2
$\begingroup$

Consider a liquid flow in which, keeping all else constant, we can vary the thermal conductance(K). Now, if we increase the thermal conductance of the fluid the Prandlt number will decrease. I read that as Prandtl number decreases the thermal boundary layer thickness becomes larger than the velocity boundary layer.

What if we have to compare the boundary layer thickness of liquid with lower K and higher K?Will the liquid with higher K( lower Pr) have thinner TBL than the liquid with lower K (higher Pr)?

I am asking this because I think higher value of K should cause heat to diffuse more easily. If heat diffuse easily then in the TBL the temperature should reach the free stream temperature in a shorter distance. So shouldn't the thermal boundary layer be thinner for higher K.

$\endgroup$
2
$\begingroup$

The definition of the thermal boundary layer is that of the distance across a boundary layer from the wall to a point where the flow temperature has essentially reached the 'free stream' temperature, $T_{0}$. This distance is defined normal to the wall in the y-direction

The Prandtl number is defined as the ratio of momentum diffusivity to thermal diffusivity. The equation which mostly reflects that (there are other forms) is probably:

$$Pr = \frac{c_p \mu}{k}$$

So the smaller the Prandtl number the higher thermal diffusivity there is in the material compared to the momentum diffusivity. However greater diffusivity means that the heat from the wall penetrates easier in the fluid flow and therefore the temperature of the wall affects a larger area in the liquid. So, the greater the heat conductivity the further, inside the material the $T_w$ will have an affect.

Essentially the TBL, shows at which distance the flow is thermally independant of the boundary walls. In that respect you might be able to understand that, increasing heat conductivity (and smaller Prandtl numbers) results in thicker boundary layer.

Relation to velocity boundary layer

You have to keep in mind that when you change the Prandtl number by changing the $k$ heat conductivity, you don't alter the kinematic behaviour. I.e. the velocity boundary layer remains the same, and its determined by the following equation:

$$\delta _{v} = 5.0 \sqrt\frac{v\cdot x}{u_0} $$ where:

  • $\delta _{v} $ is the thickness of the velocity boundary layer thickness
  • $u_{0}$ is the freestream velocity
  • $x$ is the distance downstream from the start of the boundary layer
  • $\nu$ is the kinematic viscosity

The velocity boundary layer thickness is mainly dependent on the viscosity, in a similar manner to the thermal boundary layer. In a similar manner because there is an analogy that:

  • greater viscosity means that higher forces can be transfer through the material, while
  • greater heat conductivity means that more heat can be transferred through the material.

So the higher the viscosity, the greater the effect of the boundary conditions (velocity equals to zero) on the flow.

The thickness of the thermal boundary layer is given by:

$$\delta_T = 5.0 \sqrt\frac{v\cdot x}{u_0} Pr^{-1/3}$$

where:

  • $\mathrm {Pr}$ is the Prandtl Number
  • $u_{0}$ is the freestream velocity
  • $x$ is the distance downstream from the start of the boundary layer
  • $\nu$ is the kinematic viscosity

Therefore, the thermal and velocity boundary are related by :

$$\delta_T = \delta _{v} Pr^{-1/3}$$

You can write the above equation in the following form: $$Pr= \left(\frac{\delta_v }{\delta_T} \right)^3$$

$\endgroup$
1
$\begingroup$

A higher thermal conductivity ratio in the flow causes the heat diffusion to catch up with farther layers of fleeing molecules of the flowing liquid that otherwise would have washed past unaffected. If we compare k to the speed of heat diffusion, a higher k means reaching more in less time.

For example, if the k converges to infinity the TBL will be the complete thickness of the liquid because it will heat up immediately before getting a chance to be carried away cold by the momentum.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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