Radiation exchange between surfaces

Lesson 10 described how single surfaces emit and absorb. Engineering problems involve several surfaces that see each other: the burner floor, water-tube walls and refractory of a furnace; a pipe inside a protective sleeve; a thermocouple inside a hot duct; the panes of a window. How much radiation leaving one surface reaches another depends on their geometry, and how much of it is absorbed depends on their properties. This lesson develops the view factor, which captures the geometry, and the radiation network method, which combines geometry and properties for gray, diffuse surfaces. It finishes with radiation shields, the thermocouple error in hot gas, and a brief look at radiating gases.

The view factor

The view factor \(F_{ij}\) is the fraction of the radiation leaving surface \(i\) that strikes surface \(j\) directly. It depends only on geometry, assuming diffuse emission and reflection. For two small areas, it follows from the cosine law of diffuse emission; for finite surfaces it is an integral over both areas, evaluated once and tabulated or given by formulas for common shapes. Three rules make view factors much easier to handle in practice.

The reciprocity relation, \(A_iF_{ij} = A_jF_{ji}\), lets one view factor be found from another. The summation rule, \(\sum_j F_{ij} = 1\) over all surfaces of an enclosure, follows because all radiation leaving a surface must go somewhere. A flat or convex surface cannot see itself, so \(F_{ii} = 0\); a concave surface can, so \(F_{ii} > 0\). With these rules, a few known view factors often give all the others. For example, a small convex object fully enclosed by a large surface has \(F_{12} = 1\), and reciprocity then gives \(F_{21} = A_1/A_2\).

For long two-dimensional geometries, such as ducts and channels, Hottel's crossed-strings method gives view factors from lengths alone: imagine strings stretched between the edges of the two surfaces, and \(F_{12} = (\text{sum of crossed strings} - \text{sum of uncrossed strings})/(2 \times \text{length of surface 1})\). For three-dimensional shapes, formulas exist for parallel coaxial disks, aligned parallel rectangles, perpendicular rectangles with a common edge, concentric cylinders and spheres and many more, and charts present the most useful.

Blackbody exchange

If all surfaces are black, radiation leaving surface \(i\) is simply its emission, and the net exchange between two surfaces is \(q_{ij} = A_iF_{ij}\sigma(T_i^4 - T_j^4)\). In an enclosure of black surfaces, each surface's net rate is the sum of its exchanges with all the others. Black surfaces are a useful limit and a good model for furnace walls covered in soot or deposits.

Gray, diffuse surfaces and radiosity

Real surfaces reflect part of what arrives, so radiation leaving a surface consists of its emission plus reflected irradiation. The total is the radiosity, \(J = \varepsilon E_b + \rho G\). For an opaque, gray, diffuse surface, with \(\rho = 1 - \varepsilon\), the net radiation leaving the surface can be written in two ways: as emission minus absorption, or as radiosity minus irradiation. Eliminating \(G\) gives

The first equation describes a surface resistance between the blackbody potential and the radiosity; the second a space resistance between the radiosities of two surfaces. Together they turn any enclosure of gray, diffuse, isothermal surfaces into a network that can be solved like an electrical circuit. A black surface has zero surface resistance, so its radiosity equals \(E_b\). An insulated surface that loses no net heat by conduction or convection, such as refractory with good insulation behind it, is called reradiating: its net radiation is zero, so its radiosity floats at whatever value the network gives, and it has no surface resistance in the circuit. Reradiating surfaces matter greatly in furnaces because they redirect radiation toward the heated surface.

Two-surface enclosures

When two surfaces form an enclosure, the network reduces to three resistances in series, giving

For infinite parallel plates this becomes \(q'' = \sigma(T_1^4 - T_2^4)/(1/\varepsilon_1 + 1/\varepsilon_2 - 1)\). For long concentric cylinders or concentric spheres, with \(F_{12} = 1\), it becomes \(q = \sigma A_1(T_1^4 - T_2^4)/[1/\varepsilon_1 + (1 - \varepsilon_2)/\varepsilon_2 \cdot (r_1/r_2)^n]\), with \(n = 1\) for cylinders and 2 for spheres. For a small object in a large enclosure, \(A_1/A_2 \to 0\), and the result reduces to \(\varepsilon_1\sigma A_1(T_1^4 - T_2^4)\), the expression used since Lesson 1.

# Radiation exchange: view factors, two-surface enclosures, shields, a thermocouple in flue gas and a radiosity solver
import math
sigma = 5.670e-8

# 1. View factor between coaxial parallel disks (radius r1 to r2, spacing L)
def F_disks(r1, r2, L):
    R1, R2 = r1 / L, r2 / L
    S = 1 + (1 + R2**2) / R1**2
    return 0.5 * (S - math.sqrt(S**2 - 4 * (R2 / R1)**2))
print(f"disk to disk, both 0.5 m radius, 1 m apart: F12 = {F_disks(0.5, 0.5, 1.0):.3f}; 0.25 m apart: {F_disks(0.5, 0.5, 0.25):.3f}")

# 2. Crossed strings: a long duct 0.4 m wide and 0.3 m deep; floor to one side wall and floor to roof
w, d = 0.4, 0.3
diag = math.hypot(w, d)
F_floor_side = (w + d - diag) / (2 * w)
F_floor_roof = (2 * diag - 2 * d) / (2 * w)
print(f"long duct: F(floor to one side) = {F_floor_side:.3f}, F(floor to roof) = {F_floor_roof:.3f}, sum with two sides = {2*F_floor_side + F_floor_roof:.3f}")

# 3. Two-surface enclosure: a 175 C steam pipe (OD 219 mm) inside a sleeve (ID 300 mm) at 60 C, evacuated or still air ignored
def q_concentric(T1, T2, D1, D2, e1, e2):
    return sigma * math.pi * D1 * (T1**4 - T2**4) / (1 / e1 + (1 - e2) / e2 * D1 / D2)
T1, T2 = 175 + 273.15, 60 + 273.15
for lab, e1, e2 in (("oxidized steel both", 0.80, 0.80), ("pipe wrapped in aluminum foil", 0.05, 0.80), ("both foil", 0.05, 0.05)):
    print(f"pipe in sleeve, {lab:30s}: radiation {q_concentric(T1, T2, 0.219, 0.300, e1, e2):6.1f} W/m")

# 4. Radiation shields between large parallel plates (all emissivities 0.8, shields 0.05 on both faces)
T1, T2 = 600.0, 300.0
q0 = sigma * (T1**4 - T2**4) / (1/0.8 + 1/0.8 - 1)
for N in (0, 1, 2, 4):
    R = (1/0.8 + 1/0.8 - 1) + N * (1/0.05 + 1/0.05 - 1)
    print(f"{N} shields: q'' = {sigma*(T1**4 - T2**4)/R:7.1f} W/m2 ({sigma*(T1**4 - T2**4)/R/q0:5.1%} of unshielded)")

# 5. Thermocouple in a flue duct: true gas temperature from the reading (energy balance on the junction)
h, eps_tc, T_read, T_wall = 80.0, 0.6, 500.0, 400.0
Tg = T_read + eps_tc * sigma * (T_read**4 - T_wall**4) / h
print(f"thermocouple reads {T_read:.0f} K with walls at {T_wall:.0f} K: true gas about {Tg:.0f} K (error {Tg - T_read:.0f} K)")
# with a thin cylindrical radiation shield (emissivity 0.3) around it, the shield sits between gas and wall
eps_s, h_s = 0.3, 80.0
Ts = T_read
for _ in range(200):                     # junction: h(Tg - Tj) = eps_tc sigma (Tj^4 - Ts^4), shield: 2h(Tg - Ts) = eps_s sigma (Ts^4 - Tw^4) (+ junction term ignored)
    Ts = Tg - eps_s * sigma * (Ts**4 - T_wall**4) / (2 * h_s)
Tj = Tg
for _ in range(200):
    Tj = Tg - eps_tc * sigma * (Tj**4 - Ts**4) / h
print(f"  with a shield the junction reads {Tj:.0f} K (error {Tg - Tj:.0f} K); shield at {Ts:.0f} K")

# 6. Radiosity method for a long furnace of equilateral triangular section (each view factor 0.5)
#    surface 1: burner floor 1500 K, eps 0.9; surface 2: water-tube wall 600 K, eps 0.8; surface 3: refractory, reradiating
row1, row2, row3 = [0, .5, .5], [.5, 0, .5], [.5, .5, 0]
F = [row1, row2, row3]
Tsurf, eps, A = [1500.0, 600.0, None], [0.9, 0.8, None], [1.0, 1.0, 1.0]     # areas per 1 m of length (m2)
# unknowns J1, J2, J3; equations: (Eb_i - J_i) eps_i/(1-eps_i) = sum_j F_ij (J_i - J_j) for 1, 2; sum_j F_3j (J3 - Jj) = 0
Eb = [sigma * T**4 if T else 0 for T in Tsurf]
M, b = [list([0.0] * 3) for _ in range(3)], [0.0] * 3
for i in range(2):
    c = eps[i] / (1 - eps[i])
    M[i][i] = c + sum(F[i]); b[i] = c * Eb[i]
    for j in range(3):
        if j != i: M[i][j] -= F[i][j]
M[2] = [-F[2][0], -F[2][1], F[2][0] + F[2][1]]
for col in range(3):                                   # Gaussian elimination
    for r in range(col + 1, 3):
        f = M[r][col] / M[col][col]
        M[r] = [a - f*bb for a, bb in zip(M[r], M[col])]; b[r] -= f * b[col]
J = [0.0]*3
for r in (2, 1, 0):
    J[r] = (b[r] - sum(M[r][c]*J[c] for c in range(r+1, 3))) / M[r][r]
q1 = (Eb[0] - J[0]) * eps[0] / (1 - eps[0]) * A[0]
T3 = (J[2] / sigma) ** 0.25
print(f"furnace: floor delivers {q1/1000:.0f} kW/m, tube wall receives {-(Eb[1]-J[1])*eps[1]/(1-eps[1])/1000:.0f} kW/m, refractory settles at {T3:.0f} K")
q_no_refr = sigma*(1500**4 - 600**4) / ((1-0.9)/0.9 + 1/0.5 + (1-0.8)/0.8)
print(f"  if the third wall were open to cold surroundings instead, floor-to-tube exchange alone would be {q_no_refr/1000:.0f} kW/m")

disk to disk, both 0.5 m radius, 1 m apart: F12 = 0.172; 0.25 m apart: 0.610
long duct: F(floor to one side) = 0.250, F(floor to roof) = 0.500, sum with two sides = 1.000
pipe in sleeve, oxidized steel both           : radiation  763.0 W/m
pipe in sleeve, pipe wrapped in aluminum foil : radiation   54.2 W/m
pipe in sleeve, both foil                     : radiation   32.3 W/m
0 shields: q'' =  4592.7 W/m2 (100.0% of unshielded)
1 shields: q'' =   170.1 W/m2 ( 3.7% of unshielded)
2 shields: q'' =    86.7 W/m2 ( 1.9% of unshielded)
4 shields: q'' =    43.7 W/m2 ( 1.0% of unshielded)
thermocouple reads 500 K with walls at 400 K: true gas about 516 K (error 16 K)
  with a shield the junction reads 515 K (error 1 K); shield at 511 K
furnace: floor delivers 165 kW/m, tube wall receives 165 kW/m, refractory settles at 1293 K
  if the third wall were open to cold surroundings instead, floor-to-tube exchange alone would be 118 kW/m

Radiation shields

A radiation shield is a thin sheet of low-emissivity material placed between two surfaces. It adds two surface resistances and one space resistance to the network without conducting heat significantly. Between large parallel plates, each shield with emissivity \(\varepsilon_s\) on both faces adds \(2/\varepsilon_s - 1\) to the denominator. With plates of emissivity 0.8 at 600 K and 300 K, one shield of emissivity 0.05 reduces the flux from 4.59 kW/m² to 170 W/m², under 4 percent of the unshielded value; four shields reduce it to 1 percent. Multilayer insulation on cryogenic tanks and spacecraft, made of dozens of aluminized films in a vacuum, applies this principle.

Radiation error in gas temperature measurement

A thermocouple inserted in a hot gas flowing through a duct with cooler walls gains heat by convection from the gas and loses it by radiation to the walls. At steady state its junction settles below the true gas temperature. An energy balance on the junction gives

A furnace with a reradiating wall

For enclosures of three or more surfaces, the network equations are written at each radiosity node and solved together: at a surface with known temperature, \((E_{b,i} - J_i)/R_{\text{surf},i} = \sum_j (J_i - J_j)A_iF_{ij}\); at a reradiating surface, the right side equals zero. The result is a set of linear equations in the radiosities.

Radiating gases

The gases in a furnace are not transparent. Carbon dioxide and water vapour, the main products of combustion, absorb and emit strongly in certain infrared bands, and soot particles in luminous flames emit across the whole spectrum. Air, nitrogen, oxygen and hydrogen are essentially transparent at ordinary temperatures. Gas radiation depends on the gas temperature, the partial pressures of the radiating species and the path length through the gas, expressed through a mean beam length for the enclosure's shape. Hottel's charts give the emissivities of carbon dioxide and water vapour as functions of temperature and the product of partial pressure and path length, with a correction for their overlapping bands. In a large utility boiler furnace, gas and soot radiation deliver most of the heat absorbed by the water walls; in a small duct, gas radiation is often negligible.

Canadian boiler and fired-heater design follows the same methods used throughout North America, and furnace performance checks during commissioning are carried out under the provincial boiler safety authority's requirements. For Northgate, Kofi's furnace model is a check on the boiler manufacturer's guaranteed outlet temperature rather than a design calculation, since the radiant section is supplied as a packaged unit.