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Heat Transfer

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Heat Transfer Heat is the total internal kinetic energy of the atoms and molecules that make up a substance. Since heat is a form of energy, it is measured in Joules. 1 Joule = 1 N*m = 1 kg m/s 2 * m 1 calorie is the heat energy needed to raise 1 gram of water by 1 degree Celsius. 1 calorie = 4.186 Joules. 1 Calorie = 1000 calories. Two liters of boiling water has more heat (energy) that one liter of boiling water. Heat will not flow between two objects of the same temperature. Heat is really energy in the process of being transferred from one object to another because of the temperature difference between them. Method of Heat Transfer The transfer of heat is normally from a high temperature object to a lower temperature object. Heat transfer changes the internal energy of both systems involved according to the First Law of Thermodynamics. Heat can be transferred by: Conduction Convection Advection Radiation
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Page 1: Heat Transfer

Heat Transfer

Heat is the total internal kinetic energy of the atoms and molecules that make up a substance. Since heat is a form of energy, it is measured in Joules.

1 Joule = 1 N*m = 1 kg m/s2 * m 1 calorie is the heat energy needed to raise 1 gram of water by 1 degree Celsius. 1 calorie = 4.186 Joules. 1 Calorie = 1000 calories.

Two liters of boiling water has more heat (energy) that one liter of boiling water.Heat will not flow between two objects of the same temperature. Heat is really energy in the process of being transferred from one object to another because of the temperature difference between them. 

Method of Heat Transfer

The transfer of heat is normally from a high temperature object to a lower temperature object. Heat transfer changes the internal energy of both systems involved according to the First Law of Thermodynamics.

Heat can be transferred by:

Conduction Convection Advection Radiation

Page 2: Heat Transfer

Conduction

                   

 

Conduction is the transfer of heat within a substance, molecule by molecule. If you put one end of a metal rod over a fire, that end will absorb the energy from the flame (this is radiation transferring energy). The molecules at this end of the rod will gain energy and begin to vibrate faster. As they do their temperature increases and they begin to bump into the molecules next to them. The heat is being transfered from the warm end to the cold end.

                          

The measurement of how well a material can conduct heat depends on how it's molecules are structurally bonded together. This is a listing of the heat conductivity of various substances:

Substance Heat Conductivity

Still air at 20 °C 0.023

Dry Soil 0.25

Water at 20 °C 0.60

Snow 0.63

Wet Soil 2.1

Ice 2.1

Granite 2.7

Iron 80

Silver 427

Page 3: Heat Transfer

As you can see air does not conduct heat very well. This is the idea behind a styrofoam coolers, the air pockets between the styrofoam beads do not conduct heat very well. On the other hand, metals do conduct heat very well. This is why metal seems cold when you touch it. The metal molecules are conducting your body heat away from your hand quickly. 

Convection

Convection is heat transfer by the mass movement of a fluid in the vertical (up/down) direction. This type of heat transfer takes place in liquids and gases. This occurs naturally in our atmosphere.

Warm air is less dense than cold air, making cold air heavier than warm air. On a sunny day, the surface of the Earth is heated by radiation from the Sun. The thin layer of molecules touching the surface are heated byconduction. We know air is a poor conductor of heat, so this warm mass of air near the surface can not immediately transfer its heat away from the surface by conduction. This warm air mass is buoyant and wants to rise upward because it is less dense, the heavy cold air takes the place of the warm bubble. This rising warm light air is called a thermal in meteorology. 

In lecture you learned that the pressure of the air decrease with height. This is an important fact for the life of these rising thermals. Recall, P/ρT = R, from the ideal gas law. For a rising thermal, you can think of this thermal as a parcel or a balloon, there will be less pressure exerted on its wall as it rises. Therefore, the parcel will expand and the temperature within the parcel will decrease. 

These warm thermals cool as they rise. In fact, the cooling rate as a parcel rises can be calculated:

If the thermal consists of dry air, it cools at a rate of 9.8°C/km as it rises.

Page 4: Heat Transfer

Advection

Advection is the transfer of heat in the horizontal (north/east/south/west) direction. In meteorology, the wind transports heat by advection. This happens all the time on Earth, heat is transported in many ways. For example, wind blowing over a body of water will pick up evaporated water molecules and carry them elsewhere, when the air with these water molecules cools, the water will condense and release latent heat. The heat is being transfered by the wind. 

Advection is very similar to Convection, however, it is in the horizontal and not vertical.

Radiation

               

Radiation allows heat to be transfered through wave energy. These waves are called Electromagnetic Waves, because the energy travels in a combination of electric and magnetic waves. This energy is released when these waves are absorbed by an object. For example, energy traveling from the sun to your skin, you can feel your skin getting warmer as energy is absorbed.

The energy a wave carries is related to its wavelength (measured from crest to crest). Shorter wavelengths carry more energy than longer wavelengths. Wavelengths are measured in terms of meters:

1 (millimeter) mm = .001 m = 10-3 m 1 (micrometer) μm = .000001 m = 10-6 m 1 μm is 1 millionth of a meter. One-hundredth the diameter of a human hair. 1 (nanometer) nm = .000000001 m = 10-9 m

Page 5: Heat Transfer

When talking about electromagnetic waves it is sometimes easier to give them characteristics of particles, we call these particles photons. A photon of x-ray radiation carries more energy than a photon of visible light.           

All things with a temperature above absolute zero emit radiation. Everything, your body, your desk, your house, grass, snow, the atmosphere, the moon, they all emit a wide range of radiation. The source of this electromagnetic radiation are vibrating electrons that exist in every atom that makes an object.

Emitted radiation can be:

Absorbed

Increasing the internal energy of the gas molecules.

Reflected

Radiation is not absorbed or emitted from an object but it reaches the object and is sent backward. The Albedo represents the reflectivity of an object and describes the percentage of light that is went back.

Scattered

Scattered light is deflected in all directions, forward, backward, sideways. It is also called diffused light.

Transmitted

Radiation not absorbed, reflected, or scattered by a gas, the radiation passes through the gas unchanged

Page 6: Heat Transfer

Summary

Describe what is going on and how each method of heat transfer works in this example:

                     

Page 7: Heat Transfer

Law of Fourier

Conduction is the transfer of heat through a solid medium. The energy transfer depends on:

Temperature differences in the solid body The material’s thickness The material's thermal conductivity The material’s cross-sectional area

The basic heat conduction equation is also known as the Fourier law of heat conduction. Below is the conduction equation for a one-dimensional system, i.e., considering heat conduction in only one direction. (See Figure 1-12.) Two-and three-dimensional problems are beyond the scope of this course.

Page 8: Heat Transfer

Equation 1-1 applies only after the system has reached steady state. This means the temperature on the cold side is not going to increase as long as the temperature on the hot side doesn’t change. If a steady state is assumed, then the temperature gradient is constant. By temperature gradient we mean the temperature across a body. Generally, in conduction problems, we’ll assume steady state has been achieved. Otherwise the problems become exceedingly complex.

Finding properties of materials is essential in solving some fire dynamic problems. Table 1-2 lists the thermal properties of some common materials. Some properties are in the Appendix, Handbook of Fire Protection. We’ll also show other tables in later course modules. The following website is a database of thermal properties of many materials: 

Page 9: Heat Transfer

Example 1-1 After investigating a fire and in preparation for a trial you want to determine the heat transfer through a sheet of gypsum wallboard at a given stage of the fire. (See Figure 1-13.) You determined that the temperature on the hot side, T2, was 900°C and the temperature on the cold side was 100°C. The conductivity of gypsum wallboard is 4.8 x 10-4 kW/m°C (Table 1-2) and the thickness was ½-inch (0.0127 m.) To simplify calculations, assume the area is 1 m2.

There are several classes of conduction heat transfer problems important in fire dynamics solvable without computers. The first class of problems is an object being heated that has a high surface area to mass ratio and a high thermal conductivity. The fusible link of a sprinkler head has these characteristics. The rate of temperature rise of these elements is independent of the heat conduction through the element. The heat conduction is assumed to occur infinitely fast, so that the link or detector element is at the same temperature throughout at any given time. This type of problem is often called a "liquid heat capacity" or "infinite thermal conductivity" problem.

The second type of problem occurs when the object is relatively thick and a good insulator, and we are interested in the surface temperature early in the heating process. These types of problems are called "semi-infinite" or "infinite slab" problems and often occur in ignition and flame spread studies.

Law of Newton

Convection is the heat transfer occurring when a fluid (liquid or gas) heats a solid material. An example is the cooling of room air at the surface of a window during winter. The basic law of convection, otherwise known as Newton's law, states:

 

Where: q = heat transfer to a solid (kW, kJ/s or Btu/s)h = convective heat transfer coefficient (kW/m2°C)A = solid surface area (m2) or (ft2)Tf = temperature of fluid (°C or °K) (°F or °R)Ts = temperature of a solid (°C or °K) (°F or °R).

Page 10: Heat Transfer

To solve convection problems we must know the convective heat transfer coefficient. In general, it is a complex function of the geometry, fluid properties, fluid velocity, and film and surface temperatures. The solution of the convective heat transfer problem is generally beyond the scope of this course. However, we’ll cover a solution for it in Module III. This will be for a specific application.

Typical heat transfer coefficient from flames is 0.005-0.01 kW/m2°C. However for a fire plume impinging on a ceiling the heat transfer coefficient is 0.0005-0.05 kW/m2°C. (See Figure 1-14.)

Fluid flow (gases and liquids) may be either laminar or turbulent. Below are a couple of simple experiments to illustrate laminar and turbulent flow.

Page 11: Heat Transfer

Law of Stefan-Boltzmann

Radiation heat transfer occurs by electromagnetic waves, such as when the sun warms the earth. The heat energy radiated from an object is proportional to the fourth power of the absolute temperature. Radiation is transmitted in straight lines.

The fundamental radiation relationship, known as the Stefan-Boltzmann Law, is:

Radiation impinging on a body may be absorbed, reflected, or transmitted, depending on the nature of the material. The amount of energy absorbed is called the material’s absorptivity while the amount reflected is called reflectivity, and the amount transmitted is called transmissivity. The sum of the reflectivity, absorptivity, and transmissivity of a surface is equal to unity.

An ideal radiator absorbs all energy striking upon it. It reflects no radiation and is often called a blackbody radiator. So, a blackbody reflects no radiation but absorbs all radiation hitting it. A blackbody emits the maximum amount of radiant energy possible at any given temperature. For the case of most solid bodies, the transmissivity is zero. The emissivity of a real radiator (e.g., a flame) is the ratio of radiation emitted by the radiator to the amount of radiation emitted by a blackbody.

Page 12: Heat Transfer

An object's absorptivity is equal to its emissivity. Figure 1-15 illustrates these concepts. A blackbody would absorb all the radiant energy. But this is a real body so only part of the incident energy is absorbed. This amount is the total incident energy less the reflected energy (assuming the transmitted energy equals zero). A blackbody that has absorbed all the energy has emissivity =100% = 1.0. This real body absorbs only 5% so its emissivity = 5% = 0.05.

The emissivity (E) of an object relates to a real versus an ideal radiator. The emissivity of an ideal radiator, blackbody, is unity. Values for emissivity vary from 0 to 1. As stated above, emissivity of an object is the ratio between the radiation of a real body and an ideal blackbody at the same temperature. Since emissivity depends on wavelength, values for use in engineering analysis are generally average values over a wavelength range. A graybody represents a real body with an emissivity between 0 and 1. The emissivity will partially determine the amount of radiation reflected or absorbed.

The emissivity of a solid object is a strong function of the object's color and physical surface condition, i.e., rough or smooth. Rough materials radiate more heat than smooth materials. In a like fashion, dark colored materials radiate more heat than light colored materials. Therfore dark rough materials absorb more heat than light smooth materials.

Emissivities for solid objects vary from 0.1 for a smooth, highly polished metallic surface to 0.9 or higher for charcoal or carbon black. The emissivity of a structural steel member, for example, varies from 0.7 to 0.9. The emissivity of a 0.18-meter-thick flame of burning kerosene is approximately 0.37 (Drysdale, p. 165).

The emissivity of a flame is directly related to its thickness. This also partially explains the increase in radiation heat transfer of a very thick flame front as opposed to a line fire. In general, the emissivity of a compartment fire is considered to be unity. The compartment fire is assumed to be a cavity radiator. A cavity radiator when viewed from the environment approximates a blackbody; thus its emissivity is equal to 1.

When two objects of different temperatures are in close proximity, radiation heat transfer occurs. Radiation heat transfer between a flame and a structural is one example. The temperature differences between the two bodies as well as their respective emissivities affect the amount of heat transferred.

Radiation from a point source is emitted in all directions and travels in a straight line. Depending on the configuration of the bodies, not all radiation from one body will reach the second. To compensate for this, a configuration factor is introduced into the radiation heat transfer equation. The mechanism of determining configuration factors is essentially a problem in solid geometry involving the areas of the surfaces and at what angles they are to each other. These calculations can be complex and for the most part we will not cover them in this course. Later however, we’ll use some simple equations to determine the radiation from flames.

Ignoring the configuration factor, the equation for radiation heat transfer between two bodies (assuming both are blackbodies, i.e., both emissivities are one, is:

Page 13: Heat Transfer

Be sure to use absolute temperature °K rather than °C.

Example 1-2 You are inspecting an industrial plant so you can make firefight plans for the facility. They have an open tank of kerosene within 5 feet (1.52 m) of a wooden wall. You want to know if the wall would ignite from burning kerosene if the keronsene ignites. To avoid configuration factors you assume both the wall and flames are fairly large. The kerosene flames are 990° C (990 + 273 = 1,263°K). You assume the wall temperature is 21°C (294°K). To simplify calculations you assume the area is 1 m2. If the minimum heat flux for the ignition of the wood is 10 kW/m2, will the wall ignite? 

No question about it. The wood wall would ignite.

Page 14: Heat Transfer

Sample Radiation Levels & Their Effects

To provide some physical meaning to the concept of thermal radiation, values for the effects of certain radiation levels are given in Table 1-3. For comparison purposes, a heat flux of 10 kW/m2would exist at a distance of 25 feet from a 10 foot square plane blackbody radiator at 2,160°R (1,200°K). Heat flux is the amount of energy transferred per unit area to the object in the left column (DiNenno, p. 32).

 

 

 

 

 

 

 

 

 

Page 15: Heat Transfer

Heat Transfer Through Composite Wall & Electrical Analogy

In practice different materials are constructed in layers to form a composite wall , e.g. wall of a building, furnace, pressure vessel. Consider a general case:

Heat transfer through composite wall

Page 16: Heat Transfer
Page 17: Heat Transfer

Heat Transfer Through A Wall

Heat transfer by conduction through a simple plane wall

A good way to start is by looking at the simplest possible case, a metal wall with uniform thermal properties and specified surface temperatures.

Fig. 2.5.5Conductive heat transfer through a plane wall

T 1 and T 2 are the surface temperatures either side of the metal wall, of thickness L; and the temperature difference between the two surfaces is ΔT.

Ignoring the possible resistance to heat flow at the two surfaces, the process of heat flow through the wall can be derived from Fourier's law of conduction as shown in Equation 2.5.1.

The term 'barrier' refers to a heat resistive film or the metal wall of a heat exchanger.

Equation 2.5.1

Where:

= Heat transferred per unit time (W)A = Heat transfer area (m²)k = Thermal conductivity of the barrier (W / m K or W / m°C)ΔT = Temperature difference across the barrier (K or °C)

= Barrier thickness (m)

It is possible to rearrange Equation 2.5.1 into Equation 2.5.6.

Page 18: Heat Transfer

Equation 2.5.6

Where:

= Heat transferred per unit time (W)A = Heat transfer area (m²)ΔT = Temperature difference across the barrier (°C)

= Barrier thickness / material thermal conductivity

It can be seen from their definitions in Equation 2.5.6 that   is the thickness of the barrier divided by its inherent property of thermal conductivity. Simple arithmetic dictates that if the length (Χ) of the barrier increases, the value   will increase, and if the value of the barrier conductivity (k) increases, then the value of   will decrease. A characteristic that would behave in this fashion is that of thermal resistance. If the length of the barrier increases, the resistance to heat flow increases; and if the conductivity of the barrier material increases the resistance to heat flow decreases. It can be concluded that the term   in Equation 2.5.6 relates to the thermal resistance of a barrier of known length.

The results of simple electrical theory parallel the equations appertaining to heat flow. In particular, the concept of adding resistances in series is possible, and is a useful tool when analysing heat transfer through a multi-layer barrier, as will be seen in a later section of this tutorial.

Equation 2.5.6 can now be restated in terms of thermal resistance, where:

as shown in Equation 2.5.7

Equation 2.5.7

Where:

Page 19: Heat Transfer

= Heat transferred per unit time (W)A = Heat transfer area (m²)ΔT = Temperature difference across the barrier (°C)R = Thermal resistance of the barrier (m² °C / W)

Thermal resistance denotes a characteristic of a particular barrier, and will change in accordance to its thickness and conductivity.

In contrast, the barrier's ability to resist heat flow does not change, as this is a physical property of the barrier material. This property is called 'thermal resistivity'; it is the inverse of thermal conductivity and is shown in Equation 2.5.8.

Equation 2.5.8

Where:

r = Thermal resistivity (m°C / W)k = Thermal conductivity (W / m°C)

Relating the overall resistance to the overall U value

Page 20: Heat Transfer

The usual problem that has to be solved in heat transfer applications is the rate of heat transfer, and this can be seen from the general heat transfer formula, Equation 2.5.3.

Equation 2.5.3

Where:U = The overall thermal transmittance (W / m °C)

By comparing Equations 2.5.3 and 2.5.7, it must be true that:

and therefore,

Equation 2.5.9

Therefore, U value (thermal transmittance) is the inverse of resistance.

Heat flow through a multi-layer barrier

As seen in Figure 2.5.4, a practical application would be the metal wall of a heat exchanger tube or plate which uses steam on one side to heat water on its other. It can also be seen that various other barriers are present slowing down the heat flow, such as an air film, a condensate film, a scale film, and a stationary film of secondary water immediately adjacent to the heating surface. 

These films can be thought of as 'fouling' the flow of heat through the barrier, and consequently these resistances are considered by heat exchanger designers as 'fouling factors'.

All of these films, in addition to the resistance of the metal wall, constitute a resistance to heat flow and, as in an electrical circuit, these resistances can be added to form an overall resistance.

Therefore:

Equation 2.5.10

Where:

R1 = Resistance of the air film

Page 21: Heat Transfer

R2 = Resistance of the condensate filmR3 = Resistance of the scale film on the steam sideR4 = Resistance of the of the metal wallR5 = Resistance of the scale film on the water sideR6 = Resistance of the product film

As resistance is   as shown in Equation 2.5.6, then Equation 2.5.10 can be rewritten as Equation 2.5.11:

Equation 2.5.11

Table 2.5.2Typical thermal conductivities of various materials

The thermal conductivities will alter depending on the film material (and temperature). For instance, air roughly has thirty times greater resistance to heat flow than water. For this reason, it is relatively more important to remove air from the steam supply before it reaches the heat exchanger, than to remove water in the form of wet steam. Of course, it is still sensible to remove wet steam at the same time.

The resistance of air to steel is roughly two thousand times more, and the resistance of air to copper is roughly twenty thousand times more. Because of the high resistances of air and

Page 22: Heat Transfer

water to that of steel and copper, the effect of small thicknesses of air and water on the overall resistance to heat flow can be relatively large.

There is no point in changing a steel heat transfer system to copper if air and water films are still present; there will be little improvement in performance, as will be proven in Example 2.5.5.

Air and water films on the steam side can be eradicated by good engineering practice simply by installing a separator and float trap set in the steam supply prior the control valve. Scale films on the steam side can also be reduced by fitting strainers in the same line.

Scale on the product side is a little more difficult to treat, but regular cleaning of heat exchangers is sometimes one solution to this problem. Another way to reduce scaling is to run heat exchangers at lower steam pressures; this reduces the steam temperature and the tendency for scale to form from the product, especially if the product is a solution like milk.


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