The Otto Cycle - Power

The Otto Cycle
Presented by: Jason Rako, Mike Nee and Ralph Spolander
Our Agenda
➢ A brief history of Nicolaus Otto
➢ The Otto Engine
➢ The Otto Cycle Explained
➢ Formulas to Calculate an Otto Cycle
➢ Critical Error
➢ An Otto Cycle Problem
A brief history of Nicolaus Otto
➢ Nicolaus Otto is credited
as the inventor of the
compressed-charge
engine, commonly known
as the four-stroke engine.
➢ Otto was a travelling
merchant selling tea,
coffee, rice and sugar
across Germany.
A brief history of Nicolaus Otto
➢ Inspired by Frenchman Jean-Joseph Etienne Lenoir’s
successful experiments with internal combustion
engines, Otto began pursuing his own engine designs.
➢ The two men established N.A. Otto & Cie., the first
engine company in the world - and the forerunner of
today’s Deutz AG.
➢ Otto built the first the first practical high-compression
engine with an ignition device in 1876.
➢ The design was based on principles patented in 1862
by Alphonse Beau de Rochas.
The Otto Engine
➢ Otto went into great detail when describing his engine.
From his American patent submission in 1877, the
operation can be described as follows:
The Otto Engine
➢ First, an intimate mixture of combustible gas or vapour
and air is introduced into the cylinder. At the same
time, it is combined with a separate charge of air or
other gas that may or may not support combustion.
➢ These two mixtures are combined in such a manner
and in such proportions that the particles of the
particles of the combustible gaseous mixture are more
or less dispersed in an isolated condition in the air.
The Otto Engine
➢ By doing this, it ensures, upon ignition, the flame will
be communicated gradually from one combustible
particle to another.
➢ This will cause a gradual development of heat and a
corresponding gradual expansion of the gases.
➢ Otto believed that his engine would utilize the motive
power produced by combustion in the most effective
manner.
The Otto Cycle Explained
➢ According to “Applied Thermodynamics for
Engineering Technologists” (Eastop and McConkey,
third edition) the Otto Cycle is not a true heat engine.
➢ The reason for this is that fuel is burned directly in the
working fluid and thus this is an internal combustion
cycle.
➢ The high temperatures obtained by the working fluid is
the main advantage of these cycles.
➢ The working fluid may reach as high as 2750 degrees
Celsius for an instant; the cylinders requires air or
liquid cooling.
The Otto Cycle Explained
The steps in an Otto Cycle
➢ The steps of the Otto cycle are summarized as
intake, compression, power and exhaust.
The Otto Cycle Explained
➢ The intake step (stroke) takes
place at (atmospheric) pressure.
The volume of the combustion
chamber increases as the piston,
originally at the height of its
stroke, moves down. Air is
drawn through the intake valve.
The Otto Cycle Explained
➢ The compression stroke is
isentropic compression. The
volume of the combustion
chamber will decrease and there
is no heat flow during this
process. Note that both valves
are closed and the piston is
moving upwards
The Otto Cycle Explained
➢ The combustion and power
stroke is a combination of a
reversible isochoric heating
process and an isentropic
expansion process. The air is
heated from the ignition of the
fuel. As a result pressure
increases; this takes place very
quickly. The air is then expands,
pushing the piston down and
performing work on the
crankshaft.
The Otto Cycle Explained
➢ The final step is the exhaust
stroke where the spent fuel and
air is pushed out of the
combustion chamber. This is
done by the piston moving back
up towards the top of the
cylinder. This is known as a
isochoric cooling process.
The Otto Cycle Explained
The PV Diagram
The Otto Cycle Explained
The TS Diagram
The Formulas to Calculate the Otto Cycle
Calculating the ideal air-standard thermal
efficiency of the Otto Cycle
➢ Calculating the thermal efficiency of the Otto Cycle
hinges on the ratio of the amount of heat supplied to
the process ( Q1 during combustion) to the amount of
heat lost ( Q2 during exhaust).
The Formulas to Calculate the Otto Cycle
Calculating the ideal air-standard thermal
efficiency of the Otto Cycle
➢ This is derived as follows, based on the Second Law of
Thermodynamics:
Eff. = W / Q1
= [ Q 1 - Q2 ] / Q 1
= 1 - Q2 / Q1
The Formulas to Calculate the Otto Cycle
➢ Now, given Q = Cv ΔT for a constant volume process:
Eff. = W / Q1
= [ Q1 - Q2 ] / Q1
= 1 - Q2 / Q1
Eff. = 1 - [ ( T4 - T1 ) / ( T3- T2 ) ]
The Formulas to Calculate the Otto Cycle
Relating the Isentropic Processes
➢ Since both the compression and expansion stages of
the Otto Cycle are isentropic, we can derive a
relationship between ( T3 - T2 ) and ( T4 - T1 ).
T2 / T1 = ( V1 / V2 )γ -1
T3 / T4 = ( V4 / V3 )γ -1
The Formulas to Calculate the Otto Cycle
➢ Now, referring to our PV diagram on page 8, we note
the following relationships:
V1 = V4
( V1 / V2 )
γ -1
And V3 = V2
= ( V 4 / V3 )
γ -1
T2 / T1 = T3 / T4 = A Ratio
= A Ratio
The Formulas to Calculate the Otto Cycle
The Ratio Behind it All
➢ Establishing the relationships between the two
isentropic processes resulted in a ratio. Looking at this
ratio in simplest form, without considering Gamma, it
can be written as follows:
( V1 / V2 ) = A Ratio (simplified)
( V1 / V2 ) = rv
The Formulas to Calculate the Otto Cycle
➢ In terms of the isentropic relationship:
( V1 / V2 )
γ -1
= ( V4 / V3 )
γ -1
v
T2 / T1 = T3 / T4 = r
γ -1
=r
γ -1
v
The Formulas to Calculate the Otto Cycle
➢ Thus we can establish:
T2 = ( rv
T3 = ( rv
γ -1
) ( T1 )
γ -1
) ( T4 )
The Formulas to Calculate the Otto Cycle
➢ Relating it to the ideal air-standard efficiency
equation:
Eff. = 1 - [ ( T4 - T1 ) / ( T3- T2 ) ]
Eff. = 1 - [ ( T4 - T1 ) / ( T4 - T1 )( rv γ -1 ) ]
Eff. = 1 - 1 / ( rv γ -1 )
The Formulas to Calculate the Otto Cycle
➢ Thus the ratio is truly behind it all.
➢ That also means that the engine efficiency will increase
with a higher compression ratio.
CRITICAL ERROR
Listen Very Carefully
➢ There is a major difference between the air-standard
efficiency and the actual efficiency.
➢ The values we have been using for air (like R=0.287)
are considered nominal values. These are not constant
when temperatures fluctuate wildly!
➢ For example, air at 850K has a γ of 1.349; at 450 K,
1.391.
CRITICAL ERROR
➢ Truly, only at temperatures below 300K are you able to
get away with using a γ of 1.4!
➢ That being said, get ready to interpolate and find
average temperatures to make sure your calculations
are accurate.
CRITICAL ERROR
Here's an Example:
➢ Take an adiabatic process where air enters at 100 kPa
and at 27 degrees Celsius at a volume of 1 m3. It is
compressed to 0.125 m3.
γ -1
Using:
T
/
T
=
(
V
/
V
)
➢
2
1
1
2
➢ We find that T2 = 689.219 K.
This is not correct!!
CRITICAL ERROR
➢ Looking at page 16, we interpolate to find that γ at this
temperature is 1.365. This is a significant difference
and it is apparent that γ is not constant.
➢ To remedy this situation, the best course of action is to
add both T1 and T2 and find the average temperature
(divide by 2).
➢ The average would be 494.60K or roughly 500K and
corresponds to a γ of 1.387. This γ value is the average
that should be used for this process at these specified
temperatures.
CRITICAL ERROR
➢ This is the temperature averaging technique.
➢ In reality, a temperature of 670.837 K is more realistic.
➢ You can ignore this factor but your “real” calculations,
realistically, may be completely incorrect.
Example Problem
Put what you just learned to use
➢ A highly-experimental Otto-cycle based engine has a
compression ratio of 12 with a clearance volume of 0.2
m3 and is fueled by a Nitropropane-air mixture
(providing 1.685 MJ/kg of heat once ignited).
➢ The fuel enters the combustion chamber at atmospheric
pressure (100 kPa) and at 50 degrees Celsius. R is
0.283 kJ/kgK and γ is 1.4 INITIALLY.
Example Problem
Determine the following:
➢ a) The volume of the combustion chamber when the
piston is all the way down.
➢ b) The ideal air standard thermal efficiency.
➢ c) The realistic temperature and pressure at points 1
through 4.
➢ d) The realistic net work of the cycle.
Example Problem
Determine the following:
➢ e) The realistic thermal efficiency of the engine.
➢ f) The realistic net change of entropy in the cycle.
Please provide proof.
➢ g) Why are the compression and expansion cycles
considered isentropic? (Hint: think about RPMs.)
Example Problem
Solution for part a
➢ V1 / V2 = rv
➢ rv = (displacement volume + clearance volume) / clearance volume)
➢ V1 = displacement volume + clearance volume
➢ V2 = clearance volume
3
V
=
(
r
1
)
V
=
2.2
m
➢ 1
v
2
Example Problem
Solution for part b
γ -1
➢ Eff. = 1 - [ 1 / ( rv ) ]
➢ rv = 12
➢ γ = 1.4 for air
➢ Eff. = 1 - [ 1 / (12 1.4 -1 ) ]
➢ Eff. = 0.629 or 63%
Example Problem
Solution for part c
➢ For T2 (through isentropic compression):
γ -1
T
=
(
r
) ( T1 )
➢ 2i
v
1.4 -1
T
=
(
12
) ( 323 K) = 872.72 K
➢ 2i
➢ Tavg = ( 872.72 K + 323 K ) / 2 = 597.86 ( about 600 K )
➢ γavg = ( at 600 K) = 1.376
1.376 -1
T
=
(using
γ
)
=
(
12
) ( 323 K) = 822.19 K
➢ 2
avg
Example Problem
Solution for part c (part deux)
➢ For P2 (through isentropic compression):
γ
P
=
P
(V
/
V
)
➢ 2 1 1 2
γ
P
=P
(
r
)
➢ 2 1 v
1.376
➢ P2 = 100 kPa ( 12 )
➢ P2 = 3054.59 kPa
Example Problem
Solution for part c (part three)
➢ For T3 (through constant volume heating):
➢ Qsupplied = Cv ( T3 - T2 )
➢ 1685 kJ/kg = Cv ( T3 - 822.19 K)
➢ Cv = ( at 822.19 K ) = ( roughly 800 K) = 0.8116 kJ/kgK
➢ T3i = ( 1685 + 667.372 ) / 0.8117 = 2898.08 K
➢ Tavg = ( 2898.08 K + 822.19 K ) / 2 = 1860.14 K
Example Problem
Solution for part c (part three)
➢ Cvavg = ( at 1860.14 K) = 0.9541
➢ T3 = ( 1685 + 784.46 ) / 0.9541 = 2588.26 K
➢ P3 = P2 ( T3 / T2 ) = 3054.59 ( 3.148 ) = 9615.87 kPa
Example Problem
Solution for part d
➢ Net heat supplied = Net work done
➢ Q1 - Q2 = Net Work
➢ 1685 kJ - 762.24 kJ = 922.76 kJ
Example Problem
Solution for part e
➢ Eff. = Net Work / Heat Supplied
➢ Eff. = 927.83 kJ / 1685 kJ
➢ Eff. = 0.550 or 55.0%
Example Problem
Solution for part f
➢ Δs = Cv (dT/T)
➢ Ignition stage: Δsi = 0.95366 Ln ( T3 / T2 )
➢ Δsi = 0.95366 Ln ( 2588.26 K
/ 822.19 K ) = 1.09363 kJ/kg K
➢ Exhaust stage: Δse = 0.8062 Ln ( T1 / T4 )
➢ Δse =0.8062 Ln ( 323 K / 1262.189 K ) = 1.09881 kJ/kg K
Example Problem
Solution for part f
➢ Δse - Δsi = Δsnet
➢ 1.09881 - 1.09363 = 0.00518 kJ/kgK
Questions and Comments?
Any Questions?