Transient conduction

Steady-state analysis tells Northgate how much heat its pipes lose in normal operation. Some of the most important questions, though, are about time. If a pump trips on a −30 °C night, how many hours before the water in a stopped service line freezes? How fast must a temperature sensor respond for a control loop to stay stable? How long does a concrete valve chamber take to cool when its heating fails, and how deep will frost reach under a cleared parking lot? These are transient conduction problems, in which temperatures change with time as well as position. This lesson presents three standard approaches: the lumped capacitance method, the one-term series solutions for walls, cylinders and spheres, and the semi-infinite solid.

The lumped capacitance method

The simplest transient model assumes that the temperature inside a body is uniform at every instant, so it changes only with time. An energy balance on the whole body, losing heat by convection to a fluid at \(T_\infty\), gives \(-hA_s(T - T_\infty) = \rho V c\,dT/dt\), whose solution is an exponential decay:

The time constant \(\tau\) is the product of a thermal resistance, \(1/(hA_s)\), and a thermal capacitance, \(\rho V c\). After one time constant the body has completed 63 percent of its temperature change; after three, 95 percent; after about 4.6, 99 percent. The analogy with a resistor-capacitor circuit is exact, and it extends to bodies cooled through insulation, where the resistance is that of the insulation rather than a convection film.

When is lumping valid?

The lumped model is accurate when conduction inside the body is fast compared with convection at its surface, so internal temperature differences stay small. The ratio of the two resistances is the Biot number:

where \(k\) is the conductivity of the solid, not of the fluid, and \(L_c\) is a characteristic length: half the thickness for a plane wall cooled on both faces, \(r_o/2\) for a long cylinder and \(r_o/3\) for a sphere. When Bi is less than about 0.1, the error of the lumped model is under about 5 percent. The time scale is described by the Fourier number, \(\text{Fo} = \alpha t/L_c^2\), a dimensionless time, and the lumped solution can be written compactly as \(\theta/\theta_i = \exp(-\text{Bi}\cdot\text{Fo})\).

# Transient conduction: lumped bodies, a plane wall by the one-term solution and frost in the ground
import math

# 1. Lumped capacitance: a bead thermocouple dropped into a hot-water stream
d, rho, c, k = 0.0010, 8500.0, 400.0, 25.0    # bead diameter (m) and properties of the junction
h = 1200.0                                    # water-side coefficient, W/m2.K
Lc = d / 6                                    # V/A for a sphere
Bi = h * Lc / k
tau = rho * c * Lc / h
print(f"thermocouple: Bi = {Bi:.4f}, time constant {tau:.2f} s; reaches 99 % of a step in {tau*math.log(100):.1f} s")

# 2. A stopped service pipe in winter: how long until the water reaches 0 C?
#    Water plus steel treated as one lumped mass; the foam is the resistance to the cold soil.
L_svc, Di = 1.0, 0.0545                       # per 1 m of DN50 service pipe (inside diameter, m)
m_w = 1000 * math.pi / 4 * Di**2 * L_svc       # water mass, kg
m_st = 7850 * math.pi / 4 * (0.0603**2 - Di**2) * L_svc
C = m_w * 4200 + m_st * 470                    # J/K
R = math.log(0.0625 / 0.03015) / (2 * math.pi * 0.027) + 0.5   # foam to a 125 mm casing plus soil, m.K/W (soil value assumed)
T_i, T_soil = 60.0, -8.0
tau_p = R * C
t0 = tau_p * math.log((T_i - T_soil) / (0 - T_soil))
print(f"service pipe: C = {C/1000:.1f} kJ/K, R' = {R:.2f} m.K/W, time constant {tau_p/3600:.1f} h, reaches 0 C after {t0/3600:.0f} h")

# 3. One-term solution: a 0.60 m concrete wall of a valve chamber, both faces suddenly at -25 C
def zeta1_plane(Bi):
    lo, hi = 1e-9, math.pi / 2 - 1e-12
    for _ in range(200):
        m = (lo + hi) / 2
        lo, hi = (m, hi) if m * math.tan(m) < Bi else (lo, m)
    return (lo + hi) / 2
kc, alpha, Lw, hc = 1.4, 6.9e-7, 0.30, 25.0   # concrete; half-thickness (m); windy outside coefficient
Bi_w = hc * Lw / kc
z1 = zeta1_plane(Bi_w)
C1 = 4 * math.sin(z1) / (2 * z1 + math.sin(2 * z1))
T0, Tinf = 10.0, -25.0
for hours in (12, 24, 48, 96):
    Fo = alpha * hours * 3600 / Lw**2
    th = C1 * math.exp(-z1**2 * Fo)
    flag = "" if Fo > 0.2 else "  (Fo < 0.2: one-term solution only approximate)"
    print(f"wall midplane after {hours:3d} h: Fo = {Fo:.3f}, T = {Tinf + th*(T0 - Tinf):6.1f} C{flag}")
print(f"  Bi = {Bi_w:.2f}, zeta1 = {z1:.4f}, C1 = {C1:.4f}")

# 4. Semi-infinite solid: depth reached by 0 C after a long cold spell (no latent heat, so an upper bound)
alpha_s, Ti_g, Ts = 5.5e-7, 5.0, -15.0       # frozen-soil diffusivity (assumed), initial ground, surface
def erfinv(y):
    lo, hi = 0.0, 5.0
    for _ in range(100):
        m = (lo + hi) / 2
        lo, hi = (m, hi) if math.erf(m) < y else (lo, m)
    return (lo + hi) / 2
eta0 = erfinv((0 - Ts) / (Ti_g - Ts))
for days in (30, 60, 100):
    t = days * 86400
    print(f"after {days:3d} days at the surface temperature: 0 C isotherm at {2*eta0*math.sqrt(alpha_s*t):.2f} m")

thermocouple: Bi = 0.0080, time constant 0.47 s; reaches 99 % of a step in 2.2 s
service pipe: C = 11.7 kJ/K, R' = 4.80 m.K/W, time constant 15.6 h, reaches 0 C after 33 h
wall midplane after  12 h: Fo = 0.331, T =   -0.7 C
wall midplane after  24 h: Fo = 0.662, T =  -11.5 C
wall midplane after  48 h: Fo = 1.325, T =  -20.8 C
wall midplane after  96 h: Fo = 2.650, T =  -24.6 C
  Bi = 5.36, zeta1 = 1.3278, C1 = 1.2433
after  30 days at the surface temperature: 0 C isotherm at 1.94 m
after  60 days at the surface temperature: 0 C isotherm at 2.75 m
after 100 days at the surface temperature: 0 C isotherm at 3.55 m

Spatial effects: the one-term solutions

When the Biot number exceeds about 0.1, the temperature inside the body varies significantly, and the heat equation must be solved in space and time. For a plane wall of thickness \(2L\), a long cylinder or a sphere, initially at \(T_i\) and suddenly exposed to fluid at \(T_\infty\) on all surfaces, the exact solutions are infinite series. For Fourier numbers above about 0.2, all terms but the first are negligible, and the midplane or center temperature is

where \(\zeta_1\) and \(C_1\) depend only on the Biot number, here defined with \(L\) (the half-thickness) or \(r_o\). For a plane wall, \(\zeta_1\) is the first root of \(\zeta\tan\zeta = \text{Bi}\) and \(C_1 = 4\sin\zeta_1/(2\zeta_1 + \sin 2\zeta_1)\); for a cylinder, \(\zeta_1 J_1(\zeta_1)/J_0(\zeta_1) = \text{Bi}\), with Bessel functions; for a sphere, \(1 - \zeta_1\cot\zeta_1 = \text{Bi}\). Tables of \(\zeta_1\) and \(C_1\) appear in every heat transfer text. The temperature at any position is the center value multiplied by a shape function, \(\cos(\zeta_1 x/L)\) for a wall, and the energy removed up to time \(t\) as a fraction of the maximum possible is also given in closed form.

The semi-infinite solid

A body that extends far enough from its surface that the far side never feels a surface change behaves as a semi-infinite solid. The ground beneath a surface, a thick wall early in a transient, and a metal part touched briefly by a hot tool all fit this model. For a sudden change of surface temperature from \(T_i\) to \(T_s\), the solution is

where erf is the Gauss error function, available in every calculator library. The penetration depth of a temperature change grows with \(\sqrt{\alpha t}\), so a disturbance that reaches 1 m in 30 days reaches only 2 m after 120 days. Related solutions exist for a constant surface heat flux and for convection at the surface, and for two semi-infinite bodies suddenly brought into contact, which explains why steel feels colder to the touch than wood at the same temperature: the contact temperature is weighted toward the body with the larger \(\sqrt{k\rho c}\).

US utilities in the northern states design buried services for frost depths set by local codes and field experience; Minnesota requirements are among the deepest in the country. Insulation boards above shallow lines are a common alternative to deep burial. In US units, the stopped service line's time constant of 15.6 hours is unchanged, and the 0 °C freeze point is 32 °F.