MECH3610 · Advanced Thermofluids
Transient Conduction & Lumped Capacitance
When a body's temperature changes with time, the simplest model is lumped capacitance, which treats the body as spatially uniform and gives an exponential approach to ambient. This chapter derives that response, the time constant, and the Biot-number test (Bi <= 0.1) that validates it. Quenching, annealing and thermocouple-response problems (both forward-for-time and inverse-for-h) are standard Week-2 tutorials and appear on the 30% open-book mid-term.
What this chapter covers
- 01Energy balance h*A_s*(T-inf - T)dt = rho*V*c_p*dT giving the exponential response
- 02Solution (T(t) - T-inf)/(T_i - T-inf) = e^{-bt} with b = h*A_s/(rho*V*c_p) [1/s]
- 03Time constant tau = 1/b = rho*V*c_p/(h*A_s); time to a target ratio by inverting the exponential
- 04Characteristic length L_c = V/A_s (sphere D/6, long cylinder r/2, plane wall of half-thickness L)
- 05Biot number Bi = h*L_c/k and the lumped-validity criterion Bi <= 0.1
- 06Instantaneous rate Q(t) = h*A_s*(T(t) - T-inf) and total energy transferred rho*V*c_p*(T(t) - T_i)
- 07Inverse problems: back out h from a measured temperature-time history, then verify Bi after the fact
- 08When Bi is not small: internal gradients matter and a series/Heisler-chart solution is needed
Quench time of a steel sphere, with a Biot-number validity check
- +1Characteristic length L_c = V/A_s = D/6 = 0.010/6 = 1.667e-3 m. Biot number Bi = h*L_c/k = 120*1.667e-3/50 = 0.004, which is well below 0.1, so the sphere is nearly isothermal and lumped capacitance is valid.
- +1b = h*A_s/(rho*V*c_p) = h/(rho*c_p*L_c) = 120/(7800*480*1.667e-3) = 120/6240 = 0.01923 /s (time constant tau = 1/b = 52 s).
- +1Invert the exponential with the temperature ratio (100 - 30)/(500 - 30) = 70/470 = 0.1489: -b*t = ln(0.1489) = -1.904, so t = 1.904/0.01923 = 99 s.
- +1Volume V = pi*D^3/6 = pi*(0.010)^3/6 = 5.236e-7 m3. Energy removed = rho*V*c_p*(T_i - T) = 7800*5.236e-7*480*(500 - 100) = 784 J. (Positive, i.e. released by the sphere to the oil.)
Key terms
- Lumped capacitance method
- A transient model that assumes a spatially uniform body temperature T(t), valid when internal conduction is fast compared with surface convection (Bi <= 0.1). It gives an exponential approach to ambient.
- Biot number (Bi)
- Bi = h*L_c/k, the ratio of internal conduction resistance to surface convection resistance. Bi <= 0.1 justifies the lumped model; larger Bi means significant internal gradients.
- Characteristic length (L_c)
- L_c = V/A_s: the volume-to-surface-area ratio used in both Bi and b. It equals D/6 for a sphere, r/2 for a long cylinder and the half-thickness for a symmetrically cooled plane wall.
- Time constant (tau)
- tau = 1/b = rho*V*c_p/(h*A_s) [s], the e-folding time of the exponential cooling. After one tau the excess temperature has fallen to 37% of its initial value.
- Rate coefficient (b)
- b = h*A_s/(rho*V*c_p) = h/(rho*c_p*L_c) [1/s], the decay rate in (T - T-inf)/(T_i - T-inf) = e^{-bt}. Larger b (small body, high h, low heat capacity) cools faster.
- Total energy transferred
- The heat exchanged from time 0 to t, rho*V*c_p*(T(t) - T_i) [J]. Negative when the body cools (it releases energy to the fluid).
Transient Conduction & Lumped Capacitance FAQ
How small does Bi have to be for lumped capacitance?
Bi <= 0.1 is the standard criterion; below it the temperature is uniform to within about 5% and the lumped model is accurate. Once Bi exceeds 0.1 the internal gradient matters and you must use a one-term-series or Heisler-chart solution instead of the simple exponential; by Bi around 1 or larger the gradient dominates.
What characteristic length do I use for a cylinder or wall?
L_c = V/A_s always. For a sphere that is D/6, for a long cylinder r/2 (the ends are ignored), and for a plane wall cooled on both faces it is the half-thickness L. Using the full radius or full thickness by mistake throws both Bi and the time off.
How do I solve an inverse problem that asks for h?
Rearrange the exponential: from a measured time and temperature ratio, b = -ln(ratio)/t, then h = b*rho*c_p*L_c. This is the 'assume lumped, verify afterwards' pattern — after finding h, compute Bi and confirm it is <= 0.1 so the assumption was legitimate.
How is transient conduction examined in MECH3610?
As Week-2 tutorial quenching, annealing and thermocouple-response problems and on the open-book mid-term, in both forward (find the time) and inverse (find h) forms, usually with an explicit Biot check for a mark. Confirm the mid-term coverage on Moodle.
Exam move
Automate the lumped-capacitance routine: compute L_c = V/A_s, check Bi = h*L_c/k <= 0.1, then use (T - T-inf)/(T_i - T-inf) = e^{-bt} with b = h/(rho*c_p*L_c). Practise inverting the exponential for time and, in reverse, solving for h from a temperature history and then verifying Bi. Memorise the three characteristic lengths (sphere D/6, long cylinder r/2, wall half-thickness). Keep a clear head about sign when computing energy transferred. Because the Biot check is where easy marks and easy mistakes both live, always write it explicitly. Confirm the mid-term format on Moodle.
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