I might be wrong but I will give a little background leading to my question. I am trying to calculate the heat generation in the advection-diffusion equation for heat generation due to friction by loss in mechanical energy in groundwater flow. Water can be considered incompressible and therefore the volume, $V$, is constant for a given mass unit of flow. This loss in energy can be related to the source term, $R$, from the advection-diffusion equation. On a per time basis this comes to power production of groundwater flow:

delta pressure*volumetric flow rate (cubic meters per second)=change in energy per second

By multiplying the change in pressure between two points by the volumetric flow rate of water, the power due to friction (total energy lost per second) can be found. The power of friction is also the amount of heat gained per time interval, which can be used for $R$.

My question is how can I relate this energy gain due to friction for $R$? I am trying to convert the change in energy per second to units of $\partial T/\partial t$ which is the units for the terms in the advection-diffusion equation.

The advection-diffusion equation for heat transfer due to conduction, advection, and generation is

$$ \frac{\partial T}{\partial t}=\nabla\cdot\left(D\nabla T\right)-\nabla\cdot\left(\mathbf vT\right)+R $$

  • $\nabla\cdot\left(D\nabla T\right)$ is the term for conduction (diffusion)
  • $\nabla\cdot(\mathbf vT)$ is the term for advection
  • $R$ is the term I am trying to quantify for heat generation due to mechanical energy loss (pressure change)

1 Answer 1


The rate of temperature change will be the power per unit mass times the specific heat.

So if you have a certain mass of water $M$ flowing per second, at a velocity $v$, losing $\Delta P$ pressure per second, then work done is $v\Delta P A$ and $A = \frac{M}{\rho v}$ . Then with a heat capacity $c$ (about 4.2 kJ/kg/K for water), and the relationship between rate of heating $E$ of a mass $M$ and the power:

$$\frac{dT}{dt}= \frac{E}{Mc}$$

You should have everything you need.

Make sure that the dimensions are correct as you manipulate your equations. It's late here and an error may have snuck in.

  • 2
    $\begingroup$ Everything is right I think except A should be inversed, A=M/(rho*v) and the heat capacity should be in the denominator. Thanks $\endgroup$ Nov 22, 2014 at 19:10
  • $\begingroup$ @tyciani thanks for pointing out my mistake. Fixed now. $\endgroup$
    – Floris
    Nov 22, 2014 at 21:09

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