BENEMÉRITA UNIVERSIDAD AUTÓNOMA

DE PUEBLA

FACULTAD DE INGENIERÍA QUÍMICA

LABORATORIO DE INGENIERÍA II

PRÁCTICA 1: Conducción Lineal de Calor

EQUIPO 4:

Brenda García Hernández

Brenda Moreno Lazcano

Jocelyn Trejo Gallardo

Amairani Vázquez Pérez

PROFESOR: FERNANDO HUMBERTO DEL VALLE SOTO

Heroica Puebla de Zaragoza a 10 de septiembre de 2015

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LINEAR CONDUCTIVITY

OBJECTIVES

GENERAL:

Determine the linear coefficient of thermal expansion given the length alteration that accompanies a specified temperature change.

SPECIFIC:

To verify the phenomenon of heat transfer for linear conduction without not knowing the presence of other forms of heat losses.

To calculate with experimental information the thermal conductivity of different materials.

To check that the thermal conductivity is a property independent from the geometry.

Analyzes the effect of the thermal conductivity on the temperature distribution.

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INDEX

OBJECTIVES...................................................................................................................................2

INTRODUCTION.............................................................................................................................3

THEORICAL FRAMEWORK..........................................................................................................4

METHODOLOGY............................................................................................................................6

DISCUSSION AND RESULTS......................................................................................................9

CONCLUSIONS............................................................................................................................18

BIBLIOGRAPHY............................................................................................................................19

INTRODUCTION

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Thermal Conductivity The transport of thermal energy from high- to low-temperature regions of a material is termed thermal conduction. For steady-state heat transport, the flux is proportional to the temperature gradient along the direction of flow; the proportionality constant is the thermal conductivity. For solid materials, heat is transported by free electrons and by vibrational lattice waves, or phonons. The high thermal conductivities for relatively pure metals are due to the large numbers of free electrons, and also the efficiency with which these electrons transport thermal energy. By way of contrast, ceramics and polymers are poor thermal conductors because free-electron concentrations are low and phonon conduction predominates.

Thermal conduction is the phenomenon by which heat is transported from high to low-temperature regions of a substance. The property that characterizes the ability of a material to transfer heat is the thermal conductivity. It is best defined 720 in terms of the expression where q denotes the heat flux, or heat flow, per unit time per unit area (area being taken as that perpendicular to the flow direction), k is the thermal conductivity, and dT/dx is the temperature gradient through the conducting medium. The units of q and k are W/m2 (Btu/ft2 -h) and W/m-K (Btu/ft-h-F), respectively. is valid only for steady-state heat flow, that is, for situations in which the heat flux does not change with time. Also, the minus sign in the expression indicates that the direction of heat flow is from hot to cold, or down the temperature gradients similar in form to Fick’s first law for atomic diffusion. For these expressions, k is analogous to the diffusion coefficient D, and the temperature gradient parallels the concentration gradient, dC/dx.

Mechanisms of Heat Conduction

Heat is transported in solid materials by both lattice vibration waves (phonons) and free electrons. A thermal conductivity is associated with each of these mechanisms, and the total conductivity is the sum of the two contributions, or where kl and ke represent the lattice vibration and electron thermal conductivities, respectively; usually one or the other predominates. The thermal energy associated with phonons or lattice waves is transported in the direction of their motion. The kl contribution results from a net movement of phonons from high to low-temperature regions of a body across which a temperature gradient exists. Free or conducting electrons participate in electronic thermal conduction. To the free electrons in a hot region of the specimen is imparted a gain in kinetic energy. They then migrate to colder areas, where some of this kinetic energy is transferred to the atoms themselves (as vibrational energy) as a consequence of collisions with phonons or other imperfections in the crystal. The relative contribution of k to the total thermal conductivity increases with increasing free electron

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concentrations, since more electrons are available to participate in this heat transference process.

THEORICAL FRAMEWORK

What is a thermocouple sensor?

A thermocouple is a sensor for measuring temperature, consisting of two different metals joined at one end. When the junction of the two metals is heated or cooled a voltage that can be correlated with temperature occurs. Thermocouple alloys are available for normal wire form.

What are the different thermocouple types?

A thermocouple is available in different combinations of metals or calibrations. The four most common calibrations are J, K , T and E. There are high temperature calibrations are R, S, C and GB. Each calibration has a different temperature range and environment, although the maximum temperature varies with the diameter of the wire used in the thermocouple. Although the thermocouple calibration dictates the temperature range, the maximum range is also limited by the diameter of the thermocouple wire.

How I choose a thermocouple type?

Because a thermocouple measures in wide temperature ranges and can be relatively rugged, thermocouples are used extensively in the industry. The following criteria were used to select a thermocouple:

Temperature range Chemical resistance of the thermocouple or sheath material Abrasion resistance and vibration Installation requirements ( it may be necessary to be compatible with

existing equipment ; existing holes may determine probe diameter )

Fourier´s law

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This law states that the flow of heat between two bodies is directly proportional to the temperature difference between them, and can only go in one direction: the heat can only flow from the hotter to the cooler body. The mechanical paths, however, are reversible: can ever imagine the reverse process. In his Analytical Theory of Heat, Fourier says: “There are a variety of phenomena that are not produced by mechanical forces, but result exclusively from the presence and heat buildup. This part of the Natural Philosophy cannot be explained under the

dynamic theories, but has particular his principles, using a similar method to the other sciences.

Suppose we put a substance in contact with two heaters at temperatures T1 and T2 and thermally isolated pockets of the rest of the universe, as indicated below:

There will be a flow of heat from the hot source to the cold source (T2 > T1). If we maintain a constant temperature of outbreaks, a stationary regime is reached. Experimentally it is found that the density of heat flux through any plane perpendicular to the z axis is proportional to the temperature gradient.

1dqAdt

=−k x=dTdz

Expression is known as Fourier Act and the coefficient of proportionality κ (with units Jm -1s -1K -1 in the international system) is known as thermal conductivity.Fourier's law applies to gases, solids and liquids, provided that the heat transfer occurs only by conduction (collisions between molecules or atoms forming substance), rather than radiation or convection (macroscopic movements due to density differences, as occurs in the ascent of hot air in the atmosphere). Clearly, conductivity coefficient values are very different in solids, liquids and gases due to density differences.The amount of heat transferred by conduction in a given direction is proportional to the area to be heat flow and how temperature decreases in said direction. That is, the amount of heat that will be lost depends on the material surface in contact with the flow and how much or little to lower the temperature in that direction. The constant of proportionality is the thermal conductivity.

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Figure 1. Equipment for lineal heat conduction practice.HT11 linear heat conduction HT11c.

METHODOLOGY

1. Connect thermocouple 1, 2, 3 corresponding to the hot zone and 6, 7,8 correspond to the cold zone of the team.

2. Determine the percentage of voltage heat.3. Allow the computer to stabilize to take readings in each of4. Thermocouples.5. Determine experimentally the distance between thermocouples.6. Take the readings corresponding to the values of each thermocouple, the7. Percentage of heat intensity, I and V.8. Increase the percentage of heat intensity , twice as9. growing and take appropriate readings

Heat transfer and thermodynamics HT11 linear heat conduction HT11c computer controlled linear heat conduction issue.

The armfield linear heat conduction accessory has been designed to demonstrate the application of the Fourier rate equation to simple steady-state conduction in one dimension. We can appreciate the equipment we worked with on the lab on Figure 1.

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The units can be configured as a simple plane wall of uniform material and constant cross sectional area or composite plane walls with different materials or changes in cross sectional area to allow the principles of heat flow by linear conduction to be investigated. Measurement of the heat flow and temperature gradient allows the thermal conductivity of the material to be calculated. The design allows the conductivity of thin samples of insulating material to be determined. (Figure2).

On the HT11c the heater power and the cooling water flow rate are controlled via the Ht10xc, either from the front panel or from the computer software. On the ht11 these are controlled manually.

Figure 2.Brass disc and thermocouples (on the equipment).

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Technical details

The accessory comprises a heating section and cooling section which can be simply clamped together or clamped whit interchangeable intermediate sections between them, as required. The temperature a difference created by the application of heat to one end of the resulting wall an cooling at the other end results in the flow of heat linearly throught the wall by conduction.

The heating section is manufactured from 25 mm diameter cylindrical brass bar with cartridge type electric heating element installed at one end. The heating element is operated at low voltage for increased operator safety and is pretected by a thermostat to prevent damage bye overheating.

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DISCUSSION AND RESULTS

Thermal Conduction calculation for Brass:

During the experiment, we observed several changes on the temperature values, but we waited until the equipment registered temperatures without variation which indicated we have reached steady state, then we proceed to take data for our calculations.

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The following data table shows the results of the laboratory practice:

Voltage 10 J/C

Intensity 0.35 C/s

Thermocouple Temperature (°C) Temperature (K) Distance1 16.0156 289.1656 0.0152 14.8437 287.9937 0.033 14.0625 287.2125 0.0456 10.7421 283.8921 0.067 9.375 282.525 0.0758 8.6914 281.8414 0.09

Now, making a graphic of Temperature versus distance with the obtained data we can note that the curve doesn’t describe a line, therefore we cannot calculate the slope directly.

So, the next step is to make a linear regression, obtaining the results below:

0 0.02 0.04 0.06 0.08 0.1278

280

282

284

286

288

290f(x) = − 107.328571428571 x + 291.073133333333

Temperature (K) vs Distance (m)

Temperature (°F)Linear (Temperature (°F))

Distance (m)

Tem

pera

ture

(K)

With this information, we can apply the next formula:

T1−T 2

x2−x1

k=QA

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Making a clearance

m= QAk

But first, we need to calculate Q and A, knowing that:

Q=V∗I

Then:

Q=(10JC

∗0.35CS )=3.5

JS

And for A, we know that:

A=π∗r2=π∗(0.0125m )2=4.9087x 10−4m2

Using the last equation of the line we obtained from the practice, we can take the slope as 107.33 positive for convenience. Then, making a clearance for K, we obtain:

k= QAm

And replacing values we have:

k=3.5JS

( 4.9087 x10−4m2 )(107.33° Km

)=66.43247372

Wm° K

The above result is not a real value; because we can say that in the transference area exists a heat loss, because the two surfaces are not in direct contact. Therefore, we proceed to calculate the heat loss value, we start making a Temperature vs Distance graphic with the first three points:

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0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05286

286.5

287

287.5

288

288.5

289

289.5

f(x) = − 65.1033333333335 x + 290.077033333333

Temperature (°F)

Linear (Tem-perature (°F))

Distance (m)

Tem

pera

ture

(K)

As the last time, we use the next formula:

m= QAk

With the clearances we have already made, and knowing that m=65.103 (positive for convenience) from the linear regression and that

Q=3.5JS

A=4.9087 x 10−4m2

Then we have:

k=3.5JS

( 4.9087 x10−4m2 )(65.103° Km

)=109.5217948

Wm° K

For the next three points we made the same steps. Making a graphic of Temperature versus distance of the others three points we obtain:

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0.055 0.06 0.065 0.07 0.075 0.08 0.085 0.09 0.095280.5

281

281.5

282

282.5

283

283.5

284

284.5

f(x) = − 68.3566666666669 x + 287.879583333333

Series2

Linear (Series2)

Distance (m)

Tem

pera

ture

(K)

Applying the formula again:

m= QAk

Also using Q and A (values obtained before):

Q=3.5JS

A=4.9087 x 10−4m2

Making a clearance, this time for Q, we obtain:

m∗A∗k=Q

Replacing the values in the formula

Q= (68.357 ° K )∗(4.9087 x 10−4m

2 )∗(109.5217948W

m2° K )

Q=3.674938176

Making the difference between these two numbers:

3.674938176W−3.5W=0.174938176W

The value we obtained represents the heat loss during the process.

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Thermal Conduction calculation for Stainless Steel:

Voltage 10 J/C

Intensity 0.33 C/s

Thermocouple Temperature (°C) Temperature (K) Distance1 22.1679 295.3179 0.0152 21.1914 294.3414 0.033 20.7031 293.8531 0.0456 9.375 282.525 0.097 8.39843 281.54843 0.1058 7.8125 280.9625 0.12

And making a graphic Temperature vs Distance we obtain:

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14270

275

280

285

290

295

300

Temperature (°K) vs Distance (m)

Temperature (°F)

Distance (m)

Tem

pera

ture

(K)

In the first graphic we can see that there is not thermocouple in the line where the specimen is placed and that’s the temperature we need to calculate K, we need the temperature in the beginning of the specimen and the ending; so, we made a linear regression of the first three points and after, for the other three points too.

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Now, we can calculate the distance from the first thermocouple till the first line of the specimen adding all distance, obtaining: 0.0525 m. replacing this value in the first linear regression equation as follows:

y=−48.827 x+295.97 y (° K )=−48.827 (0.0525 )+295.58y (° K )=293.01658 ° K

0.085 0.09 0.095 0.1 0.105 0.11 0.115 0.12 0.125280

280.5

281

281.5

282

282.5

283

f(x) = − 52.0833333333333 x + 287.147393333333

Distance (m)

Tem

pera

ture

(K)

The last point in the end of specimen is -0.0075 and the temperature in that point is calculated as follows:

y=−52.083x+287.15 y (° K )=−52.083 (−0.0075 )+287.15 y (° K )=287.5406K

And now we can put those points in a graphic.

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0.05 0.055 0.06 0.065 0.07 0.075 0.08 0.085284

285

286

287

288

289

290

291

292

293

294

f(x) = − 182.532666666666 x + 302.599545

Series2Linear (Series2)

With this information we can apply the next formula:

m= QA K

But first we need to calculate the Q and A like the next way:

Q=V∗I=(10JC

∗0.33CS )=3.3

JS

The area is calculated:

A=4.9087 x 10−4m2

Using the last formula:

K=3.3JS

( 4.9087 x10−4m2) (182.53° Km

)=36.8309

Wm°K

We can compare the results of our operations with the next table, which shows several materials and their thermal conductivity.

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As we can observe the values we calculate for Brass and Stainless steel are in the range the table shows.

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CONCLUSIONSWith the results it is found that the heat transfer by conduction is a process in which heat flows from a region of high temperature to a low temperature region within a medium or between different media in direct physical contact.

The thermal conductivity values depend on the material and temperature. A material will be a better driver of heat while major it is his conductivity of the same one. Due to the fact that the values of temperature were verified, is observed that a stationary condition exists since in the same one variation was not observed with regard to the time. With the results of the graphs obtained there was verified that the heat transfer is better to minor length.

BIBLIOGRAPHY

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Cengel, Y., “Transferencia de calor”, 3ª Edición, McGraw Hill, México (2004).

Incoprera, F. & Dewitt D., “Fundamentos de transferencia de calor”, Prentice-Hall, México (1999).

Mc Cabe, W., Smith, J. &Harriot, P., “Operaciones unitarias en ingeniería química, 6ª Edición, McGraw-Hill, México (2002).

P. A. Hilton LTD. “Experimental Operating and Maintenance Manual”. Heat Transfer Service Unit. November, 2000.