Methods of establishing minimum allowable thickness range from the highly complex to the simple. With the average heater, the operating pressure and temperature are known only for the heater inlet and outlet. The pressure and temperature at intermediate points is typically estimated by calculations or measured with pressure gauges and thermocouples installed at appropriate locations.

The metal temperature governs the allowable working stress for tube materials. Therefore, for a given tube size and a given operating pressure, the minimum allowable thickness varies with the tube temperature. Tube temperature is an important parameter to know, especially at the highest levels of the normal operating condition.

API Std. 530 gives extensive information on the calculation of required wall thickness of new tubes (carbon steel and alloy tubes) for petroleum refinery heaters. The procedures given are appropriate for designing tubes or checking existing tubes in both corrosive and non-corrosive services.

Many methods, including those involving tube skin thermocouples, infrared cameras, infrared pyrometers, and optical pyrometers, are available to determine the metal temperature of a tube. A simple method is to estimate the metal temperature from the operating fluid temperature and then adjust the temperature estimate based on the location of the tube in the heater—the skin temperatures on a tube closer to the flame or nearer the heater outlet will be hotter than one at the heater inlet.

Under certain conditions, the methods described in the preceding text may result in a thickness that is too small for practical purposes. The minimum allowable thickness must be great enough to give the tube sufficient structural strength to prevent sagging between supports and to withstand upset operating conditions. For this reason, it may be appropriate to add some amount based on experience to the calculated minimum allowable thickness and to use this greater thickness as the limit at which a tube should be replaced. Generally, this would be about 0.125 in. (0.32 cm) in these cases.

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