Heat transfer boundary condition

5 replies · 4K views · last reply
EastWindBreaks
Messages
128
Reaction score
3

Homework Statement


Slide titled Convection Boundary Condition showing two boxed equations: -k dT(0,t)/dx = h1[T∞1 - T(0,t)] and -k dT(L,t)/dx = h2[T(L,t) - T∞2], with diagrams of a slab with convection and conduction arrows at each face.


I am confused on how it's using the surrounding temperature minus the surface temperature as its the other way around in the Newton's law of cooling, Doing that would change the sign of convection right? I don't see the reason to do that, since if left side is hotter, then conduction is positive going from left to right.

Homework Equations



Slide titled Newton's law of cooling showing the equation Q-dot conv = hA_s (T_s - T_inf) in watts, with definitions: h convection heat transfer coefficient (W/m^2 K), A_s surface area convection takes place, T_s surface temperature, T_inf temperature of fluid sufficiently far from surface.

The Attempt at a Solution

 
Attachments
  • Slide titled Convection Boundary Condition showing two boxed equations: -k dT(0,t)/dx = h1[T∞1 - T(0,t)] and -k dT(L,t)/dx = h2[T(L,t) - T∞2], with diagrams of a slab with convection and conduction arrows at each face.
    upload_2018-5-13_8-29-33.png
    38.2 KB · Views: 1,813
  • Slide titled Newton's law of cooling showing the equation Q-dot conv = hA_s (T_s - T_inf) in watts, with definitions: h convection heat transfer coefficient (W/m^2 K), A_s surface area convection takes place, T_s surface temperature, T_inf temperature of fluid sufficiently far from surface.
    upload_2018-5-13_8-31-3.png
    11.5 KB · Views: 1,087
What is the context for the form of Newton's Law that you quote? Seems to me that T∞ here refers to the ambient temperature in the heat sink. In the image you post, T∞1 is the distal temperature in the heat source.
 
haruspex said:
What is the context for the form of Newton's Law that you quote? Seems to me that T∞ here refers to the ambient temperature in the heat sink. In the image you post, T∞1 is the distal temperature in the heat source.
it's from here :
Slide titled CONVECTION showing a velocity profile and temperature profile of air flowing over a hot block, with Newton's law of cooling Q_conv = hA_s(T_s - T_inf) and definitions of h, A_s, T_s and T_inf.

isn't distal temperature same as ambient temperature in this case? so I guess if surface is cooling, then we have to take surface temperature minus ambient temperature, if the surface is being heated, then we have to take the ambient temperature minus surface temperature? but sometimes even when the surface temperature is not given, the book still uses" ambient temperature minus surface temperature" on the left side of the wall. like this example from the book:
Schematic of a south wall of thickness L with inner surface at T∞1 and outer surface facing sun, sky at Tsky, and ambient T∞2, showing convection, radiation, solar and conduction heat flows, with x-axis from 0 at inner surface to L at outer surface.
 
Attachments
  • Slide titled CONVECTION showing a velocity profile and temperature profile of air flowing over a hot block, with Newton's law of cooling Q_conv = hA_s(T_s - T_inf) and definitions of h, A_s, T_s and T_inf.
    upload_2018-5-13_22-26-1.png
    48.7 KB · Views: 859
  • Schematic of a south wall of thickness L with inner surface at T∞1 and outer surface facing sun, sky at Tsky, and ambient T∞2, showing convection, radiation, solar and conduction heat flows, with x-axis from 0 at inner surface to L at outer surface.
    upload_2018-5-13_23-17-58.png
    72.8 KB · Views: 650
EastWindBreaks said:
isn't distal temperature same as ambient temperature in this case?
Yes, but as I posted, one equation is looking at the ambient temperature in the heat source, while the other, Newton, is looking at the ambient temperature in the heat sink.
 
  • Like
Likes   Reactions: EastWindBreaks
Your book isn't doing a very good job of explaining this. It has led you think in terms of which temperature is higher in a particular problem and which temperature is lower. This is not the proper way of analyzing this. You need to focus rather on which temperature is situated at lower x, and which temperature is situated at higher x. This is because the heat flux is really a vector quantity. In reality, it should be accompanied by a unit vector in the x direction. If the ambient temperature is at lower x and the wall is at higher x (like say adjacent to the wall at x = 0), then the heat flux vector is $$\mathbf{q}=h(T_{ambient}-T_{wall})\mathbf{i_x}\tag{1}$$It doesn't matter which temperature is higher and which temperature is lower. The temperature at smaller x comes first and the temperature at higher x comes last. Similarly, at the boundary x = L, the wall temperature is situated at lower x than then ambient. Therefore, at that boundary, $$\mathbf{q}=h(T_{wall}-T_{ambient})\mathbf{i_x}\tag{2}$$It doesn't matter which temperature is higher and which temperature is lower.

In Eqn. 1, if the wall temperature is higher than the ambient temperature, all that means is the the heat flux is in the negative x direction (rather than the positive x direction).
 
  • Like
Likes   Reactions: EastWindBreaks
Chestermiller said:
Your book isn't doing a very good job of explaining this. It has led you think in terms of which temperature is higher in a particular problem and which temperature is lower. This is not the proper way of analyzing this. You need to focus rather on which temperature is situated at lower x, and which temperature is situated at higher x. This is because the heat flux is really a vector quantity. In reality, it should be accompanied by a unit vector in the x direction. If the ambient temperature is at lower x and the wall is at higher x (like say adjacent to the wall at x = 0), then the heat flux vector is $$\mathbf{q}=h(T_{ambient}-T_{wall})\mathbf{i_x}\tag{1}$$It doesn't matter which temperature is higher and which temperature is lower. The temperature at smaller x comes first and the temperature at higher x comes last. Similarly, at the boundary x = L, the wall temperature is situated at lower x than then ambient. Therefore, at that boundary, $$\mathbf{q}=h(T_{wall}-T_{ambient})\mathbf{i_x}\tag{2}$$It doesn't matter which temperature is higher and which temperature is lower.

In Eqn. 1, if the wall temperature is higher than the ambient temperature, all that means is the the heat flux is in the negative x direction (rather than the positive x direction).
wow, you are right, I got stuck on trying to write that term base on which temperature is higher and which is lower, thank you very much for the awesome explanation!