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M1A2 Abrams, with current and proposed 1500 hp turbine engine. Photos courtesy of United States Military Academy
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2. Piston-Cylinder Engines
4 Stroke Engine Process
*Intake Stroke
Compression Stroke
Power Stroke (Expansion)
Exhaust Stroke
* For Spark Ignition engines, intake is of an air/fuel mixture. For Diesel engines, intake is air only.
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3. Air Standard Analysis The following assumptions are made:
Air, an Ideal Gas, is the working fluid
Combustion is replaced with Heat Addition (see Chap 13 for details)
No exhaust and intake strokes – constant volume heat rejection
All processes are internally reversible
For Cold-Air Standard, Specific Heats are also assumed constant
For hot cycle, Cp is not constant, so EES will give direct property.
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4. Otto Cycle *Cycle Analysis: 4 Internally Reversible Processes:
Isentropic Compression
Constant Volume Heat Addition
Isentropic Expansion
Constant Volume Heat Rejection
* Sign Conventions (Work in negative, etc.) are sometimes changed for cycle applications
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5. Example problem on Otto cycle
The compression ration of an Otto cycle is 8. At the beginning of compression stroke
P=.1MPa, t=15C. Heat transfer to the air per cycle is 1800kJ/kg. Find the mep, efficiency
Net work output per cycle.
Example problem on Otto cycle
p1=100; t1=15; q23=1800
v1=volume(air,p=p1,t=t1)
v2=v1/8
s1=entropy(air,p=p1,t=t1)
p2=pressure(air,s=s1,v=v2)
wc=integral(pressure(air,s=s1,v=v),v,v1,v2)
t2=temperature(air,v=v2,s=s1)
u2=intenergy(air,t=t2)
q23=u3-u2; q14=u1-u4; u4=intenergy(air, t=t4); u1=intenergy(air, t=t1)
t3=temperature(air,u=u3); t4=temperature(air,s=s3,v=v1)
v3=v2; v4=v1
p3=pressure(air,t=t3,v=v3)
s3=entropy(air,v=v3,t=t3)
wt=-integral(pressure(air,s=s3,v=v),v,v3,v4)
p4=pressure(air,v=v1,s=s3)
wnet=wt+wc;
eff=wnet/q23;
mep=wnet/(v1-v2)
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6. Analysis & Performance Parameters
Air Standard Analysis
Cold Air Standard Analysis
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7. Diesel Cycles *Cycle Analysis: 4 Internally Reversible Processes:
Isentropic Compression
Constant Pressure Heat Addition
Isentropic Expansion
Constant Volume Heat Rejection

8. Analysis & Performance Parameters
Air Standard Analysis
Cold Air Standard Analysis
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9. Example problem An air standard Diesel cycle has compression
ratio of 18, and the heat transfer to the working
fluid is 1800kJ/kg. At the beginning of the
Compression process, the pressure is 0.1mPa,
And temp=15C. Find the net work output, mep and
Efficiency .
module work(s3,v3,v4 : w)
w=integral(pressure(air,s=s3,v=v),v,v3,v4)
end
p1=100; t1=15; q23=1800
v1=volume(air,p=p1,t=t1)
v2=v1/18
s1=entropy(air,p=p1,t=t1)
p2=pressure(air,s=s1,v=v2)
wc=integral(pressure(air,s=s1,v=v),v,v1,v2)
t2=temperature(air,v=v2,s=s1)
u2=intenergy(air,t=t2)
q23=w23+u3-u2; w23=p2*(v3-v2)
t3=temperature(air,u=u3);
p3=p2; v4=v1
v3=volume(air,p=p3,t=t3)
s3=entropy(air,v=v3,t=t3)
call work(s3,v3,v4 : wt)
wnet=wt+w23+wc;
eff=wnet/q23
mep=wnet/(v1-v2)
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10. Dual Cycle *Cycle Analysis: 5 Internally Reversible Processes:
Isentropic Compression
Constant Volume Heat Addition
Constant Pressure Heat Addition
Isentropic Expansion
Constant Volume Heat Rejection
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11. Gas Turbines Assumptions for the basic gas turbine cycle:
Air, as an ideal gas, is the working fluid throughout
Combustion is replaced with heat transfer from an external source
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12. The Brayton Cycle
Cycle Analysis:
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13. Ideal Brayton Cycle
Using Constant Specific Heats, the cycle thermal efficiency is:
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14. s1=entropy(air,p=p1,t=t1) h1=enthalpy(air, t=t1)
Example problem
To a gas turbine air enter the compressor at 0.1MPa, 15C and leaves the compressor
At 1MPa. Max cycle temp is 1100C. Comp efficiency is 0.8 and turbine efficiency is 0.85.
Pressure drop between comp and turbine is 15kpa. Determine comp work, turbine work
and cycle efficiency.
p1=100; t1=15; p2=1000; p2t= ; t3=1100
s1=entropy(air,p=p1,t=t1)
h1=enthalpy(air, t=t1)
h2s=enthalpy(air,p=p2,s=s1);
(-h1+h2s)/(-h1+h2)=0.8
h3=enthalpy(air,t=t3)
s3=entropy(air,p=p2t ,t=t3)
h4s=enthalpy(air,p=p1,s=s3)
(h3-h4)/(h3-h4s)=.85
wt=h3-h4
wc=h2-h1
eff=(wt-wc)/qh
qh=h3-h2
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15. Regenerative Gas Turbines
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16. Reheat and Intercooling
Reheat, regeneration, and intercooling are most effective when used in combination with one another. However, weight limitations (e.g. aircraft applications) often limit their usage.
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17. Aircraft Applications
This clip showing a T-62T-40-1 Auxiliary Power Unit from a US Army helicopter is provided courtesy of the U.S. Military Academy at West Point.
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18. Aircraft Applications
Turbo Jet Applications, with and without afterburner
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19. Aircraft Applications
Turboprop, Turbofan, and Ramjet applications
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20. Compressible Flows: The Basics
1-D Steady Flow Momentum Equation:
Velocity of Sound, isentropic wave, ideal gas:
Mach Number:
Stagnation Enthalpy:
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21. Nozzle & Diffuser Types
Cases 1 & 2 are Nozzles, Cases 3 and 4 are Diffusers
Converging-Diverging Nozzle
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