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Thermodynamics: Control Volume analysis using energy

  1. Jan 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Steam flows through an uninsulated pipe at 0.5 kg/s, entering at 5 bar, 440 C and exiting at 3 bar, 320 C. How much energy is lost from the steam per hour?

    2. Relevant equations

    Which terms in the energy eqn can drop out? I'm having trouble with that.

    Is there only one W?

    3. The attempt at a solution

    Starting w/ the full energy eqn:

    de/dt = Q - W_ + (m_i)[u_i + (p_i)(v_i) + (v_i)^2 + g*z_i] - (m_e)[u_v + (p_e)(v_) + (v_e)^2 + g*z_e]


    where Q = vol. flow rate
    W = work rate
    m = mass flow rate
    subscript e = out
    subscript i = in
    v = velocity
    u = specific internal energy
    p = pressure

    So z_i = z_e = 0

    Q = mv = (0.5)(0.6548) = 0.3274?
     
  2. jcsd
  3. Jan 17, 2009 #2
    I would treat this as a steady state problem. Meaning your mass flow rate in is going to equal your mass flow rate out. and de/dt is zero.

    Since we're not given any sorts of elevation, the potential energy will be drop out. No work is happening, so W will also drop out.

    You will be left with
    Q/m=hb-ha+(V^2/2)b-(V^2/2)a


    h is of course equal to u+pv

    a=in
    b=out
    m=mass flow rate
    Q=energy rate
    v=specific volume
    h=enthalpy
    V=velocity

    I can't check any of your values because I don't have a thermo book with me.
     
  4. Jan 17, 2009 #3
    ok that makes sense, but how do i find velocity?
     
  5. Jan 17, 2009 #4
    You know what, based on the information given, I would let velocity drop out as well. In a basic thermo class most prevalent place you're going to see velocity -not- being negligible is in nozzle and diffuser problems, or where it is expressively given to you in the problem statement.
     
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