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
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