Ask Question
27 October, 03:49

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.

+3
Answers (1)
  1. 27 October, 05:53
    0
    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%
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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) ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers