1. Internal Combustion Engines

2. Ideal Diesel Cycle Ideal Diesel Cycle

3. Internal Combustion Engines Ideal Diesel Cycle Ideal Diesel Cycle –Ideal Gas Laws  pV = mRT wherep = absolute pressure (kPa) V = volume (m 3 ) V = volume (m 3 ) m = mass (kg) m = mass (kg) R = air gas constant [kJ/(kg∙K)] R = air gas constant [kJ/(kg∙K)] = 8.314/29 = 8.314/29 T = absolute temperature (K) T = absolute temperature (K)

4. Internal Combustion Engines  p 1 V 1 n = p 2 V 2 n where n = 1.4 for ideal process = 1.3 for practical processes = 1.3 for practical processes  p 1 V 1 /T 1 = p 2 V 2 /T 2  T 2 = T 1 (V 1 /V 2 ) n-1  Useful relationships: –r = V 1 /V 2 = compression ratio –Displacement = (  ∙bore 2 /4)stroke = V 1 - V 2 = V 1 - V 2

5. Internal Combustion Engines –First Law of Thermodynamics 1 Q 2 = U 2 – U 1 + 1 W 2 Where 1 Q 2 = heat transfer = mc (p or v) (T 2 – T 1 ) 1 U 2 = internal energy = mc v (T 2 – T 1 ) 1 U 2 = internal energy = mc v (T 2 – T 1 ) 1 W 2 = work = (p 1 V 1 – p 2 V 2 /(n-1) = ∫pdV 1 W 2 = work = (p 1 V 1 – p 2 V 2 /(n-1) = ∫pdV c v = air specific heat @ constant volume c v = air specific heat @ constant volume = 0.718 kJ/(kg∙K) = 0.718 kJ/(kg∙K) c p = air specific heat @constant pressure c p = air specific heat @constant pressure = 1.005 kJ/(kg∙K) = 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 = 30 o C, and pressure = atmospheric ? What is work done during compression stroke of a diesel engine with r=16.5, intake temperature = 30 o C, and pressure = atmospheric ? 1 W 2 = work = (p 1 V 1 – p 2 V 2 /(n-1) T 2 = T 1 (V 1 /V 2 ) n-1 1 W 2 = work = (p 1 V 1 – p 2 V 2 /(n-1) T 2 = T 1 (V 1 /V 2 ) n-1 [Don’t have V’s. Can use the fact that for ideal adiabatic compression process, 1 W 2 = - 1 U 2 = - mc v (T 2 – T 1 )] [Don’t have V’s. Can use the fact that for ideal adiabatic compression process, 1 W 2 = - 1 U 2 = - mc v (T 2 – T 1 )] 1 W 2 = -1.0×0.718[(30+273.15) ×16.5 (1.4-1) – 303.15] 1 W 2 = -1.0×0.718[(30+273.15) ×16.5 (1.4-1) – 303.15] = - 450.3 kJ/kg = - 450.3 kJ/kg Also, p 2 = 101.3kPa×16.5 1.4 = 5129.4 kPa Also, p 2 = 101.3kPa×16.5 1.4 = 5129.4 kPa

7. Internal Combustion Engines Practical Power Production Practical Power Production –P fe = (HV∙ṁ f )/3600 (kW) = (HV∙ṁ f )/2545 (hp) = (HV∙ṁ f )/2545 (hp) where P fe = fuel equivalent power where P fe = fuel equivalent power HV = heating value of fuel HV = heating value of fuel = 45,500 kJ/kg = 19,560 BTU/lb = 45,500 kJ/kg = 19,560 BTU/lb ṁ f = mass fuel consumption (?/h) ṁ f = mass fuel consumption (?/h)

8. Internal Combustion Engines –P i = (imep D e N e )/(2×60,000) (kW) = (imep D e N e )/(2×396,000) (hp) = (imep D e N e )/(2×396,000) (hp) where imep = indicated mean effective pressure or mean pressure during compression and power strokes (kPa or psi) where imep = indicated mean effective pressure or mean pressure during compression and power strokes (kPa or psi) D e = engine displacement (L or in 3 ) D e = engine displacement (L or in 3 ) N e = engine speed (rpm) N e = engine speed (rpm) 396,000 = 33,000 ft∙lb/min∙hp x 12 in./ft 396,000 = 33,000 ft∙lb/min∙hp x 12 in./ft

9. Internal Combustion Engines –P b =(2  T e N e )/60,000 (kW) =2  T e N e /33,000 (hp) =2  T e N e /33,000 (hp) where P b = brake (engine) power where P b = brake (engine) power T e = engine torque (kJ or lb∙ft) T e = engine torque (kJ or lb∙ft) N e = engine speed (rpm) N e = engine speed (rpm)

10. Internal Combustion Engines  Efficiencies –Brake thermal efficiency e bt = (P b / P fe) ×100 e bt = (P b / P fe) ×100 –Brake specific fuel consumption BSFC = ṁ f /P b (kg/kW∙h) or (lb/hp∙h) BSFC = ṁ f /P b (kg/kW∙h) or (lb/hp∙h) (The above two efficiencies can be extended to PTO power by substituting PTO for Brake.) –Mechanical efficiency e m = (P b / P i ) × 100 e m = (P b / P i ) × 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. 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. P i = (imep D e N e )/(2×396,000) e m = P b /P i × 100 P i = (imep D e N e )/(2×396,000) e m = P b /P i × 100 P i = (125psi×505in 3 ×2200rpm)/(2×396,000)=175.3 hp P i = (125psi×505in 3 ×2200rpm)/(2×396,000)=175.3 hp P b = (2  ×358.1 lb∙ft×2200 rpm)/(33000 ft∙lb/min∙hp) P b = (2  ×358.1 lb∙ft×2200 rpm)/(33000 ft∙lb/min∙hp) = 150 hp = 150 hp e m = (150 hp /175.3 hp)×100 = 85.6% e m = (150 hp /175.3 hp)×100 = 85.6%

12. Problem 112 in practice problems: 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? 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? e bt = (P b / P fe) × 100 P fe = (HV∙ṁ f )/3600 e bt = (P b / P fe) × 100 P fe = (HV∙ṁ f )/3600 P fe = (45,400 kJ/kg × 37 l/h × 0.847 kg/l)/3600 P fe = (45,400 kJ/kg × 37 l/h × 0.847 kg/l)/3600 = 395.2 kW = 395.2 kW e bt = (135 kW/395.2 kW) × 100 e bt = (135 kW/395.2 kW) × 100 = 34.2% = 34.2%

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

14. Practice Problem: P b =(2  T e N e )/60,000 (kW) P b =(2  T e N e )/60,000 (kW) T e = 60,000 x 135/(2  x 2200) T e = 60,000 x 135/(2  x 2200) = 586 kJ = 586 kJ BSFC = (37 l/h × 0.847 kg/l)/135 kW BSFC = (37 l/h × 0.847 kg/l)/135 kW = 0.232 kg/kW = 0.232 kg/kW