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Work done and total heat transfer for piston-cylinder device

  1. Jun 25, 2011 #1
    Hi everybody, this question is on an old exam paper given to us for exam prep purposes. No answers have been supplied however. Am I on the right track?

    Question:
    A mass of 0.2kg of saturated refrigerant-134a is contained in a piston-cylinder device at 200kPa. Initially 75% of the mass is in the liquid phase. Now heat is transferred to the refrigerant at constant pressure until the cylinder contains only vapor. Determine:
    a) the work done, and
    b) the total heat transfer

    Assumptions:
    Piston-cylinder device is frictionless, constant pressure, quasi-equilibrium, no heat lost to surroundings

    a) the work done
    Formula:
    W = P (V2-V1) where,

    V2 = Volume at state 2
    V1 = Volume at state 1
    P = Pressure
    W = Boundary work

    Properties:
    P = 200 kPa
    V2 = 0.2 x vg where vg = 0.099867 (gas table) = 0.0199734 m3/kg
    V1 = 0.2 x (0.75 x vf + 0.25 x vg) = 0.005106345 m3/kg

    Analysis:
    W = P(V2-V1)
    = 200 x (0.0199734 - 0.005106345)
    = 3.89 kJ boundary work

    b) total heat transfer
    Formula:
    Q = h2 - h1 where,

    Q = total heat transferred
    h2 = Enthalpy state 2
    h1 = Enthalpy state 1

    Properties:
    h2 = hg = 244.46 kJ/kg (from gas tables)
    h1 = .75 x hf + .25 x hg = .75 x 38.43 + .25 x 244.46 = 89.9375 kJ/kg

    Analysis:
    Q = h2 - h1
    = 244.46 - 89.9375
    = 154.52 kJ heat
     
  2. jcsd
  3. Jun 26, 2011 #2

    rock.freak667

    User Avatar
    Homework Helper

    I think that is all correct.
     
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