1 Topic 4 Power Cycles Gas power cycles Vapour and combined power cycles Refrigeration cycles 2 Modelling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy, as in the example to the right. Our study of power cycles will explore ideal cycles that provide insights into the performance of real devices. They will, however, not include some features of real cycles. We are, therefore, establishing the foundations that you will build on later. Introduction to power cycles Common assumptions for analysis of power cycles in this course include: 1) The cycle does not involve any friction. 2) All expansion and compression processes are quasi-equilibrium processes. 3) The pipes connecting the various components of a system are well insulated, i.e. heat transfer through them is negligible. 4) Changes in kinetic and potential energies of the working fluid can be neglected.
Transcript
1
Topic 4
Power Cycles
Gas power cyclesVapour and combined power cycles
Refrigeration cycles
2
Modelling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy, as in the example to the right.
Our study of power cycles will explore ideal cycles that provide insights into the performance of real devices. They will, however, not include some features of real cycles. We are, therefore, establishing the foundations that you will build on later.
Introduction to power cycles
Common assumptions for analysis of power cycles in this course include:
1) The cycle does not involve any friction.
2) All expansion and compression processes are quasi-equilibrium processes.
3) The pipes connecting the various components of a system are well insulated, i.e. heat transfer through them is negligible.
4) Changes in kinetic and potential energies of the working fluid can be neglected.
3
The areas under the process curves on the P-v diagram represent the work done for closed systems. The net cycle work done is the area enclosed by the cycle on the P-v diagram.
The areas under the process curves on the T-s diagram represent the heat transfer for the processes, when the processes are internally reversible. The net heat added to the cycle is the area that is enclosed by the cycle on the T-s diagram.
For a cycle we know Wnet = Qnet; therefore, the areas enclosed on the P-v and T-s diagrams are equal.
! thnet
in
W
Q=
Net cycle efficiencyOur analysis will focus on understanding the major parameters affecting the cycle affect the performance of heat engines. This performance is often measured in terms of the cycle efficiency.
4
Carnot CycleThe Carnot cycle is the most efficient heat engine that can operate between two fixed temperatures TH and TL.
We often use the Carnot efficiency as a means to think about ways to improve the cycle efficiency of other cycles. For example, thermal efficiency increases with an increase in the average temperature at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system.
! th CarnotL
H
T
T, = "1
Note: Although Topic 3 typically focused on closed systems, the Carnot cycle can be executed in a closed system (piston-cylinder device) or a steady-flow system with two turbines and two compressors.
5
Our study of gas power cycles will involve the study of those heat engines in which the working fluid remains in the gaseous state throughout the cycle. We often study the ideal cycle in which internal irreversibilities and complexities (the actual intake of air and fuel, the actual combustion process, and the exhaust of products of combustion among others) are removed. Before studying some particular cycles, we will introduce some standard assumptions and terminology.
4.1 Gas power cycles
6
4.1.1 Air-Standard AssumptionsIn our study of gas power cycles, we assume that the working fluid is air, and the air undergoes a thermodynamic cycle even though the working fluid in the actual power system does not undergo a cycle (due to the combustion process the composition of the working fluid changes during the real cycle).
To simplify the analysis, we approximate the cycles with the following assumptions: • The air continuously circulates in a closed loop and always behaves as an ideal gas.
• All the processes that make up the cycle are internally reversible.
• A heat rejection process that restores the working fluid to its initial state replaces the exhaust process.
• The cold-air-standard assumptions apply when the working fluid is air and has constant specific heat evaluated at room temperature (25oC or 77oF).
• The combustion process is replaced by a heat-addition process from an external source.
7
4.1.2 Terminology for Reciprocating Devices
The top dead centre (TDC) and bottom dead centre (BDC) can be used to find the change in volume during a cycle. The bore and stroke can be used to determine this displacement volume.
In real systems, the cylinder will also contain a small clearance volume. Air and fuel will enter the cylinder through the intake valve and combustion products will exit through the exhaust valve.
8
The compression ratio (r) of an engine is the ratio of the maximum volume to the minimum volume formed in the cylinder.
rV
V
V
VBDC
TDC
= =max
min
The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the real cycle.
MEPW
V V
w
v vnet net="
="max min max min
9
4.1.3 Otto CycleThe Otto cycle is the air-standard ideal cycle approximation for spark-ignition reciprocating engines. It is named after Nikolaus Otto, who built a successful 4-stroke engine in 1876 in Germany. In most spark-ignition engines the piston executes four complete strokes (two mechanical cycles) within the cylinder, and the crankshaft completes two revolutions for each thermodynamic cycle. These engines are called 4-stroke internal combustion engines. The Otto cycle can also occur in 2-stroke engines, but these are less efficient than their 4-stroke counterparts because of the incomplete expulsion of the exhaust gases and the partial expulsion of the fresh air-fuel mixture with the exhaust gases
Four-stroke engines
10
Two-stroke enginesIn two-stroke engines all four functions described above are executed in just 2 strokes: 1) the power stroke and 2) the compression stroke. In these engines the crankcase is sealed and the outward motion of the piston is used to slightly pressurise the air-fuel mixture in the crankcase. Also the intake and exhaust valves are replaced by openings in the lower portion of the cylinder wall.
During the latter part of the power stroke, the piston uncovers first the exhaust port, allowing the exhaust gases to be partially expelled, and then the intake port, allowing the fresh air-fuel mixture to rush in and drive most of the remaining gases out of the cylinder. This mixture is then compressed as the piston moves upward during the compression stroke and is subsequently ignited by a spark plug.
C. Cylinder Intake/ Cylinder Exhaust
A. Crankcase intake/ Cylinder Compression
B. Ignition/ Power/ Crankcase pressure
11
Process Description for Ideal Otto Cycle1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection
12
13
Example 4-1: Net thermal efficiency of an ideal Otto cycle by calculation
! thnet
in
net
in
in out
in
out
in
W
Q
Q
Q
Q Q
Q
Q
Q= = =
"= "1
So, need to find Qin and Qout.
To find Qin
Apply first law closed system to process 2-3, V = constant.
Thus, for constant specific heats,
Q U
Q Q mC T T
net
net in v
,
, ( )
23 23
23 3 2
=
= = "
#
Numerical example:T2 = 700K and P2 = 20 barMaximum pressure (P3) = 50 bar1kg of air (Cv = 0.718 kJ/kgK)
P2 V2$
T2
P3 V3$
T3
T3 T2
P3
P2
$:= T3 1750K=
Q23 1kg 0.718$kJ
kg K$1750K 700K"( )$:=
Q23 754kJ=
14
To find Qout
Apply first law closed system to process 4-1, V = constant. Thus, for constant specific heats
Q U
Q Q mC T T
Q mC T T mC T T
net
net out v
out v v
,
, ( )
( ) ( )
41 41
41 1 4
1 4 4 1
=
= " = "
= " " = "
#
The net thermal efficiency can then be calculated ! th Ottoout
in
Q
Q, = "
"
1
Numerical example:As above plus V1/V2 = 8.5, k=1.4
V1
V2
8.5T2
T1
V1
V2
!"#
$%&
k 1"
V1 8.5V2:= V1 0.9m3
=
T1
T2
V1
V2
!"#
$%&
k 1":= T1 297K=
T4 T3
V3
V4
!"#
$%&
k 1"
$T4
T3
V3
V4
!"#
$%&
k 1"
V4 V1 0.854m3
=:= V3 V2 0.1m3
=:=
T4 1750K0.1m
3
0.854m3
!""#
$%%&
1.4 1"
$:=
T4 742K=
Q41 1kg 0.718$kJ
kg K$297K 742K"( )$:=
Q41 320" kJ=
!th_Otto 1320kJ
754kJ" 57.6%=:=
15
! th Otto
T T
T T,
( )
( )= "
"
"
"
1 4 1
3 2
Recall processes 1-2 and 3-4 are isentropic, so
Since V3 = V2 and V4 = V1, we see that or
or
T
T
T
T4
1
3
2
=
! th Ottoout
in
v
v
Q
Q
mC T T
mC T T
,
( )
( )
= "
= ""
"
1
1 4 1
3 2
Example 4-2: Net thermal efficiency of an ideal Otto cycle as a function of temperatures in the cycleWe can also improve our understanding of Otto cycle efficiency by further consideration of the relationship between temperature of the working fluid at different states and net cycle efficiency.
T
T
T
T2
1
3
4
=
T T T
T T T
( / )
( / )= "
"
"1
1
11 4 1
2 3 2
The Otto cycle efficiency becomes ! th Otto
T
T, = "1 1
2
16
Compression ratioAs shown in Example 4-2, the Otto cycle efficiency is ! th Otto
T
T, = "1 1
2
Process 1-2 is isentropic, so
where the compression ratio is r = V1/V2
! th Otto kr, = "
"1
11And, hence
We see that increasing the compression ratio (r) increases the thermal efficiency (since k=1.4 for air-standard assumptions). However, there is a limit on r depending upon the fuel. Fuels under high temperature resulting from high compression ratios will prematurely ignite, causing knock.
For a given compression ratio the thermal efficiency of an actual spark-ignition engine is less than that of an ideal Otto cycle because of irreversibilities, such as friction, incomplete combustion, etc.
In compression-ignition engines the air is compressed to a temperature that is abovethe autoignition temperature of the fuel and combustion starts as the fuel is injected into this hot air. In CI engines only air is compressed during the compression stroke eliminating the possibility of autoignition. Therefore, diesel engines can be designed to operate at much higher compression ratios, typically between 12 and 24 (compared to 7-10 for spark-ignition). 17
4.1.4 Diesel Cycle
The Diesel cycle is the air-standard ideal cycle approximation for compression-ignition reciprocating engines. Before considering the cycle itself, it is useful to note some important differences between spark-ignition and petrol-ignition engines. These differences occur since in diesel engines the spark plug is replaced by a fuel injector and only air is compressed during the (real) compression cycle.
In spark-ignition engines the air-fuel mixture is compressed to a temperature that is below the autoignition temperature of the fuel and the combustion process is initiated by firing a spark plug.
18
Not having to deal with the problem of autoignition has another benefit: many of the stringent requirements placed on the petrol can now be removed, and fuels that are less refined (thus less expensive) can be used in diesel engines.
Processes in an air-standard Diesel cycleThe fuel injection process in Diesel engines starts when the piston approaches TDC and continues during the first part of the power stroke. Therefore, the combustion process in these engines takes place over a longer time interval. Because of this longer duration the combustion process in the ideal diesel cycle is approximated as a constant-pressure heat-addition process. This is the onlyprocess where the Otto and the Diesel cycles differ.
The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine is, therefore:
Process Description
1-2 Isentropic compression
2-3 Constant pressure heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection
The P-v and T-s diagrams for a Diesel cycle are shown on the next slide.
19
P-v and T-s diagrams for the ideal Diesel cycle
P-v and T-s diagrams for the ideal Otto cycle
20
Example 4-3: Net thermal efficiency of an ideal Diesel cycle by calculation
! thnet
in
net
in
in out
in
out
in
W
Q
Q
Q
Q Q
Q
Q
Q= = =
"= "1
So, need to find Qin and Qout.
To find Qin
Apply first law closed system to process 2-3, P = constant.
Thus, for constant specific heats,
Q U P V V
Q Q mC T T mR T T
Q mC T T
net
net in v
in p
,
,
( )
( ) ( )
( )
23 23 2 3 2
23 3 2 3 2
3 2
= + "
= = " + "
= "
#
21
To find Qout
Apply first law closed system to process 4-1, V = constant as in the Otto cycle.
Q mC T T mC T T
net out v
out v v
,
( ) ( )
41 1 4
1 4 4 1= " " = "
The net thermal efficiency can then be calculated
! th Dieselout
in
Q
Q
mC T T
,
( )
= "
"
1
Numerical example: m=1kg, r=15, rc=2, T1=297K, P1=1bar, cold air standard conds.
V1
mair R$ T1$
P1
0.85m3
=:=
V2
V1
r0.06m
3=:=
V3 rc V2$ 0.11m3
=:=
V4 V1 0.85m3
=:=
T2 T1
V1
V2
!"#
$%&
k 1"
$ 877K=:=
P2 V2$
T2
P3 V3$
T3
P3 P2
T3 T2
V3
V2
!"#
$%&
$ 1755K=:=
T4 T3
V3
V4
!"#
$%&
k 1"
$ 784K=:=
Qin 1kg 1.005$kJ
kg K$1755K 877K"( )$ 882 kJ$=:=
Qout 1kg 0.718$kJ
kg K$784K 297K"( )$ 350 kJ$=:=
!th_Diesel 1Qout
Qin
" 0.604=:=
22
Example 4-4: Net thermal efficiency of an ideal Diesel cycle as a function of temperatures in the cycleWe can also improve our understanding of Diesel cycle efficiency by further consideration of the relationship between temperature of the working fluid at different states and net cycle efficiency.
! th Dieselout
in
v
p
Q
Q
mC T T
mC T T
,
( )
( )
= "
= ""
"
1
1 4 1
3 2
! th Dieselv
p
C T T
C T T
k
T T T
T T T
,
( )
( )
( / )
( / )
= ""
"
= ""
"
1
11 1
1
4 1
3 2
1 4 1
2 3 2
As in Example 4-2, we can examine the temperature ratios further. In particular, we introduce the concept of the cutoff ratio (rc), defined as the ratio of the cylinder volumes after and before the combustion process (V3 /V2). This is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder.
23
Since we are using air-standard assumptions, we can assume that air behaves like an ideal gas.The combined gas law can, therefore, be applied:
PV
T
PV
TP P
T
T
V
Vrc
3 3
3
2 2
2
3 2
3
2
3
2
= =
= =
where
PV
T
PV
TV V
T P
4 4
4
1 1
1
4 1= = where AlsoT
T
P
P
4 1
4
1
4
1
=so
Recall processes 1-2 and 3-4 are isentropic, so PV PV PV PVk k k k1 1 2 2 4 4 3 3= = and
Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain
24
! th Diesel
ck
c
k
ck
c
k
T T T
T T T
k
T
T
r
r
r
r
k r
,
( / )
( / )
( )
( )
= ""
"
= ""
"
= ""
""
11 1
1
11 1
1
11 1
1
1 4 1
2 3 2
1
2
1Note: as rc tends to 1 (i.e. towards V3 = V2), the final term in the equation to the right tends towards unity and the efficiencies of the Otto and Diesel cycles become identical for the same compression ratio (r).
Remember though that diesel engines operate at much higher compression ratios and thus are more efficient than the spark-ignition engines (as shown to the right with k=1.4 for air-standard assumptions).
Diesel engines also burn fuel more completely since they operate at lower revs and the air-fuel mass ratio is much higher than in petrol engines.
Therefore,T
T
V
Vrc
3 2
3
2
3
2
= =
T
T
P
P
4 1
4
1
4
1
=
25
Dual CycleApproximating the combustion process in internal combustion engines as constant-volume (Otto cycle) or constant-pressure (Diesel cycle) heat-addition process is overly simplistic and not quite realistic. Probably a better but slightly more complex approach would be to model the combustion process in both petrol and diesel engines as a combination of two heat-transfer processes – one at constant volume and the other at constant pressure. The ideal cycle based on this concept is called the dual cycle. The P-v diagram is sketched below and this cycle is considered as a tutorial sheet example.
26
1) Fresh air at ambient conditions is drawn into the compressor where its temperature and pressure are raised.
2) The high-P air proceeds into the combustion chamber, where fuel is burned at constant pressure.
3) The resulting high-T gases then enter the turbine where they expand to atmospheric pressure while producing power.
4) The exhaust gases leave the turbine and are vented to atmosphere.
Note: in this cycle air has two main functions: a) it supplies the necessary oxidant for the combustion of the fuel, and b) it serves as a coolant to keep the temperature of various components within safe limits.
4.1.5 Brayton CycleThe Brayton cycle is the air-standard ideal cycle approximation for gas-turbine engines. As in previous section, before considering the cycle itself, it is useful to review the real cycle that is being modelled.
27
It is also sometimes useful to calculate the fraction of the turbine work used to drive the compressor. This is called the back work ratio.
Open Brayton cycle
Closed Brayton cycle
Some definitions used in Brayton cycle modelling
28
The open cycle gas turbine can be modelled as a closed cycle when air-standard assumptions are used. This the approach taken in this course. The Brayton cycle is the air-standard ideal cycle approximation for the gas-turbine engine. The ideal cycle that the working fluid undergoes in this closed loop is made up of four internally reversible processes.
The P-v and T-s diagramsare shown on the next slide.
Process Description 1-2 Isentropic compression (in a compressor)2-3 Constant pressure heat addition (in a heat exchanger) 3-4 Isentropic expansion (in a turbine)4-1 Constant pressure heat rejection (in a heat exchanger)
Example 4-5: Net thermal efficiency of an ideal closed Brayton cycle by calculation
An important difference between this cycle and the Otto and Diesel cycles is that the processes are open systems or control volumes. Therefore, an open system, steady-flow analysis is used to determine the heat transfer and work for the cycle.
We will consider the cold-air-standard cycle, with the working fluid as air and constant specific heats.
! th Braytonnet
in
out
in
W
Q
Q
Q, = = "1 Brayton
out
in
out
in
Q
Q
q
q
!
!= " = "
"
1 1
31
! !
! ! !
E E
m h Q m h
in out
in
=
+ =2 2 3 3
So, need to find Qin and Qout.
To find Qin
Apply the conservation of energy to process 2-3 for P = constant (no work), steady-flow, and neglect changes in kinetic and potential energies.
The conservation of mass gives
! !
! ! !
m m
m m m
in out=
= =2 3
For constant specific heats, the heat added per unit mass flow is
! ! ( )
! ! ( )
!
!( )
Q m h h
Q mC T T
qQ
mC T T
in
in p
inin
p
= "
= "
= = "
3 2
3 2
3 2
Numerical example:P1 = 1 bar and T1 = 297KMaximum temperature (T3) = 1400K1kg/s of air (Cp = 1.005 kJ/kgK, k=1.4)
P2 rp P1$ 12bar=:=
T2 T1
P2
P1
!"#
$%&
k 1"
k
$ 604K=:=
T3 1400K= given in the data
qin 1.005kJ
kg K$1400K 604K"( ):=
qin 800kJ
kg=
32
To find Qout
The conservation of energy for process 4-1 with constant specific heats gives: The net thermal efficiency can
then be calculated
Numerical example continued:
! ! ( )
! ! ( )
!
!( )
Q m h h
Q mC T T
qQ
mC T T
out
out p
outout
p
= "
= "
= = "
4 1
4 1
4 1
! th Braytonout
in
out
in
Q
Q
q
q,
!
!= " = "
"
1 1
T4 T3
P4
P3
!"#
$%&
k 1"
k
$
P4 P1 1bar=:= T4 1400K1bar
12bar
!"#
$%&
1.4 1"
1.4
$:=
P3 P2 12bar=:=T4 688K=
qout 1.005kJ
kg K$688K 297K"( )$:=
qout 393kJ
kg=
!th_Brayton 1
393kJ
kg
800kJ
kg
":=
!th_Brayton 0.509=
33
Recall processes 1-2 and 3-4 are isentropic, so
or
T
T
T
T4
1
3
2
=
Example 4-6: Net thermal efficiency of an ideal closed Brayton cycle as a function of temperatures in the cycleWe can also improve our understanding of Brayton cycle efficiency by further consideration of the relationship between temperature of the working fluid at different states and net cycle efficiency.
T
T
T
T2
1
3
4
=
T T T
T T T
( / )
( / )= "
"
"1
1
11 4 1
2 3 2
The Brayton cycle efficiency becomes
! th Braytonout
in
out
in
p
p
Q
Q
q
q
C T T
C T T
,
!
!
( )
( )
= " = "
= ""
"
1 1
14 1
3 2
! th Brayton
T T
T T,
( )
( )= "
"
"1 4 1
3 2
Since P3 = P2 and P4 = P1, we see that
! th Brayton
T
T, = "1 1
2
34
Pressure ratioAs shown in Example 4-6, the closed Brayton cycle efficiency is
Process 1-2 is isentropic, so where the pressure ratio is rp = P2/P1
And, hence
! th Brayton
T
T, = "1 1
2
!th Brayton
p
k kr
, ( )/= "
"1
11
35
4.1.6 Gas power cycles summaryAir-standard ideal cycle approximations can be used to gain useful insights into the efficiency of cycles where the working fluid remains gaseous throughout the whole cycle. Three common cycles are the Otto, Diesel and Brayton cycles. The Dual cycle can also be considered for analysing internal combustion engines.
Diesel Cycle
Compression-ignition engine model
Closed processes
Otto Cycle
Spark-ignition engine model
Closed processes
Brayton Cycle
Open or closed cycle gas turbine model
Open processes
36
4.2 Cycles including phase changesWe can also consider cycles where the working fluid undergoes a phase change. A common example is the steam power cycle where water/steam is the working fluid. Refrigeration cycles typically also include a phase change (as in the lab).
This section will introduce the Rankine cycle for power generation (with water/steam as the working fluid) and the vapour compression cycle, which is demonstrated in the lab. To understand these cycles, it is first useful to consider practical problems that are associated with attempting to implement a Carnot vapour cycle.
The discussion that follows will consider a steady-flow Carnot cycle executed within the saturation dome of a pure substance.
The ideal cycle consists of four processes:Process Description 1-2 Isentropic compression (compressor) 2-3 Isothermal heating (boiler) 3-4 Isentropic expansion (turbine)4-1 Isothermal cooling (condenser)
37
Problems with a real Carnot vapour cycle
1) Processes 2-3 and 4-1 can be approached closely in actual boilers and condensers (maintaining constant pressure fixes the temperature when the working fluid is wet). However, limiting the heat transfer to 2-phase systems severely limits the maximum temperature that can be used in the cycle as it has to remain under the critical-point value (374ºC for water).
2) Process 3-4 can be approached closely by a well-designed turbine. However, the quality of the steam decreases during this process. Thus the turbine has to handle steam with low quality, i.e. steam with a high moisture content. Liquid droplets would cause erosion of turbine blades would cause erosion and be a major source of wear. The quality of the steam in the operation of power plants should be 90% or higher. This problem can be eliminated by using a fluid with a very steep saturated vapor line.
3) Process 1-2 involves the compression of a liquid-vapour mixture to a saturated liquid. But, (a) it is not easy to control the condensation process so precisely as to end up with the desired quality at state 4; (b) 2-phase flow in compressors should be avoided.
38
Some of these difficulties can be eliminated by executing the Carnot cycle so that it is not contained within saturation dome of a pure substance. However, this cycle presents other problems such as isentropic compression to extremely high pressures and isothermal heat transfer at variable pressures.
Thus we conclude that the Carnot cycle cannot be approximated in real devices and is not a realistic model for vapour power cycles. We can, however, use it as standard against which real vapour power cycles are compared.
In particular, remember that the thermal efficiency of the Carnot cycle is given as
! th Carnotnet
in
out
in
W
Q
Q
Q, = = "1
Thus, thermal cycle efficiency will be increased by increasing TH and by decreasing TL. For a power cycle, this means that to improve efficiency we need to increase the average temperature at which heat is transferred to the working fluid in the boiler, and/or decrease the average temperature at which heat is rejected from the working fluid in the condenser.
L
H
T
T= "1
39
4.2.1 Rankine CycleThe simple Rankine cycle has the same component layout as the Carnot cycle, but continues the condensation process 4-1 until the saturated liquid line is reached.
Process Description 1-2 Isentropic compression in pump 2-3 Constant pressure heat addition in boiler 3-4 Isentropic expansion in turbine 4-1 Constant pressure heat rejection in condenser
Note: The vertical distance between 1 and 2 on the T-s diagram is greatly exaggerated for clarity. (If water were truly incompressible –there would not be any temperature change at all during 1-2 process.)
Additionally, we can recall the property relation:
40
Example 4-7What is the thermal efficiency of an ideal Rankine cycle where steam leaves the boiler as superheated vapour at 6 MPa, 500oC, and is condensed at 75 kPa?
! th =Desired Result
Required Input! th
net out
in
W
Q=
,
0 2 4 6 8 10 12120
100
200
300
400
500
s [kJ/kg-K]
T [
C]
6000 kPa
10 kPa
1
2
3
4
Pump (1-2) The pump work is obtained from the conservation of mass and energy for steady-flow, assuming that potential and kinetic energy changes and negligible, and that the pump is adiabatic and reversible.
! ! !
! ! !
! ! ( )
m m m
m h W m h
W m h h
pump
pump
1 2
1 1 2 2
2 1
= =
+ =
= "
dh = T ds + v dP
Since the ideal pumping process 1-2 is isentropic, ds = 0, so:
41
The incompressible liquid assumption then allows
v v const
h h v P P
% =
" % "
1
2 1 1 2 1
.
( )
The pump work is calculated from
! ! ( ) ! ( )
!
!( )
W m h h mv P P
wW
mv P P
pump
pump
pump
= " % "
= = "
2 1 1 2 1
1 2 1
Additionally, h2 can then be found from
2 1pumph w h
kJ kJ
= +
P1 75kPa 0.75bar=:=
v1 vf v1 0.001037m
3
kg:=
h1 hf h1 384.3kJ
kg:=
wpump v1 P2 P1"( )$
wpump 0.001037m
3
kg6000 75"( )kPa$:=
wpump 6.1kJ
kg=
h2 wpump h1+:=
h2 390.4kJ
kg=
= (3422.9 – 390.4) kJ/kg
= 3032.5 kJ/kg
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Boiler (2-3)To find the heat supplied in the boiler, we apply the steady-flow conservation of mass and energy to the boiler. Again, we assume that changes in potential and kinetic energies are negligible. Additionally, we note that no work is done on the steam in the boiler:
! ! !
! ! !
! ! ( )
m m m
m h Q m h
Q m h h
in
in
2 3
2 2 3 3
3 2
= =
+ =
= "
3 2in
in
Qq h h
m
kJ
= = "!
!
P3 = 6 MPa = 60 bar
T3 = 500°C
so h3 = 3422.9 kJ/kg
s3 = 6.882 kJ/kgK
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Turbine (3-4)The turbine work is again obtained from the application of the conservation of mass and energy for steady flow. We assume the process is adiabatic and reversible and that changes in kinetic and potential energies are negligible.
Since the turbine is isentropic, we know that s4 = s3 = 6.882 kJ/kg-K.
Additionally, we have been told that the condenser pressure (P4) is 0.75 bar.
sf = 1.2130 kJ/kgK and sg = 7.456 kJ/kgK
and, hence, the steam is wet.! ! !
! ! !
! ! ( )
m m m
m h W m h
W m h h
turb
turb
3 4
3 3 4 4
3 4
= =
= +
= "
s4 sf x4 sg sf"( )$+
x4
s4 sf"
sg sf"0.908=:=
h4 hf x4 hfg$+
h4 384.37kJ
kg0.9082278.0$
kJ
kg+:=
h4 2453kJ
kg$=
wturb h3 h4" 970kJ
kg$=:=
44
Finally, the net work done by the cycle is
The thermal efficiency is
Options to improve the simple Rankine cycle efficiency include
• Increase boiler pressure (for fixed maximum temperature) (2-3) Availability of steam is higher at higher pressures.Moisture is increased at turbine exit.
• Superheat the vapor (2-3)Average temperature is higher during heat addition.Moisture is reduced at turbine exit (we want x4 > 90%).
•Lower condenser pressure (4-1)Less energy is lost to surroundings.Moisture is increased at turbine exit.
wnet wturb wpump"
wnet 970kJ
kg6
kJ
kg":=
wnet 964kJ
kg=
!th
wnet
qin
!th
964kJ
kg
3032kJ
kg
31.8%=:=
RefrigeratorThe purpose of a refrigerator is the removal of heat, called the cooling load, from a low-temperature medium.
Heat pumpThe purpose of a heat pump is the transfer of heat to a high-temperature medium, called the heating load.
In general, the term heat pump is used to describe the cycle as heat energy is removed from the low-temperature space and rejected to the high-temperature space.
4.2.2 Refrigeration Cycle
The vapor compression refrigeration cycle is a common method for transferring heat from a low temperature to a high temperature. It is introduced in the lab for this course.
45
46
REMINDER – The Reversed Carnot CycleRecall that the Carnot cycle is a totally reversible cycle that consists of two reversible isothermal and two isentropic processes. It has maximum efficiency for given temperature limits and it serves as a standard against which actual power cycles can be compared. Since it is a reversible cycle, all four processes that comprise the Carnot cycle can be reversed. Reversing the cycle also reverses the directions of any heat and work interactions.
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REMINDER – Coefficient of performanceThe performance of refrigerators and heat pumps is expressed in terms of coefficient of performance (COP), defined as
COPQ
W
COPQ
W
RL
net in
HPH
net in
= = =
= = =
Desired output
Required input
Cooling effect
Work input
Desired output
Required input
Heating effect
Work input
,
,
Both COPR and COPHP can be larger than 1. Under the same operating conditions, the COPs are related by
COP COPHP R= +1
Refrigerators, air conditioners, and heat pumps are rated with an SEER number or seasonal adjusted energy efficiency ratio. The SEER is defined as the Btu/hr of heat transferred per watt of work energy input. The Btu is the British thermal unit and is equivalent to 778 ft-lbf of work (1 W = 3.4122 Btu/hr).
Refrigeration systems are also rated in terms of tons of refrigeration. The capacity of a refrigeration system that can freeze 1 ton of liquid water at 0ºC into ice at 0ºC in 24 hours is said to be 1 ton.
48
A refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnotheat pump, and their Coefficients of Performance are
COPT T
T
T T
COPT T
T
T T
R Carnot
H L
L
H L
HP Carnot
L H
H
H L
,
,
/
/
="
="
="
="
1
1
1
1
Also since a turbine is used for the expansion process between the high and low-temperatures, some of the work produced by the turbine can be used to supply work to the compressor.
Note: both COPs increase as the difference between the two temperatures decreases, that is as TL rises or TH falls.
49
Processes 1-2 and 3-4 can be approached closely in evaporators and condensers. Processes 2-3 and 4-1 cannot, however, be approximated closely in practice. Process 2-3 involves the compression of a liquid-vapour mixture which would require compression of two phases. Process 4-1 would involve the expansion of high-moisture refrigerant in a turbine. It is also cheaper to have irreversible expansion through an expansion valve rather than in a turbine
Challenges for using a reversed Carnot cycle in reality
We can consider executing the reversed Carnot cycle outside the saturation region to avoid these problems. But this would then make it difficult to maintain isothermal conditions during the heat-rejection (and possibly also heat-absorption) processes. Therefore we conclude that the reversed Carnot cycle cannot be approximated in real devices and is not a realistic model for refrigeration cycles. We can, however, compare real refrigeration cycles to it.
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The most widely used refrigeration cycle is the vapour-compression refrigeration cycle. This cycle has four components: evaporator, compressor, condenser, and expansion (or throttle) valve.
It removes many of the impracticalities associated with the reversed Carnot cycle by (1) vapourising the refrigerant completely before it is compressed and (2) replacing the turbine with a throttling device.
In an ideal vapour-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapour and is cooled to the saturated liquid state in the condenser. It is then throttled to the evaporator pressure and vapourises as it absorbs heat from the refrigerated space.
The Vapour-Compression Refrigeration Cycle
Reminder: Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid.
51
Process Description 1-2 Isentropic compression 2-3 Constant pressure heat rejection in the condenser3-4 Throttling in an expansion valve4-1 Constant pressure heat addition in the evaporator
52
The area under the process curve in T-s diagram represents the heat transfer for internally reversible processes.
The area under 1-4 curve represents the heat absorbed by the refrigerant in the evaporator, and the area under the 2-3 curve represents the heat rejected in the condenser.
If the throttling device were replaced by an isentropic turbine, the refrigerant would enter the evaporator at state 4’ (not 4). As a result the refrigeration capacity would increase by the area under the 4-4’ curve. A turbine is not used in reality, however, due to complications and costs that this would add.
As illustrated in the lab, the P-h diagram is another convenient diagram often used to illustrate the refrigeration cycle.
53
Example 4-8: The ordinary household refrigerator
COPQ
W
h h
h h
COPQ
W
h h
h h
RL
net in
HPH
net in
= ="
"
= ="
"
!
!
!
!
,
,
1 4
2 1
2 3
2 1
See Tutorial Sheet 4B
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Real Vapour-Compression Refrigeration CycleA real vapour-compression cycle differs from the ideal one in several ways including:
1) It is hard to control the state of the refrigerant precisely and normally the refrigerant is slightly superheated as it leaves the evaporator. This slight overdesign ensures that the refrigerant is completely vapourised when it enters the compressor.
2) Real compression processes are not reversible and adiabatic. Where practical and economical, the refrigerant should be cooled during compression. This minimises specific volume of refrigerant and, hence, work input.
3) The refrigerant is subcooled before it enters the throttling valve (i.e. not a saturated liquid) since it is difficult to execute very precisely.
4) Pressure drops can be expected throughout the cycle, although this will be more significant in some parts of the cycle than others (e.g. the throttling valve and the evaporator are located very close to each other, so the pressure drop in the connecting line between them is expected to be small).
See the next slide for diagrams that illustrate these points. Within this course we will not undertake quantitative analysis of this real cycle. You will, however, analyse real cycles quantitatively in Thermo 3.
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4.2.3 SummaryThe Carnot cycle cannot be implemented in reality, so modifications are needed for ideal (and real) cycle analysis. This section has introduced two ideal cycles that will be considered further in later courses.
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