Page 182 - WCS - Electrical

8
Load
15000 =
8

Load
2π ∗
30
MA =
2π ∗
Effort
Distance
moved
by

8∗
15000 ∗
V.R. =

Distance
moved

by

15000 ∗
load

Distance
V.R. = by moved
3008
by
Distance
load

moved
2∗
22∗
3008
Output
work

100%
Efficiency =
×
Output

work
Distance
moved

load
by
Iutput
work
Efficiency =
×
100%
=
Distance

moved
by
Distance
load
work
= by d
move
Iutput
Distance
load
by
moved

8
work Load
Output

=
15000 =
MA =
Efficiency =

work
300
2π ∗
D
πD
Iutput
Efficiency = Effort
=
D
πD
Iutput
d
πd
=
=
d
πd
moved

by
effort

Distance
7
8∗
15000 ∗
Distance

by
moved

load
Load×
D
=
Distance
Load× Distance
moved oved
by load
by
=
m
d
load
3008
2∗
22∗
= moved e
effort
Distanc
Effort ×

by
d
by
moved


effort
Distance


Effort ×
2
1
=
of
 πd



2
work
Output
1
of
2
= 
− πd
 πd

100%
Efficiency =
×

move
load

2 Distance
moved

load
Load


Distance
by

work
1
Iutput
2 =
=
×
 1
Distance
load
Load

by

moved
− d
d
=
π
by
moved
load
Distance

1
= move e
Effort MA = Effort V.R. = work effort Output work effort 1 = = 2∗ D 22∗ = − πd    7 1 = =0 15000 = 8∗  7    by d 300 load
effort
× by d
Distanc
2
 1

Effort Distance moved by effort 2 = 2 π d − d  


Output
work
2 
1
Efficiency =
= Mechanical Advantage = 1 Iutput work π   d − d    1 πD 2  D

π =
d − =
d 
=
Ve
Advantage =
= Mechanicallocity Ratio 1 WORKSHOP CALCULATION & SCIENCE - CITS
 d
 πd
Velocity Ratio 2 =
= Velocity Ratio = Distance moved by 2 Load
D

Loadad×
MA = = Eff Distance moved by load = Distance Ratio = by moved Distance moved by Load
Lo
Load
Velocity

Load
by
d
Distance
Load
moved
17 (x) = 100 x 50
moved
effort
Effortort ×
Distance
by
MA =
1
of
− πd
=
100 x 50 17 X = Effort = πD 1    πd πD 2  
1200 Distance moved by load π  1 - d 2   2 = 
d
= Load
x =100x50\17= 294.1 cm
1200
=
d1
d 2
300 × Distance moved by effort   1 π  1 - − d 2    
π
=
d
=

Effort

= 294.1 cm 300  1 2  2  
π  d - d   d 1 d  

2 
: Distance between the load and point of force
d
 d −

π
Mechanical Advantage Advantage = 1 = πD/ 2 = πD/   π 1 - 2   
Mechanical
=
Mechanical
Advantage Velocity
= 294.1-50
Velocity Ratio Velocity Ratio Ratio 2 = 2 2

= 244.1 cm = πD = ∗ Veloci 2 Ra tio ∗ Distan 2 ce moved by Load
 1ty
= πD =
4 Load π. d - d   Distan - d 2  moved by Load
 1ce

=
π.
MA
4
5 = 2.4410 m Ans = Effort 2D d  

=
5
3 Screw Jack = πD
2D
Load d 1 - d 2 = =
MA = Load  1 2 
1200
2
1
- A screw jack consists of a threaded screw and a nut that moves along the screw. dπ
Effort MA = = 300  d - - d d 
15
Effort
D
- When the screw is turned, it causes the nut to move up or down, lifting or lowering the load.
=
=

15 2 

D 1
= d
75 75 d 6 = πD/ π  d = d- 6  
Simple screw Jack
=
25 Mechanical
Advantage
2
150
=
Load
Working principle of simple screw jack is similar to that of an inclined plane. In consists of a screw fitted in a nut.
=
=
25
150
Load
Ratio
Velocity
x
Effect
Simple screw jack is shown in the figure below. For lifting the load, an effort is applied at the end of the lever l.e
=
= 2

Effect
3
x
= πD
effort arm. 4 3 75 15 15 ∗ π.  1 - d 2  
=
d
4
=
=
/ Length of lever or effort arm = 100 × 6 = 8 = 75  × 15 = 15 
4
5

=
moved

by

Distance
P = Pitch of screw effort 15 100 2D 6 8
= Velocity Ratio = Load moved by effort d 1 15 2
Distance

-
d
MA
= Distance Ratio = by moved = Effort Load moved by Load 8 8
Velocity
Distance
2π l 150 = 15 D 15
15
150==
= 2π l x 8 =
p = 75 x d 6 8
= p
25 150x8 Load 150

by
Distance moved by effort =2 πl / = Effect = x
2ππ
Distance
150x8

effort
moved
= Velocity Ratio = Distance = by effort 2ππ 8

moved
3
8
Load
= Distance Ratio = by moved Distance moved 2 by Load = 1 2 75 15 15
1
Velocity
p
Distance moved by load = p p −
=
4
Differential screw Jack p − p = 100 × 6 = 8
15
by
moved

screw Distance

= F × / = This is an improved form of a simple screw jack in which the velocity ratio is intensified with the help of a
of
Pitch
effort

Ratio =
Velocity
differential screw as shown in the figure.
of

Pitch

2 Circumfere F × nce / screw of Distance screw by Load 8
moved
=
2
L = length of level or effort arm screw of nceCircumfere 150 15
2π
l
=
P = Pitch of the screw A x = 8
1 p
P = Pitch of the screw 2
2
= Velocity Ratio = Distance moved by effort = 2ππ 150x8
8
1
Distance moved by Load p − p 2
Examples
1.m15 kN Pitch (P) 8 mm R30 cm F N

screw
Pitch

of
=
F ×
Solution 2 / Circumfere nce of screw
Length of lever R = 30 cm = 300mm
Load F = 15 kN = 15000 newtons
2
Effort F = To find
1
Pitch of screw P = 8 mm
Circumference of screw
= 2πR= 2π × 300 mm
F1=F2. tan a
300 mm 15000 newtons
169
CITS : WCS - Electrical - Exercise 17