Unit IV Testing of IC Engines & Supercharging

Syllabus : IC Engines Unit-IV : testing of ic engines Objective of testing, various performance parameters for IC engines - indicated power, brake power, friction power, SFC, AF ratio etc, methods to determine various performance parameters, characteristic curves, heat balance sheet SUPERCHARGING Supercharging and turbo-charging methods and their limitations

Lecture No 23 Learning Objectives: To understand objectives of engine testing To learn about engine performance parameters

Objectives of Testing Engine performance during development Engine performance after development/ sample testing after production by manufacturers Engine performance testing by Govt Testing Agencies for certification like ARAI

Objectives of Testing Whether engine is performing as per design? Indicated Power (IP) Brake Power (BP) Frictional Power (FP) Mechanical Efficiency Thermal Efficiency Brake Specific Fuel Consumption (bsfc) Heat Balance Sheet Whether engine is meeting emission norms? CO HC NOx PM Soot

Objectives of Testing Leakage in engine? Smooth running of engine? Engine performance after overhaul/repairs?

Performance Parameters Brake Power (BP) Power available at output shaft/ crank shaft Mechanical Losses/Frictional Power (FP) Sum of frictional losses and pumping losses incl power required to operate engine accessories like water pump, dynamo etc Indicated Power (IP)/ Theoretical Power Power produced within eng cylinder IP = BP + FP

Tests Performed on IC Engines Indicated Power (IP) Brake Power (BP) Frictional Power (FP) Mechanical Efficiency Air-Fuel Ratio Brake Specific Fuel Consumption (bsfc) Thermal Efficiency Working out Heat Balance Sheet

Engine/ Mechanical Indicator Indicator Paper Wrapped Drum p-V diagram Stylus Rope connected To Piston Rod Pulleys Weight Piston Coupling Nut To Combustion Chamber

Measurement of IP on Mech/Eng Indicator To determine IP, p-V diagram is required, the area of which represents work developed by engine per cycle Apparatus used for drawing actual p-V diagram is called Mechanical/ Engine Indicator Eng indicator consists of a cylinder, piston, piston rod coupling nut, straight line linkage with stylus, spring of required stiffness, indicator card wrapped drum, pulley, rope and weights . Vertical movement of stylus and horizontal movement of the cord combines to produce a closed figure called Indicator diagram Area of indicator diagram can be measured by Planimeter to a definite scale giving work developed

Mean Effective Pressure Indicated Mean Effective Pressure (imep) imep is the average pressure, which if acted over the entire stroke length, would produce the same work done by the piston as is actually produced by the engine during a cycle Let ‘a’ be the net area of indicator diagram (cm2) ‘l’ be length of diagram (cm) and ‘k’ be spring stiffness N/cm2/cm Hence, mean height of diagram = a/l

Indicated Power (IP) Let pm= imep (N/cm2); L= Length of stroke (m) A= Piston top area (cm2)=Лd2/4; N= RPM n= Power stroke /min(=N/2 for 4 S eng as one power stroke per 2 rev & N for 2S eng) Hence, Force on Piston= Pm x A (Newton) WD per Cycle= pm x A x L (Nm) Hence, IP = pm x A x L x n (Nm/min)/Cycle/Cylinder

Lecture No 24 Learning Objectives: To learn working out/ measurement of Brake Power (BP) & Friction Power (FP)

Measurement of BP 1. Rope Brake Friction Dynamometer S W Spring Balance Rope Flywheel/Brake Drum W Weight

Brake Mean Effective Pressure (bmep) Rope Brake Friction Dynamometer (Contd.) Let W=Dead Weight (mg) in Newton (N) S=Spring Balance Reading (N) Rb=Radius of Brake Drum (D+d)/2 (m) D=Brake Drum dia and ‘d’ rope dia N=Engine RPM Hence, net Brake Load= (W – S) Braking Torque = (W-S) x Rb Hence, Brake Power (BP) = Brake Mean Effective Pressure (bmep)

2. Prony Brake Dynamometer Load Arm Flywheel/ Brake Drum Brake Shoes Length, L Weight W

Prony Brake Dynamometer (Contd.) Let W (=mg) be the weight (N) Let L be the distance from centre of brake drum to hanger, called load arm (m) Then, Torque=W x L (Nm)

Frictional Power (FP) Difference between IP and BP is called FP FP includes: - Pumping losses due to intake & exhaust processes - Frictional losses in bearings, rotary/sliding parts Power required to drive auxiliaries like governor, water, lub oil, fuel pumps, alternator/dynamo, valve operating mechanism etc FP increases as square of N but practically FP ∞ N1.6 Higher FP results in: - Reduced power output - Decreased mech efficiency - Increased bsfc - Increased requirement of cooling

Methods of Measurement of FP By measurement of IP and BP Willan’s Line Method Morse Test Motoring Test

FP by Willan’s Line Method ( Fuel Rate Extrapolation Method ) At Constant Eng Speed, say 1500 RPM 4 3 Fuel Flow Rate (kg/h) 2 1 A - 8 - 4 0 4 8 12 16 20 BP (kW)

FP by Willan’s Line Method A graph between fuel consumption rate (kg/h) taken on y-axis and BP (kW) on x-axis is drawn, while engine is made to run at some constant speed, say 1500 RPM The graph is extrapolated back to zero fuel consumption, which cuts on –ve x-axis at point ‘A’ The –ve intercept on x-axis represents FP at that speed of the engine Although when BP=0, some fuel consumption is there. This fuel is consumed to overcome engine friction Only for CI engine to be run at constant speed as Fuel consumption rate v/s BP plot is almost straight line in case of diesel engine, hence can be extrapolated

FP by Morse Test Morse Test can be used for determining FP/IP of multi-cylinder IC engines, generally 3 cyl and more by cutting off each cylinder in turn In SI engines, each cylinder is rendered in-operative by short-circuiting the SP or cutting off fuel supply in MPFI systems. In CI engines, fuel supply is cut off Consider 4 stroke, 4 cylinder SI engine coupled with dynamometer Engine is run at constant speed N throughout one set of test parameters, as FP ∞ N2 It is assumed that pumping & mech losses are same whether a cylinder is working or not Throttle position is kept fixed, however, to attain same speed N, load is decreased by dynamometer

FP by Morse Test Let B=BP of eng when all cylinders are working B1=BP of eng when Cylinder No 1 is cut off Similarly, B2=BP of eng when Cylinder No 2 is cut off B3=BP of eng when Cylinder No 3 is cut off B4=BP of eng when Cylinder No 4 is cut off Let I1, I2, I3 & I4 be the IPs developed by Cylinder Nos 1, 2, 3 & 4 respectively and their corresponding FPs be F1, F2, F3 & F4 Total BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4)

FP by Morse Test BP(B) = (I1+I2+I3+I4) - (F1+F2+F3+F4) Hence, B1=(I2+I3+I4) – (F1+F2+F3+F4) On subtracting; B – B1 = I1 Similarly, B – B2 = I2 B – B3 = I3 B – B4 = I4 On adding; IP = I1 + I2 + I3 + I4 = 4B – (B1+B2+B3+B4) B, B1, B2, B3 & B4 can be measured by Dynamometer, Hence IP can be calculated Therefore, FP = IP - BP

Lecture No 25 Learning Objectives: To understand working out of heat balance sheet To learn measurement of air/fuel consumption

Theoretical/ Air Std Efficiencies Otto Cycle: Diesel Cycle: Dual Cycle:

Some Definitions Thermal Efficiencies: i) Indicated Thermal Efficiency ii) Brake/Overall Thermal Efficiency Where mf is fuel consumed in kg/sec CV is Calorific Value of fuel in kJ/kg IP/BP is in kW

Some Definitions Mechanical Efficiency: Relative Efficiency : Defined as the ratio of Brake Thermal Efficiency to Air Standard Efficiency at same Compression Ratio(CR)

Some Definitions Volumetric Efficiency: Ratio of actual mass of charge inducted during suction stroke to mass of charge corresponding to swept volume of the engine at atm pr & temp Reduced Volumetric Efficiency causes reduction in Power Output Volumetric Efficiency puts a limit on the amt of fuel that can be burnt and hence on its power, since output of eng depends on amt of air inducted

Some Definitions (Brake) Specific Fuel Consumption (bsfc/sfc): bsfc is defined as the amount of fuel required to be supplied to eng to develop 1kW of power per hour at crankshaft Specific Output: BP per unit of piston displacement

Heat Balance Sheet Heat Balance Sheet is an account of heat released on combustion of fuel in the combustion chamber and its utilization in the engine To draw heat balance sheet, tests are carried out on engine, while it is run at some constant speed Heat Supplied: Heat Supplied = mf x CV (kJ/min) Where mf = mass flow rate of fuel (kg/min) CV = Calorific Value of fuel (kJ/kg)

Heat Balance Sheet Heat Expenditure/Utilization: 1) Heat Equivalent to BP: Heat Equivalent to BP = BP x 60 (kJ/min) 2) Heat Rejected to Cooling Water: Heat carried away by water =mwCpw(Two – Twi) kJ/min Where mw=cooling water circulation kg/min & Cpw=4.187 kJ/kgK 3) Heat carried away by Exhaust gases Heat carried by exh gases=mgCpg(Tge – Tsa) kJ/min Where mg=(ma+mf) =flue gases flow rate (kg/min) 4) Unaccounted Heat: By difference

Heat Balance Sheet Heat Supplied kJ/ min Heat Utilization 100 % Heat Utilization 100 a) Heat to BP=BPx60 Heat Supplied by comb of Fuel =mf x CV b) Heat to water =mwxCpw(Two – Twi) c) Heat carried away by exhaust gases =mgxCpg(Tge – Tsa) d) Heat Unaccounted (By difference) 100 Total 100

Volumetric Fuel Flow Meter(Burette Type) Fuel from Tank 3-Way Cock Start Start 200cc 100cc Stop Stop 3-Way Cock Fuel to Eng

Fuel Measurement Time required to supply given volume of fuel is noted Mass Flow Rate of Fuel Supply: Density of Fuel = Sp Gravity of fuel x Density of water This method does not give very accurate mass flow rate due to variation in density with temp

Gravimetric Fuel Flow Meter Fuel Tank A Fuel to Engine Valves B Flask Weighing Machine

Air Flow Meter Thermometer Air Surge Orifice Plate (A, Cd) Tank Manometer ΔH Air Intake to Eng

Measurement of Air Consumption by Air Flowmeter Surge tank is connected to intake side of the engine Manometer measures the pressure difference Vol Flow Rate Cd – Coeff of discharge for given orifice A – Orifice cross sectional area ΔHw-Head of water to be converted to air head

Lecture No 26 Learning Objectives: To learn about engine characteristic curves

SI Engine Characteristic Curves IP BP FP POWER (kW) Speed

SI Engine Characteristic Curves Lab tests carried out to determine eng performance During tests, throttle is kept full (full /rated load, max fuel consumption) and speed is varied by adjusting the brake load IP, BP, FP, bsfc, mechanical & volumetric efficiencies etc are worked out Same tests can be repeated at half load At rated output, max p-V diagram area, hence max imep; For given torque; power ∞ N

SI Engine Characteristic Curves IP increases when imep or speed or both increase IP initially increases faster with speed, if inlet conditions are kept constant However, after certain limit, rate of increase of IP reduces with speed due to reduction in vol efficiency as air/charge velocity increase results in inlet pr drop Mech losses increase with increase in speed(FP∞ N2) due to which increase in IP is off-set by steep increase in FP

SI Engine Characteristic Curves IP IP BP bsfc Mech Eff BP bsfc x Mech Efficiency Speed

SI Engine Characteristic Curves As FP ∞ N2, mech efficiency reduces due to steep increase in FP At lower speeds, due to lower charge velocity because of low piston speed, bsfc reduces since volumetric efficiency increases and mech efficiency also increases After certain speed, bsfc increases due to reduction in volumetric efficiency and increase in mech losses x Point x represents economical speed of eng for min fuel consumption

SI Engine Characteristic Curve Vol Efficiency Speed

SI Engine Characteristic Curve Volumetric Efficiency reduces with increase in speed due to increase in intake velocity resulting in drop of suction pressure Higher the speed, lesser the time available for induction of charge Suction valve fully opens only when pressure inside cylinder slightly below the surrounding pressure, thus reducing effective suction stroke

CI Engine Characteristic Curves BP Power bsfc bsfc Speed

CI Engine Characteristic Curves IP and BP increase with speed but due to steep increase in FP, IP and BP start coming down For bsfc curve, same reasons as in SI engine

Engine Characteristic Curves SI bsfc CI BP

CI Engine Characteristic Curve Stoichiometric Mixture Brake/ Overall Efficiency Lean Mixture Rich Mixture A/F Ratio

Lecture No 27 Learning Objectives: To understand working out of engine parameters through numerical problems

Q1. Obtain cylinder dimensions of a twin-cylinder, 2-S IC engine from the following data: Engine speed=4000RPM; Volumetric efficiency=77%; Mech Efficiency=75%; Fuel consumption=10 lit/hr; Sp. Gr. Of fuel=0.73; A/F ratio=18; Piston speed= 600m/min; imep=5 bar. Also, determine power output at STP conditions (p=101325 N/m2; Ta=25˚C; R for air=0.287 kJ/kgK) Solution: Cylinder Dimensions=? D & L Piston speed=2LN Since speed & N are given, L=? Now, to find out D=?

Solution(Contd):

Solution(Contd):

Solution(Contd):

Q2. A 6 cylinder gasoline engine operates on 4 stroke cycle. Bore of cylinder is 80mm and stroke 100mm. Clearance volume per cylinder is 70CC. At 4000 RPM, Fuel consumption is 20kg/hr and the torque developed is 150Nm. Calculate:- BP (b) Brake mean effective pressure (c) Brake thermal efficiency If CV of the fuel is 43000kJ/kg, find relative efficiency on brake power basis, assuming engine works on Constant volume cycle and gamma for air =1.4. Solution:

Lecture No 28 Learning Objectives: To understand working out of engine parameters and heat balance sheet through numerical problems

Q3. During trial of a single cylinder, 4 stroke oil engine the following results were obtained: Cyl bore=200mm, Stroke=400mm, mep=6 bar, Torque=407Nm, speed=250 RPM, Oil consumption=4kg/hr, CV of fuel=43MJ/kg, Cooling water rate=4.5kg/min, Air used per kg of fuel= 30kg, Rise in cooling water temp=45°C, Temp of Exhaust gases=420°C, Room temp=20°C, mean sp. heat of exhaust gases=1kJ/kgK, Sp. Heat of water= 4.18kJ/kgK, Barometric pressure=1.01325 bar Find IP, BP and draw up heat balance sheet in kJ/hr. Solution:

Heat Balance Sheet 1. Heat supplied by fuel to eng =mfxCV =4x43000kJ/hr=172,000kJ/h

2. Heat utilized (i) Heat to Power Output=BPx3600kJ/h =10.65x3600=38,358kJ/h (ii) Heat to cooling water=mw x Cpw x ∆Tw =4.5x60x4.18x45 = 50,787 kJ/h (iii) Heat to exhaust gases=mg x Cpg x (Te-Ta ) To find mg : ma=30kg/kg of fuel; hence mg=31kg Since fuel consumption is 4kg/h; mg=31x4kg/h Hence, Heat to exhaust gases=31x4x1(420-20) =49,600kJ/h (iv) Unaccounted Heat=33,255kJ/h (by difference)

Heat Balance Sheet Heat Supplied kJ/hr % Heat Utilized Heat supplied by fuel= mfxCV 172,000 100 To BP= BPx3600 38,358 22.3 To Cooling Water =mw.Cpw.∆Tw 50,787 29.5 To exhaust gases =mg.Cpg.∆Tg 49,600 28.8 Unaccounted Heat 33,255 19.3 Total

Q4. During a test on a 4 stroke oil engine, the following data were obtained: Mean height of indicator diagram = 21mm Indicator spring number/stiffness=27kN/m2/mm Swept volume=14 lit, effective brake load=77kg, Effective brake radius= 0.7m, speed=6.6 rev/s, fuel consumption=0.002kg/s, CV of fuel=44MJ/kg, Cooling water rate=0.15kg/s, water inlet temp=38°C, cooling water outlet temp=71°C, Sp. heat of water= 4.18kJ/kgK, energy carried by exhaust gases=33.6kJ/s Determine IP, BP and mech efficiency and draw up heat balance sheet in kJ/s and %. Solution:

Heat Balance Sheet Heat Supplied kJ/s % Heat Utilized Heat supplied by fuel= mf x CV = 0.002x44000 (=88kJ/s) 88 100 To BP 21.92 24.9 To Cooling Water =mw.Cpw.∆Tw 0.15x4.18x (71-38) 20.69 23.5 To exhaust gases (given) 33.6 38.2 Unaccounted Heat 11.79 13.4 Total

Q5. A 4 cylinder 4 stroke SI engine has a bore of 5.7cm and stroke 9cm. Its rated speed is 2800 RPM and it is tested at this speed against a brake which has a torque arm of 356mm. The net brake load is 155N and fuel consumption is 6.74 lit/hr. Sp gravity of petrol is 0.735 and CV is 44200kJ/kg. A Morse test is carried out and cylinders are cut off in order of 1, 2, 3 & 4 with corresponding brake loads of 111, 106.5, 104.2 and 111N. Determine engine torque, bmep, brake thermal efficiency, sfc, mech efficiency and imep. Solution: Engine torque= (W-S).Rb=WxL = 155x0.356=55.18Nm

We know that IP = 4BP - (BP1+BP2+BP3+BP4)

Q6. During trial of a 4 cylinder 4 stroke SI engine running at 50 rev/s, the brake load was 267N when all cylinders were working. When each cyl was cut off in turn and speed returned to same 50 rev/s, brake readings were 178N, 187N, 182N and 182N. Determine BP, IP and mech efficiency of the engine. For brake, BP=F.N/455(kW), where F is brake load in Newtons and N rev/s. The following results were obtained: Fuel consumption=0.568lit in 30 seconds, SG of fuel=0.72, CV=43000kJ/kg, A/F ratio=14:1, Exh temp=760°C, Cpg=1.015kJ/kg, Water inlet temp= 18°C and outlet temp=56°C, water flow rate=0.28kg/s, Ambient temp=21°C. Draw heat balance sheet in kJ/s. Solution:

We know that IP = 4BP - (BP1+BP2+BP3+BP4) Therefore IP= 4x29.34 - (19.56+20.55+20+20) =37.25kW

Heat Balance Sheet Heat supplied =mfxCV Heat utilized (i) Heat to BP=BP= 29.34kJ/s (5%) (ii) Heat to cooling water=mw x Cpw x ∆Tw =0.28x4.187x(56-18) =44.55 kJ/s (7.6%) (iii) Heat to exhaust gases=mg x Cpg x (Te-Ta ) To find mg :(ma +1)xmf=(14+1)x0.01363=0.204kg/s Hence, Heat to exhaust gases=0.204x1.015(760-21) =153kJ/s (26.1%) (iv) Unaccounted Heat=356kJ/s (61%) (by difference)

Supercharging & Turbocharging

Syllabus : IC Engines SUPERCHARGING Supercharging and turbo-charging methods and their limitations

Lecture No 29 Learning Objectives: To learn effects of supercharging /turbo-charging and limitations

How can engine power be increased? Increasing Eng speed (BP= T x 2πN) (FP ∞ N2 & Volumetric η ↓ ) Higher CR (Peak Pr increases; Thermal Load increases; Weight to Power ratio increases) (HUCR limited due to knocking/detonation in SI engines and heat load in CI engines) Utilization of exh energy in gas turbine, thus ↑ BP Use of 2-stroke cycle; but cooling, emission problems, lower volumetric & thermal efficiency Increasing charge density by - Lowering charge temp (Cooling) and / or - Increasing induction pressure

Objectives of Supercharging To increase power output To reduce bulk size of engine To increase power to weight ratio To compensate loss of power at high altitude

Supercharging Supplying air /Air-Fuel mixture at higher pressure than the pressure, at which the engine naturally aspirates, by a boosting device is called supercharging The device which boosts the pressure is called supercharger. Purpose of supercharging to have small displacement engines but developing more power and to meet emission legislation on fuel consumption for emission control More power is achieved by raising density of charge, thus more mass of air making available more oxygen for combustion

Supercharging & Turbo Charging Systems

Effects of Supercharging Increased Eng Output (p-V diagrams) Turbulence Effect (Higher BP) Power required to drive supercharger, thus ↓ BP Mech Efficiency increases bsfc ↑ for SI (due to reduced delay) but ↓ for CI engines due to better combustion & higher mechanical efficiency Better scavenging; Increase in power output SI Engines→ Knocking tendency as ign delay ↓ For CI Engines→ Smoother running, low F/A ratio, ↑ durability & reliability and lower bsfc

Effects of Supercharging Better atomization Better mixing of air and fuel Reduced exhaust smoke Better torque characteristic over whole speed range Better and smoother combustion Increased thermal stresses Increased gas load Increased valve overlap period of 60° to 160° of crank angle Increased cooling requirements of piston and valves

P-V Diagrams of Naturally Aspirated & Supercharged Engines 4 3 p p +(c) 4 3 5 6 2 +(a) 5 6 2 7 1 +(d) patm patm 7 1 -(b) V V Naturally Aspirated Engine Supercharged Engine

Limitations of Supercharging Power o/p limited by knock, thermal & mech loads For SI engines, knocking reached earlier In CI Engs, thermal & mech loads reached earlier Increase in intake pr increases peak pr leading to increase in weight of cylinder (limitation on peak pr) Increase in peak pr→ ↑ tendency to detonate (SI) Increase in peak pr increases friction losses Increase in peak pr, increases bearing loads ↑ peak pr → ↑ peak T →Reqmt of better cooling sys ↑T → ↑ exh gas temp →overheating of exh valves Due to the above reasons, supercharging generally limited to 2.5 bar

Limitations of Supercharging In SI Engs. Detonation is the limitation as it increases with ↑ pr, ↑ T, ↑ CR, ↑ density of charge Strongest detonation at stoichiometric A/F ratio CR limited due to detonation for given Octane Rating of fuel used Detonation can be reduced by reducing CR but BP & thermal efficiency decreases & bsfc increases For the above reasons, SI Engines are normally NOT supercharged except for aircraft, high altitude compensation or higher power of aero engines required at the time of take off of aircraft at the expense of higher fuel consumption

Limitations of Supercharging In CI Engs. Increased induction pr helps in suppressing knocking tendency, improve combustion, higher power output & thermal efficiency and hence can use lower Cetane fuels Supercharging is limited by: - peak pressure →mechanical loading - peak temp →thermal loading - thermal stresses developed - mean temp of cylinder walls - loads on bearings

Lecture No 30 Learning Objectives: To understand methods of supercharging To learn various methods of turbocharging

Types of Superchargers Centrifugal Type Supercharger Root’s Type Supercharger Vane Type Supercharger

Centrifugal Type Supercharger

Root’s Type Supercharger

Vane Type Supercharger

Vane Type Supercharger

Arrangements of Supercharging After Cooler Air outlet from Compressor Inlet to Engine Exhaust from Engine Compressor Gears Engine Load Air inlet to Compressor

Arrangements of Turbocharging Air inlet to Compressor Exhaust from Turbine Compressor Turbine After Cooler Air outlet from Compressor Exhaust from Engine Inlet to Engine Engine Load

Method of Super/Turbocharging After Cooler Air outlet from Compressor Exhaust from Engine Inlet to Engine Turbine Compressor Load Engine Exhaust from Turbine Air inlet to Compressor

Method of Super/Turbocharging Exhaust from Turbine After Cooler Load Turbine Air outlet from Compressor Inlet to Engine Exhaust from Engine Compressor Gears Engine Load Air inlet to Compressor

Turbochargers generated in the eng cylinder Exhaust gases carry about 1/3 of the total energy generated in the eng cylinder In order to utilize this energy, hot gases can be allowed to expand further in a gas turbine and its work output can be utilized to drive a supercharger. This system of supercharger coupled to Turbine is called Turbocharger Due to cyclic fluctuations of the pressure in exhaust pipe, turbo charging is not employed in single cylinder eng, however, system is suitable for engines having 4 or more cylinders

Turbo Charger Advantages Disadvantages Turbocharger does not consume eng power No gearing required between turbine and compressor as both are connected by single shaft Gain in power at nominal cost Exhaust energy, which is 1/3 of total energy generated in the engine, is gainfully utilized Exhaust noise level reduces Suitable for high speed engines Disadvantages Increase in fuel consumption at low power output Total cost of unit increases

End of Unit - IV