor (film, wall, contact, insulation), decide series vs parallel, then compute R_tot and Q = deltaT/R_tot. Practise both the 'per unit area' (q'') and 'per unit length' (q') conventions and be deliberate about which radius each convection resistor uses. For generation problems, split the two skills — surface temperature from the overall energy balance, and centreline rise from the g_dot*r^2/(nk) formulae — and remember the profile is parabolic. Rehearse peeling off one layer's temperature drop as deltaT_i = Q*R_i. Confirm the mid-term format on Moodle.
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