# Examples of Convection

We express the examples of convection heat transfer with definition, mathematical model of convection, coefficient of convection and characteristics of convection.. Lets Read Physics…

## Examples of Convection examples of convection heat transfer

TheÂ **Convection**Â is the heat transfer mechanism in which the kinetic energy is transported from one point to another through the movement of a fluid.

Depending on the movement of the fluid, it can beÂ **Natural Convection**Â , which is produced only by the differences in densities of matter;Â orÂ **Forced Convection**Â , in which matter is forced to move from one place to another, for example air turned into wind with a fan or water turned into a strong current by means of a centrifugal pump.

Convection only occurs in liquids and gases where atoms and molecules are free to move in the medium.Â In nature, most of the heat gained by the atmosphere by Conduction and Radiation near the surface is transported to other layers of the atmosphere by Convection.

## Mathematical Model of Convection

A model of heat transfer by Convection H, calledÂ **Newton’s Law of Cooling**Â , is:

**H = h * A * (TÂ _{A}Â – T)Â **

Where “h” is the Coefficient of Convection, in W / (mÂ ^{2}Â K), A is the surface that delivers heat with a temperature TÂ _{A}Â to the adjacent fluid, which is at Temperature T.

If the fluid gains heat from the surface, the convective heat flux is positive (H> 0).Â This would happen in the case where the Surface Temperature TÂ _{A}Â was higher than that of the fluid (TÂ _{A}Â > T).

If the fluid loses heat to the surface, the heat flux by convection is negative (H <0).Â This would happen in the case where the Surface Temperature TÂ _{A}Â was lower than that of the fluid (TÂ _{A}Â <T).

The value of the Coefficient of Convection hÂ **varies depending on the type of fluid and the type of convection.**

## Coefficient of Convection examples of convection heat transfer

When there isÂ **Natural Convection**Â , there will beÂ convection coefficientsÂ for theÂ **Gases**Â with valuesÂ **between h = 2 W / (mÂ ^{2}Â K) and h = 25 W / (mÂ ^{2}Â K)**Â .

When there isÂ **Natural Convection**Â , there will beÂ convection coefficientsÂ forÂ **Liquids**Â with valuesÂ **between h = 50 W / (mÂ ^{2}Â K) and h = 1000 W / (mÂ ^{2}Â K)**Â .

When there isÂ **Forced Convection**Â , there will beÂ convection coefficientsÂ forÂ **Gases**Â with valuesÂ **between h = 25 W / (mÂ ^{2}Â K) and h = 250 W / (mÂ ^{2}Â K)**Â .

When there isÂ **Forced Convection**Â , there will beÂ convection coefficientsÂ forÂ **Liquids**Â with valuesÂ **between h = 50 W / (mÂ ^{2}Â K) and h = 20000 W / (mÂ ^{2}Â K)**Â .

## Characteristics of Convection

The fluid that is in contact with the surface of the solidÂ **can be in laminar movement or in turbulent movement**Â , and this movement can beÂ **caused by external forces**Â , that is to say, be forced convection;Â **or by density gradients**Â induced by temperature differences, and it will be natural convection.Â In addition, it may be changing phase (boiling or condensation).

Convection is, by concept, energy transferred from a solid surface to a fluid in motion.Â Convection is the mechanism of heat transfer byÂ **mass movement or circulation within the substance**Â , it is a combination of energy transfer by random molecular movement (conduction) and volumetric movement of the fluid (advection).

## Examples of Convection

1.- The glass of a window is at 10 Â° C and its area is 1.2mÂ ^{2}Â .Â If the outside air temperature is 0 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 4 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 10 Â° C = 283 K

T = 0 Â° C = 273 K

A = 1.2 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [4 W / (mÂ ^{2}Â K)] * (1.2 mÂ ^{2}Â ) * (283 K – 273 K)

**H = 48 W = 48 J / s**

2.- The glass of a window is at 20 Â° C and its area is 1.5mÂ ^{2}Â .Â If the outside air temperature is 14 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 4.2 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 20 Â° C = 293 K

T = 14 Â° C = 287 K

A = 1.5 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [4.2 W / (mÂ ^{2}Â K)] * (1.5 mÂ ^{2}Â ) * (293 K – 287 K)

**H = 37.8 W = 37.8 J / s**

3.- The glass of a window is at 40 Â° C and its area is 2.0 mÂ ^{2}Â .Â If the outside air temperature is 10 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 5 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 40 Â° C = 313 K

T = 10 Â° C = 283 K

A = 2.0 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [5 W / (mÂ ^{2}Â K)] * (2.0 mÂ ^{2}Â ) * (313 K – 283 K)

**H = 300 W = 300 J / s**

4.- The glass of a window is at 60 Â° C and its area is 1.9 mÂ ^{2}Â .Â If the outside air temperature is 45 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 2 W / (mÂ ^{2}Â K).

The data are: examples of convection heat transfer

TÂ _{A}Â = 60 Â° C = 333 K

T = 45 Â° C = 318 K

A = 1.9 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [2 W / (mÂ ^{2}Â K)] * (1.9 mÂ ^{2}Â ) * (333 K – 318 K)

**H = 57 W = 57 J / s**

5.- The glass of a window is at 80 Â° C and its area is 1.55mÂ ^{2}Â .Â If the outside air temperature is 40 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 3 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 80 Â° C = 353 K

T = 40 Â° C = 313 K

A = 1.55 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [3 W / (mÂ ^{2}Â K)] * (1.55 mÂ ^{2}Â ) * (353 K – 313 K)

**H = 186 W = 186 J / s**

6.- The glass of a window is at 35 Â° C and its area is 1.8mÂ ^{2}Â .Â If the outside air temperature is 20 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 5.4 W / (mÂ ^{2}Â K).

The data are: xamples of convection heat transfer

TÂ _{A}Â = 35 Â° C = 308 K

T = 20 Â° C = 293 K

A = 1.8 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [5.4 W / (mÂ ^{2}Â K)] * (1.8 mÂ ^{2}Â ) * (308 K – 293 K)

**H = 145.8 W = 145.8 J / s**

7.- The glass of a window is at 65 Â° C and its area is 2.3mÂ ^{2}Â .Â If the outside air temperature is 15 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 4.8 W / (mÂ ^{2}Â K).

The data are: examples of convection heat transfer

TÂ _{A}Â = 65 Â° C = 338 K

T = 15 Â° C = 288 K

A = 2.3 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [4.8 W / (mÂ ^{2}Â K)] * (2.3 mÂ ^{2}Â ) * (338 K – 288 K)

**H = 552 W = 552 J / s**

8.- The glass of a window is at 75 Â° C and its area is 0.80 mÂ ^{2}Â .Â If the outside air temperature is 30 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 4.3 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 75 Â° C = 348 K

T = 30 Â° C = 303 K

A = 0.80 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [4.3 W / (mÂ ^{2}Â K)] * (0.80 mÂ ^{2}Â ) * (348 K – 303 K)

**H = 154.8 W = 154.8 J / s**

9.- The glass of a window is at 25 Â° C and its area is 0.60 mÂ ^{2}Â .Â If the outside air temperature is 0 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 3.9 W / (mÂ ^{2}Â K).

The data are:

TÂ _{A}Â = 25 Â° C = 298 K

T = 0 Â° C = 273 K

A = 0.60 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [3.9 W / (mÂ ^{2}Â K)] * (0.60 mÂ ^{2}Â ) * (298 K – 273 K)

**H = 58.5 W = 58.5 J / s**

10.- The glass of a window is at 10 Â° C and its area is 1.3mÂ ^{2}Â .Â If the outside air temperature is -10 Â° C, calculate the energy that is lost through convection every second.Â Consider that the coefficient is h = 4.7 W / (mÂ ^{2}Â K).

The data are: examples of convection heat transfer

TÂ _{A}Â = 10 Â° C = 283 K

T = -10 Â° C = 263 K

A = 1.3 mÂ ^{2}

Using Newton’s law of cooling:

**H = h * A * (TÂ _{A}Â – T)**

H = [4.7 W / (mÂ ^{2}Â K)] * (1.3 mÂ ^{2}Â ) * (283 K – 263 K)

**H = 122.20 W = 122.20 J / s**