27156908 concentric tube heat exchanger

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  • TABLE OF CONTENTS

    No. Topics Page No.1. Title 22. Objective 23. Theory 24. Equipment 35. Procedure 56. Data and Results 67. Sample Of Calculation 78. Analysis and Discussion 99. Conclusion 1010. References 10

    TITLE: CONCENTRIC TUBE HEAT EXCHANGER

    OBJECTIVE: To demonstrate the effect of flow rate variation on the performance

    characteristics of a counter-flow concentric tube heat exchanger.

    1

  • THEORY:

    The equations for calculating the performance characteristics: power emitted, power

    absorbed, power lost, efficiency (), logarithmic mean temperature (), and overall heat

    transfer coefficient (U).

    The efficiency for the cold medium is:

    c = (Tc,out Tc,in) / (Th,in Tc,in) 100

    The efficiency for the hot medium is:

    h = (Th,in Th,out) / (Th,in Tc,in) 100

    The mean temperature efficiency is:

    mean = ( c + h ) / 2

    The power emitted is given below (where Vh is the volumetric flow rate of the hot

    fluid):

    Power Emitted = Vh h Cph ( Th,in Th,out )

    The power absorbed is given below (where Vc is the volumetric flow rate of the cold

    fluid):

    Power Absorbed = Vc c Cpc ( Tc,out Tc,in )

    2

  • The power lost is therefore:

    Power Lost = Power Emitted Power Absorbed

    The overall efficiency () is:

    =(Power Absorbed / Power Emitted) 100

    The logarithmic mean temperature difference (Tm) is:

    Tm = (T1 T2) / ln (T1/T2)

    = [ (Th,in Tc,out) (Th,out Tc,in) ] / ln [(Th,in Tc,out) / (Th,out Tc,in)]

    The overall heat transfer coefficient (U) is:

    U = Power Absorbed / As . Tm

    Where the surface area (As) for this heat exchanger is 0.067 m

    EQUIPMENT: The experiment set-up consists of:a) Set of tube heat exchanger.b) Cold fluids supply and hot fluids supply.c) Digital stopwatch.

    3

  • Figure 1: Set of tube heat exchanger. Figure 2: Volumetric Flow Rate.

    Figure 3: Decade Switch. Figure 4: Flow Diagrams.

    This experiment can be made using either parallel or counter flow operation. This

    experiment was conducted as counter flow operation.

    4

  • PROCEDURE:

    1. Configure the experiment for counter flow heat exchanger operation such as turn

    ON the heating elements to heat the fluids.

    2. Set the required hot water inlet temperature to Th,in = 60 with the decade switch

    and set the cold water volumetric flow rate (Vc) to run at a constant 2000

    cm/min.

    3. Initially set the hot water volumetric flow rate Vh to 1000 cm/min. Wait until 5

    minutes before the six temperature readings are records.

    4. Repeat this for volumetric flow rate,Vh of 2000, 3000 and 4000 cm/min for hot

    water. Record the temperature readings in the table.

    5. After finish up the experiment, turn OFF the heating elements, close the valve for

    hot and cold water.

    5

  • DATA AND RESULTS:

    Vh Th,in Th,mid Th,out Tc,in Tc,mid Tc,out(cm/min

    )

    (m/s) (C) (C) (C) (C) (C) (C)

    1000 1.667E-

    5

    60 53 47 30 31 35

    2000 3.333E-

    5

    60 55 50.5 30 33 38

    3000 5E-5 60 57 53 30 34 403900 6.5E-5 60 58 54 30 35 41

    Vh Power

    Emitted

    Power

    Absorbed

    Power

    Lost

    Efficiency T1 T2 Tm U

    cm/

    min

    m/s (W) (W) (W) (%) (C) (C) (C) W/(m. C)

    1000 1.667E

    -5

    891.786 693.479 198.307 77.76 25 17 20.74 499.06

    2000 3.333E

    -5

    1302.99 1109.57 193.42 85.16 22 20.5 21.24 779.69

    3000 5E-5 1440.29 1386.96 53.33 96.3 20 23 21.46 964.634000 6.5E-5 1604.89 1525.65 79.24 95.02 19 24 21.4 1064.07

    SAMPLE CALCULATION:

    From table A-9 (Properties of saturated water):

    6

  • At Tc,in = 30 C.

    Vc = 2000 cm/min = 2000 cm/min 1 min/60 s 1 m/100 cm

    =3.333E-5 m/ s

    c = 996 kg / m

    Cpc = 4178 J/kg.K

    At Th,in = 60 C.

    h = 983.3 kg / m

    Cph = 4185 J/kg.K

    a) Power Emitted = Vh h Cph ( Th,in Th,out )

    = (1.667E-5 m)(983.3 kg / m)(4185 J/kg.K)(333K 320K)

    = 891.786 W

    b) Power Absorbed = Vc c Cpc ( Tc,out Tc,in )

    = (3.333E-5 m/ s)( 996 kg / m)(4178 J/kg.K)(308K 303K)

    = 693.479 W

    c) Power Lost = Power Emitted Power Absorbed

    = 891.786 W 693.479 W

    = 198.307 W

    d) Overall Efficiency, =(Power Absorbed / Power Emitted) 100

    = (693.479 W / 891.786 W) 100

    = 77.76 %

    e) Logrithmic Mean Temperature Difference, Tm= (T1 T2) / ln (T1/T2)

    T1 = Th,in Tc,out= 60 C - 35 C

    7

  • = 25 C

    T2 = Th,out Tc,in= 47 C 30 C

    = 17 C

    Tm = (T1 T2) / ln (T1/T2)

    = (25 C 17 C) / ln [(25 C/17 C)

    = 20.74 C

    f) Overall Heat Transfer Coefficient, U = Power Absorbed / As . Tm

    U = Power Absorbed / As . Tm

    = 693.479 W / (0.067 m20.74 C)

    = 499.06 W/ m .C

    ANALYSIS AND DISCUSSION

    8

  • Heat exchanger are commonly used in practice, and an engineer often finds

    himself or herself in a position to select a heat exchanger that will achieve a specified

    temperature change in a fluid stream of known flow rate, or to predict the outlet

    temperatures of the hot and cold fluids stream in a specified heat exchanger.

    The variation of temperature of hot and cold fluids in a counter-flow heat

    exchanger is given in figure 4. Note that the hot and cold fluids enter the heat exchanger

    from opposite ends, and the outlet temperature of the cold fluid in this case may exceed

    the outlet temperature of the hot fluid. In the limiting case, the cold fluid will be heated to

    the inlet temperature of the hot fluid. However, the outlet temperature of the cold fluid can

    never exceed the inlet temperature of the hot fluid, since this would be a violation of the

    second law of thermodynamics.

    CONCLUSION

    9

  • It can be concluded that the experiment is done successfully. The power emitted

    and power absorbed are increased when we compared the effect of changing the

    volumetric flow rate of the hot fluid. Besides, the power lost that we get shows decreasing

    value unless the last reading give some increased value. This is maybe because of the

    error while doing the experiment that may cause by conduction and convection between

    hot and cold fluid while doing counter flow operation. The overall efficiency are

    reasonable and doesnt exceed the 100%. From our experiment, the overall heat transfer

    coefficient will increase when the volumetric flow rate of the hot fluid are increase. So

    that, the conclusion that can be done is the overall heat transfer coefficient, the power

    emitted and power absorbed are influenced by the changing of volumetric flow rate of the

    hot fluid.

    REFERENCE

    1. Heat and Mass Transfer (A Practical Approach) 3rd EditionYunus A. CengelMcGraw Hill (2006).

    2. Class Note KJM531-Heat Transfer.

    10