Copyright © John Wiley & Sons Ltd. M1A2 Abrams, with current and proposed 1500 hp turbine engine. Photos courtesy of United States Military Academy Copyright © John Wiley & Sons Ltd.

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. Copyright © John Wiley & Sons Ltd.

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. Copyright © John Wiley & Sons Ltd.

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 Copyright © John Wiley & Sons Ltd.

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) Copyright © John Wiley & Sons Ltd.

Analysis & Performance Parameters Air Standard Analysis Cold Air Standard Analysis Copyright © John Wiley & Sons Ltd.

Diesel Cycles *Cycle Analysis: 4 Internally Reversible Processes: Isentropic Compression Constant Pressure Heat Addition Isentropic Expansion Constant Volume Heat Rejection

Analysis & Performance Parameters Air Standard Analysis Cold Air Standard Analysis Copyright © John Wiley & Sons Ltd.

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) Copyright © John Wiley & Sons Ltd.

Dual Cycle *Cycle Analysis: 5 Internally Reversible Processes: Isentropic Compression Constant Volume Heat Addition Constant Pressure Heat Addition Isentropic Expansion Constant Volume Heat Rejection Copyright © John Wiley & Sons Ltd.

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 Copyright © John Wiley & Sons Ltd.

The Brayton Cycle Cycle Analysis: Copyright © John Wiley & Sons Ltd.

Ideal Brayton Cycle Using Constant Specific Heats, the cycle thermal efficiency is: Copyright © John Wiley & Sons Ltd.

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=1000-15; 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 Copyright © John Wiley & Sons Ltd.

Regenerative Gas Turbines Copyright © John Wiley & Sons Ltd.

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. Copyright © John Wiley & Sons Ltd.

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. Copyright © John Wiley & Sons Ltd.

Aircraft Applications Turbo Jet Applications, with and without afterburner Copyright © John Wiley & Sons Ltd.

Aircraft Applications Turboprop, Turbofan, and Ramjet applications Copyright © John Wiley & Sons Ltd.

Compressible Flows: The Basics 1-D Steady Flow Momentum Equation: Velocity of Sound, isentropic wave, ideal gas: Mach Number: Stagnation Enthalpy: Copyright © John Wiley & Sons Ltd.

Nozzle & Diffuser Types Cases 1 & 2 are Nozzles, Cases 3 and 4 are Diffusers Converging-Diverging Nozzle Copyright © John Wiley & Sons Ltd.