Thermodynamics
Gas Power Cycles
Gas Power Cycles
Otto Cycle
Single Cylinder Four
Stroke Spark Ignition
Engine
Multi-Cylinder Spark Ignition Engine
THE OTTO CYCLE
COMPRESSION STROKE
Air and fuel are mixed and compressed so rapidly that there is no time for heat
to be lost. (Figure A) In other words the compression is adiabatic. Work must be
done to compress the gas.
IGNITION
Just before the point of maximum compression, the air is hot and a spark
ignites the mixture causing an explosion (Figure B). This produces a rapid rise
in the pressure and temperature. The process is idealized as a constant volume
process in the Otto cycle.
EXPANSION OR WORKING STROKE
The explosion is followed by an adiabatic expansion pushing the piston and
giving out work. (Figure C)
EXHAUST
At the end of the working stroke, there is still some pressure in the cylinder.
This is released suddenly by the opening of an exhaust valve. (Figure D) This is
idealized by a constant volume drop in pressure in the Otto cycle. In 4 stroke
engines a second cycle is performed to push out the products of combustion
and draw in fresh air and fuel. It is only the power cycle that we are concerned
with The four ideal processes that make up the Otto cycle are as follows.
The Ideal Cycle for Spark Ignition Engine
( the Otto Cycle )
Assumptions:
- The working medium is a thermally perfect gas with constant specific heats ( taken as those of air at atmospheric conditions)
- The compression and expansion strokes are isentropic- Heat addition from external source and heat rejection to external sink and both take place at constant volume
WORKED EXAMPLE :
An Otto cycle is conducted as follows. Air at 100 kPa and 20oC is
compressed reversibly and adiabatically. The air is then heated at constant
volume to 1500oC. The air then expands reversibly and adiabatically back
to the original volume and is cooled at constant volume back to the
original pressure and temperature. The volume compression ratio is 8.
Calculate the following.
i. The thermal efficiency.
ii. The heat input per kg of air.
iii. The net work output per kg of air.
iv. The maximum cycle pressure.
cv = 718 kJ/kg K
R = 287 J/kg K
SOLUTION
Remember to use absolute temperatures throughout. Solve for a mass of 1
kg.
T1=20 +273=293K T3=1500+273=1773K rc=8
Diesel – Engine
The gasoline-engine (Otto-engine) wasn't very efficient, that is why 1892
Rudolf Diesel had developed the engine with great efficiency named after
him. It is like the Otto-engine a 4-stroke engine. The main differences are:
1st stroke (Intake):
Only air is sucked in.
2nd stroke (Compression): The air is powerfully compressed.
3rd stroke (Combustion): Diesel is directly injected into the compressed air
and ignites spontaneously.
4th stroke (Exhaust): Like the Otto-engine
DUAL COMBUSTION CYCLE
This is the air standard cycle for a modern fast running diesel engine.
First the air is compressed isentropically making it hot.
Fuel injection starts before the point of maximum compression. After a
short delay in which fuel accumulates in the cylinder, the fuel warms up
to the air temperature and detonates causing a sudden rise in
pressure. This is ideally a constant volume heating process.
Further injection keeps the fuel burning as the volume increases and
produces a constant pressure heating process.
After cut off, the hot air expands isentropically
At the end of the stroke, the exhaust valve opens producing a sudden
drop in pressure. This is ideally a constant volume cooling process. The
ideal cycle is shown in figure
The analysis of the cycle is as follows.
The heat is supplied in two stages hence Qin = mcp(T4 - T3) + mcv (T3 - T2)
The heat rejected is Qout = mcv (T5 - T1)
The thermal efficiency may be found as follows.
THE DIESEL CYCLE
The Diesel Cycle proceeded the dual combustion cycle.
The Diesel cycle is a reasonable approximation of what happens in slow
running engines such as large marine diesels.
The initial accumulation of fuel and sharp detonation does not occur and the
heat input is idealized as a constant pressure process only.
Again consider this cycle as being carried out inside a cylinder fitted with a
piston.
The p-V and T-s cycles diagrams are shown in figure
WORKED EXAMPLE
An engine using the Diesel Cycle has a compression ratio of 20/1 and a cut off
ratio of 2. At the start of the compression stroke the air is at 1 bar and 15oC.
Calculate the following.
i. The air standard efficiency of the cycle.
ii. The maximum temperature in the cycle.
iii. The heat input.
iv. The net work output.
SOLUTION
Gas Turbine Power Plants
Two simple gas turbines:
The Air-Standard Analysis
Based on 2 assumptions:
· Working fluid is air and it behaves as an ideal gas.
· The temperature rise that would actually be brought about by combustion
is modeled as being accomplished by heat transfer from an external source
(this simplifies the analysis).
The Air-Standard Brayton Cycle
· Based on the two air-standard analysis assumptions.
· The turbine exhaust air is restored to the compressor inlet state by being
passed through a heat exchanger which affects Qout to the surroundings
The Air-Standard Brayton Cycle
Analysis
Gas Turbines With Reheat
Due to metallurgical considerations, the temperature of gaseous combustion
products must be limited. This is done by providing air in excess of the
amount required to burn the fuel in the combustor.
Therefore, gases leaving the combustor (and the turbine) have sufficient air for
further combustion. This is exploited in a multistage turbine with additional
combustors between the individual stage
Thermodynamics
Vapor Power Cycles
Example
Saturated vapour steam enters the turbine at 8 MPa and saturated
liquid leaves the condenser at 0.008 MPa. Find: h, m , Q in (for boiler),
Q out (for condenser) if W net =100 MW
In actuality, there are irreversibilities in vapour power cycles
– most notably those associated with the turbine and pump.
Effects of Boiler and Condenser Pressures on the Rankine Cycle
Consider the standard IDEAL Rankine Cycle:
The area marked in red (1 – b – c
– 4 – a – 1) represents the
heat transfer into the working
fluid per unit mass passing
through the boiler, i.e.,
In the figure above, T in is higher
for the cycle with the larger
boiler pressure (1’ – 2 ’ – 3 ’ – 4 ’ 1’) than for the lower boiler
pressure (1 – 2 – 3 – 4 – 1). Thus
as the boiler pressure increases,
so does the thermal efficiency.
In the figure above, Tout is lower for the
cycle with a lower condenser pressure (P
< Patm) (cycle 1 – 2” – 3” – 4” – 1) than
for the cycle with the higher pressure (P
= Patm). Thus as the condenser pressure
decreases, the thermal efficiency
increases.
Improvements to Rankine Cycle
It is common practice to ensure that the quality of the steam
(or other fluid) at the turbine exit is not less than 0.9 .
1. Superheat
After the boiler, further energy is added to the steam in a superheater to bring
it to a superheated vapour state (boiler + superheater = “steam generator”).
2. Reheat
In a reheat cycle, the steam expands to the condenser in more than one stage.
For example, the steam is passed through the 1st stage of a turbine, is reheated in the steam generator to the same T as for the inlet of the 1st stage +
passed through a 2nd stage of the turbine to the condenser. More than two
stages for a turbine are possible.
A reheat cycle allows for taking advantage of the increase thermal
efficiency associated with a higher boiler pressure while avoiding low
quality steam at the turbine exhaust.
Thermodynamics
Some Refrigeration Cycles
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 P-h diagram is another convenient diagram often used to illustrate the
refrigeration cycle.
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
Process
Description
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
Major Components in Vapour Compression Cycle
Refrigeration Cycle: PH diagram
Evaporator

A diagram of a typical vapor-compression refrigeration cycle is
superimposed on a pressure-enthalpy (P-h) chart to demonstrate the
function of each component in the system.

The pressure-enthalpy chart plots the properties of a refrigerant—
refrigerant pressure (vertical axis) versus enthalpy (horizontal axis).

The cycle starts with a cool, low-pressure mixture of liquid and
vapor refrigerant entering the evaporator where it absorbs heat
from the relatively warm air, water, or other fluid that is being
cooled.

This transfer of heat boils the liquid refrigerant in the evaporator,
and this superheated refrigerant vapor is drawn to the compressor.
Refrigeration Cycle: PH diagram
Compressor

The compressor draws in the superheated refrigerant vapor
and compresses it to a pressure and temperature high
enough that it can reject heat to another fluid.

This hot, high-pressure refrigerant vapor then travels to the
condenser.
Refrigeration Cycle: PH diagram
Condenser

Within the condenser, heat is transferred from the hot
refrigerant vapor to relatively cool ambient air or cooling
water.

This reduction in the heat content of the refrigerant
vapor causes it to desuperheat, condense into liquid, and
further subcool before leaving the condenser for the
expansion device.
Refrigeration Cycle: PH diagram
Expansion Device

The high-pressure liquid refrigerant flows through the
expansion device, causing a large pressure drop that reduces
the pressure of the refrigerant to that of the evaporator.

This pressure reduction causes a small portion of the liquid to
boil off, or flash, cooling the remaining refrigerant to the
desired evaporator temperature.

The cooled mixture of liquid and vapor refrigerant then enters
the evaporator to repeat the cycle.
The ordinary household refrigerator is a good example of the application of this cycle.
Q L
C O PR 
W

Q H

W

net , in
C O PH P
net , in
h1  h 4
h 2  h1
h2  h3
h 2  h1
Example 11-1
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

kJ
h

238.41
 1
C om pressor inlet  
kg

o
T1   2 0 C
  s  0.9456 kJ
1
 
kg  K
x1  1.0



kJ
C om pressor exit

  h 2 s  278.23
kg
P2 s  P2  9 0 0 kP a

  T  43.79 o C
kJ
  2s
s 2 s  s1  0.9456
kg  K 

kJ
h

101.61
 3
C ondenser exit  
kg

P3  9 0 0 kP a
  s  0.3738 kJ
3
 
kg  K
x 3  0 .0


  x 4  0.358
T hrottle exit

kJ

o
T 4  T1   2 0 C   s 4  0.4053
kg  K


h 4  h3

State 1
State 3
State 2
State 4
C O PR 
QL

W net , in
m ( h1  h 4 )
m ( h 2  h1 )
(238.41  101.61)

(278.23  238.41)

h1  h 4
h 2  h1
kJ
kg
kJ
kg
 3.44
The tons of refrigeration, often called the cooling load or refrigeration effect, are
Q L  m ( h1  h 4 )
3
kg
(238.41  101.61)
m in
kJ
kg
1 T on
kJ
211
m in
 1.94 T on
C O PR , C arnot 

TL
TH  TL
(  20  273) K
(43.79  (  20)) K
 3.97
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.
W net , in

QL
4.715
C O PR

4.715 hp
3.44 T on
 1.37
hp
T on
Superheating

Superheating occurs inside the final length of tubes at which the
temperature difference between refrigerant and air is highest

Such large temperature difference increases the rate of heat
transfer and the refrigerant vapor absorbs much heat.

Liquid refrigerant completely evaporated

Superheating shifts from the liquid/vapor region to vapor

It ensures the refrigerant vapor is completely free liquid before
entering the compressor.
Heat Pump Systems