Essential Physics

Section 2 review

The power of heat flowing through conduction varies directly as the thermal conductivity, the area through which the heat flows, and the temperature difference and inversely as the length the heat flows through. Examples are developed for calculating heat conduction through rods and walls. Convection is more complex than conduction, but convection can transfer much more heat in circumstances where fluid such as air or water is free to flow. For example, the heat flow next to a window via conduction may be a few watts per square meter, whereas the heat flow through convection may be 100 W or more. The equation for convection is a model in which many complexities are wrapped up in the heat transfer coefficient h, which can vary from less than 10 to more than a million depending on the exact situation. For this reason, convection calculations are best treated as approximations subject to experimental verification. Read the text aloud

Read the text aloud

heat transfer coefficient

P = \frac{κ A}{L} Δ T

P = h A Δ T

Review problems and questions

How much heat can flow through a solid, round aluminum rod that is 0.50 m in length with a diameter of 1.0 cm? The temperature difference is 50°C between the two ends of the rod.

Answer: 1.8 W

P = \frac{κ A}{L} Δ T = \frac{(226 {W m}^{- 1} ° C^{- 1}) π {(0.005 m)}^{2}}{0.50 m} (50 ° C) = 1.77 W

Rod and tube dimensions for the problem

Repeat the calculation of the first problem if the rod is replaced by a hollow tube with a wall thickness of 0.50 mm. The outside diameter and length are the same and the temperatures are the same.

Answer: 0.34 W

P = \frac{κ A}{L} Δ T = \frac{(226 {W m}^{- 1} ° C^{- 1}) [π {(0.005 m)}^{2} - π {(0.0045 m)}^{2}]}{0.50 m} (50 ° C) = 0.337 W

Less heat can flow per unit time because there is less cross-sectional area for the tube than for the solid cylinder.

An insulated wall has an average thermal conductivity of 0.080 W/(m °C). How much heat passes through the wall if the area is 20 m² and the thickness is 0.2 m? The temperature difference is 20°C between the inside surface of the wall and the outside surface.

Answer: 160 W

P = \frac{κ A}{L} Δ T = \frac{(0 {.08 W m}^{- 1} {°C}^{- 1}) (20 m^{2})}{0.2 m} (20 °C) = 160 W

A cooling surface for a computer chip has a heat transfer coefficient of 50 W/(m² °C). The outside air temperature is 25°C, the chip temperature is 100°C, and when the chip is operating it must dissipate 100 W or else it will overheat. How much surface area does the heat sink need?

Answer: 270 cm²

Start with the equation

P = h A Δ T

and solve for the area:

A = \frac{P}{h Δ T} = \frac{100 W}{(50 {W m}^{- 1} ° C^{- 1}) (100 ° C - 25 ° C)} = 0.027 m^{2} = 270 {cm}^{2}


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