Conduction Heat Transfer

Abstract

This chapter covers the basic concepts and applications of conduction heat transfer. It starts with the introduction and definition of conduction heat transfer (Sect. 9.1). Fourier’s law of heat conduction and its application is explained in Sect. 9.2. Conduction through rectangular entities and R-values of insulating and building materials are described in Sect. 9.3. Sections 9.4 and 9.5 examine conduction through cylindrical and spherical walls, respectively.

Appendices

Practice Problems

Practice Problem 9.1

Use the data in Example 9.1, and assume constant, steady-state heat flow through the system. The temperature of the inside surface of the insulation is 680 °F. Determine in inches the thickness of fiberglass insulation (k = 0.023 Btu / hr – ft –°F) required to keep the temperature of the outside surface of the insulation below 70 °F.

Practice Problem 9.2

A window in a home consists of a double glass pane each with a thickness of ¼ in. The air gap between the window panes is (1/16) in. The thermal conductivity of glass is 0.61 Btu / hr – ft – °F and that of air is 0.015 Btu / hr – ft – °F. Compare the heat loss through the double-pane glass window with the heat loss from a single glass pane window of ½ in thickness, and calculate the percentage reduction in heat loss due to the use of the double-pane window.

Practice Problem 9.3

The insulation surface in Example 9.3 can be a source of heat hazard at 150 °F. Calculate the thickness of an additional layer of calcium silicate insulation (k = 0.035 Btu / hr – ft – °F) required to limit the surface temperature to 80 °F.

Practice Problem 9.4

The heat transfer to liquid ammonia in Example 9.4 will cause the heating of liquid ammonia to its boiling point −33 °C and subsequently vaporization of liquid ammonia. The relevant physical properties of liquid ammonia are density = 686 kg/m³; specific heat = 4.744 kJ/kg.K; and enthalpy of vaporization = 1369 kJ/kg. Determine the time required in hours for liquid ammonia to reach its boiling point.

Solutions to Practice Problems

Practice Problem 9.1

• (Solution)

Under steady-state conditions, the heat flow through the insulation layer will be the same as the heat flow through the insulating brick. From Example 9.1, the heat flow per unit area through the insulating brick is 110.34 Btu / hr – ft² – °F. Calculate the required thermal resistance per unit area of the fiberglass insulation by using Eq. 9.5.

$$ {R}_{\mathrm{cond},\mathrm{ua}}=\frac{\Delta T}{q/A}=\frac{680\kern0.5em {}^{\circ}\mathrm{F}-70\kern0.5em {}^{\circ}\mathrm{F}}{110.34\ \frac{\mathrm{Btu}}{\mathrm{hr}-{\mathrm{ft}}^2}}=5.53\ \mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}/\mathrm{Btu} $$

Calculate the required thickness of fiberglass insulation by using Eq. 9.4.

$$ {\displaystyle \begin{array}{l}\Delta X=\left({R}_{\mathrm{cond},\mathrm{ua}}\right)(k)=\left(5.53\frac{\mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}}{\mathrm{Btu}}\right)\left(0.023\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}-{}^{\circ}\mathrm{F}}\right)\\ {}\kern7.8em =0.1272\ \mathrm{ft}\end{array}} $$

Convert the insulation thickness to inches.

$$ \Delta X=\left(0.1272\ \mathrm{ft}\right)\left(\frac{12\ \mathrm{in}}{\mathrm{ft}}\right)=1.53\ \mathrm{in} $$

Practice Problem 9.2

• (Solution)

The thermal circuits for the double-pane and single-pane windows are shown in the figure. The overall temperature will be the same in both cases, and any convenient value, say, 20 °F, can be taken for calculation purposes. The air is treated as a conductive layer.

Convert all thicknesses from inch to feet.

$$ \Delta {X}_{\mathrm{P}1}=\Delta {X}_{\mathrm{P}2}=\left(0.25\ \mathrm{in}\right)\left(\frac{1\ \mathrm{ft}}{12\ \mathrm{in}}\right)=0.0208\ \mathrm{ft} $$

$$ \Delta {X}_{\mathrm{Air}}=\left(0.0625\ \mathrm{in}\right)\left(\frac{1\ \mathrm{ft}}{12\ \mathrm{in}}\right)=0.0052\ \mathrm{ft} $$

$$ \Delta {X}_{\mathrm{SP}}=\left(0.50\ \mathrm{in}\right)\left(\frac{1\ \mathrm{ft}}{12\ \mathrm{in}}\right)=0.0417\ \mathrm{ft} $$

Calculate the heat loss per unit area for the double-pane window by using Eq. 9.6. Include conduction resistance of air.

$$ {\displaystyle \begin{array}{l}\frac{q}{A}=\frac{\Delta {T}_{\mathrm{overall}}}{\sum {R}_{\mathrm{cond},\mathrm{ua}}}=\frac{\Delta {T}_{\mathrm{overall}}}{\frac{\Delta {X}_{\mathrm{P}1}}{k_{\mathrm{G}}}+\frac{\Delta {X}_{\mathrm{Air}}}{k_{\mathrm{Air}}}+\frac{\Delta {X}_{\mathrm{P}2}}{k_{\mathrm{G}}}}\\ {}\kern0.8em =\frac{20\kern0.5em {}^{\circ}\mathrm{F}}{2\left(\frac{0.0208\ \mathrm{ft}}{0.61\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}-{}^{\circ}\mathrm{F}}}\right)+\left(\frac{0.0052\ \mathrm{ft}}{0.015\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}-{}^{\circ}\mathrm{F}}}\right)}\\ {}\kern0.8em =48.21\ \mathrm{Btu}/\mathrm{hr}-{\mathrm{ft}}^2\end{array}} $$

Calculate the heat loss per unit area for the single-pane window by using Eq. 9.6.

$$ {\displaystyle \begin{array}{l}\frac{q}{A}=\frac{\Delta {T}_{\mathrm{overall}}}{R_{\mathrm{SP},\mathrm{ua}}}=\frac{\Delta {T}_{\mathrm{overall}}}{\frac{\Delta {X}_{\mathrm{SP}}}{k_{\mathrm{G}}}}\\ {}\kern5.45em =\frac{20\kern0.5em {}^{\circ}\mathrm{F}}{\left(\frac{0.0417\ \mathrm{ft}}{0.61\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}-{}^{\circ}\mathrm{F}}}\right)}\\ {}\kern5.35em =292.57\ \mathrm{Btu}/\mathrm{hr}-{\mathrm{ft}}^2\end{array}} $$

Calculate the percentage reduction in heat loss as shown.

$$ {\displaystyle \begin{array}{l}\%\kern0.5em \mathrm{reduction}\ \mathrm{in}\ \mathrm{heat}\ \mathrm{loss}=\left(\frac{{\left(q/A\right)}_{\mathrm{SP}}-{\left(q/A\right)}_{\mathrm{DP}}}{{\left(q/A\right)}_{\mathrm{SP}}}\right)\times 100\\ {}\kern10.3em =\left(\frac{292.57\frac{\mathrm{Btu}}{\mathrm{hr}-{\mathrm{ft}}^2}-48.21\frac{\mathrm{Btu}}{\mathrm{hr}-{\mathrm{ft}}^2}}{292.57\frac{\mathrm{Btu}}{\mathrm{hr}-{\mathrm{ft}}^2}}\right)\times 100\\ {}\kern10.3em =83.52\%\end{array}} $$

• Conclusion: 83% reduction in heat loss is achieved by using a double-pane window, and the air gap makes a significant contribution to the thermal resistance of the system.

Practice Problem 9.3

• (Solution)

At steady state, the heat flow through the different layers will be constant at 73.37 Btu / hr – ft (calculated in Example 9.3). Let R_cs,ul be the thermal resistance of the calcium silicate insulation per unit length of the pipe. From Eq. 9.12 and using the results from Example 9.3:

$$ {\displaystyle \begin{array}{l}\frac{q_{\mathrm{cyl}}}{L}=73.37\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}}=\frac{\Delta {T}_{\mathrm{overall}}}{R_{\mathrm{total},\mathrm{ul}}}=\frac{\Delta {T}_{\mathrm{overall}}}{R_{\mathrm{pipe},\mathrm{ul}}+{R}_{\mathrm{ins}1,\mathrm{ul}}+{R}_{\mathrm{cs},\mathrm{ul}}}\\ {}\kern1.3em =\frac{400^{\circ}\mathrm{F}-{80}^{\circ}\mathrm{F}}{3.4075\frac{\mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}}{\mathrm{Btu}-\mathrm{ft}\ \mathrm{length}}+{R}_{\mathrm{cs},\mathrm{ul}}}\end{array}} $$

Solve the preceding equation for the thermal resistance of calcium silicate insulation.

$$ {\displaystyle \begin{array}{l}\left(73.37\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}}\right)\left(3.4075\frac{\mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}}{\mathrm{Btu}-\mathrm{ft}\ \mathrm{length}}\right)\\ {}\kern4.5em +{R}_{\mathrm{cs},\mathrm{ul}}\left(73.37\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}}\right)=320\kern0.5em {}^{\circ}\mathrm{F}\end{array}} $$

$$ {R}_{\mathrm{cs},\mathrm{ul}}=0.9540\ \mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}/\mathrm{Btu}-\mathrm{ft}\ \mathrm{length} $$

Let r₄ be the outer radius with the calcium silicate insulation in place. Calculate r₄ by using Eq. 9.13.

$$ {R}_{\mathrm{cs},\mathrm{ul}}=0.9540\frac{\mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}}{\mathrm{Btu}-\mathrm{ft}\ \mathrm{length}}=\frac{\ln \left(\frac{r_4}{r_3}\right)}{2\pi {k}_{\mathrm{cs}}} $$

$$ \ln \left(\frac{r_4}{5.315\ \mathrm{in}}\right)=\left(0.9540\frac{\mathrm{hr}-{\mathrm{ft}}^2-{}^{\circ}\mathrm{F}}{\mathrm{Btu}-\mathrm{ft}\ \mathrm{length}}\right)\ \left(2\pi \right)\left(0.035\frac{\mathrm{Btu}}{\mathrm{hr}-\mathrm{ft}-{}^{\circ}\mathrm{F}}\right) $$

Solve the preceding equation for r₄, r₄ = 6.556 in

Therefore, the thickness of calcium silicate insulation required is:

$$ {\displaystyle \begin{array}{l}\Delta {r}_{\mathrm{cs}}={r}_4-{r}_3=6.556\ \mathrm{in}-5.315\ \mathrm{in}\\ {}\kern4.75em =1.241\ \mathrm{in}\end{array}} $$

Practice Problem 9.4

• (Solution)

Calculate the volume of the spherical tank in Example 9.4.

$$ {\displaystyle \begin{array}{l}V=\left(\frac{4}{3}\right)\pi {r_1}^3=\left(\frac{4}{3}\right)\left(\pi \right){\left(0.89\ \mathrm{m}\right)}^3\\ {}\kern5.7em =2.953\ {\mathrm{m}}^3\end{array}} $$

Calculate the mass of liquid ammonia in the tank using the definition of density (mass per unit volume).

$$ {\displaystyle \begin{array}{l}m=\rho \times V=\left(686\frac{\mathrm{kg}}{{\mathrm{m}}^3}\right)\left(2.593\ {\mathrm{m}}^3\right)\\ {}\kern4.75em =1779\ \mathrm{kg}\end{array}} $$

Calculate the heat required to raise the temperature of liquid ammonia to its boiling point.

$$ {\displaystyle \begin{array}{l}Q={mc}_{pl}\left({T}_b-T\right)\\ {}\kern1.1em =\left(1779\ \mathrm{kg}\right)\left(4.744\frac{\mathrm{kJ}}{\mathrm{kg}\cdot \mathrm{K}}\right)\left(-33\kern0.5em {}^{\circ}\mathrm{C}-\left(-60\kern0.5em {}^{\circ}\mathrm{C}\right)\right)\\ {}\kern1.1em =227869\ \mathrm{kJ}\ \left(2.279\times {10}^8\ \mathrm{J}\right)\end{array}} $$

The time required to reach the boiling point is the heat required divided by the rate of heat transfer.

$$ {\displaystyle \begin{array}{l}{t}_{bp}=\frac{Q}{q}=\frac{2.279\times {10}^8\mathrm{J}}{100\ \mathrm{W}}\\ {}\kern1.4em =2.279\times {10}^6\ \mathrm{s}\ \left(633\ \mathrm{hrs}\right)\end{array}} $$