Video 1: Rankine
Cycle
Overview
Network cycle:
waste-wout. will hittipt=
I
din-But
Thermal Efficiency:
y Waycle/m
=
①in/in
1
=
-
(Port/in
Din/r
in
Vikco2:Rankine
Cycle
components
enthalpy
decrease
enthalpy decrease
enthalpy
increase
enthalpy increase
Video
3:Rankine
Cycle
Ideal Processes
I
A.
⑤I
V
T Tc
=
↳
%
②
P Pc
=
P Pb
·
=
⑤
⑭
-
T Tc
Wine
=
·
B
-
⑪
Wort
↳
furt
②
P Pc
=
Video 4:
Fixing State 4
*
not
necessary
Videos:Ideal
As the
As
the
Cycle Performance
averageit tintin
Pout
we are
limited by
ambient
temperature
Tin
<
*
AT
is
in
not
boiler, Nth
for
but
is
good performance
x
a
at outlet of
tubine which
Video 6: Rankine
internal
*
->
Cycle
within
the
Principle
Irreversibilities
system (we seek to minimize)
Isentropic Processes
*
are
adiabatic Breversible
Internal Irreversibilities
·
pump
&
(constant entropy)*
Bankine Cycle:
turbine -> friction
moving parts (mechanical
expansion
in
in
that
we
dont restrain
completely
=He
fluid friction),
ideal
=
actual
External Irreversibilities
·
beattrusten- system
-
condensers
fluid
↳
to
Gate:
smondings
I
to
system
surrounding
thid
of
boiler-Temp ofcombustion products Temp working
*there is
-
Rankine
in
a
finite
there
temperature difference
mustbe finite
Bronking third
spontaneous
nuclear reaction
trap
diff between
cooling
Video:Rankine Gale Superbeat
change boil
A
steam
to
generator A
↳ same functionality
* we
a
can use a
higher quality
boiler and
superbeat
higher
at the outlet ofturbine
order to
in
a
-isentropin
....
-steam
1
generator
4
3
2
ersen
E 2s
4s
and 2
lie
on
same
Pressureline
and
lie
on
Same
Pressure line
4
3
actual
produce
Video
8:Rankiee
*
Assume
Bring
Cycle Rebeat
turbine inlet
back fluid
stage
through
is
superheated steam genenter
chamber with combustion
through portion of expansion
a
in
gases
E
to reheat it after its been
turbine
pressure
condenser
in
3
I
2
p P,
↳
-
S
③
~
88
I
p P2 Ps
=
↳
6
S
=
high pressure
~ reheat
3
.
⑧
p Pc
=
·
A
turbine
1
&iorpasmetbin
=
T,
⑧
m
process
4th Wycle/m
=
Qin/m
0
4th Wupt
in with
=
-
this is
atire because
reg
its
&
m
bQ,
in
*
y
is
I
z
Qs
in
improved because
the work
outputproduced by the 2 turbines
than the aditional Heat transfer thatcomes
in
during reheat
is more
Video:Bankine Cycle
Supercritical
p
~
goes
from supercritical liquid
supercritical vapor
to
Pcrit
=
Video
10:Rankine
Gale Regenerative Open Feedwater Heater
(1
-
x)
(1)
Y
Open Feedwater Heater
into
-> increases
Temp of feedwater
boiler) by mixingwith High Temp Steam ofTurbine
(water thatgoes
*
we
take
Pump
↳
a
portion of flow from
mix it with
t-bine
middle of
low
temp flow coming from
R
6
since
higher than 16 without feedwater heaten,
is
-x3
(
(1 y)
-
Theseindicate
the fraction
Qin at steam
generator
ofthe
mass
is
flow thatis
alongthatparticular path
reduced
flowing
Ibleed
y
fraction of
mass
=
flow extracted bleed flow
=
(ty) fraction ofremainder
=
through
*mpsort ant
-
->
all
-
mass
flow
condenser pump
to note
x
paths my only
mass
mass flow
one
flow rate
Ex. m3 my=is
=
i, m>=m
=
way
for fluid to flow will have constant
exiting
MPT and
going
*
Pressurein OFM
must have same pressure
=> all inputs Boukuts
mixing together
if
my
diferectflow
Pressure were
that direction
in
world
go
since
theyre all
need
we
a
pump
·
M
·
for this
(I-y) reaches states it mixes in
when
I
OFH with
-
7
Tzt, Is' until
-2
6
if
*
we
flow to
↳
back to condenser
some need
Instatewe needto
P6*P,
3
connected stateto
the OFH
go
·
⑧
·
->
Pa=Ps=P6
reaching
T6
58
Pump ->
fluid at state B:
We
then
so we
increase
P6beause
addanother
Higher Pressure at state I
would
Not What We Want
so
flow is baked and
goes
in
direction
want
we
pump
cause
the
State
w/o
Sa
OF
is
process
Video
Calculations
II:DEW
To calculate
y:
Mass Rate Balance &
I
-6 .......
i
OFM(
Energy
Rate Balance
Assumption:
-
S
Adabatic
-No menk
i
done on
being
ev
.
Mass Balance:
0 m2 +
=
*
=
-
m2 mg r,
=
+
-
Energy Balance:
O
=
mehz+mghs -m,h6
=
hi,
mis-n6
is m3
0
y2
I
my M,
=
around
of
O
mehz+mghs -m,h6
=
I
m,
0
yhz
=
0
+
((-y)h) hx
-
yhz hs
=
+
-hsy
-
bb
h6-hs yhz -yh5
=
ho-bs
y= b6-hs
he -hs
The
*
ra
3
bleed
energy
ha-hs
y
=
fraction
rate balances for work
are
normalized
by
the
He (ii)
highest
flow
mass
Work:
Turbines:
MPI
-
S
......
I
wou
b
........
WE=m,(h, -he)
-
it,p=ms(ha-by)
>
WE WE, MP WE, LP
=
hiz:
mi
+
hi-hat
(1-y)ha-hy
neuseing
becausey has already left
e
Pumps
:
"
up,=ms(huh t
ip, m,(h6 -ha)
6
q
z
=
D
Wp,
Up
z
lip, hip.2
=
it
Wp m,(hy -hy) m3/hn hs)
=
+
-
p ha-b (r-y) (by-h)
+
=
Heat Transfers:
Steam Generator!
①in mi(h, -ha)
=
Gain
...
"
.. .
Die=
-
:
-
t
hi-ha
7
y Wcycle/mi
ain/m,
=
1
(h, hz) ( y)(ha hz) (h6 x (1 y)(hn b5)
-
+
-
-
+
=
h, -by
-
+
-
-
Video
Regenerative
12:
Cycle
Closed Feedwater
~Massflowrate
reduces
Din
1
-
Y
(1)
wit
2
3
I -Y
[Y)
E
steam
generator
S
Condenser
Port
[I]
(1)
Pump
I
4
S
6
heater
Trap
7
doesnt
mix
S
between CFM &OFM
with flow
5-6
8(Y)
X
LY)
Difference
*
↑
Up
closed
feedwater
A
is
that
in
CFM
the flow from
2-L
A
steam
trap
->
operates
these
I
we
as
throttle, lowers pressure, no power input,
↳ constant enthelpy
pressures
don't haveto
don't have to
worry
be
equal
since
about direction
it is not
no
moving parts
mixing
Ps P6 P,
=
=
In
F
y8
8)
S
Qin
to
I
advantage
O
from s-x6
of
energy
is
reduced because
from
2.77
me
are
taking
Video
Calculations
13:CFM
y i2
=
mi;
To calculate
W
-
!
I
b
31
y:
Mass Rate Balance &
·------
↓
-
Energy
Rate Balance
around
Mass Balance
↳
I
S
↓I
..........
mz
in
m6 Ms i, my
=
=
=
=
of
Energy
Rate Balance
mahy-m>hytinsts-myho
8
=
8
=
mahy-m>hytinsts-myho
i,
0
yhz -yh>
=
hs
+
-
h6
hy-bs yhz-yh>
=
y
ho-hs
ha-bs
I
↳2 hy
ha-hy
Y
=
-
Work:
HPT
·
sier
it,Mp
m, (h,
=
-ha)
T
-
I
swe
3
Lit, (p ms(h2 -hy)
=
its is because
*
at LPT inlet
the bleed flow has
been extracted
already
ip=in (hn-hs)
si
->Y
Heat Transfer:
steam gen
⑨in m,(h,-h6)
I
Qin
=
-
->
↑
6
y
waycle/in
=
①in/in,
h
y itMP,lt,P+
in,
=
mi,
Din/m,
2 (h, hz) (( y)(ha h3)
+
=
-
-
-
+hy-hs
hi-h6
Heat Transfer Out
Condenser:
↓
Gout-mzhz mehe-myhy
+
Qurt
->
nb48
⑳ut=(y)
hy+
yhs h
Video 14:
Multiple
Feedwater Heaters
Videos:Multiple
Feedwater
Heater Gakulations
(1 y' y")
-
y
-
y"
lisentropic)
HP
&
*
EWH
every
* reheat
is
multiple
CEN
FWH
should have atleast 1 OFM
section between MT $
LPT
*
y'
By"are
defined
y':
i
in
ms
=
y
in
Closed FWM:
After
"F--;
"
:
terms of in,
Mrs.
doingEB:
=hi, his
Y
No
ha-hiz
↓ ...
-
-
- -
12
Open
FWH:
S
........
"
V
After
I
↳
I
i.
-
is
-
!
8
"
=
doing EB:
y (1-y)h8+
y'his-hp
ho-hs
it,
itz
weare
Wt3
Dinn
I
Why
Qin3th
state l
T, 480°
=
Pi
=
8 MPa
hi
3i
state 2
Se S,
=
Pa 2 MPa
=
3h
State 3
5
P3
=
se
0.7
=
MPa
3hg=
State 4
Tn=440°
Py P3
=
3
state s
55 S4
=
Ps
0.3 MPa
=
ihs
state 6
346=
56 Ss
=
P=0.008MB
state 7
3
=
e
state 8
P8 P5 0.3MPa h8
3
=
=
58
57
=
=
state o
Pg Ps 0.3 MPa
=
=
Xy
0
=
3
hq
Sp
=
=
State 10:
3
Sio Sq
=
Pio Pr Pi 8MP
=
=
a
=
hio=
state II
Pi D1o=P,
=
8MPa
=
T1 205
=
State
]
12
Piz =P2 MPa
2
=
h
e
hin
State
B
his hie
:
Pis PS Py 98
=
=
=
it
0.344.3
his
=
he
=
its=/1-y)
wizz ((
=
-
hasa
y)(hy hs)
-
in,
ity=(ky'-y")h5-ho
in,
up,:(ly'-y) ha-ho
in,
*
for and
upo
in,
pump
allmass flow rates
mixed
are
together
*
ha-hio
=
ain=h,-h,+(ry) hm-h3
i,
Wade
i,
Wtz +WEn+ tz lizy
+
+
p, ripe
e
are
thesevalues
negative
A Wade/mi
=
L
Din/i
in
Wcycle
=
>
IOOMW
Wayce/m, Wtz +W2+ity
hity
+
+
p, ripe
Video
Choosing Working Fluid
16:
a
Criteria:
-
-
inexpensive
abundant
-nontoxic
-chemically stable
-
chemical reactions
in
flow
-chemical reactions
-
relatively
we
-large
damage to components
of irreversibility
strong
cause
are a
source
non corrosive
dont wortham in components
enthalpy of vaporization
we want it
giveoff alotof energy during expansion
-
to
in
turbine
large specific volume change during vaporization
-
Pr. Wort 4
ne
water hits
all these
spots
↳ cuban
hydrogen band
Topping
Ist
Cycle Boiler-takes high temp
source
to
turbine
HeatTransfer of
-
topping
i
S
·
I
fluid
to
bottoming flid
Toppingcycle
↓ Qut
↓
.
a
,
↓
⑤
->
Bottoming cycle
b
C
steam atinlet of
for
↳ Qin
Qin
-
generate
buttoning cycle
Water
3
stre
S
*
improved
of
because you have high
heat addition
average temp
to environment
have reduced the
but
best
of
you
average temp
y
is
a
rejection
Video 17:
Cogeneration
Cycles:
Video
H:
Vapor Compression Refrigeration
3
3
energy
Refrigeration cycle-transfer
hot
B from
system
into
reservoir
into
system
outofsystem
from cold reservoir
reservoir
(superbeated vapor phase)
(constath)(4s)
cold
e
into
system
Trh->
I
which
1
·
3.
Tr,h-
v
·
Tr,c->
S
into
> TM
into hot reservoir
Rejection)
Tn -> Terop of
Ty -----------------
Tr,h
goes
(Heat
2
Tc.-------------
Tr, -
of
high temp Refrigerant
Not Reservo:
Temp of Refigment
in
energy is
working
graph
on
TS
of Meat
Transfered
Diagram for remusible
processes
slid of
tells us
amount
↳
In
Counter Clockwise
Meaning its
->
->
a
net out
Waste,net,out=Qayele,
-
Transfer
(Refrigeration Cycle) thatMeat
in
Wagels,ort=-Quale, in
Wayce, Qcycle, out
in:
location
transfered from cold reservoir
Tc<Tr,c
Area inside
cold
is
Negative
resing
Video HI:
Cycle Components
Vapor Compression
F
T3
TH
1
~
·
-
-
M
3
S
that
Exchange theen
I exchang
hot reservoir
Heut
I
a
3
-
X
4
Rout
to
M
L
-
fiz
Condenser
of win
-
V
xX
↳
I
-
I
>
&
Din
Heat
↑
Exchange
that
exchang
Qin from cold
reservoir into
es
system
(compressa)
nic=m/h, -h2)
Diort=m/ha-ha)
Typically, X3
Gendersel
0
=
hy hy
(expansion value)
=
Ideally, X, 1.0
=
* For
a
practical
reasons
avoid
little to
we
droplets
superheat
in
compressor
3
a
ain=m/h, -hu)
For instance:
1Th 20°C,
=
Tn 25
=
T3 (25 20)
=
+
Greficient
of Performance:
B Qin hihy
Iwyclel/hi-hal
=
=
(evaporator)
The
So
Explaining
bullet
first
point:
point
of
TySTc
process
/Temp of
w -frdp
=
3-4
cold
is
just
a
blockage
to reduce
reservoir)
compression is
butif
had
you
phase
value
is
done w/
a
so
mixture
vapor
turbine at
pressure
so wis
expansion,
high
so
it would be
the would be too low
tempt
expanding liquid
a
or
a
2
Video
42:
p-h
Fiagrams
Video 43:Cascade
Refrigeration Cycles
Refrigerants
Video 44:
selection
seal
easier to
Video
4S:Heat
Pump
Cycles
~
Heattransfer
the
reserve
not
Video 46:Gas
Refrigeration Gale
* 400-sooK
-300K
Video
S3:
Ideal GasMixtures
to caracterize state A
*In mixtures
we need extra info
Ways
to characterize mixture:
moles, mass,fructions, average molecular weight
n
#moles
=
m ma55
=
in
component
component
of
M molarmass/Mobeclan neight
=
Ratios
are
used more
commonly
n =#ofmoles
n total
=
component
of given
of molesin mixture
Bitmassofrepareinte
the
Yi&mfi
·
2waysto
are
intensive
properties
->
dont
depends
size
on
of
system
calculate:
M
Yi
molar mass
mixture
average
componenti
mole fraction of
=
=
Mi=Mola Mass
I
msi=mass
component:
of
fraction of
componenti
Mi=Molar mass
component:
f
o
Video
p-v-T
SH:
How to
relate
Mixtures
Pressure, Volume Itemp
to each other
mixtures of ideal
for
gases:
unstert
R universal gas
(5/molK) (Btw/IbmolR
=
=8.314
=
unigas
constant
average mixture mole
mass
*
We
can
aracterize behavior of each
component
Pi=pantial pressure
comparenti
of
p:total pressure
in
terms
ofideal
gas
law w
Partial Pressure
will have:
component
same total extensive volume V
each
->
·
same
·
We
also relate
can
Di
temperature
I
P
of
a
T
partial
ni
pressure
Di
V
-
is a
property
particular component
to mole functions:
Yi
=
n
nRT
V
·
Di
p
Yi
=
only
*
defined in terms ofmole fraction
defined
not
*
↳ Because
So
I
in
R
terms ofmass fractions
is
different
for each
would not
cancel
in
equation
*
componentIR for
overall mixture
i
Video SS:Mixture
Properties
M total
=
To
put it
emeralize this
component:
in
terms of mass
or
of
components
molar
in
mixture
specific
propertiesof each
Video
Psychrometric Principles
56:
↳n
Yaz
motefraction
total
either solid or lig
of a
Pressure
D:
·
D
my
air
Air->
particular agreed upon ideal specification for
treat
that involves
pure component
thatwe can
Nitrogen, Oxygen
Both water
·
as
some
a
and alittlebit of
-
-
same
Some
volume
Temp
or
composition of
mixture of
Argen
will fill the entire volume
rapar dry
system temperature they will both:
$
the
of
the
container at the
Other ways
to characterize mole fractions than
justby mole functions
-
partialpressure
Mu
18
=
Ma
of
air
0.622
=
28.2
~
specified
in
weather
3
reports
tempPene
keep Icons but add
Me
mass
water
of
vapor
more
into mix
[HumidityRatio 4, pantial pressure of
water
By
T
·
-
Pu
..--
-
-
-
vapor
4)
Video
S7:Moist
remember
*
Air
->
Properties
moist at=mix ofdry air water
vapor
were
both are treated
ideal
as
gases
calculate thetotal
We
↳ internal
of the moist
can
energy
air mixture
ji:total intenergy
mixture
totalintenergy
of
energy
(an approximation)
Las
W
⑤
·
ofideal
gasdepends
gasdepends
total int
a
vapor
m
=
approximation)
internal energy ofan ideal dependson
gas
mideal
of
on
enthalpy
entropy
water
of
the sun of
air
ma
b
(approximation)
·
as
of
on
Temp (u)
Temp (h)
Temp, Pressure (st
↳
partial
pressure of
component
particular
P
=
+,P
Video 58:Dew Point
So
Temperature
far behave treated
rap
dry
to when theis
a condensed
get
exchange
well
water
mass
with the
Dry
&
independent
ideal
phase of water(hig
or
but
gases
solid) which
vapor phase
Hir
WaterVapor
some
moleculesof
liquidwill
I some
Mee
liquid
air as
become we par
moleculesof
repor
will become
liguid
lig: satliquid
sat
vapor:
vapor
3
the rol e at
-
which
is
this
occurs
balanced
now
is
able to
start:
①
Dry Air
hatVup
Tsat=Dem Point Temp
water
of
T
Pr= YvP
P
Reduce
&
-
rap
I
I
Isat
it
.
③.
·2
I ............
Du
Prz
·
Dry Air
⑬
WaterVap
-
-
W
2-33
·
·
occurs
Der Point
Temp
on
is
sat line
related to the initial
High ValuesofDPT Feels
=
to decrease
&a
quite
as
more
partial pressure
braid outside because
much for water to start
higher partial pressure
of
water
water
of
vapor
vapor
temp
the
doesnot have
condensing:More
water
rep
in air
Video S4: Adiabatic Saturation
·
·
How
to measure I
for
partial
pressure
Temperature
calculate humidity ratio use that to
ofwater
vapor in a
to characterize moist
air mixtures
2
ways
get
moist mixture
humidity ratio, partial pressure
-
Temp of adiabatic saturation
Tas:
......
T
-
-
Pi
WI
" ....
19)
-.........
Tz Ta5
=
2
Assumption:
Pa PI
=
%2 100%
=
w
-
?
=
liquid
bi
Open System
-
water
-
-
Mass Rate Balance:
mas= maz:ba
hi, + mw hve
=
nic
ma
Remember
*
W2 nive
=
ina
in a
W, mv,
=
niw
in a
We -W,
=
ma
a
*
Adiabatic
No Work
No
PE
no
KE
value
Energy Rate Balance
O
ma,hai+mvhr, nwho-mashazivahva
=
t
Dividing everything by ina:
0
ha w,hr,
=
+
Pr=
*
+
(wa-wi)hw-baz-wahve
Py(tas)
W2 0.022
=
->
sat
mix at
outlet
Pg (Tas)
4-Py(Tas]
Wi haz-hast
=
we
[hva-hm]
Gr,-hw
A
hu h5 (Ta5]
=
hav ha (T1)
haz ha(Tas]
=
hr, hr(+1)
=
huz=hr(Tas)
Drybulb
-
Wet Bulls
-
Tab:TI
Tub:Tas
.........
wrap tick
around thebulb
sulk wick in
flow
Ma
air over
wick
*
Video 60:
Psychrometric
Charts
Another way to relate propertiesofmoistair mixtures (notjustusing
equations)
·
Dry
Bulb
Temperature temperature
-
thatdoes not have
·
wehumidity ratio
partial pressure
a
metmick
-> Because
of the air that
on
we measure
with a
it
total Pressure is
constant, wis constant
->
directly related
of water
-> lines
thermometer
of constantrelative
Note:4
*
=
100%
is
humidity (b)
sat mix
*
to the
->
lines
of constantmixture
enthalpy
line used to
find metbulb
->
temperature
lines of
volume per unit mass of
dry air
V
ma
P latm
=
Lets
say
6 401
=
Tb
8
30°
=
starting
-
condition
.
·
8
We
can use
any
2
properties
so
long
as
their lines
intersect:
W&b
->
->
->
I B Tab
Tub I
->
I
->
etc
Tdb
& Tub
Tdb 30C
=
·
b 40%
=
·
·W 0.011
=
E
·
for dew point
E
water
E
·
for a constant
composition process
thup (follow sloped straight lines) (Remember
until
reach the
vapor into mixture
you
Not the
*
Turb
for
8
occurs
Tdp= 15.5
for met bulb
A
temp (Remember it
=
20
specific enthalpy (follow same
H
ma
=Sin
For volume
(strepest diagmal)
line
as
adding
you
saturation conditions)
are
steepest line
Tub butextendity
Given
particular
al
direction
Ex. Air
water
on
the
type of conditioning process,
air
psychrometric chart that process
Process
Condi
t
i
o
ni
n
g
world
that
process
r
where Im
go
determine
we can
on
the
cooling
following
the
should
ar
direction:
what
go.
by evaporation (by adding
Tb
W4
humidification by
doing
injecting
would be the
If I'm
then the
direction
steam
following:
Tab
7
Y
w
If Im
doing
dehumidification by condensation from
coding
TdbN
wi
L
If I'm
doing
changingtemp
V
a
i
desiccation
process (Silica
packets)
->
absorb water who
Video 61:Air
Conditioning Processes
Wm
=
There
are
3
air
conditioning processes:
Ma
2 ways:
&
Evaporating
③
Mix
in
some
liquid water
steam, some
Temp of sup steam Temp of
causes
we, Tab ↓
superbected vapor
moist mixture
Tdb*
w4
causes
butbecause
2
ways:
desiccation ->
·
other chemical
·
condensation
-> we
need
a
point
the wath
from
absorbing
vapor
such
->
as
reducing
silica to reduce
w
cooling coil to reduce
mixture
the
some
w
temp ofmoistair
environment it
at be cold
so
if
An
is
for
occupants
enough
moist
be added
mixture below its der
ejected to
in
that
space
where
m/lower
heating section might
constant
humidity process
heating coil to 4Tdb happening
a
*
into
a
also redrew the db
mig
air
it into
by removingwatervapor from mixture I turning
liquid
reduces but it
->
the moist
air mix
w
is
sent
to
in a
Similar to humidification
works well
in
dry
climates
process
butthe
purpose
is
to
get
some
Temp reduction
->
is constructed
System
to have a wick
Process
to
go
So
mater
sort ofstructure that can hold on to water,
liquid water out ofwick
evaporatingliquid
certain
down because
satuated with that
->
or some
I then
me flow the
of
we
require
Process will tTdb
energy
&Pw
air over that
and into moist air
input
to
liquid wate
mix
temp
causes
evaporate that water
Video 62:
Analyzing AC Processes
H20
Adding
->
wa wi
-
Assumptions:
Steady State
-
may
maz: ma
=
inv,
tniw=iva
min
We-WI
=
nic
0
Dc -Wa +ha,+whr,
=
ma
inc
Removing Hasi
+(wa-m)hw-haz-wahre
w, swe
man=in a
=
a
inv, mint
inve
=
inw W,-W2
=
in a
O
Qu_Wer+ hart wihr, -(w-wahw-haz-wahre
:
ma
ina
Constant
composition
Case
Wi W1
=
hia,=mas:man
mr=rive
0
ger-wartha, -has) w(hn eit
=
+
-
One
inlet, one outlet
Video 63:
Cooling Towers
Natural Draft
·
to
Towers-> rely
Cooling
drive the flow into the bottom of
and then out the
on
top where you typically
differences
coolingtower
see
a
up
in
past
billowing steam
density
the
ofair
following water
·
Freed draft->
small
Were
*
uses
droplets evaporate
a
drive the flow
far to
better than
droplets
big
to cool down this water inlet
trying
&remove more
energy
so
thatwe can send it back to the rankine
the
water inlet into
some of
Evaporating
->
from outside of
cooling tower
atmospheric
water droplets will
->
evaporate into atmospheric
droplets
warm
makeup
air
mater to
make sure
in
atmospheric air
somemoist
comes in
some
->
condenser
out of Rankine Cycle
->
air
cycle
of
of return water:in
& flows
->
up
Reduces
incoming
warm
through
the
temp both of
water
water
falling
droplet
air
the water