2020 PPI Practice #57

[BuffaloWings]

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.  

 

[Slay the P.E.]

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.  

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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]

Why does y equal m3/m1 and also m2/m7

 

[Slay the P.E.]

Why does y equal m3/m1 and also m2/m7

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

 

[MikeGlass1969]

This problem looks intimidating but once you do your mass and energy balances around each piece of equipment it becomes an algebra problem.