lesson 1 methods of heat transfer
DESCRIPTION
Lesson on modes of heat transfer.TRANSCRIPT
Koshu Takatsuji
1
Title: Methods of Heat Transfer Description:
One of the primary duties of a chemical engineer is to design properly working plants. As these plants depend heavily on maintaining a constant temperature, chemical engineers must understand how heat transfer works within certain materials.
Throughout this guide we will be referencing to a thing called “heat flow” or the rate at which energy goes from one place to another. It then follows that the units for heat flow to be energy/time or (J/s). With that in mind, I will now explain what the three modes of heat transfer are. Radiation: Radiation is the transfer of heat through electromagnetic waves and is inherently a form of energy released by anything that has a temperature above 0 Kelvin. You can imagine this as little heat waves emanating from anything that has a temperature.
Image from: http://enseki.or.jp/e_tokusei.php
As can be seen in the above, radiation waves are being emitted from the sun and are transferring packets of energy or heat from the sun to the earth. Mathematically, the amount of heat emitted through radiation is:
𝒒 = 𝑨𝝈𝑻𝟒 (Radiation Equation)
Here, q is the amount of heat emitted by the object and can be found by multiplying the area of the object (A) by 𝝈, the Stefan-‐Boltzman constant (!.!"!∗!"
!!!!!!!
), and the temperature (T) to the fourth power. It becomes evident from the equation that if you increase either the area of the object or the temperature of the object that the heat emitted through radiation increases.
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Conduction: Conduction is the mode of heat transfer through a substance and exists only when there is a difference in temperature within the object. You can imagine this as energy being diffused throughout an object going from a place of higher energy to lower energy.
Image from: http://www.universetoday.com/82331/what-‐is-‐conduction/
Using the above picture we say that:
𝒒 = 𝒌𝑨𝛁𝑻𝒅𝒙
(Conduction Equation General) Although we use a ∇T, to signify that heat flow is dependent on a temperature difference in the x,y, and z direction, we will mostly use it one dimension, so we can simplify it to:
𝒒 = 𝒌𝑨𝐝𝐓𝒅𝒙
(Conduction Equation Simplified) In other words, heat travels from the hotter place to the colder place dependent on the area it travels through (A), the thermal conductivity of the material (k) (how easily it allows heat to transfer), the difference in temperature from the hotter place to the colder place (dT), and the distance it has to travel (dx). Intuitively, it makes sense that the greater the thermal conductivity (k) that heat can flow more easily and the heat flow term will be greater. Likewise, a greater area allows for more heat to flow too. Convection: Convection is the transfer of heat due to movement within a fluid and is most easily imagined by wind moving heat from an area of high temperature throughout the rest of the system.
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Image from: http://www.me.rochester.edu/courses/ME223.jcl/modes/convection.htm To better ingrain this concept into your head, imagine a block of metal that is extremely hot (1000 degrees Celsius). This block is left outside which has an ambient temperature of 25 degrees Celsius. Because the block is hotter than the ambient temperature, heat is released from the block and exits into the environment where wind can then move the heat and disperse it throughout the environment. The idea where this happens is what convection is and is mathematically written in the form of:
𝒒 = 𝒉𝑨∆𝑻 (Convection Equation)
Here, like always, q is the heat flow, A is the area, ∆𝑻 is the difference in temperature between the environment and the solid, and h is the convective heat transfer coefficient. H, the convective heat transfer coefficient, is basically a measure that tells us how quickly the fluid (or in this case wind) is able to take away heat from the solid object. It is important to understand that the value of h is dependent on the fluid and not the solid. Types of questions: Now that we have a general idea of what the methods of heat transfer are, it becomes important to know how to apply them in real life situations. Below lists a couple of the main methods an engineer should know how to manipulate the equations and the general intuition that is needed to understand how the equations were manipulated.
1. Single Mode of Heat Transfer In single mode of heat transfer type problems, heat is transferred from one
area to another through one of the methods mentioned above: radiation, conduction, or convection. And the hard part of this type of question is identifying which method of heat transfer there is.
Examples: Question: You have an object with a surface area of 1m2 and a temperature of 200K, how much heat is it releasing. Solution: Identify that this is a radiation heat transfer question and refer to the radiation equation: 𝑞 = 𝐴𝜎𝑇!. From this equation, you know what the Area,
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Temperature, and 𝝈 (boltzman constant) is. You then plug and chug your values to get the heat flow. Question: You have a 1m by 1m by 1m metal box that has a temperature of 1000K on one side and 100K on the other side. The box is insulated on all sides expect for the side that is at 1000K and the side that is at 100K. The thermal conductivity of the metal box is 5W/mK. What is the heat flow? Solution: Identify that this is heat flow through a solid object and will therefore be a conduction question. Use the simplified conduction equation because it is easier to use: 𝑞 = 𝑘𝐴 !"
!". Here we know what the thermal conductivity is, the area the heat
flow will be traveling through (1m2), the change in temperature (dT) and the distance the heat has to travel through (dx). We can then find heat flow by plugging and chugging.
2. Multiple Heat Transfer Methods In Multiple Layers of Heat Transfer problems, we have to consider that heat flow dependent on multiple methods of heat transfer.
Example: Question:
Given the ambient temperature, the cold temperature, and the thermal conductivity of plastic, cork, and copper, the convection heat transfer coefficient, and the surface area find the heat flow through each material assuming steady state. Solution: It is important to first understand that in questions that involve multiple methods of heat transfer have only one uniform heat flow throughout the entire thing. That means the heat flow taken to the environment from the plastic through convection is the same as the heat flow from through conduction in the plastic, the heat flow through conduction in the cork and the heat flow through conduction in the copper. If heat flow was varying from material to material then the temperature
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of the material would be changing and that would mean that the system was not at steady state. Mathematically speaking, we can then say that:
𝑞 = 𝑞!"#$%!&'"# = 𝑞!"#$%&' = 𝑞!"#$ = 𝑞!"##$% where:
𝑞!"#$%!&'"# = ℎ𝐴(𝑇0− 𝑇1)
𝑞!"#$%&' = 𝑘!𝐴(T1− T2)(2𝑐𝑚)
𝑞!"#$ = 𝑘!𝐴(T2− T3)(2𝑐𝑚)
𝑞!"##$% = 𝑘!𝐴(T3− 𝑇!"#$)(2𝑐𝑚)
Here, you have four unknowns (q, T1, T2, and T3) and four equations so you would then be able to solve for the unknown value and therefore find the heat flow. Another way to solve this equation would be to use a pre-‐derived formula where hereon out we shall call “Ohm’s Law for Heat Transfer.” This equation is:
𝒒 = ∆𝑻
𝑹𝒕𝒉𝒆𝒓𝒎𝒂𝒍
(Ohm’s Law for Heat Transfer) Here, ∆𝑻 is the change in temperature of the final and initial place of heat flow and 𝑹𝒕𝒉𝒆𝒓𝒎𝒂𝒍 is the sum of all Rconvection, Rconduction, and Rradiation. Rconvection is always equal to !!! where h is the convection thermal coefficient and A is the area. Rconduction for a rectangular object is equal to !!" where L is the length the heat has to flow through the material, A is the surface area that the heat flow is able to travel through, and k is the thermal conductivity of the material. However, in pipes
Rconduction is equal to !"(
!!"#$%&'!!"#!$%
)
!!"# where k is the thermal conductivity coefficient, L is
the length the heat flow has to transfer through, routside is the outer radius, and rinside is the inner radius. Using this method, we can find the heat flow to be equal to:
𝒒 = ∆𝑻
𝑹𝒕𝒉𝒆𝒓𝒎𝒂𝒍=
𝑻𝟎− 𝑻𝒄𝒐𝒍𝒅𝑹𝒄𝒐𝒏𝒗𝒆𝒄𝒕𝒊𝒐𝒏 + 𝑹𝒑𝒍𝒂𝒔𝒕𝒊𝒄 + 𝑹𝒄𝒐𝒓𝒌 + 𝑹𝒄𝒐𝒑𝒑𝒆𝒓
where:
𝑹𝒄𝒐𝒏𝒗𝒆𝒄𝒕𝒊𝒐𝒏 = 𝟏𝒉𝑨
𝑹𝒑𝒍𝒂𝒔𝒕𝒊𝒄 = 𝟐𝒄𝒎𝒌𝟏𝑨
𝑹𝒄𝒐𝒓𝒌 = 𝟐𝒄𝒎𝒌𝟐𝑨
𝑹𝒄𝒐𝒑𝒑𝒆𝒓 = 𝟐𝒄𝒎𝒌𝟑𝑨
And because we know what h, k1, k2, k3, A, and T0 and Tcold are (because they will be given in the problem), we can find the heat flow.
Koshu Takatsuji
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3. Conduction: When the Area Changes