Read e-book online An Introduction to Heat Transfer Principles and Calculations PDF

By A. J. Ede

ISBN-10: 008013517X

ISBN-13: 9780080135175

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Since the pipe will probably become dirty eventually, a high value of 70 per cent is chosen. Heat transfer is directly proportional to emissivity in a simple problem such as this, so that it now becomes 4 = 0-577x^ = 0-425 kW. (b) The foil is wound on tightly, so that its temperature may be taken as 200°C. A conservative value for the emissivity of aluminium is 20 per cent, which gives 20 4 = 0-577 x — = 0-12kW. (c) Neglecting the ends of the casing, its surface area will be 0-5 m 2 . For estimating the loss to the room, eqn.

The surface temperature of the mag­ nesia is 30°C. What is the rate of loss of heat through the insulation for a 12 m length of pipe? This problem can be solved by means of eqn. 10). 7, p. 266, the thermal conductivity of magnesia is 0·07 W/m degC. The radii are r1 =2·5 cm =0-025 m, r2 =0-045 m. Therefore 2πχ0·07χ12χ(150-30) ~ loge(0-045/0-025) q = 1080W or 1-08 kW. There is no essential difference involved in carrying out the cal­ culations in any other set of consistent units. , would be in feet, the temperatures in Fahrenheit degrees, and the thermal conductivity in Btu/ft h degF; the loss of heat would be given in Btu/h.

7, p. 266, the thermal conductivity of magnesia is 0·07 W/m degC. The radii are r1 =2·5 cm =0-025 m, r2 =0-045 m. Therefore 2πχ0·07χ12χ(150-30) ~ loge(0-045/0-025) q = 1080W or 1-08 kW. There is no essential difference involved in carrying out the cal­ culations in any other set of consistent units. , would be in feet, the temperatures in Fahrenheit degrees, and the thermal conductivity in Btu/ft h degF; the loss of heat would be given in Btu/h. Find the corresponding rates if the 2 cm layer of magnesia is replaced by: (a) a 1 cm layer of magnesia followed by a 1 cm layer of silica aerogel, (b) the same materials in the reverse order.

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An Introduction to Heat Transfer Principles and Calculations by A. J. Ede


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