Gas Power Cycles
Air-Standard Assumptions
In 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.
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.
 The combustion process is replaced by a heat-addition process from an
external source.
 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).
Terminology for Reciprocating Devices
The following is some terminology we need to understand for reciprocating
engines—typically piston-cylinder devices. Let’s look at the following figures for
the definitions of top dead center (TDC), bottom dead center (BDC), stroke, bore,
intake valve, exhaust valve, clearance volume, displacement volume,
compression ratio, and mean effective pressure.
The compression ratio r of an engine is the ratio of the maximum volume to the
minimum volume formed in the cylinder.
1
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 mount of net
work as that produced during the actual cycle.
Otto Cycle: The Ideal Cycle for
Spark-Ignition Engines
Consider the automotive sparkignition power cycle with the
following processes:
- Intake stroke
- Compression stroke
- Power (expansion) stroke
- Exhaust stroke
Often the ignition and combustion
process
begins
before
the
completion of the compression
stroke. The number of crank angle
degrees before the piston reaches
TDC on the number one piston at which the spark occurs is called the engine
timing. What are the compression ratio and timing of your engine in your car,
truck, or motorcycle?
2
The air-standard Otto cycle is the ideal cycle that approximates the spark
ignition combustion engine with the following processes:
1-2 Isentropic compression
2-3 Constant volume heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection
---------------------------------------------------------------------------------------------------------------------------------------------
Otto, Diesel, and Dual Cycles
Although most gas turbines are also internal combustion engines, the name is
usually reserved to reciprocating internal combustion engines of the type
commonly used in automobiles, trucks, and buses.
Two principal types of reciprocating internal combustion engines are the sparkignition engine and the compression-ignition engine. In a spark- ignition engine a
mixture of fuel and air is ignited by a spark plug. In a compression ignition
engine air is compressed to a high-enough pressure and temperature that
combustion occurs spontaneously when fuel is injected.
In a four-stroke internal combustion engine, a piston executes four distinct
strokes within a cylinder for every two revolutions of the crankshaft. Figure 12.7
gives a pressure-displacement diagram as it might be displayed electronically.
With the intake valve open, the piston makes an intake stroke to draw a fresh
charge into the cylinder. Next, with both valves closed, the piston undergoes a
compression stroke raising the temperature and pressure of the charge. A
combustion process is then initiated, resulting in a
high pressure, hightemperature gas mixture. A power stroke follows the compression stroke, during
which the gas mixture expands and work is done on the piston. The piston then
3
executes an exhaust stroke in which the burned gases are purged from the
cylinder through the open exhaust valve. Smaller engines operate on two-stroke
cycles. In two-stroke engines, the intake, compression, expansion, and exhaust
operations are accomplished in one revolution of the crankshaft. Although
internal combustion engines undergo mechanical cycles, the cylinder contents do
not execute a thermodynamic cycle, since matter is introduced with one
composition and is later discharged at a different composition.
A parameter used to describe the performance of reciprocating piston engines is
the mean effective pressure, or mep. The mean effective pressure is the
theoretical constant pressure that, if it acted on the piston during the power
stroke, would produce the same net work as actually developed in one cycle. That
is,
where the displacement volume is the volume swept out by the piston as it moves
from the top dead center to the bottom dead center. For two engines of equal
displacement volume, the one with a higher mean effective pressure would
produce the greater net work and, if the engines run at the same speed, greater
power.
Detailed studies of the performance of reciprocating internal combustion engines
may take into account many features, including the combustion process occurring
within the cylinder and the effects of irreversibilities associated with friction and
with pressure and temperature gradients. Heat transfer between the gases in the
cylinder and the cylinder walls and the work required to charge the cylinder and
exhaust the products of combustion also might be considered. Owing to these
complexities, accurate modeling of reciprocating internal combustion engines
normally involves computer simulation.
To conduct elementary thermodynamic analyses of internal combustion engines,
considerable simplification is required. A procedure that allows engines to be
studied qualitatively is to employ an air standard analysis having the following
elements: (1) a fixed amount of air modeled as an ideal gas is the system; (2) the
combustion process is replaced by a heat transfer from an external source and
represented in terms of elementary thermodynamic processes; (3) there are no
exhaust and intake processes as in an actual engine: the cycle is completed by a
constant-volume heat rejection process; (4) all processes are internally reversible.
The processes employed in air-standard analyses of internal combustion engines
are selected to represent the events taking place within the engine simply and
mimic the appearance of observed pressure-displacement diagrams. In addition to
the constant volume heat rejection noted previously, the compression stroke and
at least a portion of the power stroke are conventionally taken as isentropic. The
4
heat addition is normally considered to occur at constant volume, at constant
pressure, or at constant volume followed by a constant pressure process, yielding,
respectively, the Otto, Diesel, and Dual cycles shown in Table 12.7.
Reducing the closed system energy balance, gives the following expressions for
work and heat applicable in each case shown in Table 12.7:
(12.32)
Table 12.7 provides additional expressions for work, heat transfer, and thermal
efficiency identified with each case individually. All expressions for work and
heat adhere to the respective sign conventions of Eq. (12.7b). Equations (1) to (6)
of Table 12.4 apply generally to air-standard analyses. In a cold air standard
analysis the specific heat ratio k for air is taken as constant. Equations (1′) to (6′)
of Table 12.4 apply to cold air-standard analyses, as does Eq. (4′) of Table 12.5,
with n = k for the isentropic processes of these cycles.
Referring to Table 12.7, the ratio of specific volumes v1/v2 is the compression
ratio, r. For the Diesel cycle, the ratio v3/v2 is the cutoff ratio, rc. Figure 12.8
shows the variation of the thermal efficiency with compression ratio for an Otto
cycle and Diesel cycles having cutoff ratios of 2 and 3. The curves are
determined on a cold air-standard basis with k = 1.4 using the following
expression:
(12.33)
where the Otto cycle corresponds to rc = 1.
5
As all processes are internally reversible, areas on the p-v and T-s diagrams of
Table 12.7 can be interpreted, respectively, as work and heat transfer. Invoking
Eq. (12.23) and referring to the p-v diagrams, the areas under process 3-4 of the
Otto cycle, process 2-3-4 of the Diesel cycle, and process x-3-4 of the Dual cycle
represent the work done by the gas during the power stroke, per unit of mass. For
6
each cycle, the area under the isentropic process 1-2 represents the work done on
the gas during the compression stroke, per unit of mass. The enclosed area of
each cycle represents the net work done per unit mass.
With Eq. (12.25) and referring to the T-s diagrams, the areas under process 2-3 of
the Otto and Diesel cycles and under process 2-x-3 of the Dual cycle represent
the heat added per unit of mass. For each cycle, the area under the process 4-1
represents the heat rejected per unit of mass. The enclosed area of each cycle
represents the net heat added, which equals the net work done, each per unit of
mass.
---------------------------------------------------------------------------------------------Thermal Efficiency of the Otto cycle:
Apply first law closed system to process 2-3, V = constant.
Thus, for constant specific heats,
Apply first law closed system to process 4-1, V = constant.
Thus, for constant specific heats,
The thermal efficiency becomes
7
Recall processes 1-2 and 3-4 are isentropic, so
Since V3 = V2 and V4 = V1, we see that
Is this the same as the Carnot cycle efficiency?
Since process 1-2 is isentropic,
where the compression ratio is r = V1/V2 and
We see that increasing the
compression ratio increases the
thermal efficiency. 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.
Example 8-1
An Otto cycle having a compression ratio of 9:1 uses air as the working fluid.
Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heat addition
process, 7.5 kJ of heat are added. Determine all T's, P's, th, the back work ratio,
and the mean effective pressure.
8
Process Diagrams: Review the P-v and T-s diagrams given above for the Otto
cycle. Assume constant specific heats with C v = 0.718 kJ/kg K, k = 1.4. (Use
the 300 K data from Table A-2)
Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9,
The first law closed system for process 2-3 was shown to reduce to (your
homework solutions must be complete; that is, develop your equations from the
application of the first law for each process as we did in obtaining the Otto cycle
efficiency equation)
Then,
9
Using the combined gas law
Process 3-4 is isentropic; therefore,
Process 4-1 is constant volume. So the first law for the closed system gives, on a
mass basis,
For the cycle, u = 0, and the first law gives
10
The thermal efficiency is
The mean effective pressure is
The back work ratio is (can you show that this is true?)
Air-Standard Diesel Cycle
The air-standard Diesel cycle is the ideal cycle that approximates the Diesel
combustion engine
11
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 are:
Thermal efficiency of the Diesel cycle
Now to find Qin and Qout.
Apply the first law closed system to
process 2-3, P = constant.
Thus, for constant specific heats
Apply the first law closed system to process 4-1, V = constant (just like the Otto
cycle)
Thus, for constant specific heats
12
The thermal efficiency becomes
where rc is called the cutoff ratio, defined as V3 /V2, and 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.
Recall processes 1-2 and 3-4 are isentropic, so
Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and
obtain
13
Therefore,
What happens as rc goes to 1? Sketch the
P-v diagram for the Diesel cycle
and show rc approaching1 in the limit.
14
Brayton Cycle
The Brayton cycle is the air-standard ideal cycle approximation for the gasturbine
engine. This cycle differs from the Otto and Diesel cycles in that the processes
making the cycle occur in 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 assume the working fluid is air and the specific heats are constant and
will consider both open and closed gas-turbine cycles.
The closed cycle gas-turbine engine
Process Description
1-2 Isentropic compression (in a
compressor)
2-3 Constant pressure heat addition
3-4 Isentropic expansion (in a turbine)
4-1 Constant pressure heat rejection
The T-s and P-v diagrams are
Thermal efficiency of the Brayton
cycle
15
Now to find Qin and Qout.
Apply the conservation of energy to process 2-3 for P = constant (no work),
steady-flow, and neglect changes in kinetic and potential energies.
For constant specific heats, the heat added per unit mass flow is
The conservation of energy for process 4-1 yields for constant specific heats
(let’s take a minute for you to get the following result)
The thermal efficiency becomes
Recall processes 1-2 and 3-4 are isentropic, so
16
Since P3 = P2 and P4 = P1, we see that
The Brayton cycle efficiency becomes
Is this the same as the Carnot cycle efficiency?
Since process 1-2 is isentropic,
where the pressure ratio is rp = P2/P1 and
17
Extra Assignment
Evaluate the Brayton cycle efficiency by determining the net work directly from
the turbine work and the compressor work. Compare your result with the above
expression. Note that this approach does not require the closed cycle assumption.
Example 8-2
The ideal air-standard Brayton cycle operates with air entering the compressor at
95 kPa, 22oC. The pressure ratio r p is 6:1 and the air leaves the heat addition
process at 1100 K. Determine the compressor work and the turbine work per unit
mass flow, the cycle efficiency, the back work ratio, and compare the compressor
exit temperature to the turbine exit temperature. Assume constant properties.
Apply the conservation of energy for steady-flow and neglect changes in kinetic
and potential energies to process 1-2 for the compressor. Note that the
compressor is isentropic.
The conservation of mass gives
For constant specific heats, the compressor work per unit mass flow is
Since the compressor is isentropic
18
The conservation of energy for the turbine, process 3-4, yields for constant
specific heats (let’s take a minute for you to get the following result)
Since process 3-4 is isentropic
We have already shown the heat supplied to the cycle per unit mass flow in
process 2-3 is
19
The cycle efficiency becomes
The back work ratio is defined as
Note that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature is greater
than the compressor exit temperature. Can this result be used to improve the
cycle efficiency?
What happens to th, win /wout, and wnet as the pressure ratio r p is increased?
20
Let's take a closer look at the effect of the pressure ratio on the net work
done.
Note that the net work is zero when
For fixed T3 and T1, the pressure ratio that makes the work a maximum is
obtained from:
This is easier to do if we let X = r p
(k-1)/k
Solving for X
Then, the r p that makes the work a maximum for the constant property case
and fixed T3 and T1 is
21
For the ideal Brayton cycle, show that the following results are true.
When rp = rp, max work, T4 = T2
When rp < rp, max work, T4 > T2
When rp > rp, max work, T4 < T2
Regenerative Brayton Cycle
For the Brayton cycle, the turbine exhaust temperature is greater than the
compressor exit temperature. Therefore, a heat exchanger can be placed
between the hot gases leaving the turbine and the cooler gases leaving the
compressor. This heat exchanger is called a regenerator or recuperator. The
sketch of the regenerative Brayton cycle is shown below.
We define the regenerator effectiveness regen as the ratio of the heat transferred
to the compressor gases in the regenerator to the maximum possible heat transfer
to the compressor gases.
For ideal gases using the cold-air-standard assumption with constant specific
heats, the regenerator effectiveness becomes
22
Using the closed cycle analysis and treating the heat addition and heat rejection
as steady-flow processes, the regenerative cycle thermal efficiency is
Notice that the heat transfer occurring within the regenerator is not included in
the efficiency calculation because this energy is not a heat transfer across the
cycle boundary.
Assuming an ideal regenerator regen = 1 and constant specific heats, the thermal
efficiency becomes (take the time to show this on your own)
When does the efficiency of the air-standard Brayton cycle equal the efficiency
of the air-standard regenerative Brayton cycle? If we set th, Brayton = th, regen then
Recall that this is the pressure ratio that maximizes the net work for the simple
Brayton cycle and makes T4 = T2. What happens if the regenerative Brayton cycle
operates at a pressure ratio larger than this value?
For fixed T3 and T1, pressure ratios greater than this value cause T4 to be less than
T2, and the regenerator is not effective.
What happens to the net work when a regenerator is added?
What happens to the heat supplied when a regenerator is added?
The following shows a plot of the regenerative Brayton cycle efficiency as a
function of the pressure ratio and minimum to maximum temperature ratio, T1/T3.
23
Example 8-3: Regenerative Brayton Cycle
Air enters the compressor of a regenerative gas-turbine engine at 100 kPa and
300 K and is compressed to 800 kPa. The regenerator has an effectiveness of 65
percent, and the air enters the turbine at 1200 K. For a compressor efficiency of
75 percent and a turbine efficiency of 86 percent, determine
(a) The heat transfer in the regenerator.
(b) The back work ratio.
(c) The cycle thermal efficiency.
Compare the results for the above cycle with the ones listed below that have the
same common data as required.
(a) The actual cycle with no regeneration, = 0.
(b) The actual cycle with ideal regeneration, = 1.0.
(b) The ideal cycle with regeneration, = 0.65.
(d) The ideal cycle with no regeneration, = 0.
(e) The ideal cycle with ideal regeneration, = 1.0.
We assume air is an ideal gas with constant specific heats, that is, we use the
cold-air-standard assumption.
Summary of Results
24
Compressor analysis
The isentropic temperature at compressor exit is
To find the actual temperature at compressor exit, T2a, we apply the compressor
efficiency
Since the compressor is adiabatic and has steady-flow
Turbine analysis
The conservation of energy for the turbine, process 3-4, yields for constant
specific heats (let’s take a minute for you to get the following result)
25
Since P3 = P2 and P4 = P1, we can find the isentropic temperature at the turbine
exit.
To find the actual temperature at turbine exit, T4a, we apply the turbine efficiency.
The turbine work becomes
The back work ratio is defined as
26
Regenerator analysis
To find T5, we apply the regenerator effectiveness.
To find the heat transferred from the turbine exhaust gas to the compressor exit
gas, apply the steady-flow conservation of energy to the compressor gas side of
the regenerator.
Using qregen, we can determine the turbine exhaust gas temperature at the
regenerator exit.
27
Heat supplied to cycle
Apply the steady-flow conservation of energy to the heat exchanger for
process 5-3. We obtain a result similar to that for the simple Brayton cycle.
Cycle thermal efficiency
The net work done by the cycle is
The cycle efficiency becomes
You are encouraged to complete the calculations for the other values found in the
summary table.
28
Other Ways to Improve Brayton Cycle Performance
Intercooling and reheating are two important ways to improve the performance of
the Brayton cycle with regeneration.
Intercooling
When using multistage compression, cooling the working fluid between the
stages will reduce the amount of compressor work required. The compressor
work is reduced because cooling the working fluid reduces the average specific
volume of the fluid and thus reduces the amount of work on the fluid to achieve
the given pressure rise.
To determine the intermediate pressure at which intercooling should take place to
minimize the compressor work, we follow the standard approach.
For the adiabatic, steady-flow compression process, the work input to the
compressor per unit mass is
29
Can you obtain this relation another way? Hint: apply the first law to processes 14.
For two-stage compression, let’s assume that intercooling takes place at constant
pressure and the gases can be cooled to the inlet temperature for the compressor,
such that P3 = P2 and T3 = T1.
The total work supplied to the compressor becomes
]
To find the unknown pressure P2 that gives the minimum work input for fixed
compressor inlet conditions T1, P1, and exit pressure P4, we get
30
or, the pressure ratios across the two compressors are equal.
Intercooling is almost always used with regeneration. During intercooling the
compressor exit temperature is reduced; therefore, more heat must be supplied in
the heat addition process. Regeneration can make up part of the required heat
transfer.
To supply only compressed air, using intercooling requires less work input. The
next time you go to a home supply store where air compressors are sold, check
the larger air compressors to see if intercooling is used. For the larger air
compressors, the compressors are made of two piston-cylinder chambers.
The intercooling heat exchanger may be only a pipe with a few attached fins that
connects the large piston-cylinder chamber with the smaller pistoncylinder
chamber.
Extra Assignment
Obtain the expression for the compressor total work by applying conservation of
energy directly to the low- and high-pressure compressors.
Reheating
When using multistage expansion through two or more turbines, reheating
between stages will increase the net work done (it also increases the required heat
input). The regenerative Brayton cycle with reheating is shown above.
The optimum intermediate pressure for reheating is the one that maximizes the
turbine work. Following the development given above for intercooling and
assuming reheating to the high-pressure turbine inlet temperature in a constant
pressure steady-flow process, we can show the optimum reheat pressure to be
or the pressure ratios across the two turbines are equal.
31
Refrigeration Cycles
The vapor compression refrigeration cycle is a common method for transferring
heat from a low temperature to a high
temperature.
The figure shows the objectives of
refrigerators and heat pumps. The
purpose of a refrigerator is the removal
of heat, called the cooling load, from a
low-temperature medium. The purpose
of a heat pump is the transfer of heat to
a high-temperature medium, called the
heating load. When we are interested in
the heat energy removed from a lowtemperature space, the device is called
a refrigerator. When we are interested
in the heat energy supplied to the high-temperature space, the device is called a
heat pump. 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.
The performance of refrigerators and heat pumps is expressed in terms of
coefficient of performance (COP), defined as
Both COPR and COPHP can be larger than 1. Under the same operating conditions,
the COPs are related by
Refrigerators, air conditioners, and heat pumps are rated with a 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). An EER of 10
yields a COP of 2.9.
Refrigeration systems are also rated in terms of tons of refrigeration. One ton of
refrigeration is equivalent to 12,000 Btu/hr or 211 kJ/min. How did the term “ton
of cooling” originate?
32
Reversed Carnot Refrigerator and Heat Pump
Shown below are the cyclic refrigeration device operating between two constant
temperature reservoirs and the T-s diagram for the working fluid when the
reversed Carnot cycle is used. Recall that in the Carnot cycle heat transfers take
place at constant temperature. If our interest is the cooling load, the cycle is
called the Carnot refrigerator. If our interest is the heat load, the cycle is called
the Carnot heat pump.
The standard of comparison for refrigeration cycles is the reversed Carnot cycle.
A refrigerator or heat pump that operates on the reversed Carnot cycle is called a
Carnot refrigerator or a Carnot heat pump, and their COPs are
Notice that a turbine is used for the expansion process between the high and lowtemperatures. While the work interactions for the cycle are not indicated on the
figure, the work produced by the turbine helps supply some of the work required
by the compressor from external sources.
Why not use the reversed Carnot refrigeration cycle?
Easier to compress vapor only and not liquid-vapor mixture.
Cheaper to have irreversible expansion through an expansion valve.
What problems result from using the turbine?
33
The Vapor-Compression Refrigeration Cycle
The vapor-compression refrigeration cycle has four components: evaporator,
compressor, condenser, and expansion (or throttle) valve. The most widely used
refrigeration cycle is the vapor-compression refrigeration cycle. In an ideal
vapor-compression refrigeration cycle, 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 pressure and vaporizes as it absorbs heat from the
refrigerated space.
The ideal vapor-compression cycle consists of four processes.
Ideal Vapor-Compression Refrigeration Cycle
1-2 Isentropic compression
2-3 Constant pressure heat rejection in the condenser
3-4 Throttling in an expansion valve
4-1 Constant pressure heat addition in the evaporator
The P-h diagram
The ordinary household refrigerator is a good example of the application of this
cycle.
34
Example 10-1
Refrigerant-134a is the working fluid in an ideal compression refrigeration cycle.
The refrigerant leaves the evaporator at -20oC and has a condenser pressure of 0.9
MPa. The mass flow rate is 3 kg/min. Find COPR and COPR, Carnot for the same
Tmax and Tmin , and the tons of refrigeration.
Using the Refrigerant-134a Tables, we have
The tons of refrigeration, often called the cooling load or refrigeration effect, are
Another
measure
of
the
effectiveness of the refrigeration
cycle is how much input power
to
the
compressor,
in
horsepower, is required for each
ton of cooling.
The unit conversion is 4.715 hp
per ton of
cooling.
35
Actual Vapor-Compression Refrigeration Cycle
Gas Refrigeration Systems
The power cycles can be used as refrigeration cycles by simply reversing them.
Of these, the reversed Brayton cycle, which is also known as the gas
refrigeration cycle, is used to cool aircraft and to obtain very low (cryogenic)
temperatures after it is modified with regeneration. The work output of the
turbine can be used to reduce the work input requirements to the compressor.
Thus, the COP of a gas refrigeration cycle is
36
Vapor and Combined Power Cycles
We consider power cycles where the working fluid undergoes a phase change.
The best example of this cycle is the steam power cycle where water (steam) is
the working fluid. The heat engine may be composed of the following
components.
Steam Power Cycle
The working fluid, steam (water), undergoes a thermodynamic cycle from 1-2-34-1. The cycle is shown on the following T-s diagram.
Carnot Vapor Cycle Using Steam
The thermal efficiency of this
cycle is given as
Note the effect of TH and TL on th, Carnot. : The larger the TH the larger the th, Carnot
and the smaller the TL the larger the th, Carnot
To increase the thermal efficiency in any power cycle, we try to increase the
maximum temperature at which heat is added.
Reasons why the Carnot cycle is not used:
Pumping process 1-2 requires the pumping of a mixture of saturated liquid
and saturated vapor at state 1 and the delivery of a saturated liquid at state 2.
To superheat the steam to take advantage of a higher temperature, elaborate
controls are required to keep TH constant while the steam expands and does
work.
To resolve the difficulties associated with the Carnot cycle, the Rankine cycle
was devised.
37
Rankine Cycle
The simple Rankine cycle has the same component layout as the Carnot cycle
shown above. The simple Rankine cycle continues the condensation process 4-1
until the saturated liquid line is reached.
Ideal Rankine Cycle Processes
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
The T-s diagram for the Rankine cycle is
given below. Locate the processes
for heat transfer and work on the diagram.
Example 9-1
Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves
the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa.
We use the power system and T-s diagram shown above.
P2 = P3 = 6 MPa = 6000 kPa
T3 = 350oC
P1 = P4 = 10 kPa
Pump
The pump work is obtained from the conservation of mass and energy for steadyflow but neglecting potential and kinetic energy changes and assuming the pump
is adiabatic and reversible.
Since the pumping process involves an incompressible liquid, state 2 is in the
compressed liquid region, we use a second method to find the pump work or the
h across the pump.
Recall the property relation:
38
Since the ideal pumping process 1-2 is isentropic, ds = 0.
The incompressible liquid assumption allows
The pump work is calculated from
Using the steam tables
Now, h2 is found from
39
Boiler
To find the heat supplied in the boiler, we apply the steady-flow conservation of
mass and energy to the boiler. If we neglect the potential and kinetic energies,
and note that no work is done on the steam in the boiler, then
We find the properties at state 3 from the superheated tables as
The heat transfer per unit mass is
Turbine
The turbine work is obtained from the application of the conservation of mass
and energy for steady flow. We assume the process is adiabatic and reversible
and neglect changes in kinetic and potential energies.
We find the properties at state 4 from the steam tables by noting s4 = s3 and asking
three questions.
40
The turbine work per unit mass is
The net work done by the cycle is
The thermal efficiency is
41
Ways to improve the simple Rankine cycle efficiency:
Superheat the vapor
Average temperature is higher during heat addition.
Moisture is reduced at turbine exit (we want x4 in the above example >
85 percent).
Increase boiler pressure (for fixed maximum temperature)
Availability of steam is higher at higher pressures.
Moisture is increased at turbine exit.
Lower condenser pressure
Less energy is lost to surroundings.
Moisture is increased at turbine exit.
Extra Assignment
For the above example, find the heat rejected by the cycle and evaluate the
thermal efficiency from
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Reheat Cycle
As the boiler pressure is increased in the simple Rankine cycle, not only does the
thermal efficiency increase, but also the turbine exit moisture increases. The
reheat cycle allows the use of higher boiler pressures and provides a means to
keep the turbine exit moisture (x > 0.85 to 0.90) at an acceptable level.
The thermal efficiency is given by
Example 9-2
Compare the thermal efficiency and turbine-exit quality at the condenser pressure
for a simple Rankine cycle and the reheat cycle when the boiler pressure is 4
MPa, the boiler exit temperature is 400 oC, and the condenser pressure is 10 kPa.
The reheat takes place at 0.4 MPa and the steam leaves the reheater at 400oC.
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Regenerative Cycle
To improve the cycle thermal efficiency, the average temperature at which heat is
added must be increased. One way to do this is to allow the steam leaving the
boiler to expand the steam in the turbine to an intermediate pressure. A portion of
the steam is extracted from the turbine and sent to a regenerative heater to preheat
the condensate before entering the boiler. This approach increases the average
temperature at which heat is added in the boiler. However, this reduces the mass
of steam expanding in the lower- pressure stages of the turbine, and, thus, the
total work done by the turbine. The work that is done is done more efficiently.
The preheating of the condensate is done in a combination of open and closed
heaters. In the open feedwater heater, the extracted steam and the condensate are
physically mixed. In the closed feedwater heater, the extracted steam and the
condensate are not mixed.
Cycle with an open feedwater heater
Rankine Steam Power Cycle with an Open Feedwater Heater
Rankine Steam Power Cycle with an Open Feedwater Heater
Cycle with a closed feedwater heater with steam trap to condenser
Rankine Steam Power Cycle with a Closed Feedwater Heater
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Cycle with a closed feedwater heater with pump to boiler pressure
Consider the regenerative cycle with the open feedwater heater.
To find the fraction of mass to be extracted from the turbine, apply the first law
to the feedwater heater and assume, in the ideal case, that the water leaves the
feedwater heater as a saturated liquid. (In the case of the closed feedwater heater,
the feedwater leaves the heater at a temperature equal to the saturation
temperature at the extraction pressure.)
Conservation of mass for the open feedwater heater:
Let y m m ! / ! 6 5 be the fraction of mass extracted from the turbine for the
feedwater heater.
Conservation of energy for the open feedwater heater:
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Example 9-3
An ideal regenerative steam power cycle operates so that steam enters the turbine
at 3 MPa, 500oC, and exhausts at 10 kPa. A single open feedwater heater is used
and operates at 0.5 MPa. Compute the cycle thermal efficiency.
Using the software package the following data are obtained.
The work for pump 1 is calculated from
Now, h2 is found from
The fraction of mass extracted from the turbine for the open feedwater heater is
obtained from the energy balance on the open feedwater heater, as shown above.
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This means that for each kg of steam entering the turbine, 0.163 kg extracted for
the feedwater heater.
The work for pump 2 is calculated from
Now, h4 is found from the energy balance for the pump.
Apply the steady-flow conservation of energy to the isentropic turbine.
The net work done by the cycle is
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Apply the steady-flow conservation of mass and energy to the boiler.
The heat transfer per unit mass entering the turbine at the high pressure, state 5, is
The thermal efficiency is
If these data were used for a Rankine cycle with no regeneration, then th = 35.6
percent. Thus, the one open feedwater heater operating at 0.5 MPa increased the
thermal efficiency by 5.3 percent. However, note that the mass flowing through
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the lower-pressure stages has been reduced by the amount extracted for the
feedwater and the net work output for the regenerative cycle is about 10 percent
lower than the standard Rankine cycle.
Below is a plot of cycle thermal efficiency versus the open feedwater heater
pressure. The feedwater heater pressure that makes the cycle thermal efficiency a
maximum is about 400 kPa.
Below is a plot of cycle net work per unit mass flow at state 5 and the fraction of
mass y extracted for the feedwater heater versus the open feedwater heater
pressure. Clearly the net cycle work decreases and the fraction of mass extracted
increases with increasing extraction pressure.
Why does the fraction of mass extracted increase with increasing extraction
pressure?
Placement of Feedwater Heaters
The extraction pressures for multiple feedwater heaters are chosen to maximize
the cycle efficiency. As a rule of thumb, the extraction pressures for the
feedwater heaters are chosen such that the saturation temperature difference
between each component is about the same.
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Example 9-4
An ideal regenerative steam power cycle operates so that steam enters the turbine
at 3 MPa, 500oC, and exhausts at 10 kPa. Two closed feedwater heaters are to be
used. Select starting values for the feedwater heater extraction pressures.
Deviation from Actual Cycles
Piping losses--frictional effects reduce the available energy content of the
steam.
Turbine losses--turbine isentropic (or adiabatic) efficiency.
The actual enthalpy at the turbine exit (needed for the energy analysis of the next
component) is
Pump losses--pump isentropic (or adiabatic) efficiency.
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The actual enthalpy at the pump exit (needed for the energy analysis of the next
component) is
Condenser losses--relatively small losses that result from cooling the
condensate below the saturation temperature in the condenser.
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Thermodynamics II
Problems on power cycles
----------------------------------------------------------------------------------------1. An ideal Otto cycle has a compression ratio of 8 . At the beginning of the
compression process, air is at 95 kPa and 27 oC, and 750 kJ / kg of heat is
transferred to air during the constant volume heat addition process. Determine
(a) the pressure and temperature at the end of the heat addition process, (b) the
net work output, (c) the thermal efficiency, and (d) the mean effective
pressure for the cycle.
Answers: (a) 3898, 1 kPa, 1538 K, (b) 392 kJ/kg, (c) 52%, (d) 495 kPa
2. A four – cylinder spark – ignition engine has a compression ratio of 8, and
each cylinder has a maximum volume of 0.6 L. At the beginning of the
compression process, the air is at 98 kPa and 17 oC, and the maximum
temperature in the cycle is 1800 K. Assuming the engine to operate on the
ideal Otto cycle, determine (a) the amount of heat supplied per cylinder, (b)
the thermal efficiency, and 9 (c) the number of revolutions per minute
required for a net power output of 80 kW.
3. The compression ratio of an air – standard Otto cycle is 9.5. Prior to the
isentropic compression process, the air is at 100 kPa, 17 oC, and 600 cm3. The
temperature at the end of the isentropic expansion process is 800 K Determine
(a) the highest temperature and pressure in the cycle, (b) the amount of heat
added in kJ, (c) the thermal efficiency, and (d) the mean effective pressure.
Answers: (a) 1987 K, 649 kPa, (b) 0.65 kJ/kg, (c) 59%, (d) 719 kPa
4. An air – standard Diesel cycle has a compression ratio of 18 .2 Air is at 27 oC
and 0.1 MPa at the beginning of the compression process and at 2000 K at the
end of the heat addition process. Accounting for the variation of specific heats
with temperature, determine (a) the cutoff ratio (b)the heat rejection per unit
mass, and (c) the thermal efficiency.
5. An ideal diesel engine has a compression ratio of 20 and uses air as the
working fluid. The state of air at the beginning of the compression process is
95 kPa and 20 oC. If the maximum temperature in the cycle is not to exceed
2200 K, determine (a) the thermal efficiency and (b) the mean effective
pressure. Assume constant heats for air at room temperature. Answers: (a)
63.5 percent (b) 933 kPa.
6. An ideal dual cycle has a compression ratio of 12 and uses air as the working
fluid. At the beginning of the compression process, air is 100 kPa and 30 oC
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and occupies a volume of 1.2 L. During the heat addition process, 0.3 kJ of
heat is transferred to air at constant volume and 1.1 kJ at constant pressure.
Determine the thermal efficiency of the cycle.
7. The compression ratio of an ideal dual cycle is 14. Air is at 100 kPa and 300 K
at the beginning of the compression process and at 2200 K at the end of the
heat addition process. Heat transfer to air takes place partly at constant
volume and partly at constant pressure, and it amounts to 1520.4 kJ/kg.
Determine (a) the fraction of heat transferred at constant volume and (b) the
thermal efficiency of the cycle.
8. A simple Brayton cycle using air as the working fluid has a pressure ratio of 8.
The minimum and maximum temperatures in the cycle are 310 and 1160 K.
Assuming an adiabatic efficiency of 75 percent for the compressor and 82
percent for the turbine , determine (a) the air temperature at the turbine exit,
(b) the net work output, and (c) the thermal efficiency.
9. Air enters the compressor of a gas–turbine engine at 300 K and 100 kPa, where
it is compressed to 700 kPa and 580 K. Heat is transferred to air in the amount
of 950 kJ/kg before it enters the turbine. For a turbine to efficiency of 86
percent, determine (a) the fraction of the turbine work output used to drive the
compressor and (b) the thermal efficiency. Answers: (a) 64.7 percent, (b) 16.4
percent.
10. A gas turbine power plant operates on the simple Brayton cycle with air as
the working fluid and delivers 15 MW of power. The minimum and maximum
temperatures in the cycle are 310 and 900 K and the pressure of air at the
compressor exit is 8 times the value at the compressor inlet. Assuming an
adiabatic efficiency of 80 percent for the compressor and 86 percent for the
turbine, determine the mass flow rate of air through the cycle.
11. A Brayton cycle with regeneration using air as the working fluid has a
pressure ratio of 8. The minimum and maximum temperatures in the cycle are
310 and 1150 K. Assuming an adiabatic efficiency of 75 percent for the
compressor and 82 percent for the turbine and an effectiveness of 65 percent
for the regenerator; determine (a) the air temperature at the turbine exit, (b)
the net work output, and (c) the thermal efficiency. Answers: (a) 763.07 kg/kg
(b) 101.64 kg/kg, (c) 21.0 percent.
12. Air enters the compressor of a regenerative gas – turbine engine at 300 k and
100 kPa where it is compressed to 800 kPa and 580 K The regenerator has an
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effectiveness of 65 percent, and the air enters the turbine at 1200 K For a
turbine efficiency of 86 percent, determine (a) the amount of heat transfer in
the regenerator and (b) the thermal efficiency. Answers: (a) 137.7 kJ/kg (b)
35.0 percent.
13. Consider an ideal gas turbine cycle with two stages of compression and two
stages of expansion. The pressure ratio across each stage of compressor and
turbine is 3. The air enters each stage of the compressor at 300 K and each
stage of the turbine at 1200 K. Determine the back work ratio and the thermal
efficiency of the cycle, assuming (a) no regenerator is used and (b) a
regenerator with 75 percent effectiveness is used.
14. Consider a regenerative gas – turbine power plant with two stages of
compression and two stages of expansion. The overall pressure ratio of the
cycle is 9. The air enters each stage of the compressor at 300 K and each stage
of the turbine at 1200 K. Determine the minimum mass flow rate of air needed
to develop a net power output of 50 MW. Answer: 113.4 kg/s.
15. A steady-flow Carnot refrigeration cycle uses refrigerant-134a as the working
fluid. The refrigerant changes from saturated vapor to saturated liquid at 30°C
in the condenser as it rejects heat. The evaporator pressure is 160 kPa. Show
the cycle on a T-s diagram relative to saturation lines, and determine (a) the
coefficient of performance, (b) the amount of heat absorbed from the
refrigerated space, and (c) the net work input. Answers: (a) 5.64, (b) 147
kJ/kg, (c) 26.1 kJ/kg
16. A refrigerator uses refrigerant-134a as the working fluid and operates on an
ideal vapor-compression refrigeration cycle between 0.12 and 0.7 MPa. The
mass flow rate of the refrigerant is 0.05 kg/s. Show the cycle on a T-s diagram
with respect to saturation lines. Determine (a) the rate of heat removal from
the refrigerated space and the power input to the compressor, (b) the rate of
heat rejection to the environment, and (c) the coefficient of performance.
Answers: (a) 7.41 kW, 1.83 kW, (b) 9.23 kW, (c) 4.06
17. Refrigerant-134a enters the compressor of a refrigerator as superheated vapor
at 0.14 MPa and _10°C at a rate of 0.12 kg/s, and it leaves at 0.7 MPa and
50°C. The refrigerant is cooled in the condenser to 24°C and 0.65 MPa, and it
is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in
the connecting lines between the components, show the cycle on a T-s
diagram with respect to saturation lines, and determine (a) the rate of heat
removal from the refrigerated space and the power input to the compressor,
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(b) the isentropic efficiency of the compressor, and (c) the COP of the
refrigerator. Answers: (a) 19.4 kW, 5.06 kW, (b) 82.5 percent, (c) 3.83
18. A gas refrigeration system using air as the working fluid has a pressure ratio
of 4. Air enters the compressor at _7°C. The high-pressure air is cooled to
27°C by rejecting heat to the surroundings. It is further cooled to _15°C by
regenerative cooling before it enters the turbine. Assuming both the turbine
and the compressor to be isentropic and using constant specific heats at room
temperature, determine (a) the lowest temperature that can be obtained by this
cycle, (b) the coefficient of performance of the cycle, and (c) the mass flow
rate of air for a refrigeration rate of 12 kW. Answers: (a) _99.4°C, (b) 1.12, (c)
0.237 kg/s
19. A steam power plant operates on a simple ideal Rankine cycle between the
pressure limits of 3 MPa and 50 kPa. The temperature of the steam at the
turbine inlet is 400 oC, and the mass flow rate of steam through the cycle is 40
kg/s. Show the cycle on a T– s diagram with respect to saturation lines, and
determine (a) the thermal efficiency of the cycle and (b) the net power output
of the power plant.
20. Steam enters the turbine of a steam power plant which operates on a simple
ideal Rankine cycle at a pressure of 6 MPa, and it leaves as a saturated vapor
at 7.5 kPa. Heat is transferred to the steam in the boiler at a rate of 10 5 kJ/s.
Steam is cooled in the condenser by the cooling water from a nearby river
which enters the condenser at 18 oC. Show the cycle on a T-s diagram with
respect to saturation lines, and determine (a) the turbine inlet temperature (b)
the net power output and the thermal efficiency and (c) the minimum mass
flow rate of the cooling water required.
21. A steam power plant operates on a simple ideal Rankine cycle between the
pressure limits of 9 MPa and 10 kPa. The mass flow rate of steam through the
cycle is 25 kg/s. The moisture content of the steam at the turbine exit is not to
exceed 10 percent. Show the cycle on a T-s diagram with respect to saturation
lines, and determine (a) the minimum turbine inlet temperature, (b) the rate of
heat input in the boiler (c) the thermal efficiency of the cycle.
22. A steam power plant operates on the ideal reheat Rankine cycle. Steam enters
the high pressure turbine at 8 MPa and 500 oC and leaves at 3 MPa. Steam is
then reheated at constant pressure to 500 oC before it expands to 20 kPa in the
low pressure turbine. Determine the turbine work output, in kJ/kg, and the
55
thermal efficiency of the cycle. Also show the cycle on a T-s diagram with
respect to saturation lines.
23. Steam enters the high – pressure turbine of a steam power plant which
operates on the ideal reheat Rankine cycle at 6 MPa and 450 oC and leaves as
saturated vapor. Steam is then reheated to 400 oC before it expands to a
pressure of 7.5 kPa. Heat is transferred to the steam in the boiler at a rate of 10
kJ/s. Steam is cooled in the condenser by the cooling water from a nearby
river, which enters the condenser at 18 oC. Show the cycle on T-s diagram
with respect to saturation lines and determine (a) the pressure at which
reheating takes place (b) the net power output and thermal efficiency and (c)
the minimum mass flow rate of the cooling water required.
24. A steam power plant operates on an ideal regenerative Rankine cycle. Steam
enters the turbine at 6 MPa and 450 oC and is condensed in the condenser at
20 kPa. Steam is extracted from the turbine at 0.4 MPa to heat the feed water
in an open feed water heater. Water leaves the feed water heater as a saturated
liquid. Show the cycle on a T-s diagram and determine (a) the net work output
per kilogram of steam flowing through the boiler and (b) the thermal
efficiency of the cycle. Answers: (a) 1016 kJ/kg (b) 37.8 percent.
25. A steam power plant operates on an ideal reheat regenerative Rankine cycle
and has a net power output of 80 MW. Steam enters the high pressure turbine
at 10 MPa and 550 oC and leaves at 0.8 MPa. Some steam is extracted at this
pressure to heat the feed water in an open feed water heater. The rest of the
steam is reheated to 500 oC and is expanded in the low pressure turbine to the
condenser pressure of 10 kPa. Show the cycle on a T-s diagram with respect
to saturation lines, and determine (a) the mass flow rate of steam through the
boiler and (b) the thermal efficiency of the cycle. Answers: (a) 54.56 kg / s (b)
44.4 percent.
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