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Engineering LibreTexts

8.2 Problem Set

  • Page ID
    9482
  • Determine the saturation temperature, specific liquid enthalpy, specific enthalpy of evaporation and specific enthalpy of dry steam at a pressure of 2.04 MPa.

    Pressure [MN/m2] Saturation Temperature [C] hf [kJ/kg] hfg [kJ/kg] hg [kJ/kg]
    2.1 214.9 920.0 1878.2 2798.2
    2.0 212.4 908.6 1888.6 2797.2
    Answer:

    MATLAB solution is as follows; >> pressure=[2.1 2.0]; >> sat_temp=[214.9 212.4]; >> h_f=[920 908.6]; >> h_fg=[1878.2 1888.6]; >> h_g=[2798.2 2797.2]; >> sat_temp_new= interp1(pressure, sat_temp,2.04) sat_temp_new = 213.4000 >> h_f_new=interp1(pressure,h_f,2.04) h_f_new = 913.1600 >> h_fg_new=interp1(pressure,h_fg,2.04) h_fg_new = 1.8844e+003 >> h_g_new=interp1 (pressure,h_g,2.04) h_g_new = 2.7976e+003

    The following table gives data for the specific heat as it changes with temperature for a perfect gas. (Data available for download). 1

    Temperature [F] Specific Heat [BTU/lbmF]
    25 0.118
    50 0.120
    75 0.123
    100 0.125
    125 0.128
    150 0.131

    Change of specific heat with temperature

    Using interp1 function calculate the specific heat for 30 F, 70 F and 145 F.


    Answer:

    MATLAB solution is as follows: >> temperature=[25;50;75;100;125;150] temperature = 25 50 75 100 125 150 >> specific_heat=[.118;.120;.123;.125;.128;.131] specific_heat = 0.1180 0.1200 0.1230 0.1250 0.1280 0.1310 >> specific_heatAt30=interp1(temperature,specific_heat,30) specific_heatAt30 = 0.1184 >> specific_heatAt70=interp1(temperature,specific_heat,70) specific_heatAt70 = 0.1224 >> specific_heatAt145=interp1(temperature,specific_heat,145) specific_heatAt145 = 0.1304

    For the problem above, create a more detailed table in which temperature varies between 25 and 150 with 5 F increments and corresponding specific heat values.


    Answer:

    MATLAB solution is as follows: >> new_temperature=25:5:150; >>

    new_specific_heat=interp1(temperature,specific_heat,new_temperature);>> [new_temperature',new_specific_heat'] ans = 25.0000 0.1180 30.0000 0.1184 35.0000 0.1188 40.0000 0.1192 45.0000 0.1196 50.0000 0.1200 55.0000 0.1206 60.0000 0.1212 65.0000 0.1218 70.0000 0.1224 75.0000 0.1230 80.0000 0.1234 85.0000 0.1238 90.0000 0.1242 95.0000 0.1246 100.0000 0.1250 105.0000 0.1256 110.0000 0.1262 115.0000 0.1268 120.0000 0.1274 125.0000 0.1280 130.0000 0.1286 135.0000 0.1292 140.0000 0.1298 145.0000 0.1304 150.0000 0.1310

    During a 12-hour shift a fuel tank has varying levels due to consumption and transfer pump automatically cutting in and out to maintain a safe fuel level. The following table of fuel tank level versus time (Data available for download) is missing readings for 5 and 9 AM. Using linear interpolation, estimate the fuel level at those times.

    Time [hours, AM] Tank level [m]
    1:00 1.5
    2:00 1.7
    3:00 2.3
    4:00 2.9
    5:00 ?
    6:00 2.6
    7:00 2.5
    8:00 2.3
    9:00 ?
    10:00 2.0
    11:00 1.8
    12:00 1.3

    Fuel Tank Level Versus Time


    Answer:

    >> time=[1 2 3 4 6 7 8 10 11 12]; >> tank_level=[1.5 1.7 2.3 2.9 2.6 2.5 2.3 2.0 1.8 1.3]; >> tank_level_at_5=interp1(time,tank_level,5);tank_level_at_5=2.7500>>tank_level_at_9=interp1(time,tank_level,9) tank_level_at_9 = 2.1500

    Footnotes

    • 1 Thermodynamics and Heat Power by Kurt C. Rolle, Pearson Prentice Hall. © 2005, (p.19)