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It has been a while since I had thermodynamics and am attempting to solve this problem for work. I need to figure out the other two mass flow rates. I worked it out with specific heats, but don't really trust it. Seems like there should be a simple way to do it with enthalpy; the process just escapes me. Thanks for the help!

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With steam in general and absolutely with any phase changes and pressure changes, you need to use enthalpies instead of specific heat. If there are phase changes, specific heat neglects the enthalpy of vaporization. Even if there aren't, the specific heat of steam changes too much with temperature.

You just need to do mass and energy balances (isn't it always mass and energy balances?). The mass balance is $\dot m_h + \dot m_c = \dot m_l$ and the energy balance is $\dot m_h h_h + \dot m_c h_c = \dot m_l h_l$. The enthalpies should be available in any thermodynamics textbook or various software packages.

I'll let you solve the whole thing, but if you have more issues, leave a comment or update the question.

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  • $\begingroup$ Good to know on the specifc heats. I was hoping it was as easy as energy balance, but it seems im still missing something. At 200psig h=1200.1 & at 30psig h=1164.0. There is not a significant difference. When you get up to 650psig it even goes back down h=1131.8. I remember "enthalpy is conserved across a throttling valve" too. But there is obviously more energy in saturated 200psig steam than 18psig steam per lbm... $\endgroup$ – ericnutsch Feb 10 '17 at 16:24
  • $\begingroup$ The absolute values don't matter very much. The only thing that matters is that the outlet enthalpy is in between the two inlets. This ensures that you can mix them to get the middle value. Otherwise, I'm not exactly sure what your outstanding question is. $\endgroup$ – ericksonla Feb 10 '17 at 19:54
  • $\begingroup$ MH (1200.1BTU/lbm) + MC (1BTU/lbm)(190F) = ML(1164.0BTU/lbm) Would give a very low value for MC. If I were reducing steam from 650psig (which has more energy than 200psig). The equation would not work (negative condensate flow). MH (1131.8BTU/lbm) + MC (1BTU/lbm)(190F) = ML(1164.0BTU/lbm). The saturated vapor enthalpy table does not make sense because the numbers do not increase with pressure and temperature. Energy balance makes sense, the numbers just dont work out. I am missing something. $\endgroup$ – ericnutsch Feb 10 '17 at 21:26
  • $\begingroup$ Yeah, doesn't take much. The enthalpy of vaporization of water is huge compared to the cp! The steam enthalpies look right but the condensate value of 1BTU/lbm doesn't look right. I get 369 BTU/lbm. $\endgroup$ – ericksonla Feb 10 '17 at 23:20

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