Heat Conduction through
Composite Cylindrical Walls

Example 9.6-1, p.286, Transport Phenomena, Bird, Stewart, and Lightfoot

by David Bennett Smith, Apr. 25, 1999

BS&L Example 9.6-1

Double-click on the cell brackets at right to open the following subsections, or press Control + a and then Shift + Control + { to open all subgroups.

Problem Description

In industry, the most common method used to transport fluids between units and around the plant is to pump the fluid through a cylindrical pipe. In most cases, it is desired to keep the fluid at some temperature either above or below the ambient air temperature. Hence, to minimize heat flux into or out of the pipe, the pipe is covered with insulation. The insulation may be composed of several different layers, each layer having a different thicknesses and a different thermal conductivity. Given (a) the temperature of the fluid inside the pipe, (b) the ambient temperature outside, (c) information about the thickness and thermal conductivity of each layer of insulation, and (d) the heat transfer coefficients at the fluid-pipe and insulation-air interfaces, derive an expression for the heat flux per unit length of pipe across the pipe.

Diagram of Problem (BS&L Fig. 9.6-2)

The cylinder is of length

The Shell Energy Balance

In its most general form, the balance equation over some control volume CV states that for every transportable quantity Ω, the rate of accumulation of Ω inside CV equals the net rate of flux of Ω into CV plus the rate of formation of Ω inside CV. In this problem, the quantity being transported is heat energy, no heat is being generated anywhere throughout the tube, and at steady-state there is no heat accumulation. Thus, we have

\(TraditionalForm
      \`E \((accumulation)\) = E(in) - E(out) + E(formed)\)
E (in) - E(out) = 0

The heat flux q measures the per unit time per unit area flow of heat. Therefore, taking the control volume as the radial shell from radius r to r + d r,

\(TraditionalForm
      \`q \((r)\)\ \(A(r)\) - 
          \(q(r + d r)\)\ \(A(r + d r)\) = 0
        \)
\(TraditionalForm
      \`q \((r)\)\ 2  π\ r\ L - 
          \(q(r + d r)\)\ 2 \( π(r + d r)\) 
            L = 0\)

Solving the Energy Balance

Dividing the shell balance equation through by 2  π L d r (the volume of this shell),

-(d/(d r)) (r q) = 0

Integrating with respect to r from (Subscript[r, 0]) to r, and using the boundary condition that q = (Subscript[q, 0]) at r = (Subscript[r, 0]),

(Subscript[(∫_r), 0]^r) d(r q) = 0
r q(r) = (Subscript[r, 0]) (Subscript[q, 0])

Applying Fouier's Law,

\(TraditionalForm
    \`r(\(-k\) \(d T\)\/\(d r\)) = 
      r\_0\ q\_0\)

Separating variables and integrating,

\(TraditionalForm
      \`\(-k\)\ d T = 
        r\_0\ \(q\_0\) \(d r\)\/r\)
\(TraditionalForm
      \`\(-k\^i\) \(∫\_T\_\(i - 1\)\%T\_i d T\) = 
        r\_0\ \(q\_0\) 
          \(∫\_r\_\(i - 1\)\%r\_i\( d r\)\/r\)\)
\(TraditionalForm
      \`\(k\^i\)(T\_\(i - 1\) - T\_i) = 
        r\_0\ q\_0\ \(ln(r\_i/r\_\(i - 1\))\)\)
\(TraditionalForm
      \`T\_\(i - 1\) - T\_i = 
        \(r\_0\ q\_0\)\/k\^i\ \(ln(r\_i\/r\_\(i - 1\))\)\)

Note that the thermal conductivity of the (i^th) layer of insulation (the region from r = (Subscript[r, i - 1]) to r = (Subscript[r, i])) is denoted as (k^i). Finally, Newton's Law of Cooling defines the heat-transfer coefficient at the two pipe-fluid interfaces: q = h Δ T, where q and Δ T have the same sign. Therefore,

(Subscript[T, a]) - (Subscript[T, 0]) = (Subscript[q, 0])/(Subscript[h, 0])
\(TraditionalForm
      \`T\_n - T\_b = 
        \(q\_n\/h\_n = \(1\/h\_n\) \((\(\(r\_0\) q\_0\)\/r\_n)\)\)\)

Calculating (Subscript[Q, 0]) = (Subscript[q, 0]) (Subscript[A, 0])

Adding the above three equations,

\(TraditionalForm
      \`\((T\_a - T\_0)\) + ∑\_\(i = 1\)\%n\((T\_\(i - 1\) - T\_i)\) + 
          \((T\_n - T\_b)\) = 
        \(\(r\_0\) q\_0\)\/\(\(r\_0\) h\_0\) + 
          ∑\_\(i = 1\)\%n\( r\_0\ q\_0\)\/k\^i\ 
              \(ln(r\_i\/r\_\(i - 1\))\) + 
          \(\(r\_0\) q\_0\)\/\(\(r\_n\) h\_n\)\)
\(TraditionalForm
      \`T\_a - T\_b = 
        \(r\_0\) 
          \(\(q\_0\)(
            1\/\(\(r\_0\) h\_0\) + 
              ∑\_\(i = 1\)\%n 1\/k\^i\ \(ln(r\_i\/r\_\(i - 1\))\) + 
              1\/\(\(r\_n\) h\_n\))\)\)
\(TraditionalForm
      \`\(r\_0\) q\_0 = 
        \((T\_a - T\_b)\) 
          \((1\/\(\(r\_0\) h\_0\) + 
                ∑\_\(i = 1\)\%n 1\/k\^i\ \(ln(r\_i\/r\_\(i - 1\))\) + 
                1\/\(\(r\_n\) h\_n\))\)\^\(-1\)\)

Defining the total heat flow through the pipe as (Subscript[Q, 0]) = (Subscript[q, 0]) (Subscript[A, 0]),

(Subscript[Q, 0]) = 2  π (Subscript[r, 0]) L (Subscript[q, 0])
\(TraditionalForm
      \`Q\_0 = 2  π\ \(L(T\_a - T\_b)\) 
          \((1\/\(\(r\_0\) h\_0\) + 
                ∑\_\(i = 1\)\%n 1\/k\^i\ \(ln(r\_i\/r\_\(i - 1\))\) + 
                1\/\(\(r\_n\) h\_n\))\)\^\(-1\)\)