# heat generation due to viscosity in a 3D fluid flow

Consider an arbitrary 3D fluid flow:

$$\vec{\nu}=\vec{V}\left( \vec{x} ,t \right) \tag{1}$$

where velocity at each point $\vec{\nu}$ is a function $\vec{V}$ of position $\vec{x}$ and time $t$ (non-steady). Due to the viscosity $\mu$ there is heat being generated at each point in space. I know this heat should be a function of velocity, viscosity and div/curl of velocity:

$$\dot{q}=F\left(\vec{\nabla},\vec{\nu},\mu \right) \tag{2}$$

But I can't just find it out.

For a 2D problem with unidirectional horizontal velocity of:

$$\nu_x=V_x\left(x,y,t\right) \tag{3}$$

I think I can write

$$\dot{q}=\mu \frac{\partial \nu_x}{\partial y}\nu_x \tag{4}$$

Where $\tau_x=\mu \frac{\partial \nu_x}{\partial y}$ is the shear force due to viscosity. But I'm not quite sure if that's correct.

I would appreciate if you could help me know what is the correct form for equation 2.

P.S. I'm not quite sure but I think it should be something like:

$$\dot{q}=\mu \left( \vec{\nu} \times \vec{\nabla}\right).\vec{\nu} \tag{5}$$

It's not what you have written. The rate of viscous heat generation per unit volume is given by $$2\mu\mathbf{E}:\mathbf{E}$$ where $\mathbf{E}$ is the rate of deformation tensor. For the shear flow you mentioned, this reduces to $\mu (dv/dy)^2$. See Transport Phenomena by Bird, et al for the full development.
• I don't understand this notation. how should I write $E$? and what type of operator $:$ is?
• $\mathbf{E}=(\nabla \mathbf{v}+(\nabla \mathbf{v})^T)/2$ (the velocity gradient tensor plus its transpose divided by 2), and $\mathbf{E}:\mathbf{E}$ is the double contraction of the rate of deformation tensor with itself. See the reference I provided, if you are not familiar working with tensors. Commented Feb 19, 2018 at 14:30
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