If you force high-pressure gas through a packed-bed, the gas pressure drops. However, the ideal gas law states:

$$P = \frac{RT}{v}$$

so for an ideal gas, the pressure drop must be associated with a property change of the gas itself as it moves through the packed bed. My question is: how does a typical gas moving through a typical packed bed adjust its pressure: is it primarily by a decrease in temperature or by an increase in specific volume?


2 Answers 2


Frictional Heat!

In simple terms, the pressure drop that is observed is a result of the friction between the moving fluid, and the stationary walls and column packing.

This friction reduces the pressure of the fluid, and in turn heats up both the fluid, and the packing.

This topic is discussed at length in Moody's treatise on the subject:

Moody, L.F. (1944). "Friction Factors for Pipe Flow". Transactions of the ASME 66 (8): 671–684.


In packed beds that I've seen, the pressure loss across the bed is miniscule compared to the absolute pressure (on the order of 2%), and the length of the packed bed is sometimes 2 or 3 diameters (or more). Finally, the gas is usually undergoing a chemical reaction (which is generally exothermic).

Looking over the net result of all of these effects:

1.) Pressure drops by 2%, which could increase specific volume or temperature

2.) Gas passes through long, uninsulated column which maintains gas temperature at or around original temperature, resulting in temperature to return to normal while maintaining density decrease

3.) Gas absorbs substantial heat from chemical interactions, which dominates the two effects above, resulting in temperature increase and an altogether different gas state which has reduced pressure (if pressure increases, the gas won't flow), with likely higher specific volume and higher temperature.

If only for effects 1 and 2, I'd argue that a packed bed results in a change in density, because the gradual temperature change is removed from the long travel length. However the final effect changes things to a chemical engineering problem.

  • $\begingroup$ How about for a filtration system, perhaps with a membrane instead of a packed bed, where no chemical reaction occurs but significant pressure changes do occur? $\endgroup$
    – tom
    Aug 30, 2015 at 12:04
  • $\begingroup$ While I'd say that's different than a packed bed, the answer to that lies in the specific gas, and whether it has a Joule-Thomson Effect. $\endgroup$
    – Mark
    Aug 30, 2015 at 15:17
  • $\begingroup$ I suspected that much. Would it be fair to say that if we're in a temperature range where significant J-T effect occurs, the pressure change is associated with a temperature change (though also potentially some volume change), but if there's negligible J-T effect (or if we're above the inversion temperature), there must be gas-expansion instead. $\endgroup$
    – tom
    Aug 31, 2015 at 3:56
  • $\begingroup$ For long pipelines, the pressure drop means the density changes mean the flow velocity increases alongthe pipe, leading to interesting iterative formulae $\endgroup$
    – mart
    Oct 29, 2015 at 13:46

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