Transcript Internal Combustion Engines - IQSoft Software Consultants

Internal Combustion Engines
Internal Combustion Engines

Ideal Diesel Cycle
Internal Combustion Engines
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Ideal Diesel Cycle
– Ideal Gas Laws
 pV = mRT where p = absolute pressure (kPa)
V = volume (m3)
m = mass (kg)
R = air gas constant [kJ/(kg∙K)]
= 8.314/29
T = absolute temperature (K)
Internal Combustion Engines
 p1V1n = p2V2n where n = 1.4 for ideal process
= 1.3 for practical processes
 p1V1/T1 = p2V2/T2
 T2 = T1(V1/V2)n-1
 Useful relationships:
– r = V1/V2 = compression ratio
– Displacement = (p∙bore2/4)stroke
= V1 - V2
Internal Combustion Engines
– First Law of Thermodynamics
1Q2 = U2 – U1 + 1W2
Where 1Q2 = heat transfer = mc(p or v)(T2 – T1)
1 U2
= internal energy = mcv(T2 – T1)
1W2
= work = (p1V1 – p2V2/(n-1) = ∫pdV
cv = air specific heat @ constant volume
= 0.718 kJ/(kg∙K)
cp = air specific heat @constant pressure
= 1.005 kJ/(kg∙K)
Thermodynamic engine example:
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What is work done during compression stroke of a diesel
engine with r=16.5, intake temperature = 30oC, and
pressure = atmospheric ?
1W2
= work = (p1V1 – p2V2/(n-1)
T2 = T1(V1/V2)n-1
[Don’t have V’s. Can use the fact that for ideal adiabatic
compression process, 1W2 = - 1U2 = - mcv(T2 – T1)]
1W2
= -1.0×0.718[(30+273.15) ×16.5(1.4-1) – 303.15]
= - 450.3 kJ/kg
Also, p2 = 101.3kPa×16.51.4 = 5129.4 kPa
Internal Combustion Engines
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Practical Power Production
– Pfe = (HV∙ṁf )/3600 (kW)
= (HV∙ṁf )/2545 (hp)
where Pfe = fuel equivalent power
HV = heating value of fuel
= 45,500 kJ/kg = 19,560 BTU/lb
ṁf = mass fuel consumption (?/h)
Internal Combustion Engines
– Pi = (imep De Ne)/(2×60,000) (kW)
= (imep De Ne)/(2×396,000) (hp)
where imep = indicated mean effective
pressure or mean pressure
during compression and
power strokes (kPa or psi)
De = engine displacement (L or in3)
Ne = engine speed (rpm)
396,000 = 33,000 ft∙lb/min∙hp x 12 in./ft
Internal Combustion Engines
– Pb=(2p Te Ne)/60,000 (kW)
=2p Te Ne/33,000 (hp)
where Pb= brake (engine) power
Te= engine torque (kJ or lb∙ft)
Ne = engine speed (rpm)
Internal Combustion Engines
 Efficiencies
– Brake thermal efficiency
ebt = (Pb/ Pfe) ×100
– Brake specific fuel consumption
BSFC = ṁf/Pb (kg/kW∙h) or (lb/hp∙h)
(The above two efficiencies can be extended to PTO
power by substituting PTO for Brake.)
– Mechanical efficiency
em = (Pb/ Pi) × 100
Internal combustion engine example:
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Calculate the mechanical efficiency of an engine if the
indicated mean effective pressure is 125 psi,
displacement is 505 cubic inches, speed is 2200 rpm and
torque is 358.1 lb ft.
Pi = (imep De Ne)/(2×396,000)
em = Pb/Pi × 100
Pi = (125psi×505in3×2200rpm)/(2×396,000)=175.3 hp
Pb = (2p×358.1 lb∙ft×2200 rpm)/(33000 ft∙lb/min∙hp)
= 150 hp
em = (150 hp /175.3 hp)×100 = 85.6%
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Problem 112 in practice problems:
Fuel consumption = 37 l/h of #2 diesel fuel, brake
power is 135 kW, and operating speed is 2200 rpm.
What is the brake thermal efficiency?
ebt = (Pb/ Pfe) × 100
Pfe = (HV∙ṁf)/3600
Pfe = (45,400 kJ/kg × 37 l/h × 0.847 kg/l)/3600
= 395.2 kW
ebt = (135 kW/395.2 kW) × 100
= 34.2%
Practice Problem:
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Calculate the engine torque and brake specific
fuel consumption of problem 112.
Practice Problem:
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Pb=(2p Te Ne)/60,000 (kW)
Te = 60,000 x 135/(2p x 2200)
= 586 kJ
BSFC = (37 l/h × 0.847 kg/l)/135 kW
= 0.232 kg/kW