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12 May, 23:23

Determine the percent error in using the ideal gas model to determine the specific volume of (a) water vapor at 4000 lbf/in. 2, 10008F. (b) water vapor at 5 lbf/in. 2, 2508F. (c) ammonia at 40 lbf/in. 2, 608F. (d) air at 1 atm, 5608R. (e) Refrigerant 134a at 300 lbf/in. 2, 1808F.

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  1. 12 May, 23:46
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    a) 24.01%

    b) 0.35%

    c) 3.43%

    d) 0%

    e) 37.31%

    Explanation:

    A) water vapor=> H2O

    From table @4000 lbf/in^2, 1000°F v = 0.1752 ft^3/ib

    m = 18.02 ib/ibmol

    P = 4000 lbf/in^2

    From the ideal gas model

    PV/m = mRT/m

    V = RT/Pm

    Where R = 1945 (ft. ibf) / (ibmol.°R)

    T = 1000°F = 1459.67°R

    Therefore

    V = (1945 * 1459.67) : (4000 * 18.02) = 2255190.19 : 72080 = 0.21727 ft^3/ib

    Error = (V - v) / v * 100% = (0.21727 - 0.1752) : 0.1752 = 0.2401 * 100% = 24.01%

    B) water vapor=> H2O

    From table @ 5 lbf/in^2, 250°F v = 84.21 ft^3/ib

    m = 18.02 ib/ibmol

    P = 5 lbf/in^2

    From the ideal gas model

    PV/m = mRT/m

    V = RT/Pm

    Where R = 1945 (ft. ibf) / (ibmol.°R)

    T = 250°F = 709.67°R

    Therefore

    V = (1945 * 709.67) : (5 * 18.02) = 84.51 ft^3/ib

    Error = (V - v) / v * 100% = (84.51 - 84.21) : 84.21 = 0.003538 * 100% = 0.35%

    C) ammonia = >

    From table @40 lbf/in^2, 60°F v = 7.9134 ft^3/ib

    m = 17.03 ib/ibmol

    P = 40 lbf/in^2

    From the ideal gas model

    PV/m = mRT/m

    V = RT/Pm

    Where R = 1945 (ft. ibf) / (ibmol.°R)

    T = 60°F = 519.67°R

    Therefore

    V = (1945 * 519.67) : (40 * 17.03) = 8.189 ft^3/ib

    Error = (V - v) / v * 100% = (8.189 - 7.9134) : 7.9134 = 0.03432 * 100% = 3.43%

    D) Air at 1 atm = >

    No table so we assume air is ideal

    m = 28.97 ib/ibmol

    P = 14.71 lbf/in^2

    From the ideal gas model

    PV/m = mRT/m

    V = RT/Pm

    Where R = 1945 (ft. ibf) / (ibmol.°R)

    T = 560°R

    Therefore

    V = (1945 * 560) : (14.71 * 28.97) = 14.099 ft^3/ib

    Since Air is ideal error is zero = 0%

    E) refrigerant = >

    From table @300 lbf/in^2, 180°F v = 0.1633 ft^3/ib

    m = 102.03 ib/ibmol

    P = 300 lbf/in^2

    From the ideal gas model

    PV/m = mRT/m

    V = RT/Pm

    Where R = 1945 (ft. ibf) / (ibmol.°R)

    T = 180°F = 639.67°R

    Therefore

    V = (1945 * 639.67) : (300 * 102.03) = 0.22422 ft^3/ib

    Error = (V - v) / v * 100% = (0.22422 - 0.1633) : 0.1633 = 0.37306 * 100% = 37.31%
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Home » Engineering » Determine the percent error in using the ideal gas model to determine the specific volume of (a) water vapor at 4000 lbf/in. 2, 10008F. (b) water vapor at 5 lbf/in. 2, 2508F. (c) ammonia at 40 lbf/in. 2, 608F. (d) air at 1 atm, 5608R.
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