Video summary
The video presents an analysis of an ideal regenerative Rankine cycle utilizing a closed feedwater heater, where water serves as the working fluid. The system operates with a turbine inlet at 500 psi and 600 degrees Fahrenheit, while the condenser is maintained at 5 psi. Steam enters the closed feedwater heater at an intermediate pressure of 40 psi before being expanded into the condenser. A key distinction in this configuration is that the extraction steam expands down to the low pressure rather than being compressed up to the high pressure, meaning a single pump handles all the compression work for the cycle. The analysis assumes ideal conditions with 100% efficiency for both the turbine and the pump, treating expansion valves as isenthalpic devices where enthalpy remains constant during throttling.
To solve the problem, the narrator identifies eight distinct state points and determines their properties using pressure, temperature, and entropy data from steam tables. The mass flow rate through the boiler, turbine inlet, and condenser is denoted as the total cycle flow, while a fraction $y$ of this flow is extracted to feed the closed heater, leaving $(1-y)$ to proceed directly to the condenser. By performing an energy balance on the closed feedwater heater, which has no heat transfer or work interactions, a specific relationship is derived to calculate the extraction fraction $y$. This calculation relies on the enthalpies of the streams entering and leaving the heater, specifically equating the energy gained by the feedwater to the energy lost by the extracted steam.
Once the value of $y$ is determined, the analysis proceeds to calculate the specific work input from the pump, the heat added in the boiler, the total work output from the turbine accounting for extraction, and the heat rejected in the condenser. These individual components are combined to find the net work output and the total heat input, which allows for the final calculation of the cycle's thermal efficiency. The results show that the extraction fraction is approximately 12.3%, the specific pump work is about 3.1 Btu/lbm, the boiler heat addition is roughly 1060 Btu/lbm, and the overall thermal efficiency reaches 30.6%.
The video concludes by visualizing these processes on a Temperature-Entropy (T-s) diagram to provide a graphical representation of the cycle's performance. The diagram illustrates the compression from the saturated liquid state at low pressure up to the high-pressure compressed liquid region, followed by heating to the maximum temperature and subsequent expansion in the turbine. The extraction points are shown as branches splitting off to the closed feedwater heater before rejoining the main flow path. This visual aid confirms that the extracted steam condenses to a saturated liquid before mixing with the condensate, effectively demonstrating how the regenerative cycle improves efficiency by preheating the feedwater using waste heat from the turbine expansion.
Read the full video transcript
an ideal regenerative rankine cycle with
a closed feed water heater uses water as
the working fluid
the turbine inlet is operated at 500 psi
and 600 degrees fahrenheit and the
condenser is maintained at 5
psi steam is supplied to the feed water
heater at 40 psi before being expanded
into the condenser
determine and complete the following
first y
the proportion of mass flow rate that
leaves the turbine early to feed the
closed feed water heater
the specific work in relative to the
cycle as a whole
and the thermal efficiency so note that
in this problem we still have the medium
pressure occurring at seven and three
for the closed feed water heater
but instead of compressing it up to the
high pressure before mixing we are
expanding it down to the low pressure
before mixing
that means that one pump is
accomplishing all of the work
and that one two five and six all have
the mass flow rates
of the cycle let's start by identifying
our
independent intensive properties for the
state points so that we can then
determine the specific work in the
specific hue in this specific workout
and the specific cue out and then the
thermal efficiency
i will start this process by pointing
out that we still have
three pressures 500 psi 5 psi and 40 psi
which stay points have the high pressure
2 5 and 6 which state points have the
low pressure
8 4 and 1 that leaves
3 and 7 with the intermediate pressure
so i will populate one independent
intensive property for
all eight of our state points using
pressure
next i will consider the pump and the
turbine i was given no indication as to
an operational efficiency
so i will assume that they are both 100
efficient
that means that the process from one to
two is assumed to be isentropic
because an isentropic process represents
the ideal work for a turbine and a pump
and then i assume that s7 and s8 are all
equal to s6
that leaves me with state point one
three four five and six
left hanging state six is easy we have a
temperature
so state six has a temperature of 600
degrees fahrenheit
next i'll assume that the condenser just
condenses
and in this case is also a mixing
chamber but the condenser condenses and
as soon as the water has condensed it
leaves the condenser
it's just a mechanical device where
water condenses and then it falls to the
bottom and leaves
there's nothing to refrigerate it to sub
cool it to hyper compress it
there is just a condensing process and
then the water leaves so we assume
it leaves as a saturated liquid
then three four and five well i can make
the same assumption
about state three that i did in the
previous problem
where i had assumed based on the same
concept of
using latent energy for the heat
exchange
that the water that leaves at 3 has
condensed
and leaves and then because i'm assuming
that our closed feed water heater is
operating ideally because i was given no
indication
otherwise i'm going to use the same
relationship for
maximum conditions which is a
temperature at five being equal to a
temperature of three
that leaves me with state point four and
to answer that question
let me pose a question what do we know
about expansion valves
well if we were to set up an energy
balance on the expansion valve i have no
opportunities for heat transfer nor work
therefore this open system undergoing a
steady process would have its energy
balance simplify down to the sum in of m
theta is equal to the sum out of m theta
theta contains enthalpy kinetic energy
and potential energy so if i neglect
changes in kinetic and potential energy
i'm left with m.3 h3 is equal to m.4 h4
and then if the mass flow rate at 3 has
to equal state 4
because it's a steady device with one
inlet and one outlet
then h3 must equal h4
therefore we treat expansion valves as
being
isenthalpic that is of a constant
enthalpy
so i use x1 and p1 to look up s1 then i
use s2 to np2 to look up h2
i guess i also look up h1 using x1 and
p1
then i use x3 and p3 to look up h3 and
use h4 being equal to h3 to
announce that h4 is known and then i use
t5
and p5 to look up h5 so i have to go
back to state three look up the
saturation temperature corresponding to
our
40 psi then i use t6 and p6 to look up
h6 and
s6 then i use s6 to look up
h7 and h8 using the pressure at the
medium pressure and the low pressure
respectively so i have everything i need
to perform
all eight enthalpy lookups so let's just
assume that we had them
and move on
the next process would be to calculate
the specific work in
the specific queue in the specific
workout
and the specific queue out
the specific work in is going to be
occurring in the
one and only pump let me back up a
second
we're defining y as the mass flow rate
that leaves the turbine early
so i'm calling y
m.7
over m dot 6 and then the remainder 1
minus y
is m dot 8 over m.6
and i recognize that 1 2 5
and 6 are all the same and they are all
the mass flow rate through the cycle
then eight stands alone
and seven three and
four are all the same
so specific work into the cycle would be
the total power input divided by m dot
cycle
which would be m.1 times h2 minus h1
divided by m dot cycle because m.1 is
equal to m.a cycle that means my
specific work in is just
h2 minus h1
for qn i'm looking at the boiler the
boiler also has the mass flow rate
through the cycle
flowing through it which means that i'm
just going to be left with h6
minus h5
for our specific workout we have the
same exact energy balance that we have
had for the previous two
examples so i'm going to use h6 minus
yh7
minus the quantity 1 minus y h8
and if you want to see why unintended
you can go back and watch those videos
again h6 minus y
h7 minus the quantity 1 minus y h8
and then and then specific queue out is
going to be the complicated one this
time
because i have multiple inlets and
outlets so let's just
walk through that energy balance
i have entering mass flow rate at eight
and four let me double check those state
point numbers
and because i have a steady device that
is an open system
my energy balance
is going to simplify quite a bit
one to skip a few i can jump to writing
q as well as the sum
in of m dot theta is equal to
q dot out plus the sum out
of m dot theta because there's no
opportunities for heat transfer in
nor work and it's an open system
operating steadily
and then theta contains enthalpy plus
specific kinetic energy plus specific
potential energy so if we assume that
changes in kinetic and potential energy
are negligibly small
that means i'm left with the sum n of m
dot h there are two inlets so that would
be
m dot eight h eight plus m dot
4 h4
is equal to q that out plus
m 1 h1 the one and
only outlet therefore
q that out is going to be
m.8 h8 plus m dot for
h4 minus
lambda 1 h1
and then if i divide everything by m dot
cycle
i'm going to be left with m.8 divided by
m.cycle which is 1 minus y
plus m.4 divided by m that cycle which
is
y
minus y times h4
minus m dot one divided by m that cycle
which is just one
h1
so with these four equations i can
calculate my specific works and heat
transfers
from that i can determine my network and
that heat transfer
and then thermal efficiency all i need
to do that
are my eight enthalpies which you'll
remember i totally know by now because i
totally looked them up
and why so in order to be able to
complete the problem
we are going to have to determine y and
remember in order to compute y
we have to perform an energy balance on
a device about which we know everything
we can't analyze the boiler the turbine
the condenser nor the pump because i
don't know cue in
work out cue out nor work in the
expansion valve is boring there's not
much to do there
therefore i'm left with the energy
balance on the closed feed water heater
so that energy balance is something that
we've done before so i will go through
it relatively quickly
so i'm going to have the sum in of m.h
is equal to the sum out of m.h
because there's no opportunities for
heat transfer nor work and changes in
kinetic and potential energy are
negligibly small
so i'm going to write that as m.2 h2
plus m dot seven h7
that is equal to the sum out of m dot
theta which is going to be
m.5 h5
plus lambda 3 h3
and then i'm going to group together my
mass flow rates to limit how many y's i
have to keep track of so
2 and 5 will be together and that's
increasing in energy
so i'm going to want to write that as
lambda 2 times h5 minus
h2 that would be the right hand side of
my equation so i'm left with
m dot 7 times h7 minus h3
5 minus 2 7 minus 3 energy gained energy
lost that makes sense
next i will divide everything by m.cycle
m.2 divided by m.cycle is
1 because lambda 2 is equal to m.6 and
m.cycle and m.6 are the same
so note here that if you had just copied
over the energy balance on the closed
feedback heater
complete with all the algebra below it
from the previous problem
you would have an incorrect answer
because the mass flow rates are
different
seven and three are both y here so
eminent seven
divided by emitted cycle is just going
to be y times h7 minus h3
therefore y is equal to h5 minus h2
divided by h7 minus h3
so at this point all that's left to do
to finish the question
is look up our eight enthalpies
plug them into this relationship down
here to determine y
plug our enthalpies and our y values
into these relationships to determine
the work in the queue in the workout in
the queue out
and then compute a network out and a
thermal efficiency
so again because the property lookups
are supposed to be something that you
master in thermal 1
it's a little bit outside the scope of
thermo2 for me to spend a whole bunch of
time in this example problem video
looking those properties up so i'm just
going to look them up and cut that out
of this video
if you want to see where i got the
numbers i will include the work with
this pdf so that you can follow it
that will be linked in the description
below the video
but in the meantime here we go you ready
all eight enthalpies three two one
now that we have all eight enthalpies
calculating why it's just a matter of
computing some numbers together
for that we will use our calculator wake
up calculator
and i am going to take h5 minus h2
which is 237.35 minus 133.27
divided by 7 minus 3
which is 1085.02
minus 236.16
that yields 0.1226
now we can take that number and plug it
into our relationships for working queue
and workout and q
out so i'm beginning with h2 minus h1
which is 133.27 minus 1 3 0.17 which
gives us 3.1
and then h6 minus h5 would be
1298.3 minus 237.35
giving me 1060
0.95
and then h6 minus y times h7
so 1298.3 minus
y times h7
minus the quantity one minus y
times h8
yields 328
0.078
and then 1 minus y times h8
which is this quantity i already have
probably would have been faster just to
start over but here we are
1 minus y times h8 plus y
times h4
which was 236.16
minus h1 which is
130.17 and i get 735.9
and then from those quantities i can
determine a network out
a net heat transfer in
and a thermal efficiency
so 328 and change minus 3 and change
yields 324.978
and then q in minus q out
is the same number which indicates that
we built those equations correctly
and then we take that number
and we divide by q in to yield
thermal efficiency of 30.6
so part a asked for y
our b asked for specific work in
and part c asked for thermal efficiency
which means that we are done with our
calculations
what i would like to do next is
probably unsurprising i'd like to plot
this on a ts diagram
so state 1 is going to be here
low pressure assumed to be a saturated
liquid
then state 2 is directly above it
all the way on the high pressure side
and you can draw that a little bit
better also let's switch to black
so it's easier to see
one and
two
and then from state two we are
gaining a little bit of heat before we
get to state five
and state five was a compressed
liquid so i'm going to leave that
in our compressed liquid region i'm
going to call that oh i don't know right
about here
and then it goes all the way up to state
six
which is our maximum temperature and
pressure
i will call right here and then that
expands down
to seven and eight which are directly
below it
i believe 7 and 8 were both saturated
liquid vapor mixtures and they are
quality of 0.9 and 0.82 so my
graph isn't perfectly to scale here but
hopefully it gets the point across
and then 7 gives up some of its heat
going to 3 which is a saturated liquid
and then we have a line of constant
enthalpy from three to four remember
those look like this
so
encountering the low pressure line
and then that is meeting the process
from eight to one
so we end up with a process line like
this
and that's our diagram