Different Types of Heat Transfer and How to Calculate their Values

Heat transfer like the name states is the way that heat transfers from one entity to another.  Heat is defined as a motion of molecules.  So, heat is anything above the absolute temperature of 0 Kelvin (-460 deg. F or -273.15 deg. C).  Thus, heat is relative.  Now, for heat to transfer, we need to have a difference in temperatures.  Energy like heat will always travel from the higher temperatures toward the lower temperatures; and there are three major ways that this can happen; conduction, convection, and radiation.  By the first Law of Thermodynamics, energy is neither created or destroyed, only transferred.  In this blog, I will explain each type of heat transfer.

  • Heat Transfer by Conduction

Conduction is about two stationary objects that are in contact.  The vibration of the molecules of one object will affect the vibration of the molecules adjacent to it.  Examples of conduction would be the cold air outside a window pane in a warm room.  Or a hot iron sitting on your wrinkled pants.  The heat from the hotter object will flow to the cooler object.  Thus, the hot object will become cooler while the cool object will become hotter.  This can be explained in Equation 1:

Equation 1 :

Q = -k * A * (T2 – T1) / x

Q – Heat Transfer (Watts)

k – Thermal Conductivity of material (Watts/K-m)

A – Heat Transfer Area (m2)

T2 – Temperature of object 2 (Kelvin, K)

T1 – Temperature of object 1 (Kelvin, K)

x – Material Thickness (m)

 

  • Heat Transfer by Convection

Convection describes heat transfer between surfaces that are in motion. This happens by moving a fluid which can be a liquid or air across an object.  There are two types, free convection and assisted convection.  Free convection is caused by gravity or buoyancy.  The basement will be cooler than the second floor because hot air will rise.  The density of warm air is less than cold air, so it will rise.  As for assisted or forced convection, the fluid will be moved over a surface with a pump, fan, or some other type of mechanical device.  An example of forced convection would be blowing your breath over your cup of coffee to cool.  Another example is the EXAIR Super Air Amplifier.  This device uses a small amount of compressed air to amplify the volume of ambient air.  When blown across a heated surface, it can cool the object quickly.   The calculation for heat transfer by convection is shown in Equation 2.

Equation 2:

Q = h * A * (T2 – T1)

Q – Heat Transfer (Watts)

h – Convective Coefficient (Watts/K-m2)

A – Heat Transfer Area (m2)

T2 – Temperature of object 2 (Kelvin, K)

T1 – Temperature of object 1 (Kelvin, K)

 

  • Heat Transfer by Radiation

Radiation refers to the transfer of heat through electromagnetic waves. Of course, the largest radiation source is our sun.  You can feel the difference when you wear a black shirt versus a white shirt.  Any object will adsorb, reflect, and transmit the radiation at different values depending on the color, surface finish, and material type.  This is called emissivity.  Emissivity, or e, is a coefficient that determines the ability of that object to adsorb the heat from radiation.  Thus, the value of e is between zero and one, and it is unitless.  By definition, 0 < e < 1.  Thus, a black object can have an emissivity of 1.  .  This is important for the EXAIR Cabinet Cooler Systems.  If the panel is outside and in full sun, we would use the color to determine the additional heat that can be absorbed by your electrical panel.  The equation for radiation heat transfer is shown in Equation 3.

Equation 3:

Q = e * A * s * ((Th)4 – (Tc)4)

Q – Heat Transfer (Watts)

e – Emissivity Coefficient

A – Heat Transfer Area (m2)

s – Stefan-Boltzmann Constant (5.6708 * 10-8 Watts/K4 m2)

Th – Temperature of hot body (Kelvin, K)

Tc – Temperature of cold body (Kelvin, K)

Thank you for reading the blog about the three main methods for heat transfer.  If you need to cool products, or remove the heat, EXAIR has many types of products to accomplish this.  You can contact an Application Engineer to discuss any of your applications dealing with heat and heat transfer.

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

 

Image courtesy of Arman Cagle, Creative Commons License

 

Calculating Force and Pressure For Air Nozzles

I assisted with an application where logs were being shaved to make thin laminate.  Because the logs were non-concentric or entirely smooth, the beginning of the sheet was riddled with scrapes and defects until it was about 8 foot (2.4 meters) long.  This was a very quick process, and once good product was coming from a shaved log, the machine would divert the material from the scrap bin to the production feed line with a nip roll.  At the speeds that the material was traveling, they needed to kept pressure on the leading edge of the sheet so that it would not “curl” up before the nip roll closed and grabbed the sheet. The drive rolls were pushing the laminate product toward the nip roll and they needed to keep the curl pushed flat along a plate and wondered if we had a product that could accomplish this.

We suggested a series of 2” flat air nozzles, model 1122, to keep the product pressed down on the plate with the force from the airflow.  In their trial runs, they tried to find the correct amount of air pressure to keep the product flat.  Once they found the pressure required, they noticed that the thin and delicate laminate was getting damaged.  Of course, it was just at the beginning length when it was being held in place as it slid into the nip roll, approximately 3 feet (0.9 meters).  Like any company, they did not want to waste any more product and wondered if we had anything else that we could recommend.

Thus a question was presented, and a solution was needed.  In thinking about this, it took me to my Michigan days where snow was abundant.  When walking on snow, you would fall through, but if you had snow shoes, you could stay on top of the snow.  This brought me to the factors of Pressure and Force.  Like with the laminate, if a smaller area does damage to the product (boots through the snow), can we expand the area to keep it from being damaged (snow shoes on top of the snow).

Snow Shoes
Snow Shoes

With the application, we needed to apply the same force on the material.  The equation for force is F = P *A (Equation 1), where F – Force, P – Pressure, and A – Area.

We can do an equality statement from Equation 1 which shows F = P1 * A1 = P2 * A2 (Equation 2).  The amount of pressure required from other EXAIR products can be determined, i.e. if I can double the surface area, then I can reduce the pressure by ½.  For model 1122, we can determine the pressure that was generated from Equation 1 and from the catalog data:

Imperial Units of Model 1122                                                      S.I. Units of Model 1122

F = 1.4 lbf (catalog)                                                                       F = 0.624 Kg (catalog)

A1 = Length X Width                                                                    A1 = Length X Width

= 5 inches X 2 inches (catalog)                                                   = 12.7 cm X 5.1 cm (catalog)

= 10 in^2                                                                                         = 64.8 cm^2

P1 = F/A1 (Rearranging Equation 1)                                         P1 = F/A1 (Rearranging Equation 1)

= 1.4 lbf/10 in^2                                                                            = 0.624 Kg/64.8 cm^2

= 0.14 PSI (pounds per in^2)                                                     = 0.0096 Kg/cm^2

Super Air Amplifier
Super Air Amplifier

Now that we have all the information from model 1122, we can determine the pressure required for a different product to keep the force the same.  With the 2” Super Air Amplifier, model 120022, it has a much larger footprint than the 2” flat air nozzle, model 1122.  So, with Equation 2, we can determine the amount of pressure required.  We will use model 1122 for our P1 and A1, and we will use model 120022 for P2 and A2.  From the catalog data for model 120022, we get a target area as follows:

 

Imperial Units for Model 120022                                               S.I. Units for Model 120022

A2 = pi * (diameter/2)^2                                                              A2 = pi * (diameter/2)^2

= 3.14 * (5.15 in/2)^2                                                                    = 3.14 * (13.1 cm/2)^2

= 20.8 in^2                                                                                      = 134.7 cm^2

 

When we apply the information to Equation 2, we get the following information:

 

Imperial Units                                                                                  S.I. Units

P2 = P1 * A1 / A2                                                                              P2 = P1 * A1 / A2

=(0.14 PSI * 10 in^2) / 20.8 In^2                                               =(0.0096 Kg/cm^2 * 64.8cm^2) / 134.7 cm^2

= 0.067 PSI                                                                                       =0.0046 Kg/cm^2

 

Now that the area was increased like the snow shoes above, the pressure was reduced and no additional waste was incurred.  Sometimes you have to think outside the igloo.  As with any application or product, you can always contact us at EXAIR for help.

 

John Ball
Application Engineer
johnball@exair.com
twitter.com/exair_jb

 

Image courtesy of VasenkaPhotography. Creative Comment License