1 / 14

# Internal Combustion Engines

Internal Combustion Engines. Internal Combustion Engines. Ideal Diesel Cycle. Internal Combustion Engines. Ideal Diesel Cycle Ideal Gas Laws pV = mRT where p = absolute pressure (kPa) V = volume (m 3 ) m = mass (kg)

## Internal Combustion Engines

E N D

### Presentation Transcript

1. Internal Combustion Engines

2. Internal Combustion Engines • Ideal Diesel Cycle

3. Internal Combustion Engines • 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)

4. 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

5. Internal Combustion Engines • First Law of Thermodynamics 1Q2 = U2 – U1 + 1W2 Where 1Q2 = heat transfer = mc(p or v)(T2 – T1) 1U2 = 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)

6. Thermodynamic engine example: • 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

7. Internal Combustion Engines • 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)

8. 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

9. 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)

10. 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

11. Internal combustion engine example: • 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%

12. 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%

13. Practice Problem: • Calculate the engine torque and brake specific fuel consumption of problem 112.

14. Practice Problem: • 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

More Related