BuffaloWings
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Can someone help me understand this solution. I get confused in the 2nd part when they calculate y using the enthalpy at State 2, 6, and 7.
2020 PPI Practice #57
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Slay the P.E.
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Perform an energy balance on the feed water heater;
energy in = energy out
m2 h2 + m6 h6 = m7 h7
but, from a mass balance m2 + m6 = m7 so the energy balance becomes:
m2 h2 + (m7 - m2) h6 = m7 h7
Now, divide this equation on both sides by m7 and note that y=m2/m7:
y h2 + (1 - y) h6 = h7
which you can use to solve for y:
y = (h7 - h6)/(h2 - h6)
BuffaloWings
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Why does y equal m3/m1 and also m2/m7
Slay the P.E.
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y is defined as the fraction of steam that gets extracted from the high pressure turbine. Hence y = m2/m1 but m7=m1 so y = m2/m7
y is NOT also equal to m3/m7. m3 is what remains in the turbine after extraction, so m3 = m1 - m2. Divide this by m1 and you get m3/m1 = 1 - y
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MikeGlass1969
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This problem looks intimidating but once you do your mass and energy balances around each piece of equipment it becomes an algebra problem.
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