# Transport equation for temperature in a fluid: Heat capacity at constant volume or pressure

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.

Background

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?

• 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 – Drew Feb 1 at 6:16

• Ahh.. I see. But if I dont make this assumption, then we have: $$\varphi(T) = \int_0^T c_v(T) \, dT = \int_0^T c_p(T) \, dT - \int_0^V p(V) \, dV$$ ? – Frysen Oct 31 '16 at 7:08
• 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. – Chet Miller Oct 31 '16 at 12:34