ME320Experiment#2 With Analysis

8/10/2019 ME320Experiment#2 With Analysis

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Lab #2: Demonstration of Fouriers Laws

Guillermo Mata, Elias S. Lopez, Andres Chavez

October 20, 2014

Professor Nima Fathi

ME320 Wed. Group A

I. Abstract

In order to obtain the amount of heat transfer that takes place in a given material, a materials heat transfer

coefficient must be obtained. The purpose of this lab is to effectively demonstrate the direct correlation between a

materials temperature gradient versus distance, heat transfer area, and the proportionality these given parameters

have with a materials heat transfer coefficient. True theoretical values for heat transfer coefficients will not be

obtained due to contact resistance within the experimental apparatus. These concepts are fundamentally important in

order to employ engineering solutions to everyday problems that incorporate conduction and the analysis of such

studies.


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I. Introduction

Fouriers law states that heat transfer is directly proportional to the negative temperature gradient in a

material and the area normal to the direction of the temperature gradient. This is true for materials undergoing

conduction. Conduction is the process upon which internal energy is transferred by collisions of particles and

microscopic diffusion within a body undergoing a temperature gradient. Conduction can occur in different phases

like solids, liquids, and gases (as long a no bulk motion occurs in the fluid). In most cases, conduction occurs in

solids in contact with one another because typically small amounts of conduction occur in fluids.

Figure 1. Heat transfer for conduction

During conduction, the particles within a body are more actively energized on the hot end of a body, which

cause kinetic energy to transfer to other particles when collision occurs. Since solids are held in place by their

crystalline structure, an excited particle can cause energy to flow to a neighboring particle. Since atoms in a crystal

are not locked in place by rigid connections (and are more or less allowed to oscillate) the temperature difference in

a body can cause lattice vibration to form (heat transfer present).

Figure 2. Lattice vibration


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Nomenclature:

q = Heat transfer

k = Heat transfer coefficient

A = Heat transfer area

= Temperature differential with respect to distance

= Thermal resistance of conductionThe only equations used for this lab is the equation for heat transfer caused by conduction

(1)

(2)

III. Experimental Apparatus

Fouriers laws were demonstrated over a series of experiments, which included different materials and

initial conditions. The majority of experiments conducted involved the use of a brass rod that was split in two

sections fitted with a heater on one end and a heat sink on the other. The brass rods were insulated with a polymer

casing in order to provide accurate heat transfer data for brass and to reduce the amount of heat loss to the

surrounding. Heat paste was also used in order to reduce the amount of surface resistance between the two rods to

provide more adequate heat transfer. Each brass rod was press fitted with three thermocouples during the

instrumented rods creation. Three different sets of experiments where ran for this lab. The first set of experiments

conducted incorporated the basic experimental apparatus with a 25mm diameter brass insertion between the rods.

The second sets of experiments were the same as the first set but included a 25mm diameter stainless steel insertion

instead of a 25mm brass insertion. The last set of experiments was similar to the previous sets but included a

aluminum insertion instead of brass, between the brass rods. The different initial conditions set for the heat source

for every set/lab can be found in table 1.

Set # Experiment # Voltage (volts) Power (watts)

1 1 5 2.7

1 2 8 6.42

2 1 5 2.7

2 2 8 6.483 1 5 2.51

3 2 8 6.4

Table 1. Heat source initial conditions


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Each experiment was allowed to reach a steady point at which data was collected and analyzed using P.A.

Hilton software. The data was given in the form of temperatures and distances (temperatures were obtained and

recorded for all six thermocouples for every experiment).

III. Experimental Results

During the first section of the previous lab we learned how to obtain thermal conductivity of the brass rods.

Data was collected to determine the thermal conductivity of the brass rods. This was done by first obtaining

relationships between temperature and horizontal distance for known power inputs. The slopes yielded by the 8, 10,

and 14 volt inputs (relating temperature vs. distance) were then plotted as points on a graph vs. their corresponding

power for each slope relating the following relationship:

Now that the rates of change have been derived, and with the known energy inserted into the experiment, it

is easy to see that a constant value (which for ease of calculation will be called C) multiplied by the rate of change is

equal to the heat rate. First below it is demonstrated mathematically.

Figure 3: K determination for brass rods

-400 -350 -300 -250 -200 -1506

8

10

12

14

16

18

20

dT/dx (K/m)

Power(W)

dT/dx Vs. Power from experiments 1, 2 and 3

Slope of the best fit line is: -0.063642


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As seen the constant C is equal to the slope of the line, which in our case is -0.063642. Now that this value

of C has been derived, K can be obtained using the following relationship.

Knowing the thermal conductivity for the two brass rods, the thermal conductivity of any insert can be

calculated using thermal resistances. Since thermal resistances add directly in series, the following equation can be

used to relate power and thermal resistance.

( ) (

) ( )

Since different inserts with similar thicknesses and cross sectional areas were used in this lab, the previous

equation can be rearranged to solve for k given all other known parameters. This new equation can be seen below:

Using this newly formed equation, the thermal conduction for any insert can be calculated by using the

other parameters that were obtained during lab. An example calculation can be seen for how thermal resistances can

be used to solve for thermal conductivity of an insert.

The equation was used in order to calculate the thermal conductivity of stainless steel, brass, and aluminum

inserts at 5V and 8V for each. The results obtained using thermal resistance method for every insert at different

initial conditions can be seen in the following table.

Insert Voltage Kexp (using ( Stainless Steel 5V 11.41

Stainless Steel 8V 15.613

Brass 5V 14.116Brass 8V 22.864

Aluminum 5V 16.1416

Aluminum 8V 25.767

Table 2: Kexp obtained using thermal resistances

Another form of calculating the thermal conductivity is to use Fouriers law to directly solve the thermal

conductivity (k) for the insert using all known given parameters. The Insert resides between the two thermo couples

T3-T6, therefore the distance between these thermo couples will be used for in the following equation:


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A sample calculation for the thermal conductivity using Fouriers law can be seen below:

Fouriers law for conduction was used to calculate the values for thermal conductivity for every insert at

different initial conditions. The results obtained using Fouriers law can be seen in the following table:

Insert Voltage Kexp (using Fouriers law( Stainless Steel 5V 15.975


Brass 5V 19.96Brass 8V 29.4

Aluminum 5V 24.025

Aluminum 8V 34.329

Table 3: Kexp obtained using Fouriers law

The first Kexp values were obtained using thermal resistances. The equation that related power to thermal

resistance was manipulated to yield the thermal conductivity of the insert . were obtained throughexperimental means and were determined to be 129.65

(were used for the k values for the brass rods). We

recalculated the Kexp values using the theoretical value for k of the brass rods (Kreal = 125

). A sample

calculation can be seen below, followed by the table of new Kexp values determined using Kreal for the thermalconductivity of the brass rods.

Insert Voltage Kexp (using

(



Brass 5V 14.26

Brass 8V 23.245

Aluminum 5V 16.33

Aluminum 8V 26.25

Table 4: Kexp obtained using thermal resistances w/ Kreal


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After obtaining the thermal conductivity for the different inserts using different methods, the percent error

and correction factor were calculated for every Kexp that was yielded from previous equations. A sample calculation

for percent error and correction factor can be seen, followed by the table of listed correction factors and percent

error for every Kexp, below:

Table 5: correction factor and percent error for Part Aa)

Calculating Using Fouriers Law

5 Volts & 0.502 Amps 8 Volts & 0.81 Amps

Calculated k w/mk

Value

Percent Error % Calculated k w/mk

Value

Percent Erro

Aluminum 24.03 9.86 90 34.33 6.90 86

Brass 19.96 6.26 84 29.4 4.25 76

Stainless Steel 15.96 0.95 6 24.24 0.62 61

Table 6: correction factor and percent error for Part Ab)

Calculating Using Thermnal Resistance (Theoretical k = 125 w/mk)

5 Volts 8 Volts

Calculated k w/mk

Value


Value

Percent Erro

Aluminum 16.33 14.51 93 26.25 9.03 89

Brass 14.26 8.77 89 23.25 5.38 81

Stainless Steel 11.51 1.31 24 15.79 0.96 5

Table 5: correction factor and percent error for Part B

The following figures represent the data obtained during lab for temperature distribution as a function of

distance.

Calculating Using Thermal Resistance (Calculated k = 129.65 w/mk)

5 Volts 8 Volts

Calculated k w/mk

Value


Value

Percent Erro

Aluminum 16.14 14.68 93 25.76 9.20 89

Brass 14.12 8.85 89 22.86 5.47 82Stainless Steel 11.41 1.32 24 15.61 0.97 3


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Figure 4: Temperature vs Distance (stainless steel 5V)

Figure 5: Temperature vs Distance (stainless steel 8V)

y = -0.0049x + 1.4925

R = 0.9324

0

0.02

0.04

0.06

0.08

0.1

0.12

282 284 286 288 290 292 294 296 298 300 302

Distance(m)

Temp (K)

Tempereature vs. Distance

Stainless Steel 5V

y = -0.0027x + 0.8759

R = 0.955

0

0.02

0.04

0.06

0.08

0.1

0.12

280 285 290 295 300 305 310 315 320

Distance(m)

temp K

Temperature vs. Distance

Stainless Steel 8V


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Figure 6: Temperature vs Distance (Brass 5V)

Figure 7: Temperature vs Distance (Brass 8V)

y = -0.0056x + 1.6805

R = 0.9427

0

0.02

0.04

0.06

0.08

0.1

0.12

284 286 288 290 292 294 296 298 300 302 304

Distance(m)

Temp K


Brass 5V

y = -0.0034x + 1.07

R = 0.9476

0

0.02

0.04

0.06

0.08

0.1

0.12

285 290 295 300 305 310 315

Distance(m)

Temp K


Brass 8V


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Figure 8: Temperature vs Distance (Aluminum 5V)

Figure 9: Temperature vs Distance (Aluminum 8V)

y = -0.0064x + 1.934

R = 0.9564

0

0.02

0.04

0.06

0.08

0.1

0.12

284 286 288 290 292 294 296 298 300

Distance(m)

temp K


Aluminum 5V

y = -0.0038x + 1.1865

R = 0.9599

0

0.02

0.04

0.06

0.08

0.1

0.12

285 290 295 300 305 310 315

Distance(m)

temp K


Aluminum 8V


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IV. Analysis

As demonstrated in the previous section of this report the difference between the thermal conductivity values

calculated greatly differentiated from those that are used in the industry or academia as a standard. The values for

both brass and aluminum differentiated by more than 85% and the values for stainless steel only differentiated in

two of the methods by less than 5%.

Analyze thermal resistance method used (2 times 1 for calculated Kbrassrod and another for Kreal)

Analyze Fouriers law method used

Another notion that can be taken from the results of the experiment is that the values of the experimental thermal

conductivity appear to not differentiate from one another. This can be attributed to mistakes or flaws in the

equipment used for this experiment.

Further as seen in tables five, six and seven a correction factor was calculated ( . This correction error iscalculated in order to counteract those mistakes or flaws in the equipment as previously discussed. If one is to run

the experiment with the same conditions using that correction factor a theoretical value for thermal conductivity can

be calculated with the data obtained. In those same tables a percent error was also calculated. As seen this percent

error is fairly large for both aluminum and brass. Those errors can be attributed to many sources such as, the lack of

a proper connection, the amount of heat paste used, the condition of the heat paste, equipment flaws, and takingmeasurements before a stable condition was met in the experiment.

Analyze why stainless steel yielded much better results for K than any other insert.(maybe the system was

flawed and it played towards stainless steals favor)

V. Conclusion

In conclusion, our group was able to understand and apply the Thermal resistance method and the Fouriers

law to calculate the results of the experiments. However, the results of our tests had a considerable percent of error,

and we figured out that there are many variables that can change the result of our test. When dealing with real

engineering problems, we need to consider all the external variables that can have an affect on the experiments. As

in the previous report, we gave a realistic meaning to the equations that we learned in class. It was interesting to

observe the better behavior of Stainless Steel compared with Aluminum and Brass.

Finally, it was noted that is necessary to pay attention to all the mechanisms involved in any test, and is

important to have patience to collect the data. So, we will reach more accurate results.

VI. References

Fathi, N. (n.d.). Fall 2014 Handout #1 Albuquerque: UNM Mechanical Engineering Heat Transfer, Laboratory 1.

Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat Transfer and Mass Transfer, Seventh Edition. New

York: John Wiley & Sons, Inc.


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IV. Analysis (second part, you can add something to Eliass part)

The thermal conductivity of Stainless Steel, Brass, and Aluminum were calculated using the thermal

resistance method and the Fouriers law. In the experiments, these three materials passed through a voltage of 5

Volts and 8 Volts, so was possible to analyze the differences between the thermal conductivity in each case. As

showed in the figures 4 to 9, the temperature decreases in the material as the distance increases.

Using the thermal resistance method, we calculated the thermal conductivity of the materials in two times.

First, we used experimental values for K of the brass rods; after, we recalculated the Kexp using the theoretical

value for k of the brass rods. In this way, we see how much our experimental values were different from the real

values; we calculated the percent error and correction factor for every Kexp.

Another method used was the Fouriers law, wherewe solved directly the thermal conductivity (k) for the

insert, using all known given parameters.

In general, we can see that the values of K found using Thermal resistance method have a higher percent of

error than the values found using the Fouriers law, comparing with the real values. The Aluminum data has the

highest percent of error, and the Stainless Steel has the smaller percent of error. So, the correction factor is higher

for Aluminum, and is smaller for the Stainless Steel.

The percent of error in our experiments is large. It can be related to potential sources of error, such as the

amount of thermal paste used, dont let the thermo couples reach equilibrium, and read the temperature outputs too

soon. The influence of the external environment is small, but can also be considered.

As stated above, the Stainless Steel yielded much better results for K than any other insert. The reason for it

is that maybe the system was flawed and it played towards stainless steel favor, or this material was better connected

with the system, and so it showed results that are more accurate.

ME320Experiment#2 With Analysis

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