CV-2-Power Plant.pptx

1. Chapter-II Combat Vehicle Power Plants 1 CV Power Plants - Prof (Col) GC Mishra,Retd

2. Introduction This chapter discusses: 1. Requirements of power plants for tanks; 2. Peculiarities; 3. Analysis of availability of space; 4. Power to weight ratio; 5. Layout of power plants in various MBTs; 6. Efficiency and performance calculations 2 CV Power Plants - Prof (Col) GC Mishra,Retd

3. 3 CV Power Plants - Prof (Col) GC Mishra,Retd

5. 5 CV Power Plants - Prof (Col) GC Mishra,Retd REQMT OF AN AFV ENGINE  ADEQUATE POWER  COMPACTNESS  RELIABILITY & EASE OF MAINT  LOW FUEL CONSUMPTION – ENDURANCE  GOOD TORQE/SPEED CHARACTERSTICS  EASE OF STARTING  HIGH POWER TO WT RATIO  IMMUNITY FROM DUST & WATER  EXTREME INVIRONMENTAL CAPABILITY  EXTREME ANGLE OF TILT

6. 6 CV Power Plants - Prof (Col) GC Mishra,Retd REQMT OF AN AFV ENGINE contd…..  SILENCE  FREEDOM FROM TOXIC SMOKE  FREEDOM FROM RADIO INTERFERENCE  LOW FIRE RISK  EASE OF PRODUCTION  LOW THERMAL SIGNATURE  LOW INITIAL COST  STADARDISATION OF ASSEMBALIES  EASY PRESERVATION  EASIER WATER PROOFING

7. 7 CV Power Plants - Prof (Col) GC Mishra,Retd REQUIREMENTS OF AFV POWER PLANTs To Produce adequate POWER for:- 1. Steady speed on the level. 2. Steady speed uphill or downhill. 3. Required acceleration or deceleration. How much is the power required for the above requirements? How to find out the same?

8. CV Power Plants - Prof (Col) GC Mishra,Retd 8                1 , 60 2 const power same the for Therefore that implies This Power  1. At minimum load/resistance, torque required is minimum and the speed is maximum. 2. At maximum load/resistance, torque required is maximum and the speed is minimum. Speed Vs Torque Concept of mass, force, work done/energy and power

9. 9 CV Power Plants - Prof (Col) GC Mishra,Retd Power Transmission by the Shaft of a Car

11. Automotive Performance of an AFV depends upon:- 1. Power to Weight ratio. 2. Ground Pressure: NGP & MMP. 3. Steerability ratio. 4. Pitch ratio. 5. Running gear. 6. Other factors- W, H, B, L, L0, Turret ring diameter

12. Power to Weight (P/W) Ratio. 1. Gross Power. (a)Maximum power produced at the test bed with all accessories driven by outside power. (b)Called corrected power if measured at STP. 2. Net Power. Maximum power produced at the test bed with all accessories driven by own power. 3. Installed Power. Power produced at the flywheel. 4. Sprocket Power. = Installed power – Transmission losses (upto 185 KW (250 hp) ≈ 40% lesser than Gross power.

13. POWER TO WEIGHT RATIO (P/W) P/W Ratio = Gross Power/Combat weight of the vehicle (kW/Ton). Ideally it should be sprocket power to mass of the tank. # P/W ratio is an indication of :- a. Speed up the gradient (W Sin θ- max 300 in general (350 for AMX-30)). b. Speed/acceleration on level ground; a function of:- i. Drag ii. Weight iii.Deformation of tyres/tracks iv.Deformation of soil. c. Speed in steer; depends upon:- i. Friction forces encountered by track with ground in generating slewing couple. ii. Weight, length and design of the tracks.

14. POWER TO WEIGHT RATIO In General, 1. Air-cooled engines consume more power than water cooled engines. 2. Losses in hydraulic transmission are higher (≈25%) than mechanical Transmission (Tx) (≈10%). 3. Epicyclic tx losses are lesser due to less churning of oil. 4. The total number of gear trains between the engine and the sprocket would determine the tx losses for determining the tx efficiency.

15. POWER TO WEIGHT RATIO Tank Power(HP) Combat Weight (ton) P/W (HP/T) AMX40(French) 1100 43 25.58 Challenger (UK) 1500 62.5 24 Leopard-II (Germany 1500 62 24.19 Merkava-IV (Israel) 1500 65 23 M1A2 (USA) 1500 63 23.8 Leclerc (French) 1500 56.5 26.55 T-72 (Russia) 780 41.5 19.02 T-80 1200 51 21.7 T-90 1000 46.5 21.5 S-Tank (Sweden) 523 39.7 13.2 Type-90 (Japan) 1500 50 30 Arjun (India) 1400 58.5 24 Al Khalid (Pakistan) 1200 48 25

16. 16 CV Power Plants - Prof (Col) GC Mishra,Retd REQUIREMENT OF POWER FROM POWER PLANTs When a vehicle is moving at a uniform speed, the driving force, or tractive effort, at the wheels must be such as to exactly balance the sum of all variable forces tending to oppose the motion. The three forces which oppose the motion of the vehicle are: 1. Aerodynamic, or air, resistance; 2. Gradient resistance, which can be either positive or negative; and 3. Rolling resistance.

17. 17 CV Power Plants - Prof (Col) GC Mishra,Retd Air offers a resistance to the passage of bodies through it, as does water or any other fluid. The magnitude of this resistance is dependent directly upon the shape and frontal area of the body exposed to the fluid it is passing through, and to the square of its velocity. 1. Aerodynamic forces

18. 18 CV Power Plants - Prof (Col) GC Mishra,Retd 1. Aerodynamic forces Air Resitance (AR) AR = ½ ρ A V2 Cd A = Frontal Area, m2; ρ = Air Density, Kg/ m3; V = Veh Speed, m/sec Cd = Drag Coeff (function of shape/size /fitment items ; streamlining ) = 0.38 TO 0.48 for private vehicles =0.7 TO 1 for military & commercial vehs AR = 1/2 of total resistance for a pvt car at 80 km/hr on level road = 15% of total resistance for large truck = Negligible for AFVs, since densely packed & slow, (insignificant compared to RR & weight)

19. 19 CV Power Plants - Prof (Col) GC Mishra,Retd The weight of the car, acting vertically downwards, can be resolved into two components: H parallel to the slope and K perpendicular to the slope. 2. Gradient Resistance (GR) GR, H = W SIN α α = 450, Max on dry road = 300 , Wheeled Vehicles = 380 , Tracked Vehicles ≤18.50, Metal Roads  To prevent the car from rolling downwards, an additional force equal and opposite to H has to be applied by the wheels at their contact with the road surface.  It follows that the gradient resistance H is dependent solely on the steepness of the slope and is unaffected by the speed of the car, provided it is constant, either up or down the gradient.

21. 3. Rolling resistance (RR)  It is the resistance encountered by a vehicle when pushed slowly over a level ground.  Following main factors contribute to the same:- 1. Bearing friction 2. Rubbing of seals. 3. Sliding and rubbing of gears. 4. Distortion of tyre and tracks. 5. Distortion of soil (negligible for metal road).  It is usual, unless extreme accuracy is required, to take rolling resistance on a good road as being directly proportional to load.

22. Rough Guideline Values for Rolling Resistance  Military Wheeled Vehicles - 2 to 2.5% of laden wt  Tracked Vehicles( slow speed ) - 4 to 5%  Tracked Vehicle( high speed ) - 8% Note:- Add 2% for firm turf ground (neat print of grousers). Add additional 8 to 18% for ploughed fields. (In these cases vehicle is supposed to run at slow speed)

23. Total resistance = AR+GR+RR  AR is dependent on the vehicle speed; GR and RR are independent of the speed of the vehicle.  If either the gradient resistance or the rolling resistance increases or decreases then the curve would simply shift up or down by the amount of the increase or decrease.  curves A, B and C are the curve shifted up by various amounts.  Thus when the speed is OS km/h the total resistance SP is composed of the rolling resistance SR, the gradient resistance RQ and the air resistance QP.

24. CV Power Plants - Prof (Col) GC Mishra,Retd 24 EXAMPLE-1 Determine the Power required for an AFV 1. At the sprocket and, 2. At the flywheel for the following GSQR parameters:- a. Laden Weight of the AFV, W ≤ 50 Ton b. Maximum Speed desired v = 70 kmph on level metal ground c. Tank be able to negotiate firm turf ground as well as ploughed field at speed of 10 kmph. d. Maximum gradient capability = 300 at a speed of 10 kmph e. Transmission efficiency = 80% Assume Air Resistance to be negligible.

25. CV Power Plants - Prof (Col) GC Mishra,Retd 25 EXAMPLE-1 Solved (a) Power requirement for high speed of 70 kmph on level metal ground. RR = 8% of the laden weight of the vehicle = 8% of (50 Ton × g) = = 39.240 kN GR = 0 for level ground; AR = 0 negligible as given Total Resistance to motion or, Tractive effort required to overcome the same T = RR + GR + AR = 39.24 KN Therefore, power required at the sprocket; P = T × v = 39.24 kN × 70 kmph =  N 81 . 9 50000 100 8    kW W or s Nm s m N 763 , 763000 / 60 60 1000 70 1000 24 . 39       contd………

26. CV Power Plants - Prof (Col) GC Mishra,Retd 26 EXAMPLE-1 Solved (a) Power requirement for Gradient and x-country run (low speed of 10 kmph). RR = 5% of the laden weight of the vehicle + 2% of the laden weight of the vehicle for firm turf ground + 18% of the laden weight of the vehicle for ploughed field = 25% of (50 Ton × g) = GR = W Sin θ = (50 Ton × g)× Sin 300 N = AR = 0 negligible as given Total Resistance to motion or, Tractive effort required to overcome the same T = RR + GR + AR = 122.6 + 245.25 = 367.85 KN contd………   kN N N 6 . 122 122600 81 . 9 50000 100 25      kN N N 25 . 245 245250 2 1 81 . 9 1000 50           

27. CV Power Plants - Prof (Col) GC Mishra,Retd 27 EXAMPLE-1 Solved end……… Therefore, power required at the sprocket; P We need to consider the higher value of the power requirement. Therefore we take the higher of the two values i.e., 1021.5 kW- this is the Sprocket Power.   kW W s m N kmph kN v T 5 . 1021 1021500 / 60 60 1000 10 1000 85 . 367 10 85 . 367            hp power horse kW or kW on transmissi power sprocket r EnginePowe 5 . 1712 746 . 0 5 . 1277 , 5 . 1277 8 . 0 5 . 1021             

28. CV Power Plants - Prof (Col) GC Mishra,Retd 28 TUTORIAL EXERCISE-1 Determine the Power required for an AFV 1. At the sprocket and, 2. At the flywheel for the following parameters:- a. Laden Weight of the AFV, W = 41.5 Ton b. Maximum Speed desired v = 60 kmph on level metal ground c. Tank be able to negotiate firm turf ground as well as ploughed field at speed of 10 kmph. d. Maximum gradient capability = 300 at a speed of 10 kmph e. Transmission efficiency = 80% Assume Air Resistance to be negligible.

29. CV Power Plants - Prof (Col) GC Mishra,Retd 29 TUTORIAL EXERCISE-2 Determine the Power required for an Infantry Combat vehicle(ICV) 1. At the sprocket and, 2. At the flywheel for the following parameters:- a. Laden Weight of the AFV, W = 13.3 Ton b. Maximum Speed desired v = 65 kmph on level metal ground c. Tank be able to negotiate firm turf ground as well as ploughed field at speed of 10 kmph. d. Maximum gradient capability = 300 at a speed of 20 kmph e. Transmission efficiency = 80% Assume Air Resistance to be negligible.

30. CV Power Plants - Prof (Col) GC Mishra,Retd 30 Efficiency and Performance Calculation Of Power Plants

31. CV Power Plants - Prof (Col) GC Mishra,Retd 31   CV m Q W s b b b           2 Supplied Heat of Rate Power Brake efficiency Overall Efficiency Thermal Brake . 1    s i i i Q W Supplied Heat of Rate Power Indicated Efficiency Thermal Indicated . 2     ρ v NTP at drawn be should which mass l Theoritica cylinder into drawn air of mass Actual Efficiency Volumetric . 3 s  v v   Engine Efficiencies and Performance CV-calorific value of fuel

32. CV Power Plants - Prof (Col) GC Mishra,Retd 32 i b mech mech W W    power Indicated power Brake produced power Actual output Power Efficiency Mechanical . 4   vehicles diesel for hr kg/kW 0.2 vehicles petrol for hr kg/kW 0.3 hour per power brake of kW per consumed fuel of Mass bsfc bsfc n, Consumptio Fuel Specific Break . 5   

33. SPECIFIC FUEL CONSUMPTION 0 0.05 0.1 0.15 0.2 0.25 0.3

34. CV Power Plants - Prof (Col) GC Mishra,Retd 34 EXERCISE-2 What is the brake thermal efficiency in the following case? An engine develops 22.4 kW and consumes 10.25 lit of fuel per hour. The calorific value of fuel is 42000 kJ/kg and the specific gravity of the fuel is 0.72     % 75 . 23 , 2375 . 0 3 . 94 4 . 22 3 . 94 46000 1000 72 . 0 sec 3600 1 1000 25 . 10 sec sec kg CV Q supplied, heat of Rate given is W power, Brake kJ/sec kJ/sec or, Q supplied, heat of Rate W power, Brake 3 3 s s b b or kW kW kW kg kJ m kg m lit kW kJ kg kJ m kW b b                                 

35. CV Power Plants - Prof (Col) GC Mishra,Retd 35 EXERCISE-3  An engine develops its maximum rated torque 504 Nm at 2500 RPM. It consumes diesel at the rate of 36.4 liter per hour.  Diesel gives out energy of 42800 kJ/Kg and has specific gravity of 0.835.  Calculate:- 1. Power developed by the engine. 2. The overall efficiency (brake thermal efficiency). 3. Brake specific fuel consumption (bsfc)

36. CV Power Plants - Prof (Col) GC Mishra,Retd 36 Solution of EXERCISE-2       kg/kWhr 23 . 0 kW 95 131 1000 835 . 0 1 1000 / 4 . 36 hour per kW per consumed fuel of mass bsfc 3. 37% or, 37 . 0 kg kJ 42800 1000 835 . 0 sec 3600 1000 / 4 . 36 95 . 131 Efficiency Overall 2. kW 131.95 is that sec / 95 . 131 sec / , sec / 13195 60 504 2500 2 60 2 Power 1. RPM 2500 N at Nm 504 Torque 3 3 3 3                      . m kg hour m m kg m kW Q W kJ J or Nm Nm RPM T N s b b   

37. CV Power Plants - Prof (Col) GC Mishra,Retd 37 MEAN EFFECTIVE PRESSURE, mep It is the mean value of the pressure inside the engine cylinder which when multiplied by the swept volume gives the same net work done as is actually produced with the varying pressures. Net Work Done = Area under curve 3-4 minus Area under curve 1-2 = mep × swept volume (Vs) We have imep (indicated mean effective pressure) and bmep (brake mean effective pressure) based upon indicated or, brake indicator diagram.

38. POWER O/P  PRESSURES FOR A TYPICAL DIESEL ENG:- 0.14 bar BELOW ATM - SUCTION STROKE 2 bar ABOVE ATM - COMPRESSION 10 bar ABOVE ATM - POWER 0.14 bar ABOVE ATM - EXHAUST  BMEP = 7 – 8 bar FOR NA COMMERCIAL VEHICLES =10 – 13 bar TURBOCHARGED ENG = 20- 25 bar LARGE TURBOCHARGED ENG

39. CV Power Plants - Prof (Col) GC Mishra,Retd 39 cylinders of no. z v bmep engine cylinder single for sec sec rev 3 2 m N v bmep 2 Power Brake s s                  z Nm m 

40. POWER FLOW IN COMBAT VEHICLE POWER PLANTS ENGINE INTERMEDIATE GEAR BOX GEAR BOX SPROCK ET C L U T C H SS FINAL DRIVE FINAL DRIVE SPROCKET STEERING SYSTEM BRAKE SYSTEM FAN DRIVE

41. CV Power Plants - Prof (Col) GC Mishra,Retd 41

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