Temperature gradient vs thermal boundary layer thickness

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The discussion centers on the relationship between temperature gradient and thermal boundary layer thickness, emphasizing that a thicker boundary layer results in a lower temperature gradient at the wall. The temperature gradient is highest at the wall and decreases towards the edge of the boundary layer, which is mathematically represented by a parabolic temperature profile. As one moves downstream on a flat plate, the thickness of both thermal and velocity boundary layers increases due to the diffusion of momentum and heat effects. Although heat continuously diffuses into the thermal boundary layer, it reaches a point of equilibrium where growth slows significantly. The conversation highlights the complexities of boundary layer behavior in fluid dynamics, particularly under varying conditions.
Idrees Afridi
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what does the relation between the temperature gradient inside the thermal boundary and thermal boundary layer thickness i mean what will be the temperature gradient ( high or low) when the thermal boundary layer is thick relative to the thin one? Kindly explain mathematically and physically as well...
Thanks
 
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The boundary layer thickness is usually not a very precisely defined quantity. In typical problems involving boundary layers, the thickness of the boundary layer is envisioned as being small compared to the characteristic physical dimensions of the system. The temperature is envisioned as varying very rapidly from the value at the wall to the value at the "edge of the boundary layer." The temperature gradient is largest at the wall, and decreases rapidly with distance to a very low value at the "edge of the boundary layer." Suppose you approximated that temperature profile in the boundary layer by a parabolic shape, $$T=(T_w-T_{\infty})\left(1-\frac{y}{\delta}\right)^2+T_{\infty}$$where ##T_w## is the wall temperature, ##T_{\infty}## is the temperature at the edge of the boundary layer, y is the distance from the wall, and ##\delta## is the boundary layer thickness. Then from this equation, the temperature gradient would be given by:$$\frac{\partial T}{\partial y}=-\frac{(T_w-T_{\infty}}{\delta}\left(1-\frac{y}{\delta}\right)$$In this approximation, the temperature gradient at the wall would be equal to ##-(T_w-T_{\infty})/\delta## and the temperature gradient at the edge of the boundary layer would be zero. So, if the boundary layer is very thin, the magnitude of the temperature gradient at the wall is very large, and if the boundary layer is thicker, the magnitude of the temperature gradient at the wall is less.
 
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Likes Yada Nandukumar
I got the point. thanks alot
 
currently i am working on momentum and thermal boundary layer so i have few more questions..
i): why the shape of thermal and velocity boundary layer is parabolic?
ii):why the thickness of these boundary layers increasing as we move from the leading edge of the flat plate?
iii):why the thermal boundary layer is not continuously growing although heat continuously diffuses into this layer?
 
Idrees Afridi said:
i): why the shape of thermal and velocity boundary layer is parabolic?

It isn't.

Idrees Afridi said:
ii):why the thickness of these boundary layers increasing as we move from the leading edge of the flat plate?

Viscosity is basically the coefficient of diffusion pertaining to momentum transport, so as you move downstream on a flat plate (without any other influences like changes in pressure gradient) it's akin to moving later in time in a typical diffusion problem. The zero-momentum effect of the wall has simply diffused further into the free stream.

Idrees Afridi said:
iii):why the thermal boundary layer is not continuously growing although heat continuously diffuses into this layer?

In theory it could be. It just likely reaches a point where it grows very slowly as the system reaches equilibrium.
 
Idrees Afridi said:
currently i am working on momentum and thermal boundary layer so i have few more questions..
i): why the shape of thermal and velocity boundary layer is parabolic?
This is only an approximation that is used sometimes.
ii):why the thickness of these boundary layers increasing as we move from the leading edge of the flat plate?
The momentum and heat effect of the wall is penetrating more deeply into the fluid as it moves along the wall.
iii):why the thermal boundary layer is not continuously growing although heat continuously diffuses into this layer?
As boneh3ad said, the thermal boundary layer does get thicker in the case of a flat plate.
 
boneh3ad said:
It isn't.
Viscosity is basically the coefficient of diffusion pertaining to momentum transport, so as you move downstream on a flat plate (without any other influences like changes in pressure gradient) it's akin to moving later in time in a typical diffusion problem. The zero-momentum effect of the wall has simply diffused further into the free stream.
In theory it could be. It just likely reaches a point where it grows very slowly as the system reaches equilibrium.
thanx to guide me but i must ask one thing more that what type of equilibrium u mean ? does it make sense physically ? if yes then how?
 
I am not sure exactly what you are asking here. Of course it makes sense physically, or it wouldn't happen. Really, there are a few possibilities. If the surface is allowed to change temperature so that the system eventually reaches thermal equilibrium, then eventually heat will stop being transferred between the fluid and the surface and the thermal boundary layer will cease to exist as a concept since everything will be isothermal. That's a rather boring case, though.

Another possibility is that the surface is held at a constant temperature, and in that case the thermal boundary layer will always grow, but at an increasingly slow rate. As the fluid near the wall gets continually hotter (or colder as the case may be), it is nearer and nearer in temperature to the surface, so all of the temperature gradients decrease and the heat transfer rate decreases. With an infinite fluid, the growth will never truly stop, but growth rate will get so incredibly small that it doesn't really matter anyway.
 
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Likes Ravi Singh choudhary
thanks a lot .i got my point ,, stay blessed
 
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