Heat Transfer: free practice, theory and problems

1. Conduction

What is it about?

Conduction is the transport of heat through a substance without the substance itself moving – the heat "hops" from molecule to molecule. This is how heat passes through a wall, a pipe or a cooling fin. For an engineer this is about sizing insulation, calculating heat loss from buildings and pipelines, and understanding why some materials feel colder to the touch than others even at the same temperature.

Concepts and formulas

Fourier's law for a one-dimensional, plane layer is

Heat flows from high to low temperature, hence the minus sign.

• The thermal conductivity $k$ has units W/(m·K). Do not confuse it with the heat transfer coefficient $h$ (W/(m²·K)) or the specific heat capacity $c$ (J/(kg·K)).
• Thermal resistance of a plane layer: $R = L/(kA)$ (K/W), where $L$ is the thickness and $A$ the area perpendicular to the heat flow.
• Layers in series: $R_{tot} = \sum R_i = \sum L_i/(k_iA)$, exactly like electrical resistors in series. The heat flow $\dot Q$ is the same through every layer in steady state.
• Thermal resistance of a cylindrical layer (e.g. insulation on a pipe), with inner radius $r_1$ and outer radius $r_2$, length $L$: $$R = \dfrac{\ln(r_2/r_1)}{2\pi kL}$$
• Heat flow: $\dot Q = \Delta T/R_{tot}$ (W). The temperature drop is distributed in proportion to the resistance of each layer.
• U-value: $U = 1/(R_{tot}A)$, the heat loss per m² and degree of temperature difference (W/(m²·K)). A low U-value means good insulation.
• A thermal bridge is a region with much lower thermal resistance than the rest of the construction (e.g. a steel beam through the insulation), causing a large local heat loss.

How to solve the problems

1. Sketch the layers the heat passes through, and find $L$, $k$ and $A$ (or $r_1$, $r_2$, $L$ for a pipe) for each of them.
2. Compute the thermal resistance of each layer, using $R = L/(kA)$ for a plane layer or $R = \ln(r_2/r_1)/(2\pi kL)$ for a cylindrical one.
3. Add the resistances in series: $R_{tot} = \sum R_i$.
4. Find the heat flow $\dot Q = \Delta T/R_{tot}$, where $\Delta T$ is the total temperature drop.
5. If you need a temperature inside the construction, use the fact that the temperature drop is proportional to the resistance up to that point.

Example

A pipe with an inner radius of 40 mm is insulated with 30 mm of insulation ($k = 0.045$ W/(m·K)) over a length of 8 m. The inner surface of the insulation is at 120 °C and the outer surface at 20 °C. What is the heat loss?

1. Outer radius: $r_2 = 40+30 = 70$ mm.
2. $R = \dfrac{\ln(r_2/r_1)}{2\pi kL} = \dfrac{\ln(70/40)}{2\pi\cdot 0.045\cdot 8} \approx \dfrac{0.5596}{2.262} \approx 0.2474$ K/W.
3. $\dot Q = \Delta T/R = (120-20)/0.2474 \approx 404$ W.

Answer: about 404 W.

Common mistakes

• Using the plane-wall formula $R = L/(kA)$ for a pipe – cylindrical layers require the $\ln(r_2/r_1)$ formula.
• Confusing $k$ (thermal conductivity) with $h$ (heat transfer coefficient) – they have different units and describe completely different phenomena.
• Forgetting that $\dot Q$ is the same through every layer in series at steady state, even though the temperature drop differs between them.
• Using the wrong area for a layer that is not plane (e.g. using the inner area for the whole pipe).
• Mixing mm and m in the same formula.

Concepts in this part

Practise conduction in the app →

2. Convection

What is it about?

Convection is heat transfer between a surface and a fluid (air, water, oil …) flowing past it. This is how a radiator heats a room, a cooler removes heat from an engine, and the wind makes you feel colder faster. Convection is more complicated than conduction because it depends on the flow: whether the fluid is driven by a fan or pump (forced convection) or by buoyancy from temperature differences (natural convection), and whether the flow is laminar or turbulent. As an engineer you need convection to size cooling systems, calculate heat loss from buildings, and decide whether a body can be treated as uniformly heated.

Concepts and formulas

Newton's law of cooling is

where $h$ is the heat transfer coefficient (W/(m²·K)), $T_s$ the surface temperature and $T_\infty$ the fluid temperature far away.

• Thermal resistance for convection: $R_{conv} = 1/(hA)$ (K/W) – added in series with conduction resistances exactly as before.
• Typical $h$ values: natural convection in air 2–25, forced convection in air 10–200, forced convection in water 50–10,000, boiling/condensation several thousand up to over 100,000 W/(m²·K).
• The Nusselt number $Nu = hL/k$ expresses the ratio between convective and conductive heat transport in the fluid; a large $Nu$ means efficient convection.
• In forced convection, $h$ increases with the fluid velocity: a faster flow makes the boundary layer (the layer near the surface where the velocity changes) thinner, and a thinner boundary layer conducts heat away into the fluid faster.
• The Biot number $Bi = hL_c/k$ ($L_c$ = characteristic length, e.g. volume/surface area) compares the resistance to conduction inside the body with the resistance to convection at the surface. When $Bi < 0.1$ the whole body can be treated as having a single, uniform temperature ("lumped capacitance").
• Fins increase the surface area and hence the heat dissipated from a surface with limited space.

How to solve the problems

1. Decide whether the convection is forced or natural, and find (or use the given) $h$.
2. For simple convection: use $\dot Q = hA\Delta T$ directly.
3. For a wall with convection on both sides and one or more layers: add the resistances $R = 1/h_iA + \sum L_j/(k_jA) + 1/h_oA$, and find $U = 1/(R_{tot}A)$ or $\dot Q = \Delta T/R_{tot}$.
4. Check whether lumped capacitance applies: compute $Bi = hL_c/k$ and compare it with 0.1.
5. Make sure $h$ and $k$ are not mixed up – they have different units and describe different mechanisms.

Example

A metal plate with a characteristic length of 8 mm and $k = 60$ W/(m·K) is cooled by air with $h = 40$ W/(m²·K). Can lumped capacitance be used, and how large is the convective heat transfer from a 0.3 m² surface with a 45 K temperature difference?

1. $Bi = hL_c/k = 40\cdot 0.008/60 \approx 0.00533$.
2. Since $Bi < 0.1$, the whole plate can be treated as having a single temperature.
3. $\dot Q = hA\Delta T = 40\cdot 0.3\cdot 45 = 540$ W.

Answer: Yes, lumped capacitance is valid, and $\dot Q = 540$ W.

Common mistakes

• Using $k$ (thermal conductivity) instead of $h$ (heat transfer coefficient) in Newton's law of cooling, or the other way around.
• Believing $h$ is a material property. It depends on geometry, fluid and flow conditions, and often has to be found from correlations or given in the problem.
• Using lumped capacitance without checking the Biot number first.
• Forgetting the convection resistance $1/(hA)$ when combining several resistances in a wall.
• Assuming natural convection always gives lower heat loss than forced convection – that is almost always true, but it is the flow, not "natural" itself, that determines $h$.

Concepts in this part

Practise convection in the app →

3. Radiation and transient heat

What is it about?

Radiation is heat transfer by electromagnetic waves and the only form of heat transport that works in a vacuum – it is how the sun heats the earth. Every body with a temperature above absolute zero radiates energy. Transient heat is about how a body's temperature changes over time, for example when a hot object cools down in air. As an engineer you need radiation to calculate heat loss from hot surfaces (pipes, furnaces, electronics) and transient formulas to find how long things take to heat up or cool down.

Concepts and formulas

• Stefan–Boltzmann's law, power radiated from a gray surface: $\dot Q_{out} = \varepsilon\sigma AT^4$, with $\sigma = 5.67\cdot10^{-8}$ W/(m²K⁴) and $T$ in kelvin.
• Net radiation loss to surroundings at temperature $T_{surr}$: $\dot Q = \varepsilon\sigma A(T^4 - T_{surr}^4)$.
• Emissivity $\varepsilon$ (0–1) says how well a surface radiates compared with a black body ($\varepsilon = 1$). A polished, shiny metal has a low $\varepsilon$; a matte, dark surface has a high $\varepsilon$.
• Kirchhoff's law (simplified): for a gray surface in thermal equilibrium with its surroundings, the absorptivity equals the emissivity, $\alpha = \varepsilon$. A surface that radiates well also absorbs well.
• Wien's displacement law: $\lambda_{max}T \approx 2898$ µm·K – warmer bodies radiate most strongly at shorter wavelengths.
• Heat is often lost by both convection and radiation at the same time from the same surface: $\dot Q_{tot} = hA(T_s-T_\infty) + \varepsilon\sigma A(T_s^4-T_{surr}^4)$.

With lumped capacitance, transient cooling/heating follows

where $\rho$ is density, $V$ volume and $c$ specific heat capacity.

• The time to reach a given temperature is found by solving the exponential equation for $t$: $t = -\tau\ln\!\left(\dfrac{T-T_\infty}{T_0-T_\infty}\right)$.

How to solve the problems

1. For radiation alone: identify $\varepsilon$, $A$, and the temperatures in kelvin, and use Stefan–Boltzmann.
2. Combine with convection when both mechanisms act from the same surface: add the two contributions.
3. For transient problems: first check that lumped capacitance is valid ($Bi < 0.1$, see the previous unit), find the time constant $\tau$, and use the exponential formula.
4. If you need the time instead of the temperature, invert the formula and solve for $t$ using the logarithm.
5. Remember to convert °C to kelvin in $T^4$ expressions, but °C can be used directly in the differences $T-T_\infty$ in the transient formula (since the 273.15 cancels).

Example

A small metal sphere with a time constant of $\tau = 90$ s cools in air at 22 °C. It started at 260 °C. How long does it take to reach 60 °C?

1. $t = -\tau\ln\!\left(\dfrac{T-T_\infty}{T_0-T_\infty}\right) = -90\ln\!\left(\dfrac{60-22}{260-22}\right) = -90\ln\!\left(\dfrac{38}{238}\right)$.
2. $\ln(38/238) = \ln(0.1597) \approx -1.834$.
3. $t \approx -90\cdot(-1.834) \approx 165$ s.

Answer: about 165 seconds (just under 3 minutes).

Common mistakes

• Forgetting to convert to kelvin in Stefan–Boltzmann's law – $T^4$ is very sensitive to a wrong unit.
• Believing radiation and convection are mutually exclusive. From a hot surface in air, both usually happen at the same time.
• Using the wrong sign in the exponential equation, so the temperature seems to increase during cooling.
• Assuming the emissivity is the same for all surfaces. Shiny, polished surfaces often have $\varepsilon$ below 0.1, while matte surfaces can have $\varepsilon$ above 0.9.
• Using lumped capacitance without checking that the Biot number is actually small enough.

Concepts in this part

Practise radiation and transient heat in the app →