Posted by Tony Garcia on February 04, 2000 at 12:09:50:
In Reply to: Re: Steam Quality posted by Kyle Sexton on February 03, 2000 at 04:51:12:
: Sorry about that. I meant to say co-op, but I guess I must have missed it when I re-read it.
: Your explanation is good, and makes sense to me. The only new thing would be the isoenthalpic process, which I can't remember doing, but the theory does works if you can keep the enthalpy the same. Thank you for the help. The original plan was to measure the quality of steam we were purchasing from our supplier and to see how efficiently we were actually using it, but depending on the costs associated with a caliometric device we may just put one on the incoming line into the plant. Thanks again for your help.
I'm sorry that I was a little brief on the isoenthalpic process, but I thought I was going a little overboard anyway. However, to clarify, if you start with your conservation of energy equation for a control volume. Now develope the energy rate balance for the control volume which has an inlet boundry just before the throttling valve and an exit boundry just after the throttling valve. Assume steady state, 1-D flow, uniform properties, 1 inlet/ 1 outlet.
Now dEcv/dt=0 due to S.S. conditions, your device is fully insulated so Qcv dot = 0, Wcv dot = 0 (no work done), delta PE = 0, and all you have left is m dot[(h1 - h2) + delta KE] = 0. Now here we should expect (assume) that the entalpy change is much larger than the velocity change across the throttling valve. So, for these assumptions delta KE is taken to be zero. What you are left with is m dot(h1 - h2) = 0. M dot cancels out and you are left with h1 = h2 or as I stated before, h of the sys. = h of the calorimeter.
I hope that works!
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