I hope you weren't offended by my comment. In a forum like this, people with all different backgrounds ask questions, and I was unaware of yours.
I don't know who the "expert" was, but he was wrong. The cause of the temperature rise was indeed viscous heating. Viscous "friction" is not the same as dry friction between the liquid with the particles. It is analogous to the heating that occurs in a viscous fluid when it is stirred. The mechanical energy of stirring is dissipated by the viscous behavior of the fluid, and converted to internal energy of the fluid, which translates into a temperature rise. This is exactly how Joule demonstrated the equivalence between mechanical work and increase in internal energy.
In the case of flowing a liquid through a packed column, the mechanical energy loss associated with viscous dissipation in the fluid flow within the pore channels of the packing produces an increase in the internal energy of the exit fluid. This is also a well-known phenomenon in viscous polymer flow through processing equipment, such as transfer lines, spinnerets, and other dies. The rough rule of thumb is a 2 degree temperature rise for every 1000 psi in pressure drop.
In your packed column, application of the open system (control volume) version of the first law of thermodynamics to the flow through the column yields:
$$\Delta h = C\Delta T+v\Delta P=0$$where h is the enthalpy per unit mass of the fluid, C is the specific heat capacity, v is the specific volume and P is the pressure. Viscosity causes the fluid pressure to decrease between the entrance and the exit of the packed bed. The corresponding viscous temperature rise of the fluid is:
$$\Delta T=-\frac{v\Delta P}{C}$$where $\Delta P$ is negative.
In the simple case of flow in a tube, the viscous flow equations are observed to accurately predict the amount of viscous dissipation and the corresponding pressure drop. And the above equation is then observed to accurately predict the temperature rise of the fluid.
In the experiments you described, there were two phenomena occurring simultaneously: expansion cooling and viscous heating. In the case of a liquid, the viscous heating wins out over the expansion cooling because a liquid is nearly incompressible, so it can't do any expansion work to cause cooling. In the case of an ideal gas, the expansion cooling exactly cancels the viscous heating, and there is no change in internal energy or temperature. In the case of a real gas, in most cases, the expansion cooling typically wins out (by a little) over the viscous heating, so there is usually a small amount of temperature decrease.
