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API 510 Temperature difference 1

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Baby-Engg

Petroleum
Apr 2, 2024
5
code API 510:

5.8.7.2 To minimize the risk of brittle fracture during a pressure test, the metal temperature should be maintained at least 30 °F (17 °C) above the MDMT or MAT for vessels that are more than 2 in. (5 cm) thick and 10 °F (6 °C) above
the MDMT or MAT for vessels that have a thickness of 2 in. (5 cm) or less. The test temperature need not exceed
120 °F (50 °C) unless there is information on the brittle characteristics of the vessel material indicating a higher test
temperature is needed.

Question: why temp. conversion deg. F - deg. C does not match?
 
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The first two temps are differential temperatures so the formula is °F / 1.8 = °C.

The last is an actual temperature so the formula is (°F - 32) / 1.8 = °C.
 
can you please simplify/explain a little bit
 
Let's say your MDMT is -40 (°C, or °F, they are the same at that temp). For a >2" walled tank, your test should be done at a temperature at least 30°F above that, so above -10°F, which converts to -23°C. Would you look at that, it's a 17°C difference. If you "converted" the 30°F like a temperature, you would guess a -1°C difference, which is nonsensical in this application (going colder than MDMT)

Like Geoff13 said, here's the formula to convert a temperature from Celsius (C) to Fahrenheit (F):

C = ( F - 32 ) / 1.8

Now look at what happens if we were converting a temperature differential, i.e. the magnitude difference between two temperatures C[sub]1[/sub] and C[sub]2[/sub]:

ΔC = C[sub]1[/sub] - C[sub]2[/sub]
= [( F[sub]1[/sub] - 32 ) / 1.8 ] - [( F[sub]2[/sub] - 32 ) / 1.8 ]
= [ F[sub]1[/sub] - 32 - ( F[sub]2[/sub] - 32 )] / 1.8
= [ F[sub]1[/sub] - 32 - F[sub]2[/sub] + 32 ] / 1.8
= ( F[sub]1[/sub] - F[sub]2[/sub] ) / 1.8
= ΔF / 1.8

So if ΔF = 30°F, then ΔC = ( 30 / 1.8 )°C = 16.67°C , or 17°C

Another good example, a 1 Kelvin difference in temperature equates to 1°C change in temperature, but 1K as a temperature is -272.15°C.
 
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