How to Calculate Heat Loss During Pipeline Depressurization

The Joule-Thomson effect causes heat loss upon depressurization.
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When a pressurized gas pipeline is rapidly depressurized (i.e., the gas is allowed to flow rapidly through an open valve to atmosphere), a thermodynamic effect causes the gas to cool. This is called a throttling process or the Joule-Thomson effect. The loss of heat is a function of the expansion of the gas from a high pressure to a lower pressure and is adiabatic in nature (no heat is exchanged).

    Determine the gas that is compressed in the pipeline. For instance, assume that carbon dioxide gas is in a pipeline at a pressure of 294 pounds per square inch (psi) and a temperature of 212 degrees Fahrenheit. At these conditions, the Joule-Thomson coefficient is 0.6375.

    Rearrange the heat loss calculation to isolate the final temperature. The Joule-Thomson equation is μ = (T1 - T2) / (P1 - P2) where μ is the Joule-Thomson coefficient, T1 is the initial temperature, T2 is the final temperature, P1 is the initial pressure and P2 is the final pressure. Rearranging yields -μ x (P1 - P2) + T1 = T2. Assume the final pressure is 50 psi.

    Calculate the final temperature and heat loss in the system. This is done by plugging in the values as -0.6375 x (294 - 50) + 212 = T2 which calculates to be T2 = 56.45. Therefore, the heat loss during depressurization is 212 - 56.45 or approximately 155 degrees Fahrenheit.

    Things You'll Need

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