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Going by the definition of convective heat transfer coefficient from Wikipedia (which I have started to doubt; reason follows): $h=\frac{q}{\Delta T}$, where the $\Delta T$ is taken as the difference between fluid and solid surface. Generally the temperature of the fluid is taken as the free stream far field temperature, $T_{\infty}$.

But in Chapter 4 of Modern Engineering for Design of Liquid-Propellant Rocket Engines by Huzel and Huang, for calculating the convective heat transfer to the engine walls, the adiabatic wall temperature, $T_{aw}$, is used instead of $T_{\infty}$.

Also Prof.Manuel Martinez-Sanchez's MIT OCW notes describe along the same line:

The $T_{aw}$ temperature is shown dashed because, as we know it is not the actual gas temperature outside the gas boundary layer, but is the one driving heat. (page 3 of 12).

The definition of $T_{aw}$ is:

Adiabatic wall temperature is the temperature acquired by a wall in liquid or gas flow if the condition of thermal insulation is observed on it.

It is not the actual temperature of the fluid and also it is the temperature of the fluid considering adiabatic conditions.

Given this: Why is such a temperature used for calculating the convective heat transfer?

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    $\begingroup$ As with a lot of things in engineering, it’s a good estimate of reality, even if it isn’t perfect or exact. Often times, it’s convenient to assume that gas slows down adiabatically because we can derive expression for thermodynamic quantities analytically rather than computationally. $\endgroup$
    – Paul
    Nov 10, 2018 at 16:46
  • $\begingroup$ @Paul what does not make sense is the necessity to use the temperature of the gas so close to the wall for convective heat transfer. For conduction, adjacency makes sense as the transfer of material is not possible. But convection is a bulk phenomenon with considerable fluid movement. Why is Taw preferred? Is it just a overkill/conservative estimate for heat flux? $\endgroup$ Nov 11, 2018 at 10:48

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A late response but this is a topic I very much enjoy. I will do my best to clarify why the adiabatic wall temperature is the driving force for the convective heat transfer.

Any fluid/gas flowing over a body caries kinetic energy. As the fluid passes over the body, a boundary layer develops. At the edge, or outer-most part of the boundary layer, the fluid velocity is that of the free stream, generic velocity. As you travel through the boundary layer (closer to the body over which the fluid is passing) the velocity of the fluid starts to decrease, and when the fluid reaches the wall, it is stationary (this is called the no-slip condition).

As the fluid velocity decreases from the free stream velocity to zero at the wall, the fluid loses all of its kinetic energy. The kinetic energy is transferred to heat energy via friction! It is important to remember that this happens in any fluid flow over any solid body, regardless of how slow or fast the fluid is moving (within reason, sometimes this fails at absurdly fast fluid velocities, but that isnt important).

For very fast fluid velocities, there is a lot of kinetic energy that is transferred to heat energy as the fluid comes to rest at the solid boundary. So, it would be intuitive to assume, that if there is enough kinetic energy that needs to be dissipated, that the heat energy could be so large that the fluid becomes very hot, more so as you get closer to the wall!

However, how hot does it get? The adiabatic wall temperature tells us this. We start with the stagnation temperature, which by definition, is the temperature reached when ALL of the kinetic energy is transferred to heat energy, without any losses. But this is not the case within a boundary layer. As mentioned before, the fluid closer to the solid boundary (which is traveling slower) is hotter than the fluid outside of the boundary layer (which is at ambient free stream temperature). This means there is actually conductive heat transfer through the fluid in the boundary layer itself! Because of this, not all of the heat converted from kinetic energy reaches the wall, only a portion of it is recovered. Therefore, the temperature reached from the heat energy that is only recovered on the wall is what we call the adiabatic wall temperature (also called the recovery temperature). It is this fluid temperature that drives convection to the wall.

In summary:

  • In all fluid flows over a solid body, kinetic energy is converted to heat energy through the boundary layer.
  • In high speed flows, this heat energy becomes very large as there is more kinetic energy to dissipate, thus heating the fluid in the boundary layer.
  • Because the boundary layer is much much hotter than the free stream temperature, conduction occurs through the boundary layer, allowing for a fraction of the heat energy generated from the kinetic energy to be recovered on the wall.
  • The temperature reached from the recovered heat energy is the adiabatic wall temperature and is the temperature of the fluid against the wall, thus driving the convection from the fluid to the wall.
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  • $\begingroup$ Sorry to say this, but I feel this as an repetition of the already available answer. And again, this answer explains why Twa is not T_stag. But still, the reason to use the Twa for calculating the convective heat transfer is not clear! $\endgroup$ Mar 7, 2022 at 17:23
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I don't know how to make this any more concise than the authors did, so here is the relevant page from Hill & Peterson, "Mechanics and Thermodynamics of Propulsion", 3rd printing, November 1970.

Note that they say the gas stagnation temperature can be used instead at the max heat transfer point.

enter image description here

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  • $\begingroup$ A very nice and apt explanation for why Twa is not T_stag. But still, the reason to use the Twa for heat transfer is not clear! Thanks for the reference! $\endgroup$ Nov 11, 2018 at 10:46
  • $\begingroup$ Thank you for sharing. Great explanation!. Note: here T0g is the stagnation temperature. $\endgroup$
    – GRANZER
    Jun 28, 2022 at 18:35
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The reason why $T_{aw}$ is used for $T_{ref}$ is simple. your solid temperature at surface can't go over your $T_{aw}$ - that's the maximum that you will observe. In terms of $h$, it's logical since when you reach max temperature, no further flux must be applied.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
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    Mar 2, 2023 at 9:14

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