Gears Power Transmission

GEARS POWER TRANSMISSION1
POWER TRANSMISSION
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GEARS POWER TRANSMISSION2
Task 1
Design a friction clutch and drive belt system for Perkins diesel engine rated for continuous use at
175kW at 2600rpm and connected to an electrical generator. The generator is to run at 3000rpm
with full load capacity of 150kW. Give a layout diagram on the final arrangement.
Solution
Length of the belt needs to be determined:
What do we know:
Diameter is 300mm (Largest pulley) Pulley 1
Diameter is 200mm (smallest pulley) Pulley 2
C-C = 800mm
Belt length= π (Pulley 1 dia. + Pulley2 dia.) X0.5 + (2 X Centre distance) + ((Pulley 2 dia. – Pulley1
dia.)2 /4 x Centre distance)) (The Engineering Mindset, 2018)
Belt length=3.142 ( 300+200 ) X 0.5+ ( 2 X 800 ) + ( ( 200−300 ) 2
4 X 800 ) =785.39+1600+3.125: therefore BL=2388.51mm∨
Centre distance of the pulleys.
( 4 XBL−6.28 X ( PD 1+ pd 2 ) ) + √ (4 X BL−6.28 X pd 2)¿2−32 X ( Pd 1− pd 2¿) 1
16
7042.04+ √(4 X BL−6.28 X pd 2)¿2−32 X ( Pd 1− pd 2¿)2 1
16

GEARS POWER TRANSMISSION3
7042.0+78120 1
16
¿ 5322.625 mm∨5.322 m
No of belts
Belt velocity = V =ω x r
In r.p.s ω = (2600 X 2π)/60 = 272.27rps X 0.15 = 40.84m/s
Power transmitted shall be found as:
P=F1 X (1−e
−u
sinθ ) X V
The angle (θ) =40
And u is the generator speed obtained as (3000X2π)/60 = 314.16rps
Hence, P=1300 X ( 1−e
−0.3
sin 40 ) X 40.84=17.597 kW
Belt needed are 175kW/17.597 = 9.9 approx. 10belts
No of clutch
Radius of the cultch is 100mm or 0.1m
R1 = 0.58 X 0.1 = 0.058m
R2^2 – R1^2 = 0.006636
R2^3 = 0.001
R1^3 = 0.000195112
R2^3 – R1^3 = 0.000804888
N=πXPX 0.006636
2600=πXpX 0.006636
=124714.54 or 124.714kW
T = P
ω = 124714.54
272.27 =458.055 Nm
Plates required shall be obtained by: