MECH4406 Lab 1 - Austin Sandstrom

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MECH4406 Heat Transfer Experiment #1 Convection and Radiation Heat Transfer Experiment Determination of Emissivity and Verification of Stefan- Boltzmann Constant Austin Sandstrom, 101030411 austinsandstrom@cmail.carleton.ca Monday October 16 th , 2023 Section L3 – Group 1

PART A: 1. Data recorded during Part A of the experiment shown in Table 1. Heater Voltage (V) 7 Heater Current (amps) 0.72 Pressure (kPa) Element Temperature ( ℃ ) Vessel Temperature ( ℃ ) 102.04 85 18 73.72 89 18 48.57 96 18 24.97 103 19 9.93 112 19 0.045 136 20 Table 1: Data Recorded during Part A of the experiment 2. The energy balance equation for the heated element can be expressed through the use of either The Effective Power to The Heater equation and the energy balance of the heated element while considering heat transfer by means of convection and radiation, both shown below respectively as Equation 1 and Equation 2. ? ℎ?𝑎𝑡?𝑟 = 0.96?? − 0.0017(𝑇 𝐸 − 𝑇 ? ) (Eq. 1) ? ℎ?𝑎𝑡?𝑟 = ? 𝐶??𝑣 + ? 𝑅𝑎? (Eq. 2) 3. Calculations of requested results using data recorded during Part A of the experiment (One sample calculation shown per requested result) a) Calculation of the Heat Transfer Coefficient, h. Through the use of 𝑇 ? (mean temperature between the heated element and the vessel), the technique of interpolation on Table A-4 was implemented to determine the Thermal Conductivity ( k) , Specific Heat (𝐶 ? ) , and Dynamic Viscosity ( 𝜇) . For 𝑇 ? = 324.5 ? a sample calculation for interpolation to determine ? 𝑇? is shown below. Sample Calculation 1: Interpolation to Determine ? 𝑇? : ? 𝑇? = ( 𝑇 ? − 300K 350? − 300? ) (? 350𝐾 − ? 300𝐾 ) + ? 300𝐾 ? 324.5𝐾 = 28.11 ? ?? Which must be corrected as Table A-4 values are shown as ? ?𝐾 𝑥 10 3

? 324.5𝐾 = 0.02811 ? ?? Repeating the process for Specific Heat (𝐶 ? ) , and Dynamic Viscosity ( 𝜇 ), the calculation of the heat transfer coefficient was done through the following calculations. Sample Calculation 2: Thermal Expansion Coefficient ? = 1 𝑇 ? ? = 1 324.5? ? = 0.003082 Next, rearrangement of the Ideal Gas Law is completed to calculate the density. Sample Calculation 3: Rearrangement of the Ideal Gas Law to Determine Density: ?? = ??𝑇 And ? = ? ? Therefore ? = ?? ?𝑇 ? = (102.04 ??𝑎)(0.02896 ?𝑔 ??? ) (8.314 ? ??? ? ) (324.5?) ? = 1.095 ?𝑔 ? 3 Sample Calculation 4: Determination of the Thermal Diffusivity: ? = ? ?𝐶 ? ? = 0.02811 ? ?? (1.095 ?𝑔 ? 3 ) (1.0079 ?? ?𝑔? ) ? = 2.546𝑥10 −5 ? 2 ?

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Sample Calculation 5: Determination of the Prandtl Number: Pr = 𝜇𝐶 ? ? Pr = (1.969𝑥10 −5 ?? ? 2 )(1.0079 ?? ?𝑔? ) 0.02811 ? ?? Pr = 0.7033 Sample Calculation 6: Determination of the kinematic viscosity: 𝝂 = Pr 𝑎 𝝂 = (0.7033)(2.546x10 −5 ? 2 ? ) 𝝂 = 1.79x10 −5 ? 2 ? Sample Calculation 7: Determination of the Grashof Number 𝐺? = 𝑔?(𝑇 𝐸 − 𝑇 ∞ )? 3 ? 2 𝐺? = (9.81 ? ? 2 ) (0.003082)(358? − 291?)(0.00635?) 3 (1.79𝑥10 −5 ? 2 ? ) 2 𝐺? = 1617.29 Sample Calculation 8: Determination of the Nusselt Number ?? = 0.48[(𝐺?)(Pr)] 0.25 ?? = 0.48[(1617.29)(0.7033)] 0.25

?? = 2.788 Sample Calculation 9: Determination of the Convective Heat Transfer Coefficient ℎ = ?? ∙ ? ? ℎ = 2.788 ∙ (0.00635?) 0.02811 ? ?? b. Calculate the heat transfer by convection. Sample Calculation 10: Determination of the Effective Area 𝐴 ?????𝑡𝑖𝑣? = 1.13(2 ?? 2 4 + ???) 𝐴 ?????𝑡𝑖𝑣? = 0.00368 ? 2 Sample Calculation 11: Determination of the heat transfer by convection ? ???𝑣 = ℎ𝐴 ?????𝑡𝑖𝑣? (𝑇 𝐸 − 𝑇 ∞ ) ? ???𝑣 = (12.34 ? ? 2 ? ) (0.00368 ? 2 )(358? − 291?) ? ???𝑣 = 3.04 ? c. Calculate the heat transfer by radiation Sample Calculation 12: Determination of the heat transfer by radiation ? 𝑟𝑎? = ? ℎ?𝑎𝑡?𝑟 − ? ???𝑣 ? 𝑟𝑎? = [0.96(7.0?)(0.72𝐴) − 0.0017(358? − 291?)] − 3.04 ? ? 𝑟𝑎? = 1.68 ?

d. Calculate he emissivity of the heated element using the provided radiation heat transfer equation for a fully enclosed body Sample Calculation 13: Determination of the emissivity of the heated element 𝜀 = ? 𝑟𝑎? 𝜎𝐴 ?????𝑡𝑖𝑣? (𝑇 𝐸 − 𝑇 𝑣 ) 𝜀 = 1.68? (5.67𝑥10 −8 ? ? 2 ? )(0.00368? 2 )(358? − 291?) 𝜀 = 0.8719 e. Calculate the convective heat transfer as a percentage of the total heat transfer Sample Calculation 14: Determination of ? ???𝑣 as a % of ? 𝑡?𝑡𝑎? % ???𝑣 = ? ???𝑣 ? ???𝑣 + ? 𝑟𝑎? 𝑥100% % ???𝑣 = 3.04? 3.04? + 1.68? 𝑥100% % ???𝑣 = 64.38%

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4. Table containing calculated results Table 2: Part A Data (calculated and recorded results) P (kpa) 102.04 73.72 48.57 24.97 9.93 0.045 Tm (K) 51.02 36.86 24.285 12.485 4.965 0.0225 K (W/mk) 0.02811 0.02826 0.0283 0.02882 0.02915 0.03008 ? 0.003082 0.003063 0.003058 0.002994 0.002954 0.002849 ? ( ?𝑔 / ? 3) 1.0954 0.7866 0.5174 0.2604 0.1022 0.0004 v (m^2/s) 0.0000179 0.0000251 0.0000381 0.000077 0.000198 0.0467 Cp (kJ/(kgK)) 1.0079 1.0081 1.0081 1.0084 1.0085 0.974 Pr 0.7033 0.7031 0.703 0.7021 0.7016 0.6757 Gr 1617.2939 869.8151 380.2188 106.4298 17.5285 0.0003804 Nu 2.7876 2.387 1.9408 1.4113 0.89887 0.06078 h (w/m^2K) 12.3414 10.6234 8.6491 6.4043 4.1262 0.2879 Qtot (W) 4.7245 4.7177 4.716 4.6956 4.6803 4.6412 Qconv (W) 3.0415 2.7744 2.2907 1.9788 1.4115 0.1228 Qrad (W) 1.683 1.9433 2.4253 2.7168 3.2688 4.5184 Qtot/Qrad (%) 64.378 58.8093 48.5721 42.142 30.1586 2.6464 ε 0.8719 0.9316 1.1409 1.0243 1.0661 1.051 TE-TV (K) 67 71 72 84 93 116 P^.25 3.178 2.93 2.64 2.235 1.775 0.461 5. Plot of ( 𝑇 𝐸 − 𝑇 ? ) vs ? 0.25 y = -18.586x + 124.78 R² = 0.9895 0 20 40 60 80 100 120 140 0 0.5 1 1.5 2 2.5 3 3.5 ( 𝑇 e− 𝑇 v) (K) P^0.25 (kPa) Temperature Difference vs. P^ 0.25 Figure 1 - Difference in Temperature of Element and Vessel vs Corrected Pressure

From the use of the equation of the ( 𝑇 𝐸 − 𝑇 ? ) vs ? 0.25 relation shown below, the temperature difference at absolute pressure between the element and the vessel can be found by setting x=0. Y = -18.586x + 124.78 Y = -18.586(0kPa) + 124.78 Y = 124.78 K Sample Calculation 15: Determination of the emissivity of the heated element 𝜀 = ? ℎ?𝑎𝑡?𝑟 𝜎𝐴 ?????𝑡𝑖𝑣? (𝑇 𝐸 4 ) Where 𝑇 𝐸 4 = [(𝑇 𝐸 − 𝑇 𝑣 ) + 𝑇 𝑣 ] 4 = [124.77? + 291?] 4 = 2.988𝑥10 10 ? 𝜀 = 0.96?? − 0.0017(𝑇 𝐸 − 𝑇 ? ) 𝜎𝐴 ?????𝑡𝑖𝑣? (𝑇 𝐸 4 ) 𝜀 = [0.96(7.0?)(0.72𝐴) − 0.0017(415.77? − 291?)] (5.67𝑥10 −8 ? ? 2 ? ) ∗ 0.00368? 2 (2.988𝑥10 10 ?) 𝜀 = 0.7420 Part B) 1. Table with data recordings from part b of the experiment. Table 3: Part B Data (Recorded results) Heater Voltage (V) 3 5 7 9 Elemental Temperature (°C) 78 95 136 184 Vessel Temperature (°C) 20 20 20 20 Heater Current (A) 0.31 0.52 0.72 0.92 Vessel Pressure (K) 0.014 0.07 0.045 0.011

2. Plot of ? 𝑟𝑎? vs 𝑇 𝐸 4 − 𝑇 𝑣 4 With the vessel assumed to be at a vacuum state, we can also make the assumption that ? ???𝑣 will not be a mode of heat transfer and such ? ℎ?𝑎𝑡?𝑟 = ? 𝑟𝑎? . Figure 2 - Radiative Heat Transfer vs Temperature Difference (element and vessel) With an understanding of the relation shown in the graph above, the slope can be used to produce the ? 𝑟𝑎? vs 𝑇 𝐸 4 − 𝑇 𝑣 4 ratio shown here ? = ? 𝑟𝑎? 𝑇 𝐸 4 − 𝑇 𝑣 4 Now, by rearranging the radiative heat equation and assuming an emissivity of 0.96, the Stephan-Boltzmann constant can be produced. 𝜎 = ? 𝜀𝐴 ?????𝑡𝑖𝑣? 𝜎 = 2.313𝑥10 10 ? ? 4 (0.96)(0.00368? 2 ) y = 2E-10x - 0.504 R² = 0.9792 0 1 2 3 4 5 6 7 8 9 0 1E+10 2E+10 3E+10 4E+10 Qrad (W) Te^4 -Tv^4 Qrad vs. Te^4-Tv^4

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𝜎 = 6.549𝑥10 −8 ? ? 2 ? 4 Finally, with a calculated Stephan-Boltzmann Constant of 𝜎 = 6.549𝑥10 −8 ? ? 2 𝐾 4 and an expected value of 𝜎 = 5.67𝑥10 −8 ? ? 2 𝐾 4 , a percent error can be found. % ?𝑟𝑟?𝑟 = | ?𝑥???𝑖????𝑎? − ?𝑥?????? ?𝑥?????? |𝑥100% % ?𝑟𝑟?𝑟 = | 6.549𝑥10 −8 − 5.67𝑥10 −8 5.67𝑥10 −8 | 𝑥100% % ?𝑟𝑟?𝑟 = 15.5% Part C) 1. In order for heat transfer via convection to occur, fluid motion must be present, however as this medium by which convection operates is removed the % of convection which will be displayed proportionally decreases also. This is described in part A.3 of the experiment where pressure was reduced and as such the convection coefficient also decreased. Clearly displayed in results provided where the convective heat transfer as a percentage of the total heat transfer initially is roughly 64.4%, while the vessel is at a pressure of 102.04 kPa. In contrast at the point where near vacuum conditions are reached for the vessel convective heat transfer as a percentage of the total heat transfer was found to be 2.65% at a pressure of 0.045 kPa. As the medium which convection operates is removed, radiation begins to dominate as this mode of heat transfer occurs through electromagnetic waves and can be seen actually increasing as pressure decreases within the vessel. 2. While comparing emissivity values determine during Step A.3 and A.5 it must first be known that an emissivity greater than 1.0 in the real world does not occur. This value represents a perfect black body which would mean all incident radiation is completely absorbed. With this understanding the omitting of the last 4 results for emissivity in step A.3 must be done, which in turn produces the remaining results having an average emissivity = 0.902 which is slightly lower than our expected values of 0.96 and 0.97. In contrast step A.5 produced an emissivity of 074, much lower than our expected values. In step A.3 heat transfer via convection was a significant mode found in total heat transfer, where in step A.5 where only radiation was considered. This leads me to believe that with less interference from convection step A.5 provides a more reliable emissivity calculation.

3. The calculation of the Stephan-Boltzmann constant was found to be 6.549𝑥10 −8 ? ? 2 𝐾 4 , producing a % ?𝑟𝑟?𝑟 of 15.5% . A couple of assumptions could be attributed to this inaccuracy, such as the ones mentioned prior to beginning the experiment. These being, system assumed to be steady state, the medium withing the vessel (air) by which convection occurs through was assumed to be an ideal gas. Of course with our limited time within the laboratory this would be unfeasible however, if the experiment was able to be completed a number of times and more time was given to allow for system stability between trials the results produced may have been more accurate.

Appendix Part A) P (kpa) 102.04 73.72 48.57 24.97 9.93 0.045 Te 85 89 96 103 112 136 Tv ('c) 18 18 18 19 19 20 Te (K) 358.3 362.3 369.3 376.3 385.3 409.3 Tv (K) 291.3 291.3 291.3 292.3 292.3 293.3 Tm (K) 324.8 326.8 330.3 334.3 338.8 351.3 K (W/mk) 0.02811 0.02826 0.0283 0.02882 0.02915 0.03008 𝜇 ( ? ∙ ? ) × 10−7 0.0000179 0.0000251 0.0000381 0.000077 0.000198 0.0467 Cp (kJ/(kgK)) 1.0079 1.0081 1.0081 1.0084 1.0085 0.974 ? 0.003082 0.003063 0.003058 0.002994 0.002954 0.002849 ? ( ?𝑔 / ? 3) 1.0954 0.7866 0.5174 0.2604 0.1022 0.0004 ? (m^2/s) 0.00002546 0.00003564 0.00005425 0.0001097 0.0002828 0.06914 Pr 0.7033 0.7031 0.703 0.7021 0.7016 0.6757 v (m^2/s) 0.0000179 0.0000251 0.0000381 0.000077 0.000198 0.0467 Gr 1617.2939 869.8151 380.2188 106.4298 17.5285 0.0003804 Nu 2.7876 2.387 1.9408 1.4113 0.89887 0.06078 h (w/m^2K) 12.3414 10.6234 8.6491 6.4043 4.1262 0.2879 Qtot (W) 4.7245 4.7177 4.716 4.6956 4.6803 4.6412 Qconv (W) 3.0415 2.7744 2.2907 1.9788 1.4115 0.1228 Qrad (W) 1.683 1.9433 2.4253 2.7168 3.2688 4.5184 Qtot/Qrad (%) 64.378 58.8093 48.5721 42.142 30.1586 2.6464 ε 0.8719 0.9316 1.1409 1.0243 1.0661 1.051 TE-TV (K) 67 71 72 84 93 116 P^.25 3.178 2.93 2.64 2.235 1.775 0.461 Part B) Heater Voltage (V) 3 5 7 9 Elemental Temperature (°C) 78 95 136 184 Vessel Temperature (°C) 20 20 20 20 Heater Current (A) 0.31 0.52 0.72 0.92 Vessel Pressure (K) 0.014 0.07 0.045 0.011 Q ?𝑎? (W) 0.7942 2.3685 4.6412 7.67 (T E4 -T V4 ) (K 4 ) 7808000000 10970000000 20610000000 36250000000

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