Video summary
The video presents an analysis of an ideal regenerative Rankine cycle incorporating a closed feedwater heater, using water as the working fluid with specific operating pressures and temperatures at the turbine inlet and condenser. The primary objective is to determine the mass flow proportion exiting the turbine early to supply the heater, calculate specific work and heat transfer values, and evaluate the overall thermal efficiency. A key distinction made in the analysis is that a closed feedwater heater introduces additional complexity compared to an open one; it requires separate pumping stages to reach high pressure before mixing or uses a trap to drop pressure, resulting in nine distinct state points instead of the usual seven because the device functions primarily as a mixing chamber rather than a heat exchanger where streams remain separate.
To solve for the unknown properties at these nine states, the presenter establishes that pumps and turbines operate isentropically due to the lack of efficiency data, while condensers and heaters are assumed to be adiabatic with negligible kinetic and potential energy changes. The analysis relies heavily on identifying two independent intensive properties for each state: pressure is determined by tracing the isobaric processes through the condenser, closed heater, mixing chamber, and boiler, while temperature or quality assumptions define the remaining states. Specifically, the outlet of the closed feedwater heater is assumed to reach thermal equilibrium with the inlet stream, meaning the exit temperatures of both the hot and cold streams are identical, which provides the necessary condition to solve for the unknown enthalpies using energy balances on the mixing chamber and the closed heater itself.
Once all enthalpy values are determined through property lookups and algebraic manipulation of the energy balance equations, the specific work inputs for the pumps, heat addition in the boiler, work output from the turbine, and heat rejection in the condenser are calculated. The resulting thermal efficiency is found to be approximately 30.8%, which is slightly lower than that of a cycle with an open feedwater heater. Despite this reduction in efficiency, the video explains that closed feedwater heaters are still utilized because they allow the entire regenerative cycle to be completed with fewer pumps, offering potential cost savings on equipment that may offset the slight loss in thermal performance and the increased mechanical complexity.
The analysis concludes by visualizing the cycle on a T-s diagram and comparing manual calculations with MATLAB simulations, which provide accurate, to-scale diagrams and allow for parametric studies. By varying the regenerator pressure in the simulation, it is demonstrated that there is an optimal pressure—around 80 to 90 psi—that maximizes thermal efficiency, rather than simply using the lowest possible extraction pressure. Furthermore, the relationship between the mass flow fraction extracted from the turbine and the regenerator pressure is illustrated, showing how changing the extraction pressure alters the cycle's performance characteristics, ultimately providing a comprehensive understanding of optimizing regenerative Rankine cycles for maximum efficiency.
Read the full video transcript
an ideal regenerative rankine cycle with
an open 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 operated at 5 psi
steam is supplied to the feed water
heater at 40 psi before being pumped and
mixed with the feed water
determine and complete the following
first why the proportion of mass flow
rate through the turbine that exits
early to serve the closed feed water
heater
then the specific work in that's
relative to the cycle itself
then lastly the thermal efficiency like
our previous example i'm going to try to
identify
two independent intensive properties
about all nine of our state points
the presence of the closed feed water
heater means that i either have to pump
my fluid back up to the high pressure
before mixing
or use a trap or expansion valve to drop
it back down to the low pressure before
mixing
in either case i'm going to have
additional steps in the process
i have to have a separate place for the
water to join back together
and that is essentially an open feed
water heater in and of itself
it's just that the purpose isn't to
actually accomplish any sort of heat
transfer
the goal is just to allow the streams to
mix together
so i have additional devices that's why
i have nine state points instead of
seven
first i recognize that i still have
three pressures and let's
think through those together which stay
points have the low pressure
that's right eight because it's the
outlet of the turbine once it has
expanded all the way
and one because the condenser operates
isobarically
and then the high pressure would be five
and six but the high pressure
also includes four nine and
two does that make sense
because the mixing chamber has to occur
isobarically
and the heat exchange process within the
closed feed water heater
also has to be isobaric so that means
the separate streams in the closed feed
water heater
that's seven to three and two to nine
are both going to be isobaric
so two is equal to nine which is equal
to four which is equal to five
therefore the high pressure includes six
and five and four
and nine and two that leaves us with
seven and 3 at the medium pressure
so pump 1 is pumping from the low
pressure all the way up to the high
pressure
2 is just going from medium to high
so nine four five two and six are high
two four five six nine
then 8 and 1 are low
that leaves us with 7 and 3 as the
medium pressure
those pressures give me one half of my
independent intensive properties
required to get to all the rest of the
lookups
next i will consider the operation of
the pumps and turbine
because i was given no indication as to
the operating efficiency of the turbine
or the pumps
i will assume that they are occurring
isentropically
meaning that they're 100 efficient so s2
is going to equal s1
s4 is going to equal s3 and s6 is equal
to s7
and s8
that leaves me with four state points
unaccounted for
one of those is fulfilled by the
temperature i was given at the inlet to
the turbine
that would be state 0.6 so i will say
t6 is
600 degrees fahrenheit
leaving me with three unaccounted for
the next one is easy the condenser is
assumed to only condense
so i assume that the outlet of the
condenser is a saturated liquid
therefore i'm assuming x1 is zero
and similarly i can assume that the
closed feed water heater
is allowing the stream at seven to
condense and that's where it's getting
its energy
to push into the stream from two to nine
so the condensation
of the steam is actually what's
accomplishing the heating process
and i'm assuming that once it condenses
it leaves
because there's not enough temperature
difference to really do a whole lot
so t3 oh so x3 is also assumed
to be zero
and again i will indicate that that's an
assumption that i'm making
with an asterisk so i have five
and nine unaccounted for
five will come from an energy balance on
the mixing chamber
once i know everything about four and
nine knowing five is just a matter of
combining the two
if i set up an energy balance on the
mixing chamber i recognize that it's
assumed to be adiabatic
and there's no opportunities for work to
occur and if i neglect any changes in
kinetic and potential energy i will end
up with
the sum in of m.h is equal to the sum
out of m.h meaning m.9
h9 plus m f4 h4 is equal to m.5 h5
if i know the proportion of mass flow
rates at 9 and 4
i can calculate h5
so i will just write h5 comes from
energy balance on mixing chamber
and that leaves me with state nine
so the key to state nine is similar to
the
assumption that we made about the
operation of the regenerator back in the
brayton cycle
if we didn't have enough information to
deduce otherwise we assume
that regenerator operated with 100
effectiveness meaning as much heat as
could be transferred was transferred
and back in that analysis
we had a heat exchanger where the flows
were flowing in opposite directions
so a high temperature here was driven to
a lower temperature here at a high
temperature here was driven to a lower
temperature here
and if i were to plot out the position
let's call this hot in
hot out iho
cold in cold out co
seco and then i am plotting
temperatures relative to x position
and if i had the worst heat exchanger in
the world
hot input would be here bold input would
be here
and absolutely nothing would happen in
between the two
so the temperature of c out would be the
same as c in
and the temperature of hot out would be
the same as the temperature of hot in
and if they were working just a little
bit then the change in enthalpy
which for air is directly correlated to
a change in temperature
is the same in both the hot stream and
the cold stream
meaning that i end up with lines that
move up
and down by the same amount
and if i were to extrapolate this to
ideal circumstances
i end up with a line directly connecting
the two
so for a heat exchanger with flows in
opposite directions like this one
ideal operation is marked by the
temperature of the cold outlet being the
same as the temperature of the hot inlet
and the temperature of the cold inlet
being the same as the temperature of the
hot outlet
but our situation is different
in our situation
we have flows
that are going more towards each other
so if i draw this as a box
i have the cold stream entering
and then undergoing a whole bunch of
surface area
before leaving again let's stick with
our naming convention here
that was c in and see out
and then hot in is steam
so we are spraying that all over these
coils
it is allowed to condense and then leave
so if we're assuming that as much heat
as can be transferred
is transferred then heat transfer will
continue
until the temperatures are the same but
here i'm not referring to the
temperature of
the inlet of one stream and the outlet
of the other
no i'm referring to the temperature of
the two outlet streams
so in conclusion ideal operation
of a closed feed water heater like this
one
is marked by the temperature of the
outlet streams being the same
so the assumption i make about state
point nine
is that it is the same as t3
and with that i have two independent
intensive properties which theoretically
would define
all nine state points and looking up all
my properties is just a matter of
putting in the time
so in an effort to continue our analysis
of the problem
before we get bogged up in the property
lookups let's assume that we had done
that okay poof we have all nine
enthalpies or rather we have eight of
them
what do we do with those enthalpies well
i want us to determine
the specific work in the specific queue
in
the specific workout and the specific
queue out
starting with work in why don't you try
that on your own first
what do you get for an equation for the
specific work into the cycle
did you get 1 minus y times h2 minus h1
plus y
times h4 minus h3 excellent
here we are defining 7 or rather
the mass flow rate that leaves as seven
relative to six
as y and whatever remains one minus y
so the proportion of the mass flow rate
that leaves early
eminence seven over m.6 is defined as y
and whatever's left over is one minus y
so if 25 percent of the stream at 6
leaves at 7
the remaining 75 percent must leave at
8.
then i recognize that the mass flow rate
at 6 is the same as 5
and those two state points are what i'm
calling
m dot cycle the overall mass flow rate
through my cycle
and that stream is split some of it goes
into 8
1 2 and 9.
and some of it goes into seven
three and four
so in my workout equation excuse me in
my work in equation
i'm taking the total power input and
dividing by
m net cycle the total power input is
going to be the power input to both
pumps
so the power input to pump 1 plus the
power input to pump 2
and the power input to pump 1 is going
to simplify down to the power input is
equal to
the mass flow rate through pump 1 which
is either m.1 or m.2
times the quantity h2 minus h1
therefore i can write the power of pump
1 as m.1 times the quantity h2 minus h1
and then for pump 2 the power of pump 2
could be written as m dot
3 or 4 times the quantity h4 minus h3
therefore w dot in is equal to m.1 times
h2 minus h1 plus lambda 3 times h4 minus
h3
and then i'm dividing that entire
quantity by m dot cycle
and because m.1 divided by m.cycle is
the same as m.8 divided by m.6
that means i'm writing it as 1 minus y
and then because m.3 divided by m dot
cycle is the same as m.7 divided by m.6
i'm writing that as y so my equation
should be
1 minus y times the quantity h2 minus h1
plus y times the quantity h4 minus h3
then qn occurs in the boiler and the
boiler has the mass flow rate
of the cycle so q dot in divided by m
dot cycle is going to simplify to
m.5 times h6 minus h5 divided by m.cycle
which is just 1 times the quantity h6
minus h5
which i can write as h6 minus h5
and then for our workout if we set up an
energy balance in the turbine
we're going to end up with the sum in
of m.h is equal to the power output plus
the sum out of m.h
therefore m.6 h6 is equal to
w dot out plus m.7 h7
plus m.8 h8 and when i divide that
entire quantity by mdat cycle i'm left
with
h6 is equal to the specific workout plus
y
times h7 plus one minus y times h8
when i rearrange that equation to write
specific workout
i should get the entrance
h6 minus y times the first outlet
minus one minus y times the
second outlet
and then for q out i'm only analyzing
the condenser
because remember the mixing chamber in
the closed feed water heater are assumed
to be well insulated
therefore q dot out would be m.8 times
the quantity h8 minus h1
dividing that quantity by mdhat's cycle
with yield m dot
h over m dot cycle times h8 minus h1
which simplifies down to one minus y
times h8
minus h1
so once again i'm left with a
relationship
that is only a function of our
enthalpies and
y so in order to be able to finish the
problem
i have to determine why and to do that
i'm going to need to perform an energy
balance on a device
about which i know everything
because i don't know the workout that
rules out the turbine because i don't
know cue out i don't
i can't analyze the condenser because i
don't know the work in i can't analyze
either of the pumps because i don't know
q and i can't analyze the boiler
that leaves me with the mixing chamber
and the closed feed water
heater but remember at this point in the
analysis we haven't yet figured out
h5 we can do everything else but we
can't do h5
therefore the only device about which i
know everything is going to be
the closed feed water heater itself so
an energy balance on the closed feed
water heater will yield
y and then i can use an energy balance
in the mixing chamber to determine h5
and then i can determine the specific
work in the specific queue in the
specific workout the specific queue out
then i can determine the thermal
efficiency at which point i'm done with
the question
so energy balance on our closed feed
water heater
and for that i will draw a big vertical
line
so that ended up being an awfully busy
drawing but the point is
we have a cool stream at two to nine
that is being heated up
by the condensation of steam which comes
in at seven
and the result of that condensation
leaves at state eight
if i set up an energy balance on this
control volume i have steady state
operation of an open system
so i'm going to skip the first couple of
steps
and write e dot in is equal to e dot out
and then because it is an open system
the energy could cross the boundary as
heat transfer or work
and because it is assumed to be
adiabatic i can neglect the heat
transfers
i have no opportunity for work to occur
so i neglect the works
and remember that theta contains
enthalpy
specific kinetic energy and specific
potential energy
and i'm neglecting changes in kinetic
and potential energy therefore i'm left
with
the sum in of m dot h is equal to
the sum out of m dot h
and then i have entering mass flow rates
in the form of states two and seven so i
can write that as
m dot two times h2
plus m.7 times h7
and then i have exiting mass flow rates
in the form of
the other two states nine and eight is
that eight
no it's three and that's confusing
let's correct that diagram i just drew
okay 2 comes in 7 comes in
9 goes out 3 goes out we got it okay
then our sum out would be m.9 h9 plus
m.3
h3
and you're expecting me to now divide by
mdat cycle and you're right that would
work
but before i do that i can make my life
a little bit easier by recognizing that
m.2
and m.9 are the same and m.7 and m.3 are
the same
so if i bring them together say by
taking m.7
times h7 minus h3
is equal to m dot two times
h9 minus h2
now i have two fewer mass flow rates to
have to deal with
so i'm not 7 times h7 minus h3
is equal to m.2 times h9 minus h2
so i'm saying the energy absorbed by the
cool stream is equal to the energy
exiting
the warm stream multiplied by mass to
get rates
cool that all makes sense now i'm going
to divide everything by m.cycle
and then i'll have seven divided by m
dot cycle which is the same as m.7
divided by m dash six
so that simplifies to because i want to
know
for that joke to work you had to say it
alone
i just operated under the assumption
that you are all participating
audibly in those rhetorical questions
that i ask you and then m.2 divided by m
cycle is equal to m.8 divided by m dot
six
which is one minus y
and then you know it's time for algebra
to solve
for y so i'm going to
foil the right hand side first i don't
really know why i'm narrating you guys
can probably figure out what i'm doing
but
here i am minus first
outside minus h2 inside minus
y times h9 last
plus h2 times y
that's equal to y times h7
minus y times h3 now i want to get all
the y's together
so i'm going to say y times the quantity
h7 minus h3
plus h9 minus h2
y times the quantity h7 minus h3
plus h9 minus
h2
i don't know why i'm still narrating my
algebra
if i hadn't spoken i could have time
lapsed this
but you know what it's fun to do algebra
together
then solving for y yields h9
minus h2 divided by the quantity
h7 minus h3 plus h9
minus h2
and nine minus two whole stream divided
by seven minus three
plus nine minus 2. yep i'm cool with it
so at this point in our analysis if we
had looked up all of the enthalpies
except for state 5
i would have enough information to
calculate y and then armed with our new
y
i could calculate h5 by performing an
energy balance
on the mixing chamber i think i can fit
that in here
i have entering mass at state let's
double check this time as opposed to
resorting to memory
nine and four nine
and four and it's leaving at five
and the mixing chamber itself is
adiabatic
and has no opportunity for work
so if i were to set up the worst control
volume ever
perform an energy balance on that
control volume
because it's steady and it's open and
i'm neglecting changes in kinetic and
potential energy
and there are no opportunities for heat
transfer or work i'm going to do
one two skip a few all the way down to
the sum in
of m.h is equal to the sum out
of m dot h
i have one exiting mass two entering
masses
so m.9 h9 plus
lambda 4 h4
is equal to m.5 h5 therefore
i'm going to get rid of this page count
therefore h5 is equal to that's
not what i wanted red for some reason
h5 is equal to m.4
h4 divided by a m.5
plus it looks a little bit too much like
a nine so i'm gonna correct that
that's a much better four while we're at
it
yeah good force that is a good
for absolutely nothing say it again
and h5 okay and if you expected me to
have
divided by m cycle i mean what i did was
solve for h5
by dividing both sides by m.5 which is
also equal to m dot cycle so it's
essentially the same thing anyway which
is fun
so to four divided by m f5 is equal to
m dot seven divided by m.6 which is y
this is y times h4
and then num.9 is equal to m.8
and then m.9 divided by m.5 would be
equal to m.8 divided by n red cycle
which is one minus y
and that gives me everything i need to
calculate h5
so let's recap our outline for the
moment
we are going to look up h1 h2
h3h4h6h7h8h9
and then we're going to use our energy
balance on our closed feed water heater
to determine
y and then we are going to use our
energy balance in the mixing chamber to
determine
our h5 and then once we have y and all
nine h's we are going to calculate the
specific work in
the specific q in the specific workout
and the specific queue out
and again because this is a thermo 2
example problem and i'm assuming that
you guys are good at property table
lookups
i'm not going to waste any more example
problem time looking up the properties
on camera
instead we are just going to jump to
having all of the enthalpies except for
state five are you ready
here we go three two one and with that
we have eight of our nine enthalpies
again i
included my work in full detail here on
the attach pages so if you download the
pdf below this video
you can follow along with the lookups if
you want to try them on your own and i
would encourage you that you do
and then our next step is going to be to
calculate y from knowing h9h2 h7 and h3
so i'm going to take 9-2
which is going to require our calculator
friend
9 minus 2 237.35 minus
133.27
and then we are dividing that quantity
by
another quantity seven minus three
that's one
zero five no one zero eight five point
zero two minus three was
two three six point one six
and then
plus 9 minus 2 plus
237.35 minus
133.27
that gives us a y value of 0.10922
and armed with our new y we now have
everything we need to calculate h5
with the exception of a little bit more
space
so i'm going to take that y value times
h4
oh no i moved the white that i had used
to block out the page number
that's interesting anyway so y times
h4 which was
239.2 and i'm adding to that
one minus y it's not minus
calculator come on you're better than
this
times h5 which was
wait times h5 what what what
four and nine what am i doing
i remember i was making a joke about the
song war
sounding like the song four
so multiplying by h9 not h5
and that h9 value was 237.35
and that would be 237.552
it's useful at this point to sanity
check that number
we are mixing together the streams at
four and nine
so our state five should be somewhere
between states four state fours enthalpy
and state nines and there'll be so it
should be somewhere between 239.2 and
237.35
since our y value is so small i should
expect a number that is so much closer
to
h9 than it is to h4 because i'm taking a
bigger quantity times h9 plus a smaller
quantity times h4
that's why our value of 237.552
being so close to 237.35 compared to
239.2
is such a good sign and now i have all
nine of my enthalpies
so i have everything i need to be able
to continue
i will calculate the work in the q in
the workout and the queue out
which i will do first by scooching some
stuff around
what did i do how did i break okay let's
try that again
which i will do by scooching some stuff
around hopefully correctly this time
and that is probably enough space to
finish the problem
so 1 minus y times h2 minus h1 so
1 minus y times h2 minus h1
133.27 minus 130.17
plus y times
h4 minus h3
h4 minus h3 was 239.2
minus 236.16 and i get
a work in of approximately three
next is q in which is easy that's just
six minus five
so i take 1298.3 minus
237.5 and i get 1060.75
just occurs to me now how many decimal
points i used on work in
but you know what we're gonna do it
already so let's just continue
then i'm taking one two nine eight point
three minus
y times h7
and h7 was 1085.02
minus 1 minus y
times h8 which is 954.18
giving me 329.83
then after such complicated calculations
such as work in and work out q out
should be
a breeze i'm taking 1 minus y
times h8 minus h1
so 954.18
minus 130.17
which gives me 734.012
and with those quantities i have enough
to calculate
the net workout the
net heat transfer in
and then the thermal efficiency
first up i have a workout minus work in
which is
329.83 minus 3.09345
according to my work that is 326.736
and then i take 1060 and i divide by
734 and i get 326.736 as well again
those values being the same
implies that i built my equations for
work in queue and work out and queue out
correctly
which is a little bit more reassuring as
they get more complicated
and then 326.736
that leaves us with thermal efficiency
which is 326.736
divided by q in which was 1000 and
change
and i get 30.8 percent
so if i were to pose the question how
does this efficiency compare
to the previous problems efficiency well
it seems like they're pretty close to
the same right 30.83
compared to 30.80 but our thermal
efficiency in the closed feed water
heater
is reflective of the differences between
the two types
of regenerative rankine cycles
closed feed water heaters are always
going to be less
efficient than open feed water heaters
that might cause you to ask yourself
why then would anyone ever use a closed
feed water heater they're
they're more mechanically complicated
they're less effective
why would i want one well when we are
talking about different ways of
implementing a regenerative rankine
cycle
the advantage of the closed feed water
heater is that you can
accomplish the overall cycle with fewer
pumps
so if i were to expand the stream from
seven to three
back down to the low pressure and mix it
together with the stream at one minus y
and have them compressed in a single
pump i would
save money on pumps and the cost savings
might be enough to offset the slight
drop in thermal efficiency with a closed
feed water heater
especially when we start to encounter
effectiveness of the feed water heater
because it's real easy to build an
effective open feed water heater it's
not particularly easy to build an
effective
closed feed water heater to the same
level but that savings might offset
the increased cost of complexity
and the slight loss of thermal
efficiency
the last thing i want us to do is to
draw a ts diagram of this cycle
for that i'm going to start
with the same
axes and
saturation lines as the previous example
nope third time's a charm
and on this expertly drawn ts diagram
i am going to indicate
three lines of constant pressure one low
pressure
one medium pressure
and one high pressure
and of course this is hugely exaggerated
because the goal of this diagram
is just to indicate rough shape what i'm
looking for is the positioning of
our state points relative to one another
and i begin by recognizing that states 1
and 3
both had assumed a saturated liquid so 1
and 3 are going to be
here and here
that was 2.3 right yep and then 2 and 4
are going to be directly above them and
i'm going to draw this in black
hopefully draw a little bit more
contrast
and then
[Music]
do
and four
then again i have
six seven
and eight all in a nice little column
here
so then the remainder of our state
points and then for the remainder of our
state points we're just going to be
looking at
this region up here and i got a little
bit overconfident when i was drawing my
state points here
2 is not on the medium pressure 2 is in
the high pressure
because again 1 to 2 is going all the
way up to the high pressure
so i'm left with five and nine
and nine and five are both going to be
on the high pressure line
they are just going to be between
four and two i will draw that
like this
so two gains a little bit of energy to
nine
four and nine mixed together to yield
five five follows a constant pressure
process up to
six seven comes back to three
before going up to four it comes back to
one
and there's our ts diagram for this
cycle if we were to analyze this problem
in matlab instead of performing the
calculations by hand
we end up with something that looks like
this so
matlab is running through the
calculations that we had done by hand
using xteam to perform all the property
lookups
and then determining why using our
values for enthalpy
calculating a specific work in specific
cue in a specific workout and a specific
queue out
and then calculating a thermal
efficiency matlab also has the benefit
of being able to plot the ts diagram
accurately so on this ts diagram
everything is actually to scale
i will point out that in my drawing we
had exaggerated
everything in the compressed liquid
region so as to actually
indicate how the state points lie
relative to one another
but in the compressed liquid region the
lines of constant pressure appear so
close together that they are
essentially inseparable when we're
looking at an actual graph
so when we look at the matlab code it's
impossible to differentiate those
isobars in the compressed liquid region
and those state points over here
referring to
2 9 5 and 4 appear
all on top of one another
another advantage of performing this
calculation in matlab is that it allows
us to perform
changes to our given information and see
how that affects the results
i could say consider the regenerator
pressure
and ask the question what happens if the
regenerator pressure is changed from 40
to 50.
we could change it to 50 and
watch how this changes our thermal
efficiency went from 31
on the nose to 31.07 we saw that it
increased
and if i wanted to i could wrap this
code in a loop and have it perform the
calculation
oh i don't know maybe a thousand times
and graph the results
this takes a second because it has to do
a lot of calculations
here let me make that figure a little
bit bigger so hopefully it's easier to
read
bring it over so on this figure i'm
showing thermal efficiency
as a function of regenerator pressure in
the center
we can see that the optimum thermal
efficiency is going to occur at a
regenerated pressure of i don't know
maybe
80 or 90 psi and then on the left graph
i'm showing how y changes as a function
of regenerator pressure
so we can't just have a y value of 0.8
it is a function of the pressure at
which we are extracting the steam
and then on the right graph i'm just
plotting thermal efficiency as a
function of that y
value so the y value corresponding to a
regenerator pressure that yields the
highest thermal efficiency
is about 0.15