# Logical explanations of factors, Heat transfer coefficient

• Conductivity
In summary, the main factors that affect convective heat transfer coefficient include the properties of the fluid such as velocity, viscosity, specific heat, and conduction coefficient. These factors can all affect the thickness of the boundary layer at the surface, which in turn affects the heat transfer. The geometric shape of the surface, such as flat, arc, or spherical, can also play a role in the heat transfer. Additionally, the angle of the surface, whether it is vertical, horizontal, or with an angle, can also impact the heat transfer. Laminar or turbulence flow can also affect the convective heat transfer coefficient, with turbulence flow generally resulting in higher heat transfer due to increased interaction between the fluid particles. The type of convection,
Conductivity
I would really like a logical explanation for some of the factors that affect convective heat transfer coefficient. I have only found mathematical one.

For example:
1) Properties of the fluid:
1)Velocity: My professor said that the lower the velocity the more time the fluid gets to interact with the surface so higher h. But in all graphs velocity is proportional to the coeff.
2)Viscosity
3) Specific heat
4) conduction coeff: The higher the better because the fluid can spread heat more.

2) Geometric shape of the surface: Flat, arc, spherical..etc. (Question on how they diff from other and who has the highest h)

3) The angle of the surface: Vertical, Horizontal, or with an angle. (Question on how they diff from other and who has the highest h)

4) Laminar or turbulence flow: Turbulence flow has the higher h because the interaction between the fluid is higher so heat is spread faster, But in laminar the only way heat can cross is by conduction between the layers or molecules effusion.

5) Type of convection: Free or forced convection.Sorry I must be asking a lot but it would help overall to have a thread like this since I didnt find any sources online.

You are correct about the effect of the velocity and your professor is incorrect. What are your thoughts on the other factors? For all the cases, this all comes down to what affects the thickness of the boundary layer at the surface (and the penetration depth of the effective temperature variation). For a better physical feel for this, see Transport Phenomena by Bird, Stewart, and Lightfoot.

Conductivity
Chestermiller said:
You are correct about the effect of the velocity and your professor is incorrect. What are your thoughts on the other factors? For all the cases, this all comes down to what affects the thickness of the boundary layer at the surface (and the penetration depth of the effective temperature variation). For a better physical feel for this, see Transport Phenomena by Bird, Stewart, and Lightfoot.
Unfortunately, I can't start on that because we haven't taken the basics of fluids which is the next chapter.

I will base my prediction on this: The more time the fluid spends near the surface, The more heat it gets the lower the temperature gradient is then less heat it will get each sec.

Velocity: The faster it is the faster the hot air gets replaced which means more heat transfer. Proportional
Viscosity: The more it is the harder for the fluid to move which means more time at the surface = less heat transfered. Inversely proportional
Specific heat: the higher it is, The lower temperature changes which means more heat transferred per sec
conduction coeff: The higher the better because the fluid can spread heat more.
Geometric shape of the surface: If we take a flat surface with 2 sides the sum of area is equal to 1 m^2 and a sphere with area 1 m^2. I would assume the hot air stays on the bottom side of those two shapes for a longer time but the sphere is less than flat surface. So the sphere has a higher coeff?
If the flat surface is with 1 side only 1 m^2 I would assume it would have the higher coeff
The angle of the surface: If we apply the same rule with a flat surface (2 sides) The heat transfer is highest in the vertical position. If one side, I would assume the horizontal..?
Laminar or turbulence flow: Turbulence flow has the higher h because the interaction between the fluid is higher so heat is spread faster, But in laminar the only way heat can cross is by conduction between the layers or molecules effusion.
Type of convection: Free or forced convection.
free convection takes more time to replace the hot air with a cold one, it depends on the density difference. However forced convection replaces the air faster so more heat.The 1 sided and 2 sided argument seems a bit shaky.

Conductivity said:
Unfortunately, I can't start on that because we haven't taken the basics of fluids which is the next chapter.

I will base my prediction on this: The more time the fluid spends near the surface, The more heat it gets the lower the temperature gradient is then less heat it will get each sec.

Velocity: The faster it is the faster the hot air gets replaced which means more heat transfer. Proportional
Viscosity: The more it is the harder for the fluid to move which means more time at the surface = less heat transfered. Inversely proportional
Specific heat: the higher it is, The lower temperature changes which means more heat transferred per sec
conduction coeff: The higher the better because the fluid can spread heat more.
Geometric shape of the surface: If we take a flat surface with 2 sides the sum of area is equal to 1 m^2 and a sphere with area 1 m^2. I would assume the hot air stays on the bottom side of those two shapes for a longer time but the sphere is less than flat surface. So the sphere has a higher coeff?
If the flat surface is with 1 side only 1 m^2 I would assume it would have the higher coeff
The angle of the surface: If we apply the same rule with a flat surface (2 sides) The heat transfer is highest in the vertical position. If one side, I would assume the horizontal..?
Laminar or turbulence flow: Turbulence flow has the higher h because the interaction between the fluid is higher so heat is spread faster, But in laminar the only way heat can cross is by conduction between the layers or molecules effusion.
Type of convection: Free or forced convection.
free convection takes more time to replace the hot air with a cold one, it depends on the density difference. However forced convection replaces the air faster so more heat.The 1 sided and 2 sided argument seems a bit shaky.
Most of this seems pretty perceptive to me. You certainly have good intuition. So, let me try to help you with this in another way.

Suppose you have purely conductive heat transfer to a semi-infinite slab of material. The material is initially all at temperature T0, but at time t = 0, you raise the temperature at the surface to T1, and you hold it at that value for all subsequent time. What will the temperature profiles look like as a function of time.

This is just the beginning of where I want to take you. So, please bear with me.

Chestermiller said:
Most of this seems pretty perceptive to me. You certainly have good intuition. So, let me try to help you with this in another way.

Suppose you have purely conductive heat transfer to a semi-infinite slab of material. The material is initially all at temperature T0, but at time t = 0, you raise the temperature at the surface to T1, and you hold it at that value for all subsequent time. What will the temperature profiles look like as a function of time.

This is just the beginning of where I want to take you. So, please bear with me.
Is this in steady state? Because we have only taken steady state.

Or do you want me to formulate a function with specific heat to get the temperature as a function of time?

Conductivity said:
Is this in steady state? Because we have only taken steady state.

Or do you want me to formulate a function with specific heat to get the temperature as a function of time?
It's not for steady state. It's transient. But you have the right idea. I would like you to only answer the question qualitatively.

Chestermiller said:
It's not for steady state. It's transient. But you have the right idea. I would like you to only answer the question qualitatively.
Sorry for being late.
For the mathematical part: (No idea where to go after this)
##\frac {d^2T} {dx^2} \frac{K}{cp} = \frac{dT}{dt}##

Heat will start to flow in the slab and the temperature of each point will increase gradually. The increase rate is lower as time passes. So the slope of dT/dt for a specific point will decrease over time.

Last edited:
Conductivity said:
Sorry for being late.
For the mathematical part: (No idea where to go after this)
##\frac {d^2T} {dx^2} \frac{K}{cp} = \frac{dT}{dt}##

Heat will start to flow in the slab and the temperature of each point will increase gradually. The increase rate is lower as time passes. So the slope of dT/dt for a specific point will decrease over time.
The latter part is what I was looking for. So, initially you have a very sharp temperature profile (spatially) near the wall, running from T1 at the boundary to T0 farther from the boundary. As time progresses the temperature profile "accordions" out, so that it has basically the same kind of shape, but encompasses a greater region of space adjacent to the boundary. The shape of the profile is of the form: $$T=T_0+(T_1-T_0)f{\left(\frac{x}{\sqrt{\alpha t}}\right)}$$where x is the distance from the boundary, ##\alpha=\frac{K}{c\rho}##, t is time, and the function f describes the (roughly parabolic) shape of the temperature profile. So, at any time, the heat flux at the boundary is:$$q=-\frac{K}{\sqrt{\alpha t}}f'(0)(T_1-T_0)$$From this, it follows that the heat transfer coefficient at the boundary is given by:
$$h=\frac{K}{\sqrt{\alpha t}}(-f'(0))$$
So, for this system, because the "thickness of the thermal boundary layer" is increasing with time, the heat transfer coefficient at the boundary is decreasing with time. The thermal boundary layer is the effective region over which the temperature goes from T1 to T0. In this system, this occurs over a distance on the order of ##\sqrt{\alpha t} ##.

Does this make sense so far?

Chestermiller said:
The latter part is what I was looking for. So, initially you have a very sharp temperature profile (spatially) near the wall, running from T1 at the boundary to T0 farther from the boundary. As time progresses the temperature profile "accordions" out, so that it has basically the same kind of shape, but encompasses a greater region of space adjacent to the boundary. The shape of the profile is of the form: $$T=T_0+(T_1-T_0)f{\left(\frac{x}{\sqrt{\alpha t}}\right)}$$where x is the distance from the boundary, ##\alpha=\frac{K}{c\rho}##, t is time, and the function f describes the (roughly parabolic) shape of the temperature profile. So, at any time, the heat flux at the boundary is:$$q=-\frac{K}{\sqrt{\alpha t}}f'(0)(T_1-T_0)$$From this, it follows that the heat transfer coefficient at the boundary is given by:
$$h=\frac{K}{\sqrt{\alpha t}}(-f'(0))$$
So, for this system, because the "thickness of the thermal boundary layer" is increasing with time, the heat transfer coefficient at the boundary is decreasing with time. The thermal boundary layer is the effective region over which the temperature goes from T1 to T0. In this system, this occurs over a distance on the order of ##\sqrt{\alpha t} ##.

Does this make sense so far?
Just to ensure I got everything right. At the beginning if I graph T(x), I get a graph that kinda of looks like 1/x. But as time progresses we get a function that looks like a bell curve.

I just didn't understand this bit: In this system, this occurs over a distance on the order of ##\sqrt{\alpha t} ##. Everything else is fine :)

Conductivity said:
Just to ensure I got everything right. At the beginning if I graph T(x), I get a graph that kinda of looks like 1/x. But as time progresses we get a function that looks like a bell curve.
No. The temperature profile looks approximately like this:

##T=T_0+(T_1-T_0)\left(1-\frac{x}{k\sqrt{\alpha t}}\right)^2## for ##x<k\sqrt{\alpha t}##

and

##T=T_0## for ##x>k\sqrt{\alpha t}##

where k is a constant on the order of 2.
I just didn't understand this bit: In this system, this occurs over a distance on the order of ##\sqrt{\alpha t} ##. Everything else is fine :)
This should now be pretty obvious from the answer I just gave.

Chestermiller said:
No. The temperature profile looks approximately like this:

##T=T_0+(T_1-T_0)\left(1-\frac{x}{k\sqrt{\alpha t}}\right)^2## for ##x<k\sqrt{\alpha t}##

and

##T=T_0## for ##x>k\sqrt{\alpha t}##

where k is a constant on the order of 2.

This should now be pretty obvious from the answer I just gave.
I didnt mean that it is 1/x but I mean it looks like 1/x as it starts T1 and gradually falls to T0. The same applies for bell curve.

Where can I read the proof of this.

Thank you, I understand now.

Conductivity said:
I didnt mean that it is 1/x but I mean it looks like 1/x as it starts T1 and gradually falls to T0. The same applies for bell curve.

Where can I read the proof of this.

Thank you, I understand now.
As I said, this is only an approximate solution. The exact solution involves a complimentary error function, and you can read about the derivation in Transport Phenomena by Bird, Stewart, and Lightfoot.

So, for the temperature profile I gave, what do you get for the temperature gradient at the wall? What is the heat flux at the wall? What is the heat transfer coefficient?

Chestermiller said:
No. The temperature profile looks approximately like this:

##T=T_0+(T_1-T_0)\left(1-\frac{x}{k\sqrt{\alpha t}}\right)^2## for ##x<k\sqrt{\alpha t}##

and

##T=T_0## for ##x>k\sqrt{\alpha t}##

where k is a constant on the order of 2.

This should now be pretty obvious from the answer I just gave.
##\frac{dT}{dx} = 2(T_1-T_0)\left(1-\frac{x}{k\sqrt{\alpha t}}\right) \frac{-1}{k\sqrt{\alpha t}} ##

## q = 2\frac{1}{\sqrt{\alpha t}} A (T_1 -T_o) (1-\frac{x}{k\sqrt{\alpha t}}) ##

## h =2\frac{1}{\sqrt{\alpha t}} (1-\frac{x}{k\sqrt{\alpha t}})##

Conductivity said:
##\frac{dT}{dx} = 2(T_1-T_0)\left(1-\frac{x}{k\sqrt{\alpha t}}\right) \frac{-1}{k\sqrt{\alpha t}} ##

## q = 2\frac{1}{\sqrt{\alpha t}} A (T_1 -T_o) (1-\frac{x}{k\sqrt{\alpha t}}) ##

## h =2\frac{1}{\sqrt{\alpha t}} (1-\frac{x}{k\sqrt{\alpha t}})##
I asked for the values at the wall. Also, the heat transfer coefficient applies only at the wall.

I'd like to try something else with you. For the approximation ##T=T_0+(T_1-T_0)\left(1-\frac{x}{\delta}\right)^2## (where ##\delta## is the "boundary layer thickness," a function of time t), what do you get if you substitute this into your differential equation?

By the way, I still want to show you how to adapt these results to determine how the heat transfer is affected by fluid flow in a steady state fluid flow heat transfer situation.

Chestermiller said:
I asked for the values at the wall. Also, the heat transfer coefficient applies only at the wall.

I'd like to try something else with you. For the approximation ##T=T_0+(T_1-T_0)\left(1-\frac{x}{\delta}\right)^2## (where ##\delta## is the "boundary layer thickness," a function of time t), what do you get if you substitute this into your differential equation?

By the way, I still want to show you how to adapt these results to determine how the heat transfer is affected by fluid flow in a steady state fluid flow heat transfer situation.
Sorry :C
##\frac{dT}{dx} = 2(T_1-T_0)\frac{-1}{k\sqrt{\alpha t}} ##
## q = 2\frac{1}{\sqrt{\alpha t}} A (T_1 -T_o)##
## h =2\frac{1}{\sqrt{\alpha t}}##

You mean like this? ##\frac{dT}{dt} =\frac{ 2 K (T_1 - T_o)}{c p \delta^2} ##

Conductivity said:
Sorry :C
##\frac{dT}{dx} = 2(T_1-T_0)\frac{-1}{k\sqrt{\alpha t}} ##
## q = 2\frac{1}{\sqrt{\alpha t}} A (T_1 -T_o)##
## h =2\frac{1}{\sqrt{\alpha t}}##

You mean like this? ##\frac{dT}{dt} =\frac{ 2 K (T_1 - T_o)}{c p \delta^2} ##
The left side too...and you're missing a minus sign.

Chestermiller said:
The left side too...and you're missing a minus sign.
## \frac{d\delta}{dt} = \frac{K}{cp(\frac{x}{\delta} -1)} ##

Conductivity said:
## \frac{d\delta}{dt} = \frac{K}{cp(\frac{x}{\delta} -1)} ##

After substituting into the differential equation, I get the following: $$\left(1-\frac{x}{\delta}\right)\frac{x}{\delta}\left[\delta\frac{d\delta}{dt}\right]=\alpha$$
Please correct me if I'm wrong. But, if I'm right, we see that our approximation, as expected, does not satisfy the differential equation exactly. However, we can still make progress by, at least, satisfying the equation on average. This can be accomplished by integrating both sides of the equation from x = 0 to x = ##\delta## (i.e., the edge of the boundary layer), and then dividing by ##\delta##. This is referred to mathematically as a weighted residual method. What do you get when you do this?

Chestermiller said:

After substituting into the differential equation, I get the following: $$\left(1-\frac{x}{\delta}\right)\frac{x}{\delta}\left[\delta\frac{d\delta}{dt}\right]=\alpha$$
Please correct me if I'm wrong. But, if I'm right, we see that our approximation, as expected, does not satisfy the differential equation exactly. However, we can still make progress by, at least, satisfying the equation on average. This can be accomplished by integrating both sides of the equation from x = 0 to x = ##\delta## (i.e., the edge of the boundary layer), and then dividing by ##\delta##. This is referred to mathematically as a weighted residual method. What do you get when you do this?
I still get this:
##\frac{dT}{dx} = 2 ( T_1- T_0) (1-\frac{x}{\delta}) \frac{-1}{\delta} ##
##\frac{d^2T}{dx^2} = 2 ( T_1- T_0) \frac{1}{\delta^2} ##
## \frac{dT}{dt} = 2 (T_1 -T_0) (1-\frac{x}{\delta}) \frac{1}{\delta^2} \frac{d\delta}{dt} ##

## \frac{d\delta}{dt} = \frac{K}{cp (1-\frac{x}{\delta})} ##

You lost an x in your partial differentiation with respect to t.

Let's try this a little differently. The method I'm going to describe now is called the "energy integral method." If we first integrate our differential equation between x = 0 and x = infinity, we obtain: $$\frac{\partial}{\partial t}\left[\int_0^{\infty}{\rho C(T-T_0)dx}\right]=\left(-K\frac{\partial T}{\partial x}\right)_{x=0}$$The term in brackets on the left side of this equation is the total increase in internal energy of our slab (per unit area of surface) from time zero to time t. So the left hand side represents the rate of change of internal energy (per unit area of surface). The right hand side of the equation is the instantaneous heat flux at the surface at time t. So, as expected, the two sides of the equation match.

What do you get for the term in brackets if you substitute our approximate equation? What do you get for the right hand side of the equation?

Chestermiller said:
You lost an x in your partial differentiation with respect to t.

Let's try this a little differently. The method I'm going to describe now is called the "energy integral method." If we first integrate our differential equation between x = 0 and x = infinity, we obtain: $$\frac{\partial}{\partial t}\left[\int_0^{\infty}{\rho C(T-T_0)dx}\right]=\left(-K\frac{\partial T}{\partial x}\right)_{x=0}$$The term in brackets on the left side of this equation is the total increase in internal energy of our slab (per unit area of surface) from time zero to time t. So the left hand side represents the rate of change of internal energy (per unit area of surface). The right hand side of the equation is the instantaneous heat flux at the surface at time t. So, as expected, the two sides of the equation match.

What do you get for the term in brackets if you substitute our approximate equation? What do you get for the right hand side of the equation?
I am really sorry I don't know partial differentiation yet. I really appreciate the time you spent on this thread and really sorry for not knowing it.

Conductivity said:
I am really sorry I don't know partial differentiation yet. I really appreciate the time you spent on this thread and really sorry for not knowing it.
No problem. I think you have the general idea now. If you have any other questions about heat transfer coefficient, please don't hesitate to ask.

Conductivity
Chestermiller said:
No problem. I think you have the general idea now. If you have any other questions about heat transfer coefficient, please don't hesitate to ask.
Can't thank you more. Thank you so much.

## 1. What is a heat transfer coefficient?

A heat transfer coefficient is a measure of how easily heat is transferred between two substances or systems. It is typically defined as the amount of heat that is transferred per unit area per unit temperature difference.

## 2. How is the heat transfer coefficient calculated?

The heat transfer coefficient is calculated by taking into account factors such as the materials involved, the surface area of contact, and the temperature difference between the two substances. It can also be determined experimentally through various tests and measurements.

## 3. What factors affect the heat transfer coefficient?

There are several factors that can affect the heat transfer coefficient, including the properties of the materials involved, the surface roughness, the flow rate of the substances, and the temperature difference between the two substances.

## 4. Why is understanding the heat transfer coefficient important?

Understanding the heat transfer coefficient is important because it helps us predict and control the rate of heat transfer in various systems. This is crucial in many engineering and scientific applications, such as designing efficient heat exchangers and managing thermal energy in buildings and industrial processes.

## 5. How can the heat transfer coefficient be improved?

The heat transfer coefficient can be improved by optimizing the factors that affect it, such as increasing the surface area of contact, using materials with higher thermal conductivity, and reducing the temperature difference between the two substances. Additionally, the use of insulation and other heat transfer barriers can also help improve the overall heat transfer coefficient.

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