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Introduction The purpose of this experiment is to predict the temperature drop of a test bar as a function of time. The temperature drop of the test bars is predicted to vary exponentially with time as what is given by the lumped capacitance method. The prediction of the temperature drop is based on the lumped capacitance method with the fulfillment of Biot Number. The lumped capacitance method is the simplest unsteady conduction model. Thus, in order to satisfy the lumped capacitance method the Biot number must always less than 0.1. The Biot number can be obtained theoretically as well as experimentally. The generalized equation for the Biot number is given in the Analytical Approach section while the Biot number obtained experimentally is predicted to be the ratio of the change of internal temperature to the change of film temperature of the test bar. Also, the time required by the test bar to cool from its initial temperature to a certain final temperature can be obtained on rearranging the generalized form of the lumped capacitance method (see also Analytical Approach). Analytical Approach In this experiment, a dimensionless parameter known as the Biot number is introduced. The Biot number plays a very significant role in deriving and verifying the validity of the lumped capacitance method. It is also a fundamental role in conduction problems that involve surface convection effects. The figure below depicts the derivation of the Biot number . T Ts,1 Ts,2 T , h x L Schematic of the fundamental derivation of Biot number on a steady-state temperature distribution in a plane wall with surface convection. Ts ,1 Ts ,2 ( L kA) hL The Biot number is Bi Ts ,2 T (1 hA) k where L = thickness k = thermal conductivity of solid h = convective heat transfer coefficient A = surface area In this experiment, some assumptions are made before suing the lumped capacitance method to predict the temperature drop across the test material. The assumptions made are: 1. No radiation 2. Negligible internal resistance of the material 3. No heat generation 4. Temperature of the material is constant 5. Constant physical properties 6. Ambient temperature (T) is constant 7. Area average film coefficient The test materials used in this experiment are aluminum and brass bars. The lumped capacitance method is the simplest unsteady conduction model. The Biot number must be less than or equal to 0.1 in order to satisfy the lumped capacitance method. However, there is a more general definition for the Biot number which will be used throughout this experiment. Lc in the following equation is the characteristic length of the bar used to calculate the Biot number. hLc Volume V Bi where Lc k Area A Since no heat generation occurs in the test bars, one can assure that the rate of heat transfer through the bars by conduction is equal to the rate of heat transfer by convection. Thus a relationship used to predict the Biot number obtained experimentally is shown below: q conduction q convection Tint kA hAT film x Tint hx Bi T film k where Tint = internal temperature of test bars Tfilm = film temperature of test bars In order to predict the time required for the test bars to drop to certain temperature, the following equation, which is a simplified form of the lumped capacitance method, is needed. Detail derivation of this relationship will be shown in the Appendix. Vc Ti T t ln where t = time hA T T = density of test bar V = volume of test bar c = specific heat capacity of test bar h = average convective heat transfer coefficient Ti = initial temperature of the test bar T = ambient temperature T = temperature of the test bar However, in order to determine the convective heat transfer coefficient, h, other dimensionless parameters were used. These are the Nussult number, Nu and the Rayleigh number, Ra. g (Ts T ) L3 c Ra L where g = local acceleration of gravity = expansion coefficient whereas and can be obtained from handbook. However, the RaL for the vertical side and the top and bottom sides of the bar is needed to calculate the NuL. The RaL for different orientations differ from their characteristic length, Lc. For vertical side Lc will be the height of bar whereas for top and bottom sides the Lc will be the ratio of the area of the bar to its perimeter. After knowing the RaL, the Nussult number can be calculated. The Nussult number for the vertical side of the bar takes the form 0.67 Ra 1/ 4 L Nu L 0.68 for RaL < 109 [1 (0.492 / Pr) 9 /16 ]4 / 9 However, the Nussult number for the upper surface of the cooled bar take the form of the following Nu L 0.27 Ra 1/ 4 L for ( 105 < RaL < 1010 ) The Nussult number for the lower surface of the cooled bar is Nu L 0.54 Ra 1/ 4 L for ( 107 < RaL < 1011 ) With the known Nussult number for different sides of the bar, the convective heat transfer coefficient for each side can be obtained by using the following equation. k h Nu L Lc After calculating the convective coefficients associate with each side of the bar, the average value is taken to calculate the lumped capacitance. Introducing the generalized form of the lumped capacitance method: hA t mc T (t ) T (Ti T )e t T (t ) T (Ti T )e mc where = thermal time constant = hA On rearranging the above equation yields the form of the equation introduced in the previous page needed to calculate the time for the temperature drop across the test bar, i.e., Vc Ti T t ln hA T T Experimental Program Since the purpose of this experiment is to predict the temperature drop on the test bars and then verify the prediction with the results obtained experimentally, the ambient temperature and the initial temperatures of the bars are critical. These quantities should be measured as precise as possible for better precision on the outcome of the experiment. Other quantities that were measured are masses of the bars and their dimensions (length, width, and height). The instruments used are General Electric freezer, SRS Stanford Research Systems (model: SR630), 16 channel thermocouple monitor, Epson Fx-85E printer, general vernier calipers 0-6”, and OHAUS LS50900 Portable Scale. The experiment was started by measuring the masses of the bars by means of the Portogram Balance followed by measuring the lengths, width, and heights of the two test bars. The two test bars used in this experiment are aluminum and brass bars. After that, the thermocouple labeled 16 was fixed to measure the ambient temperature. On the other hand, the thermocouples labeled 3 and 4 were used to measure the internal temperature of the test bars, i.e., aluminum and brass bar respectively. Thermocouples labeled 10, 11, and 12 were attached to the surface of the test bars to measure the film temperature. In this case, thermocouple 10 was used to measure the film temperature of the aluminum bar whereas 11 and 12 were used to measure the brass bar. The starting time of the experiment was recorded after the above procedures were done. All the thermocouples were allowed to reach their steady measurements before the readings were taken. The test bars were then let to be cooled by putting them in a freezer. The thermocouple monitor was connected to a printer which printed the temperatures read by all thermocouples at every minute. These readings were needed to compare with the predicted temperature drop by using the Lumped Capacitance Method. Results Since the printer printed the temperatures of all thermocouples at one minute interval, only 30 data points are used in this section to calculate the result. The table below shows the specifications of the two test bars used in the experiment. Table 1 Raw Data For Aluminum and Brass Bars Aluminum Bar Variable Units Uncertain ty Length (inch) 0.01 1 Width (inch) 0.01 1.99 Height (inch) 0.01 6 Mass (g) 0.5 664 Brass Bar Variable Units Uncertain ty Length (inch) 0.01 1 Width (inch) 0.01 1.99 Height (inch) 0.01 5.99 Mass (g) 0.5 1662 Table 2 shows that the temperatures of the aluminum and brass bars indicated by the thermocouple monitor which were printed by the printer. Since thermocouple 11 and 12 were attached to the brass bar, the average of the temperatures indicated by these two thermocouples are taken to be used in the calculations computed in Table 3. With the calculated values in Table 3, the temperature drop on the test bars as a function of time can be plotted. Figure 1 shows the temperature drop, (t) versus time, where hA t (t ) T T (Ti T )e mc From Figure 1, it is seen that the temperatures on the test bars decreases exponentially as time goes on. This satisfied the equation defined by the lumped capacitance method. Also, as seen from the graph, the temperature drop on the aluminum bar is faster than the temperature drop on the brass bar because the aluminum bar has a higher thermal conductivity, k. Also, the curves in Figure 1 can also be accounted for the hA larger mass of the brass bar which makes the term becomes smaller. The mc exponential term would then become smaller. Thus, the curve for the brass bar seems to have less steep slope compared to the aluminum curve. Table 4 shows the calculated values for the predicted Biot number, which is the ratio of the change in internal temperature to the change in film temperature of the test bar. Figure 2 shows the Biot number versus time for both the aluminum and the brass bar. It is seen that the Biot number fluctuates with time. This could be caused by the unsteady heat loss from the bars. The time required for the test bar to cool from its initial temperature to its final temperature, say 40F, was calculated based on the lumped capacitance method since the Biot number obtained from the calculation is less than 0.1. The Biot number for aluminum bar is 1.85x10-4 whereas for brass bar is 3.98x10-4. Thus, the time required for the cooling of aluminum bar takes approximately 71.34 min while brass bar takes 75.14 min. Compare to the experimental results obtained from the thermocouple monitor, the time taken for the aluminum bar to cool to 40F is shorter than what was calculated. The results show that the aluminum bar took approximately 56 min to cool to 40F. However, the time predicted for the cooling of the brass bar is fairly close to the actual results. The predicted time required by the brass bar to cool to 40F was 75.14 min and the actual time required was approximately 70 min. This shows that the calculated value is in 7% agreement with the actual result obtained from the experiment. The calculated value for brass bar is more accurate probably because there were two thermocouples connected to it and a more steady temperature for the surface temperature was read. Also, the cooling rate of the bars could be affected by some barriers in the refrigerator. This makes the cool air inside the refrigerator could not flow evenly on the surface of the test bars. Below shows the interpolated air properties used to calculate the RaL and Nu L . Detailed calculations are shown in the Appendix. 30109 265.48 . Properties evaluated at T = 283.29 K 2 air 12394 kg m 3 . 20.29 x10 6 m 2 s c p 1006 kJ kg K . 353x10 3 K 1 . 14.4 x10 6 m 2 s g 9.81 m s 2 Pr 0.7113 k 24.96 x10 3 W m K 40.00 35.00 30.00 Temperature Drop, t) ( C) 0 25.00 20.00 15.00 10.00 5.00 Temperature Drop on Aluminum Bar Temperature Drop on Brass Bar 0.00 0 10 20 30 40 50 60 Time (minute) Figure 1: Temperature Drop on the Test Bars as a Function of Time 0.01800 Predicted Biot Number for Aluminum Bar Predicted Biot Number for Brass Bar 0.01600 0.01400 Predicted Biot Number, Bi 0.01200 0.01000 0.00800 0.00600 0.00400 0.00200 0.00000 0 10 20 30 40 50 60 Time (minute) Figure 2: Predicted Biot Number versus Time for Both Aluminum and Brass Bar Table 2 Temperature of Aluminum and Brass Bar Obtained From the Experiment Time Center Temperature Surface Temperature Ambient (min) (C) (C) Temperature (C) Aluminum Brass Aluminum Brass Brass Brass (Average) 0 81.3 83.3 81.1 83.1 82.9 83 18.2 2 80 82.4 79.8 82 81.8 81.9 31 4 77.7 80.4 77.1 80 79.8 79.9 27.7 6 75 78.2 74.8 77.9 77.7 77.8 26.3 8 72.6 76.4 72.3 76.1 75.7 75.9 25 10 70.5 74.6 70.1 74.3 73.9 74.1 24.5 12 68.1 72.6 68 72.3 72.1 72.2 23.8 14 66.2 70.8 65.8 70.7 70.3 70.5 23.8 16 64 69 63.8 68.7 68.5 68.6 23.8 18 62.2 67.6 61.8 67.2 67.1 67.15 23.6 20 60.4 65.8 60 65.4 65.4 65.4 23.4 22 58.6 64.4 58.4 64 63.8 63.9 23.6 24 57 62.9 56.6 62.6 62.4 62.5 23.2 26 55.5 61.5 55 60.9 60.8 60.85 23.4 28 53.9 60 53.7 59.7 59.5 59.6 22.5 30 52.5 58.6 52.3 58.2 58.1 58.15 22.7 32 51.2 57.3 51 57.2 57 57.1 22.7 34 49.8 56.1 49.6 55.7 55.5 55.6 22.5 36 48.5 54.8 48.3 54.5 54.5 54.5 22.1 38 47.3 53.6 47.3 53.6 53.4 53.5 21.8 40 46.2 52.7 46 52.3 52.1 52.2 21.8 42 45.1 51.4 44.9 51.2 51 51.1 21.2 44 44 50.3 43.7 50.1 50 50.05 20.5 46 42.9 49.4 42.8 49.1 49.1 49.1 21.2 48 41.9 48.3 41.9 48.2 48 48.1 21.1 50 41 47.4 40.8 47.1 47.1 47.1 20.7 Conclusion From the results calculated, the temperature on the each test bar decreases exponentially with time. The trend is analogous with the equation given by the lumped capacitance method. However, the time predicted for the cooling of the aluminum and brass bars were not too accurate due mainly to the uncertainties arise in the measurement. The result also shows that the Biot number fluctuates with time. This is probably caused by the unsteady heat loss from the surface of the test bars. Appendix All properties of the test bars were interpolated from handbook corresponding to their film temperatures, Tf, where Tf is the average temperature of the surface temperature of the bars and the ambient temperature. Ts T Tf 2 As discuss in the Analytical Approach section, in order to predict the time required for the test bars to drop from their initial temperatures to a certain final temperature, the following procedures were performed. First find the Rayleigh number, RaL. Assume that the properties of aluminum and brass bar have negligible temperature difference and their properties were evaluated at the film temperature. Average surface temperature of the bars are at 301.09K whereas the ambient temperature is at 265.48K. 30109 265.48 . Thus, T f 283.29 K 2 For the vertical sides of the bars, g (Ts T ) L3 c Ra L (9.81m / s 2 )(353x10 3 K 1 )(300.54 265.48 K )(01524m) 3 . . Ra L 6 2 6 2 (14.4 x10 m / s)(20.29 x10 m / s) Ra L 14.71x10 6 For the upper and the lower surface of the bars, (9.81m / s 2 )(353x10 3 K 1 )(300.54 265.48 K )(0.008453m) 3 . Ra L (14.4 x10 6 m 2 / s)(20.29 x10 6 m 2 / s) Ra L 4100 All the properties given above were interpolated from handbook at T = 283.29K. This is followed by calculating the Nussult number with the known RaL. Also, with the known Nussult number, the convective coefficient can be obtained. For the vertical side, 0.67 Ra 1/ 4 L Nu L 0.68 [1 (0.492 / Pr) 9 /16 ]4 / 9 0.67(14.71x10 6 ) 1/ 4 Nu L 0.68 [1 (0.492 / 0.7113) 9 /16 ]4 / 9 Nu L 32.534 Thus, Nu L k 32.534(24.9632 x10 3 W / m K ) h vertical 5.329W / m 2 K L . 01524m For the upper surface of the bar, Nu L 0.27 Ra 1/ 4 0.27(4100) 1/ 4 2.1605 L Nu L k 2.1605(24.9632 x10 3 W / m K ) Thus, h top 6.3804W / m 2 K L 0.008453m For the lower surface of the bar, Nu L 0.54 Ra 1/ 4 0.54(4100) 1/ 4 4.32 L Nu L k 4.32(24.9632 x10 3 W / m K ) Thus, h bottom 12.7577W / m 2 K L 0.008453m The total surface area of the bars is 0.02581m2. To calculate the average convective coefficient, 0.02323m 2 0.00129m 2 0.00129m 2 h average 2 5.329W / m K 2 2 6.3804W / m K 2 2 12.7577W / m K 2 0.02581m 0.02581m 0.02581m h average 5.7528W / m K 2 Take the brass bar as sample calculation. The measured temperature of the test bar as a function of time, T(t) is derived from the lumped capacitance method shown below. The lumped capacitance method is derived first by performing an energy balance on the bar. E E in E out E gen dT mc 0 hAT T 0 dx dT hA dt T T mc T dT t hA Ti T T 0 mc dt hA t mc T T (Ti T )e However, h is used instead of h because the average convective coefficient for each side of the bar is needed to take into consideration when calculating T(t). Thus, by lumped capacitance method hA t mc T (t ) T (Ti T )e ( 5.7528W / m2 K )( 0.02581m2 ) (1.662 kg )( 380 J / kg K ) T (t ) 265.48 (30165 265.48)e . 4 T (t ) 265.48 3617e ( 2.351x10 ) t . If (t) = ( T-T ), then the temperature drop of the brass bar is 4 (t ) 3617e ( 2.351x10 ) t . On the other hand, the time required for the brass bar to cool from its initial temperature of 83F(301.65K) to a certain final temperature, say 40F(277.59K), can be computed as follow: mc T1 T t ln hA T T 30165 265.48 . t (2.351x10 4 ) 1 ln 277.59 265.48 t 4508s 7514 min 125hrs . . To predict the Biot number for the bars, the following relationship is needed. q conduction q convection Tint kA hAT film x Tint hx Bi T film k The predicted Biot number for the aluminum bar in the first trial is T Ts 813 811 . . Bi center 0.00318 Ts T 811 18.2 . While the predicted Biot number for the brass bar is 83.3 83.0 Bi 0.00463 83.0 18.2 The same procedures are repeated for the rest of the trials.

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posted: | 7/5/2013 |

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