Thermodynamics: Control Volume analysis using energy

  • Thread starter aznkid310
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  • #1
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Homework Statement



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?

Homework Equations



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

Is there only one W?

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?
 

Answers and Replies

  • #2
18
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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.
 
  • #3
109
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ok that makes sense, but how do i find velocity?
 
  • #4
18
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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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