internal combustion engines. ideal diesel cycle ideal diesel cycle

Internal Combustion Internal Combustion EnginesEngines


Ideal Diesel CycleIdeal Diesel Cycle


Ideal Diesel CycleIdeal Diesel Cycle– Ideal Gas LawsIdeal Gas Laws

pV = mRT wherepV = mRT where p = absolute pressure (kPa)p = absolute pressure (kPa)

V = volume (mV = volume (m33))

m = mass (kg)m = mass (kg)

R = air gas constant R = air gas constant [kJ/(kg[kJ/(kg∙K∙K)])]

= 8.314/29= 8.314/29

T = absolute temperature (K)T = absolute temperature (K)


pp11VV11nn = p = p22VV22

nn where n = 1.4 for ideal process where n = 1.4 for ideal process = 1.3 for practical = 1.3 for practical

processesprocesses pp11VV11/T/T11 = p = p22VV22/T/T22

TT22 = T = T11(V(V11/V/V22))n-1n-1

Useful relationships:Useful relationships:

– r = Vr = V11/V/V22 = compression ratio = compression ratio

– Displacement = (Displacement = (∙∙borebore22/4)stroke/4)stroke

= = VV1 1 - V- V22


– First Law of ThermodynamicsFirst Law of Thermodynamics

11QQ22 = U = U22 – U – U11 + + 11WW22

Where Where 11QQ22 = heat transfer = mc = heat transfer = mc(p or v)(p or v)(T(T22 – T – T11))

11UU22 = internal energy = mc = internal energy = mcvv(T(T22 – T – T11))

11WW22 = work = (p = work = (p11VV11 – p – p22VV22/(n-1) = /(n-1) = ∫pdV∫pdV

ccvv = air specific heat @ constant volume = air specific heat @ constant volume

= 0.718 kJ/(kg∙K)= 0.718 kJ/(kg∙K)

ccpp = air specific heat @constant = air specific heat @constant

pressure pressure

= 1.005 kJ/(kg∙K)= 1.005 kJ/(kg∙K)

Thermodynamic engine example:Thermodynamic engine example:

What is work done during compression stroke of a What is work done during compression stroke of a diesel engine with r=16.5, intake temperature = diesel engine with r=16.5, intake temperature = 3030ooC, and pressure = atmospheric ?C, and pressure = atmospheric ?

11WW22 = work = (p = work = (p11VV11 – p – p22VV22/(n-1) T/(n-1) T22 = T = T11(V(V11/V/V22))n-1n-1

[Don’t have V’s. Can use the fact that for ideal adiabatic [Don’t have V’s. Can use the fact that for ideal adiabatic compression process, compression process, 11WW22 = - = - 11UU2 2 = - mc= - mcvv(T(T22 – T – T11)])]

11WW22 = -1.0 = -1.0××0.718[(30+273.15) 0.718[(30+273.15) ××16.516.5(1.4-1)(1.4-1) – 303.15] – 303.15] = - 450.3 kJ/kg= - 450.3 kJ/kg

Also, pAlso, p22 = 101.3kPa = 101.3kPa××16.516.51.41.4 = 5129.4 kPa = 5129.4 kPa


Practical Power ProductionPractical Power Production– PPfefe = (HV = (HV∙ṁ∙ṁff)/3600 (kW) )/3600 (kW)

= (HV= (HV∙ṁ∙ṁff)/2545 (hp))/2545 (hp)

where Pwhere Pfefe = fuel equivalent power = fuel equivalent power

HV = heating value of fuelHV = heating value of fuel

= 45,500 kJ/kg = 19,560 = 45,500 kJ/kg = 19,560 BTU/lbBTU/lb

ṁṁff = mass fuel consumption = mass fuel consumption (?/h)(?/h)


– PPii = (imep D = (imep Dee N Nee)/(2)/(2××60,000) (kW)60,000) (kW)

= (imep D= (imep Dee N Nee)/(2)/(2××396,000) (hp)396,000) (hp)

where imep = indicated mean effective where imep = indicated mean effective pressure or mean pressure or mean pressure pressure during compression during compression and and power strokes (kPa or psi) power strokes (kPa or psi)

DDee = engine displacement (L or in = engine displacement (L or in33))

NNee = engine speed (rpm) = engine speed (rpm)

396,000 = 33,000 396,000 = 33,000 ft∙lb/min∙hpft∙lb/min∙hp x 12 x 12 in./ftin./ft


– PPbb=(2=(2 T Te e NNee)/60,000 (kW) )/60,000 (kW)

==22 T Te e NNee/33,000 (hp)/33,000 (hp)

where where PPbb= brake (engine) power= brake (engine) power

TTee= engine torque (kJ or lb= engine torque (kJ or lb∙∙ft)ft)

NNee = engine speed (rpm) = engine speed (rpm)

Internal Combustion Internal Combustion EnginesEngines EfficienciesEfficiencies

– Brake thermal efficiency Brake thermal efficiency

eebtbt = (P = (Pbb/ P/ Pfe)fe) ××100100

– Brake specific fuel consumptionBrake specific fuel consumption

BSFC = BSFC = ṁṁff/P/Pbb (kg/kW (kg/kW∙∙h) or (lb/hph) or (lb/hp∙∙h)h)

(The above two efficiencies can be extended to (The above two efficiencies can be extended to PTO power by substituting PTO for Brake.)PTO power by substituting PTO for Brake.)

– Mechanical efficiencyMechanical efficiency

eemm = (P = (Pbb/ P/ Pii) ) ×× 100 100

Internal combustion engine example:Internal combustion engine example: Calculate the mechanical efficiency of an engine if the Calculate the mechanical efficiency of an engine if the

indicated mean effective pressure is 125 psi, indicated mean effective pressure is 125 psi, displacement is 505 cubic inches, speed is 2200 rpm displacement is 505 cubic inches, speed is 2200 rpm and torque is 358.1 lb ft.and torque is 358.1 lb ft.

PPi i = (imep D= (imep Dee N Nee)/(2)/(2××396,000) e396,000) emm = P = Pbb/P/Pi i ×× 100 100

PPi i = (125psi= (125psi××505in505in33××2200rpm)/(22200rpm)/(2××396,000)=175.3 396,000)=175.3 hphp

PPbb = (2 = (2×358.1 lb∙ft×2200 rpm)/(33000 ft∙lb/min∙hp)×358.1 lb∙ft×2200 rpm)/(33000 ft∙lb/min∙hp) = 150 hp= 150 hp

eemm = (150 hp /175.3 hp) = (150 hp /175.3 hp)××100 = 85.6%100 = 85.6%

Problem 112 in practice problems:Problem 112 in practice problems:

Fuel consumption = 37 l/h of #2 diesel fuel, Fuel consumption = 37 l/h of #2 diesel fuel, brake power is 135 kW, and operating speed is brake power is 135 kW, and operating speed is 2200 rpm. What is the brake thermal efficiency?2200 rpm. What is the brake thermal efficiency?

eebtbt = (P = (Pbb/ P/ Pfe) fe) ×× 100 P 100 Pfefe = (HV = (HV∙ṁ∙ṁff)/3600)/3600

PPfefe = (45,400 kJ/kg = (45,400 kJ/kg ×× 37 l/h 37 l/h ×× 0.847 kg/l)/3600 0.847 kg/l)/3600

= 395.2 kW= 395.2 kW

eebtbt = (135 kW/395.2 kW) = (135 kW/395.2 kW) ×× 100 100

= 34.2%= 34.2%

Practice Problem:Practice Problem:

Calculate the engine torque and brake Calculate the engine torque and brake specific fuel consumption of problem 112.specific fuel consumption of problem 112.

Practice Problem:Practice Problem:

PPbb=(2=(2 T Tee N Nee)/60,000 (kW))/60,000 (kW)

TTee = 60,000 x 135/(2 = 60,000 x 135/(2x 2200)x 2200)

= 586 kJ= 586 kJ

BSFC = (BSFC = (37 l/h 37 l/h ×× 0.847 kg/l)/135 kW 0.847 kg/l)/135 kW

= 0.232 kg/kW= 0.232 kg/kW

internal combustion engines. ideal diesel cycle ideal diesel cycle

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