sir the pump work in for state 1-2 should be multiplied by the fraction of (1-y) because it only handelling thet amount of fluid and your videos are still helping thsnks a lot
Remember that we are dividing through by m_dot_5, or the mass flow rate through the boiler, as shown at 4:50 in the video. If we were to evaluate the work out of the turbine as (h5-h6)+(1-y)(h5-h7) we would be accounting for the work in the high-pressure turbine twice, once at a mass fraction of 1 and the second at a mass fraction of (1-y), and this would incorrectly overestimate the work output from the high-pressure turbine. The low-pressure turbine (expansion from state 6 to 7) only has a mass flow rate of (1-y)*m_dot_5. Given that we have divided by the mass flow rate through the boiler, when examining the work output of the low-pressure turbine we need only consider expansion from 6 to 7 (h6-h7) and then multiply it by the reduced mass fraction (1-y).
In determining "y", you want to divide by the flow going through the boiler (that is why m_dot_5 is in the denominator). Why put the mass flow rate through the Open FWH in the numerator and not the mass flow rate of the flow through the Condenser? You could do that, but it would give you a slightly different equation when you divide Eq. 1 (first law applied to the Open FWH) by m_dot_5. If you solve it this way, watch how you label both your process diagram and T-s diagram as the y and (1-y) streams will change.
sir the pump work in for state 1-2 should be multiplied by the fraction of (1-y) because it only handelling thet amount of fluid and your videos are still helping thsnks a lot
In the beginning around 2:45 when you said state 1-3-5, you meant state 3-4-5. It was a little confusing for a second.
that is correct (I know im sayin this 6 years after you posted this but I guess its for other people who may have also looked at the comments)
@@XplosiveAction yeah i looked at that, thank u
at 7:32, why is it not (h5-h6)+(1-y)(h5-h7)?
Remember that we are dividing through by m_dot_5, or the mass flow rate through the boiler, as shown at 4:50 in the video. If we were to evaluate the work out of the turbine as (h5-h6)+(1-y)(h5-h7) we would be accounting for the work in the high-pressure turbine twice, once at a mass fraction of 1 and the second at a mass fraction of (1-y), and this would incorrectly overestimate the work output from the high-pressure turbine. The low-pressure turbine (expansion from state 6 to 7) only has a mass flow rate of (1-y)*m_dot_5. Given that we have divided by the mass flow rate through the boiler, when examining the work output of the low-pressure turbine we need only consider expansion from 6 to 7 (h6-h7) and then multiply it by the reduced mass fraction (1-y).
I am writing a thermodynamics exam tomorrow, this really helped even after 4 years. Thank you.
Great lesson once again!
Sir , how is implementing a deaerator diffrent to an open feedwater heater ?
sir how are u assuming (y=m_dot_6 / m_dot_5)
In determining "y", you want to divide by the flow going through the boiler (that is why m_dot_5 is in the denominator). Why put the mass flow rate through the Open FWH in the numerator and not the mass flow rate of the flow through the Condenser? You could do that, but it would give you a slightly different equation when you divide Eq. 1 (first law applied to the Open FWH) by m_dot_5. If you solve it this way, watch how you label both your process diagram and T-s diagram as the y and (1-y) streams will change.