Components of Thermodynamic Cycles


What can be the components of thermodynamic cycles?
Turbines, valves, compressors, pumps, heat exchangers
(evaporators, condensers), mixers,
BUTTERFLY VALVE
GATE VALVE
SPHERE VALVE
PIPE CRACK
GLOBE VALVE
Throttling Devices (Valves)
Typical assumptions for throttling devices
1.
2.
3.
4.
5.
No work
Potential energy changes are zero
Kinetic energy changes are usually small
Heat transfer is usually small
Two port device
Look at energy equation:
Apply assumptions from previous page:
0
V V
q  w  ( h2  h1 ) 
 g ( z 2  z1 )
2
0
0
We obtain:
20
2
2
1
( h2  h1 )  0 or h2  h1
Does the fluid temperature:
increase,
decrease, or
remain constant
as it goes through an adiabatic valve?
Look at implications: If the fluid is liquid/vapor:
h const.
T
P const.
During throttling process:
• The pressure drops,
• The temperature drops,
• Enthalpy is constant
h const.
s
Look at implications: if fluid is an ideal gas:
( h2  h1 )  C p ( T2  T1 )  0
Cp is always a positive number, thus:
T2  T1
P
T
S
P1 > P2
T
v
v
Pressão
Entalpia
Temperatura
En. Interna
Entropia
diminui
constante
gás ideal h=h(T), portanto T fica constante
gás ideal u=u(T), portanto u fica constante
diminui, Ds = Cpln(T2/T1)-Rln(P2/P1)
s
611K (340oC)
T (K)
373K (100oC)
s, kJ/(kgK)
Cold embrittlement
(dureza e rigidez mas baixa resistência a tensão)
T-1000 (Robert Patrick)
(dureza e rigidez mas baixa resistência a tensão)
Cold embrittlement
Consequences of the Temperature Drop on Material Strenght
Low temperature embrittlement does affect most materials
more or less pronounced. It causes overloaded components to
fracture spontaneously rather than accommodating the stress by
plastic deformation. The picture shows a fractured fitting whose
material was not suitable for low temperatures.
TEAMPLAY
Refrigerant 12 enters a valve as a saturated liquid at 0.9607 Mpa
and leaves at 0.1826 MPa. What is the quality and the temperature
of the refrigerant at the exit of the valve?
State (1)
Liquid saturated, x=0
Psat = 0.9607 MPa
40oC
Tsat = ?
Hliq = ? 75kJ/kg
State (2)
Liq+vap x=? 0.33
Psat = 0.1826 MPa
-15oC
Tsat = ?
Hliq = ? 22 & 180 kJ/kg
P-h Diagram of an Ideal VaporCompression Refrigeration Cycle

Throttling



Isentropic
compression (1 to 2)
Constant pressure
condensation (2 to 3)
Isenthalpic expansion
(3 to 4)
Constant pressure
evaporation (4 to 1)
The refrigerant enters the compressor as a saturated
vapor and is cooled to the saturated liquid state in the
condenser. It is then throttled to the evaporator
and vaporizes as it absorbs heat from the refrigerated
space.
Heat exchangers are used
in a variety of industries





Automotive - radiator
Refrigeration - evaporators/condensers
Power production - boilers/condensers
Power electronics - heat sinks
Chemical/petroleum industry- mixing processes
Something a little closer to
home..
Heat Exchangers
Condenser/evaporator
for heat pump
Heat Exchangers

Now, we must deal with multiple inlets and outlets:
m 4
m 2
m 1
m 3
If we have steady
flow, then:
1 m
2m
A
m
3 m
4 m
B
m
Heat Exchangers
0
0
Q  W CV


V12 V22
 A  h1  h2  
m

 g z1  z2 
2
2

A
0, (sometimes 0, (usually
negligible)
negligible)


V32 V42
 B  h3  h4  
m

 g z3  z4   0
2
2

B
0, (sometimes 0, (usually
negligible)
negligible)
Heat Exchangers

And we are left with
 A ( h1  h2 )  m
 B ( h4  h3 )
m
The energy change of fluid A is equal to the
negative of the energy change in fluid B.
mw = ? kg/s
hw = 100 kJ/kg
SC
mvap = 0.1 kg/s
hvap = 2776 kJ/kg
Resp.: mw = 2.576 kg/s
mc = ? kg/s
hc = 200 kJ/kg
Nozzles and Diffusers


Nozzle--a device which
accelerates a fluid as the
pressure is decreased.
Diffuser--a device which
decelerates a fluid and
increases the pressure.
V1, P1
V2,
P2
V1,
P1
V2,
P2
These configuration is for sub-sonic flow.
For supersonic flow, the shape
of the nozzle is reversed.
Nozzles
General shapes of
nozzles and diffusers
nozzle
diffuser
Subsonic flow
nozzle
diffuser
Supersonic flow
conservation of energy
0
V V
q  w  ( h2  h1 ) 
 g ( z2  z1 )
2
0
0
2
2
2
1
q = 0 (adiabatic)
w = 0 (these are not work producing devices;
neither is work done on them)
V V
( h2  h1 ) 
2
2
1
2
2
Sample Problem
An adiabatic diffuser is employed to reduce the velocity of a stream
of air from 250 m/s to 35 m/s. The inlet pressure is 100 kPa and the
inlet temperature is 300°C. Determine the required outlet area in
cm2 if the mass flow rate is 7 kg/s and the final pressure is 167 kPa.
INLET
OUTLET
T1 = 300C
P1 = 100 kPa
V1 = 250 m/s
m
= 7 kg/s
Diffuser
P2 = 167 kPa
V2 = 35 m/s
A2=?
Conservation of Mass: Steady State Regime
m 
V1 A1
1

V2 A2
2
solve for A2
2
m
A2 
V2
But we don’t know v2!
Remember ideal gas equation of state?
RT1
1 
P1
and
RT2
2 
P2
We know T1 and P1, so v1 is simple. We
know P2, but what about T2?
NEED ENERGY EQUATION!!!!
Energy Eqn
V V
 h2  h1   CP T2  T1  
2
2
1
2
2
If we assumed constant specific heats, we could get T2 directly
T2  T1  V  V
2
1
The ideal gas law:
2
2

2  CP  602 K
 2  RT2 P2  1.0352m kg
3
And the area:
m 2 7 1.035 4
2
A2 

10  2070cm
35
V2
Vapor Power Cycles

We’ll look specifically at the Rankine
cycle, which is a vapor power cycle.

It is the primary electrical producing
cycle in the world.

The cycle can use a variety of fuels.
We’ll simplify the power plant
3
BOILER
wout
TURBINE
qin
4
2
CONDENSER
win
1
PUMP
qout
T
compressor
and turbine
must handle
two phase
flows!
TH < TC
TH
2
4 TL
3
Carnot Vapor Cycle
1
s

The Carnot cycle is not a suitable model for vapor
power cycles because it cannot be approximated in
practice.
Ideal power plant cycle is called the
Rankine Cycle
T
qin
3
wout
2
win
1
qout
4
s

The model cycle for vapor power cycles is the Rankine cycle which
is composed of four internally reversible processes:
 1-2 reversible adiabatic (isentropic) compression in the pump
 2-3 constant pressure heat addition in the boiler.
 3-4 reversible adiabatic (isentropic) expansion through turbine
 4-1 constant pressure heat rejection in the condenser
Ideal power plant cycle is called the
Rankine Cycle
P
p=c
2
s=c
1
3
s=c
p=c
4
v

The model cycle for vapor power cycles is the Rankine cycle which
is composed of four internally reversible processes:
 1-2 reversible adiabatic (isentropic) compression in the pump
 2-3 constant pressure heat addition in the boiler.
 3-4 reversible adiabatic (isentropic) expansion through turbine
 4-1 constant pressure heat rejection in the condenser
What are the main parameters we want to
describe the cycle?
Net Power



Wout Wtur Wpump
Net Specific
Work
wout  wtur  w pump
Efficiency

Wnet
 
Q
in
or
wnet

qin
Start our analysis with the pump


 h 2  h1  DKE  DPE
Q

W
pump
Pump  m
T
qin
3
wout
2
win
1
qout
4
The pump is adiabatic,
with no kinetic or
potential energy
changes. The work per
unit mass is:
s
w pump  h1  h2   ( P1  P2 )
Boiler is the next component


 h 3  h 2  DKE  DPE
Q

W
boiler
boiler  m
T
qin
3
wout
2
win
1
qout
4
Boilers do no work. In
boilers, heat is added to
the working fluid, so the
heat transfer term is
already positive. So,
s

Q
boiler
 q boiler  h 3  h 2

m
Proceeding to the Turbine


 h4  h 3  DKE  DPE
Q
turbine  Wturbine  m
T
qin
3
wout
2
win
1
qout
4
s
Turbines are
almost always
adiabatic. In
addition, we’ll
usually ignore
kinetic and
potential energy
changes:

W
turbine
 w turb  h 3  h 4

m
Last component is the Condenser


 h1  h4  DKE  DPE
Q

W
cond
cond  m
T
qin
3
wout
2
win
1
qout
4
Condensers do no
work (they are heat
exchangers), and if
there is no DKE and
DPE,
s

Q
cond
 q cond  h1  h 4

m
Efficiency
wout

qin
h3  h4  v(P2  P1 )

h3  h2
Start an analysis:
A Rankine cycle has an exhaust pressure from the turbine of 0.1
bars. Determine the quality of the steam leaving the turbine and the
thermal efficiency of the cycle which has turbine inlet pressure of
150 bars and 600C.
T
Assumptions:
Pump and turbine are
isentropic;
P2 = P3 = 150 bars =
15 MPa
T3 = 600C
P4 = P1 = 0.1 bars =
0.01 MPa
Kinetic and potential
energy changes are
T3 = 600oC
3
120 bars
PP== 15
MPa
2
P =P =0.01
MPa
0.1 bars
1
4
s
Put together property data
State
T (C)
P(MPa)
v(m3/kg) h(kJ/kg) s(kJ/kgK)
x
1
0.01
0
2
15
n.a.
3
4
600
15
----
0.01
----
Pump (1 to 2) -> isoentropic (const. volume)
Boiler [heat exchanger] (2 to 3) -> const. pressure
Turbine (3 to 4) -> isoentropic
Condenser [heat-exchanger] (4 to 1) -> const. pressure
Property Data
v(m3/kg)
State
T (C)
P(MPa)
h(kJ/kg) s(kJ/kgK)
1
45.81
0.01
0.00101 191.83
0
2
49.42
15
0.00101 206.93
3
600
15
0.02491 3582.3
6.6776
4
45.81
0.01
12.266
6.6776
Liq.
comp
Super
aquec
0.8037
2114.9
x
T
T3 = 600oC
3
120 bars
P P==15
MPa
2
= 0.1 bars
P =P 0.01
MPa
1
4
s
Let start
with pump
work
T
T3 = 600oC
3
P =15
120 bars
P=
MPa
2
= 0.1 bars
P =P0.01
MPa
1
4
w pump   (P1  P2 )  h1  h2
w pump
s
m3
 (0.00101)
(0.01 15)MPa
kg
w pump
kJ
 15.1
kg
More calculations...
Enthalpy at pump outlet:
h2  h1  w pump
Plugging in some numbers:
kJ
h2  (191.83  15.1)
kg
kJ
h2  206.93
kg
How Can I Get The Pump
Outlet Temp?
If the Enthalpy at pump outlet is 206.93
KJ/kg, then consider the compressed
liquid a the same temperature of the
saturated liquid which has h = 206.93
KJ/kg
Interpolating from the saturated steam
table one finds: 49oC
T
Calculate
heat input
T3 = 600oC
3
P =15
120 bars
P=
MPa
2
= 0.1 bars
P =P0.01
MPa
1
qboiler
4
kJ
 h3  h2  ( 3582.3  206.93)
kg
qboiler
kJ
 3375.4
kg
s
Turbine
work
T
T3 = 600oC
3
P =15
120 bars
P=
MPa
2
= 0.1 bars
P =P0.01
MPa
1
Isentropic:
s4  s3  6.6776kJ/kg K
 x4  0.8037;
Turbine work:
4
w turbine
s
h4  2114.9kJ/kg
kJ
 h3  h4  (1467.4)
kg
Overall thermal efficiency

wturbine  w pump
qin
kJ
(1467.4 15.1)
kg
 
 0.430
kJ
3375.4
kg
Some general characteristics
of the Rankine cycle



Low condensing pressure (below
atmospheric pressure)
High vapor temperature entering the
turbine (600 to 1000C)
Small backwork ratio (bwr)
BWR 
w pump
w turbine

h1  h2
h3  h4
 0.01
Questions …

Consider the ideal Rankine cycle from 1-2-3-4:

•How would you increase its
thermal efficiency ?
•What determines the upper T
limit?
•What determines the lower T
limit?
wturbine  w pump
qin
T
qin
3
wout
2
win
1
qout
4
s

Improve the efficiency of the Rankine Cycle: decrease exhaust
pressure of turbine
• decreases condensing temperature
• increases work output
• increases heat input
• decreases quality at turbine outlet
Lowering Turbine Exit Pressure


The average
temperature during
heat rejection can be
decreased by lowering
the turbine exit
pressure.
 Consequently, the
condenser pressure of
most vapor power
plants is well below the
atmospheric pressure.
wturbine  w pump
qin
Reducing Condenser Pressure





Notice that reducing the
condenser pressure (which
will lower the temperature of
heat rejection and again
increase the efficiency) will
also reduce the quality of the
steam exiting the turbine.
Turbines do not like to see
water coming out the
exhaust.
Lower qualities mean water
droplets are forming before
the steam leaves the turbine.
Water droplets lead to
turbine blade erosion.
Efforts are made to keep the
quality > 90%.

w net
q in
Raising Boiler Pressure

The average temperature during heat
addition can be increased by raising the
boiler pressure or by superheating the fluid
to high temperatures.
 There
is a limit to the degree of
superheating, however, since the fluid
temperature is not allowed to exceed a
metallurgically safe value.
Increase Superheat (Boiler Temperature)

T
w net
q in
Work output
increases faster
than heat input, so
the cycle efficiency
increases.
3'
3
2
1
Increasing Superheat
4 4'
higher quality
s
* increases heat input
* increases work output
* increases quality at turbine outlet
* may produce material problems
if temperature gets too high
Effect of Increasing Boiler Pressure on
the Ideal Rankine cycle keeping constant
the Boiler Outlet Temperature Tmax
TEAMPLAY
T
3'

Consider a simple ideal Rankine cycle with
fixed turbine inlet temperature and
condenser pressure. What is the effect of
increasing the boiler pressure on
3
2
1
4 4'
s
Pump work input: (a) increases, (b) decreases, (c) the same
Turbine work
output:
(a) increases, (b) decreases, (c) the same
Heat supplied:
(a) increases, (b) decreases, (c) the same
Heat rejected:
(a) increases, (b) decreases, (c) the same
Cycle efficiency:
(a) increases, (b) decreases, (c) the same
Moisture content
at turbine exit:
(a) increases, (b) decreases, (c) the same
Simple schematic of
Rankine reheat cycle
qinlo
5
4
BOILER
w
Low
Pressure
TURBINE
w
outhi
outlo
3
2
High
Pressure
TURBINE
6
CONDENSER
qinhi
win
1
PUMP
qout
The Ideal Reheat
Rankine Cycle
Reheat on T-s diagram:
T
3
Note that T5 < T3.
Many systems
reheat to the
same temp
(T5=T3).
qinhi
wouthi
qinlo
5
4
2
win
woutlo
1
qout
6
s
Cogeneration
 The
production of more than one useful form of
energy (such as process heat and electric power)
from the same energy source is called
cogeneration.
 Cogeneration
plants produce electric power
while meeting the process heat requirements of
certain industrial processes. This way, more of
the energy transferred to the fluid in the boiler
is utilized for a useful purpose.
 The
fraction of energy that is used for either
process heat or power generation is called the
utilization factor of the cogeneration plant.
An Ideal Cogeneration Plant
Schematic and T-s Diagram
for Cogeneration
Combined Gas-Steam
Power Plant