Heat Transfer II -- How heat is lost from a hot surface to the surrounding air?

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Tiberious
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Homework Statement


(a) Explain how heat is lost from a hot surface to the surrounding air.

The Attempt at a Solution


We will assume as there has been no other stipulation that the effect of radiation is negligible and thus we are taking purely about conduction.

We will also assume as there has been no othe stipulation that the fluid 'air' is in motion.

The image below depicts the flow of a fluid over a hot surface.

Image shows the flow profile and Temperature profile.

Figure 2a, Teesside University, 'Overall Heat Transfer Rates', unknown, 8nd April 2018.

Firstly, the fluid in contact with the hot surface has no velocity. Seen in the diagram above the velocity of the fluid increases with distance γ until it reaches the mainstream velocity of the fluid flowing of U_f. This region is known as the Boundary Layer.

The fluid that is in contact with the surface has the same temperature as the surface itself. This then falls through the Boundary Layer to the mainstream temperature T_f.

Further to our prior stipulation that we are to assume the radiation effect is negligible. Consider the fluid in contact with the surface is stationary, heat is transferred by conduction thus the Fourier's Law applies.

ϕ=-kA(dT/dy)_s

We can see the subscript s has been attached to the base of the temperature bracket indicating that the temperature gradient at the surface determines the heat Loss ϕ.

Can anyone let me know if this is a reasonable answer?
 
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On one hand, you say that the air is not in motion. On the other hand, you say that the air is moving parallel to the plate, with a velocity profile across the boundary layer. Which is it?
 
Really don't understand, is there a perceivable difference? In both cases the fluid is in motion.
 
Tiberious said:
Figure 2a, Teesside University, 'Overall Heat Transfer Rates', unknown, 8nd April 2018.
Could you Upload this image? Use the UPLOAD button in the lower right of the Edit window to Upload a PDF or JPEG image. Thanks. :smile:
 
So your problem is forced convection from a heated flat plate. A classic problem in convection heat transfer. A boundary layer with both a velocity profile and a temperature profile. The boundary values are : U(y=0) = 0, T(y=0) = T0, U(y=δ) = U(free stream), T(y=δT) = T(free stream), The boundary layer thicknesses for velocity, δ, and temperature, δT, are seldom the same. Just as an aside, I do a demonstration in my Physics class comparing free and forced convection from two coffee cups filled with hot water, The airspeed for the forced convection is the same speed that you would blow your breath across a hot liquid. The result: the forced convection film coefficient = 3 x the free convection film coefficient. That's one data point. Forced convection heat transfer is extremely complicated since there are so many properties that effect it. It is prone to analysis using a great number of dimensionless ratios representing groups of these properties, thus it is difficult to generalize. It is like turbulent air flow, every geometry is a special case.
 
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berkeman said:
Could you Upload this image? Use the UPLOAD button in the lower right of the Edit window to Upload a PDF or JPEG image. Thanks. :smile:

Uploaded. I understand the wider topic is more complex. However, as a answer for the question posed is mine suitable?
 
By definition, heat transfer is an energy flow across a boundary which is characterized by a temperature gradient at a boundary. What happens to that energy, especially with a flowing gas is largely dependent on the properties of the gas and whether the flow is laminar or turbulent. Turbulent mixing greatly enhances the energy transfer. In my former day job I was concerned with the effect of the turbulent intensity on the heat transfer from a gas turbine engine combustor to the turbine vanes and blades. I set up an experiment to measure the turbulent intensity of the products of combustion downstream of an operating combustor at the plane corresponding to the turbine inlet using a laser anemometer. I should have received hazardous duty pay for that experiment. Our immediate application was to duplicate these turbulent intensities in the heat transfer tunnel to insure we were accurately modeling the phenomena.