Ask Question
25 September, 19:30

Determine the phase or phases in a system consisting of H2O at the following conditions, sketch p-v and T-v diagrams showing the location of each state and determine the specified property at at the indicated state.

a.) p = 5 bar, T = 151.83C; and find the values of v and u at this state if x=0.6

b.) p = 1.25 MPa, T = 200C; and find the values of v and u at this state (Interpolation may be needed)

c.) p = 5 psia, T=124F; and find the values of v and u at this state (Interpolation may be needed)

d.) p=500 psia, T = 320F; and find the values of v and u at this state (Interpolation may be needed)

+3
Answers (1)
  1. 25 September, 21:45
    0
    a) v = v_f + x * (v_g - v_f) = 0.225377 m^3 / kg

    u = u_f + x * (u_g - u_f) = 1792.592 kJ/kg

    b) v = 0.2060 + (0.1325 - 0.2060) * (1.25 - 1) / (1.5 - 1) = 0.16925 m^3/kg

    u = 2621.9 + (2598.1 - 2621.9) * (1.25 - 1) / (1.5 - 1) = 2610 kJ/kg

    C) v = 0.0162 ft^3 / lb and u = 92 Btu/lb

    D) v = 0.0176 m^3/kg, u = 289 Btu / lb

    Explanation:

    a)

    From steam properties, at p = 5 bar, we get T_sat = 151.83 deg C which is the same as given T. Hence, phase is liquid-vapor mixture.

    Further,

    v_f = 0.0010926 m^3/kg, and v_g = 0.3749 m^3/kg

    u_f = 639.68 kJ/kg, and u_g = 2561.2 kJ/kg

    v = v_f + x * (v_g - v_f) = 0.225377 m^3 / kg

    u = u_f + x * (u_g - u_f) = 1792.592 kJ/kg

    b)

    From steam properties, at T = 200 deg C, we get P_sat = 15.54 bar = 1.554 MPa which is greater than 1.25 MPa. Hence, phase is superheated vapor.

    From superheated tables, at 200 degg C and p = 1 MPa, we get v = 0.2060 m^3/kg and u = 2621.9 kJ/kg

    From superheated tables, at 200 degg C and p = 1.5 MPa, we get v = 0.1325 m^3/kg and u = 2598.1 kJ/kg

    By interpolation, At T = 200 deg C and p = 1.25 MPa we get

    v = 0.2060 + (0.1325 - 0.2060) * (1.25 - 1) / (1.5 - 1) = 0.16925 m^3/kg

    u = 2621.9 + (2598.1 - 2621.9) * (1.25 - 1) / (1.5 - 1) = 2610 kJ/kg

    c)

    From steam properties, at 5 psia we get T_sat = 162.2 F which is less than 124 F. Hence, phase is sub-cooled liquid.

    we get v = 0.0162 ft^3 / lb and u = 92 Btu/lb

    d)

    p=500 psia, T = 320F ... phase = subcooled liquid

    v = 0.0176 m^3/kg, u = 289 Btu / lb
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Determine the phase or phases in a system consisting of H2O at the following conditions, sketch p-v and T-v diagrams showing the location ...” in 📙 Engineering if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers