// CODE SNIPPET
A.6.5 Example 5: Calculating Pressure Drops Due to Temperature Changes
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A test piping system is installed on a warm
autumn afternoon when the temperature is 70°F (21°C). In
accordance with local custom, the new piping system is subjected
to an air pressure test at 20 psig (138 kPa). Overnight,
the temperature drops and when the inspector shows up first
thing in the morning the temperature is 40°F (4°C).
If the volume of the piping system is unchanged, then the formula based on Boyle's and Charles' Jaw for determining the new pressure at a reduced temperature is as follows:
Therefore, the gauge could be expected to register 18 psig (124 kPa) when the ambient temperature is 40°F (4°C).
If the volume of the piping system is unchanged, then the formula based on Boyle's and Charles' Jaw for determining the new pressure at a reduced temperature is as follows:
| where: | ||
| T1 | = | Initial temperature, absolute (T1 + 459) |
| T2 | = | Final temperature, absolute (T2 + 459) |
| P1 | = | Initial pressure, psia (P1 + 14.7) |
| P2 | = | Final pressure, psia (P2 + 14.7) |
 |
||
| P2 | = | 32.7 - 14.7 |
| P2 | = | 18 psig |
Therefore, the gauge could be expected to register 18 psig (124 kPa) when the ambient temperature is 40°F (4°C).
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