I want to understand the temperature equation for a Newtonian viscous fluid from first principles. In the following, the section "Background" describes how I arrive to the problem, and the section "The problem" describes the actual problem.


I start by considering a fixed control volume $dV$ with surface $dS$. The change of total energy $e$ [energy/(mass $\cdot$ volume)] with time in this element is given by:

$$ \frac{\partial}{\partial t}\int_{dV} \rho e \, dV= -\oint_{dS}\rho e u_in_i\, dS - \oint_{dS} q_in_i \, dS + \oint_{dS}n_j \sigma_{ji} u_i \, dS + \int_V \rho f_i u_i \, dV. $$

In addition to the above, I have introduced these notations:

  • $\rho$: density
  • $u_i$: velocity field
  • $n_i$: unit normal vector directed out of the fixed control volume
  • $q_i$: heat flux vector
  • $\sigma_{ij}$: stress tensor
  • $f_i$: body force

I apply the divergence theorem and note that the control volume is arbitrary. Therefore this must holw in every point of the fluid:

$$ \frac{\partial \rho e}{\partial t} + \frac{\partial }{\partial x_i} \rho e u_i = - \frac{\partial q_i}{\partial x_i} + \frac{\partial }{\partial x_j} \sigma_{ji} u_i + \rho f_i u_i. $$

Mass continuity and the product rule gives:

$$ \rho \frac{\partial e}{\partial t} + \rho u_i \frac{\partial }{\partial x_i} e = - \frac{\partial q_i}{\partial x_i} + \sigma_{ji} \frac{\partial }{\partial x_j} u_i + u_i \frac{\partial }{\partial x_j} \sigma_{ji} + \rho f_i u_i. $$

I subtract the kinetic energy from the total energy and find that the internal energy $\varphi$ of the fluid is governed by:

$$ \rho \frac{\partial \varphi}{\partial t} + \rho u_i \frac{\partial }{\partial x_i} \varphi = - \frac{\partial q_i}{\partial x_i} + \sigma_{ji} \frac{\partial }{\partial x_j} u_i. $$

The problem

At this point I want to translate the the governing equation for internal energy into a governing equation for temperature. But here I get stuck. My approach is this:

In every point of the fluid, we must have an internal energy given by the temperature T:

$$ \varphi = \int_0^T c_v(T) \, dT. $$

Thus, we find that:

$$ \frac{\partial \varphi }{\partial t} = \frac{\partial \varphi}{\partial T} \frac{\partial T}{\partial t} = c_v(T) \frac{\partial T}{\partial t}. $$

Similarily, we find that:

$$ \frac{\partial \varphi }{\partial x_i} = \frac{\partial \varphi}{\partial T} \frac{\partial T}{\partial x_i} = c_v(T) \frac{\partial T}{\partial x_i}. $$

Inserting this into the governing equation for internal energy, we find:

$$ \rho c_v \frac{\partial T}{\partial t} + \rho c_v u_i \frac{\partial T}{\partial x_i} = - \frac{\partial q_i}{\partial x_i} + \sigma_{ji} \frac{\partial }{\partial x_j} u_i. $$

But my problem is that I can see no reason for why we can not use $c_p$ instead of $c_v$! Can we not just as well describe the internal energy by:

$$ \varphi = \int_0^T c_p(T) \, dT $$

and then we arrive to

$$ \rho c_p \frac{\partial T}{\partial t} + \rho c_p u_i \frac{\partial T}{\partial x_i} = - \frac{\partial q_i}{\partial x_i} + \sigma_{ji} \frac{\partial }{\partial x_j} u_i, $$

which contradicts the earlier result? What am I doing wrong here? How can I tell whether it is $c_v$ or $c_p$ that should be used?

  • $\begingroup$ You're definition of internal energy is not the proper general form, you need to include volume dependence. Also this seems like a painful way to derive the thermal energy equation; I'd just subtract the mechanical energy equation from the total energy equation $\endgroup$ Commented Feb 1, 2019 at 6:16
  • $\begingroup$ What kind of strange rule is in effect? "Edits must be at least 6 characters; is there something else to improve in this post?" I was planning to edit the typo holw to hold but the system prevented me from doing it. $\endgroup$ Commented Oct 12, 2019 at 11:19

1 Answer 1


So you are assuming that the internal energy is a function only of temperature, but not specific volume? This, of course, is true in the limit of an incompressible fluid. For an incompressible fluid, the heat capacity at constant pressure is equal to the heat capacity at constant volume.

  • $\begingroup$ Ahh.. I see. But if I dont make this assumption, then we have: $$ \varphi(T) = \int_0^T c_v(T) \, dT = \int_0^T c_p(T) \, dT - \int_0^V p(V) \, dV $$ ? $\endgroup$
    – Frysen
    Commented Oct 31, 2016 at 7:08
  • 1
    $\begingroup$ Actually, that's not the equation for the change in internal energy. Look it up in a Thermo book. It is $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$I suggest you check out the detailed development in Transport Phenomena by Bird, Stewart, and Lightfoot. $\endgroup$ Commented Oct 31, 2016 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.