Internal Combustion Engines

Ideal Diesel Cycle Ideal Diesel Cycle

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)

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

Internal Combustion Engines –First Law of Thermodynamics 1 Q 2 = U 2 – U 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 constant volume c v = air specific constant volume = kJ/(kg∙K) = kJ/(kg∙K) c p = air specific pressure c p = air specific pressure = kJ/(kg∙K) = kJ/(kg∙K)

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[( ) ×16.5 (1.4-1) – ] 1 W 2 = -1.0×0.718[( ) ×16.5 (1.4-1) – ] = kJ/kg = kJ/kg Also, p 2 = 101.3kPa× = kPa Also, p 2 = 101.3kPa× = kPa

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)

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

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)

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

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 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 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%

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 × kg/l)/3600 P fe = (45,400 kJ/kg × 37 l/h × kg/l)/3600 = kW = kW e bt = (135 kW/395.2 kW) × 100 e bt = (135 kW/395.2 kW) × 100 = 34.2% = 34.2%

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.

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