Introduction pantherFILE

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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            hAT film
              x
         Tint       hx
                           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 40F, 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 40F is shorter than what was calculated. The

results show that the aluminum bar took approximately 56 min to cool to 40F.
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 40F 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  hAT  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 83F(301.65K) to a certain final temperature, say 40F(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            hAT film
             x
        Tint       hx
                          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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