CV-2-Power Plant.pptx

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

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

3

4

5
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
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
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?

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
Power Transmission by the Shaft of a Car

10

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

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.

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.

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.

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
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
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
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
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.

20

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.

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)

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.

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.

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………

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 











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














TUTORIAL EXERCISE-1
Determine the Power required for an AFV
2. At the flywheel
for the following parameters:-
a. Laden Weight of the AFV, W = 41.5 Ton
speed of 10 kmph.

TUTORIAL EXERCISE-2
Determine the Power required for an Infantry Combat vehicle(ICV)
2. At the flywheel
for the following parameters:-
a. Laden Weight of the AFV, W = 13.3 Ton
speed of 10 kmph.

Efficiency and Performance
Calculation
Of
Power Plants

 
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

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




SPECIFIC FUEL CONSUMPTION
0
0.05
0.1
0.15
0.2
0.25
0.3

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


































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)

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




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.

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

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


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

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