Steady conduction and thermal resistance networks

Northgate's expansion adds 1.8 km of new transmission main between the plant and Riverside General Hospital. Part of it runs through an existing utility tunnel and part is buried. The pipe will be pre-insulated: a steel carrier pipe, a thick layer of polyurethane foam, and a polyethylene casing, factory-bonded into one unit. Kofi needs the heat loss per metre for the tender documents, and Graham wants to know how hot the casing will run in the tunnel, where maintenance crews work. Both questions are answered by treating the pipe as a set of thermal resistances in series, the most useful single idea in heat transfer.

The heat equation and one-dimensional conduction

Fourier's law gives the flux when the temperature field is known. To find the field itself, an energy balance is applied to a small element of material. In Cartesian coordinates with constant conductivity, the result is the heat equation:

where \(\dot q\) is the volumetric generation rate (W/m³) and \(\alpha\) the thermal diffusivity (m²/s). This lesson deals with the simplest and most common case: steady state, no generation, and temperature varying in one direction only. In a plane wall the equation reduces to \(d^2T/dx^2 = 0\), so the temperature is linear through the wall. In a long cylinder, where heat flows radially, it becomes \(\frac{1}{r}\frac{d}{dr}\big(r\frac{dT}{dr}\big) = 0\), whose solution varies with \(\ln r\); in a sphere the temperature varies with \(1/r\).

Thermal resistance

In each of these cases the heat rate is proportional to the overall temperature difference, so we can define a thermal resistance by analogy with Ohm's law, with temperature difference playing the role of voltage and heat rate the role of current: \(q = \Delta T / R\). The standard resistances are:

Resistances in series add; resistances in parallel combine as reciprocals. Convection and radiation from the same surface to air and surroundings at the same temperature act in parallel, so their coefficients add. Once the network is drawn, the problem becomes arithmetic, and the relative sizes of the resistances show at once which layer controls the heat flow and where design effort should go.

For pipes it is convenient to work per metre of length, writing \(R'\) in m·K/W and \(q'\) in W/m. For walls, a resistance per unit area, \(R''\) in m²·K/W, is the familiar insulation RSI value. The overall heat transfer coefficient \(U\) is the inverse of the total resistance per unit area: \(q = UA\,\Delta T\), with \(UA = 1/R_{\text{tot}}\). For a cylinder, \(U\) must always be stated with the area it is based on, inside or outside, since the two differ.

The pre-insulated pipe

The new main uses a DN200 carrier, outside diameter 219.1 mm and wall 5.9 mm, inside a 315 mm polyethylene casing. Standards for these bonded pipe systems, such as EN 253 for directly buried hot-water networks, which is used in Canada and the United States as well as Europe, limit the aged conductivity of the foam; typical manufacturers' values are around 0.022 to 0.029 W/m·K at a mean foam temperature of 50 °C. We take 0.027 W/m·K as a conservative aged value. Water at 110 °C flows inside, and the tunnel air is at 20 °C.

# Thermal resistance networks: a pre-insulated district heating pipe in a utility tunnel
import math
def R_cyl(r1, r2, k, L=1.0):  return math.log(r2 / r1) / (2 * math.pi * k * L)
def R_conv(r, h, L=1.0):      return 1.0 / (h * 2 * math.pi * r * L)

# DN200 steel carrier pipe, polyurethane foam, polyethylene casing (per 1 m length of pipe)
r_i, r_o = 0.20730 / 2, 0.21910 / 2           # steel inside and outside radius (m)
r_f, r_c = 0.3052 / 2, 0.3150 / 2             # foam outside radius, casing outside radius (m)
k_st, k_pur, k_pe = 50.0, 0.027, 0.43         # W/m.K (foam value is a typical aged value)
h_i, h_o = 5000.0, 9.0                        # water side; tunnel air, convection plus radiation
T_w, T_air = 110.0, 20.0

parts = [("water film", R_conv(r_i, h_i)), ("steel wall", R_cyl(r_i, r_o, k_st)),
         ("PUR foam", R_cyl(r_o, r_f, k_pur)), ("PE casing", R_cyl(r_f, r_c, k_pe)),
         ("outside air", R_conv(r_c, h_o))]
R_tot = sum(R for _, R in parts)
q = (T_w - T_air) / R_tot
print(f"heat loss q' = {q:.1f} W/m   (total R' = {R_tot:.3f} m.K/W)")
T = T_w
for name, R in parts:
    print(f"  {name:12s} R' = {R:8.5f} m.K/W  ({R/R_tot:6.1%})   T after = {T - q*R:7.2f} C")
    T -= q * R
print(f"overall U based on casing area: {1/(R_tot*2*math.pi*r_c):.3f} W/m2.K")
print(f"loss over a 1.8 km supply line: {q*1800/1000:.1f} kW")

# Composite plane wall with a contact resistance: a boiler casing panel
layers = [("steel skin", 0.003, 45.0), ("mineral wool", 0.10, 0.045), ("aluminum jacket", 0.001, 200.0)]
Rc = 2e-4                                  # contact resistance between skin and wool, m2.K/W (assumed)
R_flat = 1/30.0 + sum(L/k for _, L, k in layers) + Rc + 1/10.0
print(f"\nboiler casing: U = {1/R_flat:.3f} W/m2.K, flux at 250 C inside, 24 C hall = {(250-24)/R_flat:.0f} W/m2")

# Critical radius: insulating a small tube and an electrical cable
for name, k, h, r0 in (("10 mm copper tube, elastomeric foam", 0.040, 8.0, 0.005),
                       ("cable with PVC jacket", 0.16, 10.0, 0.004)):
    r_cr = k / h
    print(f"{name}: r_cr = {r_cr*1000:.1f} mm (bare radius {r0*1000:.1f} mm)")
    for t in (0, 2, 5, 10, 20, 40):
        r2 = r0 + t / 1000
        R = (math.log(r2 / r0) / (2*math.pi*k) if t else 0) + 1/(h*2*math.pi*r2)
        print(f"    thickness {t:2d} mm -> q' = {50/R:6.2f} W/m for a 50 K difference")

heat loss q' = 43.3 W/m   (total R' = 2.078 m.K/W)
  water film   R' =  0.00031 m.K/W  (  0.0%)   T after =  109.99 C
  steel wall   R' =  0.00018 m.K/W  (  0.0%)   T after =  109.98 C
  PUR foam     R' =  1.95371 m.K/W  ( 94.0%)   T after =   25.37 C
  PE casing    R' =  0.01170 m.K/W  (  0.6%)   T after =   24.86 C
  outside air  R' =  0.11228 m.K/W  (  5.4%)   T after =   20.00 C
overall U based on casing area: 0.486 W/m2.K
loss over a 1.8 km supply line: 78.0 kW

boiler casing: U = 0.424 W/m2.K, flux at 250 C inside, 24 C hall = 96 W/m2
10 mm copper tube, elastomeric foam: r_cr = 5.0 mm (bare radius 5.0 mm)
    thickness  0 mm -> q' =  12.57 W/m for a 50 K difference
    thickness  2 mm -> q' =  11.96 W/m for a 50 K difference
    thickness  5 mm -> q' =  10.53 W/m for a 50 K difference
    thickness 10 mm -> q' =   8.78 W/m for a 50 K difference
    thickness 20 mm -> q' =   6.94 W/m for a 50 K difference
    thickness 40 mm -> q' =   5.44 W/m for a 50 K difference
cable with PVC jacket: r_cr = 16.0 mm (bare radius 4.0 mm)
    thickness  0 mm -> q' =  12.57 W/m for a 50 K difference
    thickness  2 mm -> q' =  16.36 W/m for a 50 K difference
    thickness  5 mm -> q' =  19.42 W/m for a 50 K difference
    thickness 10 mm -> q' =  20.98 W/m for a 50 K difference
    thickness 20 mm -> q' =  20.45 W/m for a 50 K difference
    thickness 40 mm -> q' =  18.20 W/m for a 50 K difference

The temperatures at each interface, printed by the program, show the same story: almost the whole 90 K drop occurs across the foam. This is typical of insulated systems, and it means that the outside coefficient, which is hard to know precisely in a tunnel, has only a modest effect on the loss. Doubling it to 18 W/m²·K would reduce the outside resistance from 0.112 to 0.056 m·K/W and raise the loss by less than 3 percent.

Composite walls and contact resistance

A boiler casing is a layered wall: an inner steel skin, mineral wool insulation and a thin aluminum outer jacket. Where two solid surfaces are pressed together, they touch only at the peaks of their roughness, and the gaps between are filled with air or a coating. The result is a thermal contact resistance \(R''_{t,c}\), typically 10⁻⁵ to 10⁻³ m²·K/W for metal joints, depending on roughness, pressure and any interstitial material. It matters most where the other resistances are small, such as electronic components bolted to heat sinks, and is usually negligible next to an insulation layer.

The critical radius of insulation

Adding insulation to a plane wall always reduces heat loss. On a cylinder it need not, because the insulation also increases the outside surface area. The conduction resistance rises with \(\ln(r_2/r_1)\) while the outside convection resistance falls as \(1/r_2\). Setting the derivative of the total resistance to zero shows that the heat loss is a maximum when the outside radius equals the critical radius

For an elastomeric foam with k = 0.04 W/m·K in still air (h ≈ 8 W/m²·K), the critical radius is 5 mm, so any insulation on a tube larger than 10 mm in diameter reduces the loss, and the effect is irrelevant for Northgate's pipes. For an electrical cable with a PVC jacket (k ≈ 0.16 W/m·K) and h ≈ 10 W/m²·K, the critical radius is 16 mm, larger than the conductor. Here the effect is useful: a jacket of the right thickness lets the cable shed more heat than a bare wire would, so it can carry more current for the same conductor temperature.

Choosing an insulation thickness

Beyond the critical radius, more insulation always helps, but with diminishing returns: the resistance of a cylindrical layer grows only with the logarithm of its outer radius. The economic thickness balances the cost of insulation, which grows faster than linearly with thickness, against the value of the heat saved over the system's life. Codes and standards set minimum thicknesses for building piping, and district energy suppliers offer pre-insulated pipe in two or three insulation series with progressively thicker foam. For a line operating year-round at 110 °C, the thicker series usually pays back, especially when heat is valued at its marginal cost in winter.

In Canada, piping insulation in buildings is governed by the National Energy Code of Canada for Buildings as adopted by each province, which tabulates minimum thicknesses by fluid temperature and pipe size, with adjustments for insulation conductivity. Buried district energy pipes are specified by the utility, usually by reference to European product standards and the manufacturer's insulation series. Manitoba Hydro and other utilities run programs that can support insulation upgrades in older plants.

The resistance concept also exposes a hidden problem: thermal bridges. A pipe support, a valve body or an anchor that passes through the insulation provides a parallel path of low resistance. In a building envelope, steel studs or slab edges do the same thing. Because parallel resistances combine as reciprocals, even a small area of bridge can carry a large share of the heat. Thermal imaging finds these bridges quickly, and Lesson 4 shows how to calculate them when the heat flow is no longer one-dimensional.