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Compressible Fluid Flow Pressurizing and Discharging of Tanks

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Compressible Fluid Flow: Pressurizing
  and Discharging of Tanks Under
Adiabatic and Isothermal Conditions




           06-363 Transport Process Laboratory
               Carnegie Mellon University
                     May 22, 2006
                        Group 6:
                       Alan Abel
                       Phil Lowe
                      Rachel Clair
                       Daqian Wu
                                                                                             2


Abstract
Models describing the isothermal or adiabatic filling and discharge of a tank were analyzed. Two
tanks were used in this experiment; the larger was 7 gallons, and the smaller was 1.1 gallons. The
smaller one was used to test the adiabatic model and the larger was used to test the isothermal
model. Due to possible malfunctions in the pressure transducers, poor response times in the
pressure transducers, and/or inadequate data-sampling rate the experimental data could not be
used to prove or disprove the models in question. However first hand observations did
qualitatively confirm the accuracy of the models tested.
                         1


Table of Contents
Introduction        1
Theory              1
Experimental        3
Results             4
Discussion          6
Conclusion          7
References          7
Appendix A          8
Appendix B          9
Appendix C          11
                                                                                               1


Introduction
        This experiment was set up to examine the relationship between pressure and discharge
rate through a needle valve. It allows us to examine the phenomena of choked flow which is
when the gas velocity becomes independent of downstream pressure. However by changing the
upstream pressure one can increase the mass flow rate out of the nozzle. This is important in
industry because allows manipulation of mass flow with the use of valves and orifice plates
under choked flow. Our setup allowed us to examine both isothermal and adiabatic discharge
conditions and data was collected using Labview software.

Theory
Discharge
        To simplify our analysis of the flow we made several assumptions about the behavior of
the gas in the tank:
    1. The gas molecules operate under isentropic thermodynamic conditions.
    2. The discharge of the gas can be either characterized as adiabatic or isothermal.
    3. There is ideal isentropic flow from the tank to the nozzle throat.
Assumption 2 relies on the time allowed for the discharge to occur. If the gas exits very rapidly
there is not enough time heat transfer to occur between the tank walls and the gas. Therefore we
modeled the adiabatic exit case with a small tank of about 5 liters. A large tank, 21 liters in our
case, provides sufficient space and discharge time for temperature to be stabilized and allows
examination of isothermal discharge.
        Under our assumptions of adiabatic and stagnant gas, we can apply isentropic
thermodynamic conditions to our analysis of the gas flow in the small tank. We can derive
pressure difference as a function of time by starting from mass flow rate and simplify it to
Fleinger’s formula.
               dm
                    vA                                                            (1)
                dt
With the stagnant gas conditions we assumed, the mass balance becomes:
                                       1
               dm                 2  1 0.5
                     APo (     (     ) )                                          (2)
                dt          RTo   1
In the case of air this can be simplified to Fliegner’s formula:
               dm                 P
                     0.04042 A o                                                  (3)
               dt                 To
This can then be substituted into a continuity equation to solve for a relationship between
pressure and time.
                 udV (4) compressible flow
                    v

                   
                      dV (5) incompressible flow
                   t V
In equations (4) and (5) u is the velocity of the gas and V is the volume of the gas. Equation (3)
gives us the initial instantaneous flow rate and by combining it with either (4) or (5), depending
on the operating conditions, we can derive a relationship between pressure and time.
                                                                                                              2

                                                  ( 1)      2
                       1  1                 2 ( 1)      1
          P  [1  (             )(          )              t ]                     (6) choked adiabatic
                         2             2
                                       ( 1)
                          1
         P   exp[ (            ) 2( 1) t  ]                                  (7) choked isothermal
                             2

                                             P
                                 P                                                  (8)
                                             Pi

                                                 t
                                  t                                                 (9)
                                            t char

                                                  V
                                 t char                                             (10)
                                                 At ai

Once enough gas has exited the nozzle, the flow becomes unchoked and we can transfer our
analysis into the unchoked region by applying a similar derivation as we did for the choked
                                                          
region. We would use the unchoking points ( Punchoke, t unchoke) as our initial conditions and occurs
when the static nozzle pressure equals atmospheric pressure. Equation (11) and (12) show the
relationship between dimensionless time and pressure over the unchoked portion for adiabatic
and isothermal conditions, respectively.
                                            ( 1)
                           2 0.5 At                 x 3 5x           3
     t   t unchoke  (
             
                              )     ( PB ) 2 {(  )( x 2  1) 0.5  ln[ x  ( x 2  1) 0.5 ]}xunchoke (11)
                           1 Ae
                                                                                               x
                                                    4    8           8

                                 2 2 0.5 At ( xunchoke  x 5 ) 2( xunchoke  x 3
                                                5                   3
                   
           t   t unchoke  (        )     [                                    )  ( xunchoke  x)] (12)
                                   1 Ae            5                 3

                                                      1
                                       P             
                                 x  [(  )                   1]0.5                 (13)
                                       PB

                                            PB (backpressure)
                                 PB                                                (14)
                                                    Pi
Charging
        The analysis in this case the almost the same as that performed for the discharging part,
expect that we can no longer assume isentropic behavior in the gas. There is some kinetic energy
in the inlet gas and it has to be accounted in our analysis. This is done by including it in the
definition of the dimensionless definitions of time and pressure.

                                            P
                                 P                                                 (15)
                                            Ps
                                                                                                                               3


                                         V
                          t char                                                                         (16)
                                        As a s

For equations (15) and (16), s denotes the conditions in the source. Choked filling can occur if
the difference in the pressure between the tank and the source is great enough.

                                         ( 1)
                               1
            P  Pi   (
                     
                                    )   2 ( 1) 
                                                    t                                              (17) choked adiabatic
                               2
                                             ( 1)
                            1          2 ( 1) 
           P  Pi  Ti (                )               t                                          (18) choked isothermal
                                2

To derive the unchoked we perform a similar procedure as mentioned for the discharging phase.
The only difference is that we now set the nozzle pressure equal to the tank pressure at
unchoking, instead of atmospheric.

                                                     1                                                      
                                                                     1             Ae  
                  
             P  {1  [(1  ( P      
                                    unchoke     )        0.5
                                                            )   (          )   0.5
                                                                                         (t  t unchoke)] 2 } 1       (19)
                                                                      2               At

                                                     1                                                            
                                                                     1 A
                  
             P  {1  [(1  ( P      
                                    unchoke     )        0.5
                                                            )   (   ) Ti e (t   t unchoke)]2 } 1
                                                                                0.5   
                                                                                                                        (20)
                                                                   2      At

Equations (19) and (20) are solutions for unchoked adiabatic and isothermal, respectively.

Experimental
      A diagram of the apparatus used for the experiment is shown below.
                                                                                                 4



The equipment used includes two vessels, the first with a volume of 7 gallons and the second
with a volume of 1.1 gallons. Each vessel is equipped with a release toggle valve on top,
connected to a thermocouple. The thermocouple allows the computer to collect and record the
temperature changes during discharging with Labview software. Also on the top of each tank, is
a safety release valve; this valve will open if the pressure inside the tank exceeds 150 psi. On the
front of each tank is a three-way valve, used for the charging experiment. The three positions of
the valve are used as follows: the first, to allow air to flow into vessel from the air compressor,
the second, to calibrate the pressure inside the vessel, and the third, to vent the gas inside the
vessel to the atmosphere. A pressure transducer is located next to the three-way valve to record
the pressure inside the vessel throughout the experiment.
        Data was collected from several trials using both the 7 gallon vessel and the 1.1 gallon
vessel. Data was collected for both charging and discharging the vessels. Several trials on each
vessel allowed for the compilation of plenty of data to create accurate conclusions. Labview
software recorded the experimental information and plotted a graph for each successful run.
Data was also recoded by hand. To ensure the most accurate accumulation of experimental
information, environmental conditions were measured and recorded to keep the data consistent.

Results
Figure 1 is a plot of tank pressure vs. time for filling of the large tank (isothermal). Similar plots
for a data set of each operation (filling and discharging for both size tanks) can be found in
Appendix B.


                                  Pressure vs. Time, Big Tank Charging

                   120


                   100


                   80
  Pressure (psi)




                   60


                   40


                   20


                    0
                         0   10   20     30       40      50      60      70       80       90
                                                   Time (s)

Figure 1: pressure vs time, isothermal charging large tank
                                                                                                5




From a mathematical analysis of the charging and discharging tanks, the tank pressure was
calculated and plotted versus time. An example of this is shown in Figure 2.


                8 10
                        5




                6 10
                        5


         P 1( t )
                       5
         P 2( t )4 10


                2 10
                        5




                        0
                            0         1           2          3           4           5

                                                        t

                    Figure 2: plot of calculated pressure vs. time for isothermal filling

Of note in the figure is the separation of flow regimes. On the above plot, the red region of
pressure values indicate the region of choked flow, whereas the green section represents the
region of unchoked flow.

Calculated pressure values were then compared to measured pressure values at the corresponding
times during the different flow regimes. It is important to note, however, that during the region
of unchoked flow for tank discharge, only the time can be calculated as an explicit function of
pressure, not vice versa. This prevents any possible pressure comparison for that section of data.
The calculated time required to charge the small tank was so short, however, that our equipment
could not effectively collect enough data points during that time. The complete results of the
pressure error comparison can be found in Appendix B. The error for the isothermal tank
charging is shown in Figure 3:
                                                                                              6




t                    pmeas       pcalc       % error
            0.521      15.375       32.08    52.07294
            1.052      15.592      49.793    68.68636
            1.222      15.652      55.463    71.77938
            1.572      16.055         67.1   76.07303
            2.093      16.142      83.465    80.66016
            2.624      16.587       97.32    82.95623
            2.964      16.965     103.943    83.67855
            3.325      17.118     108.621    84.24062


measured tfinal:        77.201
calculated tfinal:   3.767
% error:             1949.403
Figure 3: Error amounts of measured pressure versus
calculated pressure in isothermal tank charging

The most striking aspect of the measured pressure data is the significantly slower response than
expected. For the pressurizing of the large tank, the measured final time is approximately 20
times longer than the calculated time.

Discussion
The results clearly show that the models employed do not agree with our data for any of the
filling or draining experiments. There are three possible causes of this. 1) The pressure
transducers were broken. 2) The pressure transducers could not respond fast enough to the
pressure change. 3) The rate at which the computer acquired the data was too slow.

The 15 psi pressure transducer was used in the initial leak tests and was certainly broken when it
was exposed to 93 psi. This renders all of the filling and draining experiments conducted with
this sensor suspect. The 150 psi transducer is, based on the model, not providing accurate
answers either. It is not very likely that this sensor was broken during testing or any subsequent
experiment since the highest pressures used were on the order of 100 psi. It is, however, likely
that the response time of this sensor or the data-sampling rate of the computers was not fast
enough to accurately record what was taking place in the tanks during filling or draining.

The predictions of the time required for draining or filling of the tanks by the various models
previously presented was confirmed by observations made on the tanks during the experiments.
Specifically the time it took for the small tank to drain from 93 psi to atmospheric pressure was
determined by listening to the air escaping from the tank. The time required for most of the air to
escape was recorded on a stopwatch. This time was found to be on the order of 1 second, which
agrees closely with the time predicted by the adiabatic tank-draining model. This type of
observation compared well with what was predicted by the model in each case of tank filling and
draining. It is then reasonable to conclude that the model for each type of flow is accurate.
                                                                                          7


Conclusions
        The tank draining and filling models presented in this report are accurate
        The pressure transducers and data acquisition equipment need to be checked to make
         sure they are capable of operating at a high enough speed to give accurate results in
         the data

References:
  1) J. Craig Dutton and Robert E. Coverdill. “Experiments to Study Gaseous Discharge and
     the Filling of Vessels.” Department of Mechanical and Industrial Engineering,
     University of Illinois at Urbana-Champaign. Full-text available at
     http://www.ijee.dit.ie/articles/999982/experime.htm
                                                                           8



Appendix A: Nomenlature
Symbol/abbrev.            Meaning                                 Units
m                         mass                                    kg
P                         pressure                                PSI
Ps                        source pressure                         PSI
Pb                        back pressure                           PSI
Pi                        initial pressure                        PSI
ρ                         density                                 kg/m3
v                         gas velocity                            m/s
V                         tank volume                             gal
T                         temperature                             K
Ti                        initial temperature                     K
Ts                        source temperature                      K
γ                         heat coefficient ratio                  dimensionless
Ae                        nozzle exit area                        m2
At                        nozzle throat area                      m2
ai                        initial sonic velocity of gas in tank   m/s
                                                                                                9


Appendix B: Pressure-time Data
                                    Pressure vs. Time, Small Tank Discharge

                  120


                  100


                  80
 Pressure (psi)




                  60


                  40


                  20


                   0
                        0       5       10       15         20        25        30        35    40
                                                          Time (s)




                                    Pressure vs. Time, Small tank charging

                  120


                  100


                  80
 Pressure (psi)




                  60


                  40


                  20


                   0
                        0   2       4        6        8     10       12    14        16    18   20
                                                          Time (s)
                                                                                            10


                                     Pressure vs. Time, Big Tank Discharge

                 120


                 100


                  80
Pressure (psi)




                  60


                  40


                  20


                   0
                       0        10           20           30          40          50         60
                                                       Time (s)




                                     Pressure vs. Time, Big Tank Charging

                 120


                 100


                 80
Pressure (psi)




                 60


                 40


                 20


                  0
                       0   10        20      30      40        50     60     70        80    90
                                                       Time (s)
                                                                                                                                                        11


Appendix C: Sample Calculation for tend
                                                                                                                                                    J
     v  26.498 L                                                                                         M  29.1            r  287.05
                                          Ti  295.81K                              1.4                                                     kg K
                    m
     ai  340
                    s                     Ts  295.81K                                                 2                         d  .283in
                                                                                     A t   
                                                                                                   d
 t char 
                v                                                                               
             A t  ai
                                                    Ti                                         2
                                          Tiplus 
                                                    Ts                                        Ae  At

 tchar  1.92 s                                                                                                     5       2
                                                                                               At  4.058  10           m

                                                                                                   Ps  110.7psi
                                                               t                                                                                Pi
                                           t plus ( t )                                          Pi  14.7psi
                                                             t char                                                                   Piplus 
                                                                                                                                                Ps
                                                                                               Punch  Ps  .528
                                                                                                                                      Piplus  0.133
 choked isothermal charge:                                                                     Punch  58.45psi

                                                                                                                    Punch
                                    
                                    
                                                (   1) 
                                                                                                  Pplusunch 
                                       1  2(  1) 
                                                                                                                        Ps
   Pplus ( t )  Piplus  Tiplus                       tplus ( t)
                                     2                                                              Pplusunch  0.528



                                        Pplusunch  Piplus
               tplusunch 
                                         
                                         
                                                     (   1) 
                                                               
                                            1  2(  1) 
                                Tiplus                     
                                          2                                         tplusunch  0.683

                                                                                        tunch  tplusunch  tchar


             P( t)  Pplus ( t)  Ps                                                   tunch  1.312 s




                                                                                                                                                
                                                                                                                                              1
                                                                                                                      
                                                                                                                                         2
                                                                      1                                                  
                                                                     2              1                                 
                        
                                                              1 
                                                                                     2                                 
                                                                               1  T  At                           
Pplusunchoked ( t)  1   1  Pplusunch
                                                                              2  iplus  A  tplus ( t)  tplusunch  
                                                                         
                                                                
                                                                            2          e                           
                                         12



                   1



Pplusunchoked( t )0.5




                   0
                        0       5   10

                                t




    Guess

              g  3s
  Given

  Pplusunchoked ( g)        1

  x  Find ( g)

  x  3.767 s
  tend  x

				
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