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    ME 231 Gomes Whharris 2-8-18SOLUTION

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    Selo
    Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar and 80°C. The work for this process is determined by first finding the specific volumes at the initial and final states from property tables, then calculating the polytropic exponent from the relation that pv^n is constant. Plugging these values into the polytropic work equation gives a work of -69.88 kJ.

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    ME 231 Gomes Whharris 2-8-18SOLUTION

    Uploaded by

    Selo
    29 views4 pages
    Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar and 80°C. The work for this process is determined by first finding the specific volumes at the initial and final states from property tables, then calculating the polytropic exponent from the relation that pv^n is constant. Plugging these values into the polytropic work equation gives a work of -69.88 kJ.

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    ME_231_Gomes_whharris_2-8-18SOLUTION

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    Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar and 80°C. The work for this process is determined by first finding the specific volumes at the initial and final states from property tables, then calculating the polytropic exponent from the relation that pv^n is constant. Plugging these values into the polytropic work equation gives a work of -69.88 kJ.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    29 views4 pages

    ME 231 Gomes Whharris 2-8-18SOLUTION

    Uploaded by

    Selo
    Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from an initial state of saturated vapor at 2 bar to a final state of 12 bar and 80°C. The work for this process is determined by first finding the specific volumes at the initial and final states from property tables, then calculating the polytropic exponent from the relation that pv^n is constant. Plugging these values into the polytropic work equation gives a work of -69.88 kJ.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    PROBLEM 3.

    33

    Two kg of Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly from


    an initial state of saturated vapor at 2 bar to a final state of 12 bar, 80oC. Determine the work for
    the process, in kJ.

    KNOWN: Refrigerant 134A undergoes a polytropic process in a piston-cylinder assembly.

    FIND: Determine the work.

    SCHEMATIC AND GIVEN DATA:

    R-134A
    m = 2 kg .2 12 bar
    80oC pvn = constant

    ENGINEERING MODEL: 1. The refrigerant is a


    .
    1
    2 bar

    closed system. 2. The process is polytropic: pvn =


    constant. v

    ANALYSIS: The work for the polytropic process is determined using Eq. 2.17, with pvn =
    constant. Following the procedure of part (a) of Ex. 2.1

    W= (*)

    In order to evaluate this expression, we need to determine the specific volumes and the
    polytropic exponent, n.

    State 1: From Table A-11; v1 = vg1 = 0.0993 m3/kg


    State 2: From Table A-12, at 12 bar, 80oC; v2 = 0.02051 m3/kg

    The polytropic exponent is found from pvn = constant as follows.

    → → n = ln(p1/p2) / ln(v2/v1)

    n = ln(2/12) / ln(0.02051/0.0993) = 1.136

    Inserting values in Eq. (*) and converting units, we get


    PROBLEM 3.33 (CONTINUED)

    W= = -69.88 kJ (in)
    PROBLEM 3.92

    Determine the volume, in m3, occupied by 2 kg of H2O at 100 bar, 400oC, using (a) data from the
    compressibility chart, (b) data from the steam tables.

    Compare the results of parts (a) and (b) and discuss.

    H2O
    V = ??
    m = 2 kg
    p = 100 bar
    T = 400oC

    (a) Using the compressibility chart, first we need to determine the reduced pressure and
    temperature. From Table A-1:
    pc = 220.9 bar and Tc = 647.3 K

    pR = p/pc = (100)/(220.9) = 0.45


    → (Figure A-1): z ≈ 0.86
    TR = T/Tc = (400 + 273.15)/(647.3) = 1.04

    Now, we can calculate the specific volume as follows.

    v= = (0.86) = 0.0267 m3/kg

    So, the volume is: V = m v = (2)(0.0267) = 0.0534 m3

    (b) From Table A-4 at 100 bar, 400oC; v = 0.02641 m3/kg

    Thus, V = m v = (2)(0.02641) = 0.05282 m3

    Comments: The compressibility chart gives a fairly accurate value considering the relative
    imprecision of reading values from the chart. The percent difference is approximately 1.1%.

    Note also that the value of z is 0.86. Hence, the ideal gas model is not particularly applicable at
    this state. The ideal gas model would predict a volume of 0.03105 m3, which is about 15% low.
    PROBLEM 3.94

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