Page 183 - WCS - Electrical

8

Load
15000 =
=
MA =
2π ∗
300
Effort

moved

by
effort

Distance
7
8∗
15000 ∗
V.R. =
=

moved
load
Distance
by

3008
22∗
2∗

work
Output
100%
×
Efficiency =

by

moved
Distance
Iutput
work
=

Distance
moved
by

load
Output
work

Efficiency =
D
πD
work
Iutput
=
=
d
πd
D
Load×

load
Distance

moved
by

=
d
Distance

by

effort
moved
Effort ×
1


2
1
of
− πd
 πd
=


2


Distance
Load
load
by
moved
1
=
×
 1
2
− d
π
d
=
Distance
Effort
by

moved
effort
2


2 

1
1

d −
d 
π
Mechanical
=
Advantage =


=
Ratio
Velocity
2

Distance
by

Velocity
Ratio =
=
Load

Distance

Load
moved
by
MA =
Effort

πD
=

1200
2
 1
d
-
π
d

=


300

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

2
= πD
∗

 1
-
π.
d
d
4


=

5

2D
=
Load

1
2
-
d
d
MA =
Effort
D
15
=
=
d
6
75
=
25
Load
150
=
=
Effect
x
3
=
15
75

15
4
=
×
=
100
8
6
15

Distance

moved
effort
by
Ratio =
=
Velocity
8
moved

by
Distance
Load

15
150
l
2π
=

=
x
8
p
150x8 2    moved load Load
= Velocity Ratio = Distance moved by effort = 2ππ 8
WORKSHOP CALCULATION & SCIENCE - CITS
moved

Load
by

2
1
Distance
p −
p
8
MA = Load = = F × / Pitch of screw
15000 =
300
2π ∗
Effort 2 Circumfere nce of screw
MA = Load = 15000 = 8
2π ∗
Effort
V.R. = Distance moved by effort 15000 ∗ 8∗ 7 300
Distance moved by load = 2∗ 22∗ 3008
Distance moved by effort
V.R. = 15000 ∗ 8∗ 7
= 63.64 newtons Ans.
Efficiency = Output work × load = 2∗ 22∗ 3008

Distance
by

moved
100%
2 A simple screw jack having pitch = 1 cm and the effective length of the lever = 50 cm. What shall be the velocity
Iutput work Distance moved by load
=
ratio? A force of 5 kg is applied on lever and lifts a load of 1100 kg. What is its efficiency?
Efficiency = Output work × 100% Distance moved by load
Iutput work worktput Distance moved by load
Ou
Solution
=
Efficiency = Distance moved by load
D
πD
Iutput work Length of level / =50 cm
=
=
Output work πd d
Efficiency = Pitch of screw p = 1 cm
Iutput work πD = D
=
D
πd
d
Load× Distance moved by load Load lifted = 1100kg
=
d
Effort × Distance moved by effort Effort applied = 5kgs

D 
2
1
of
− πd
 πd
=
Load× Distance moved by load 1 3 Wheel and Axle

=

2
d 
moved
= Effort × DistanceLoad × Distance moved by loadby effort 1 1   1 2 2  
Simple wheel and axle are keyed to the same shaft. The shaft is mounted on ball bearings in order to reduce the
1
 πd dπ
− πd d
 
of
−
Effort Distance moved by effort = = effect of frictional resistance to minimum. A string is wound round the axle, which carries load to be lifted. Another
2 2
 
 
string is wound round the wheel in opposite direction to that of the string on axle, ie., string on wheel and axle are
Load × Distance moved by load 1  1 2  

=
1
wound in opposite directions to each other. Since the two strings are wound in opposite directions, the downward
π
Effort Distance moved by effort 1 = π  d dd − − d 2  
= Mechanical Advantage = 2    
Velocity Ratio = motion of effort will raise the load.
2
2 

1
1 Let D = Diameter of wheel
d −
π
d 
= Mechanical Advantage = Velocity Ratio MA = Load = Velocity Ratio  Distance moved by Load = 15000 = 8


= =
8
Distance
moved

by
Load Load Effort = d = Diameter of axle . Load 2π ∗ 300
2
15000 =
MA = MA =
2π

Distance moved by effort dby moved Distance ∗
Effort
Effort Load = Velocity Ratio = = 15000 = 300 8 Loa

πD
MA =
MA = Load Distance moved by effort Distance moved by effort
2π
300
moved ∗
Load
by
Distance
=
Effort
1200

2
 1
Effort
π
=
=
Distance moved by effort V.R. = Distance moved by load =πD d - d   15000 ∗ 8∗ 7
πD
300
V.R. = 15000 ∗ 8∗ 7 2∗ 22∗ 3008
=
=
by
Dist
V.R. = Distanceved ance mo moved load effort 2∗ 2  1 1 --0 2  2 

by
Distance moved by load

1200
 78∗
d d∗
d d30082∗
π π 1500
=
=


work
load
by

∗
3008
= πd 222∗
Advantage
Mechanical
Output
Efficiency = Dista 300 moved nce work × 100% Efficiency = Output work × 100% = πD/ π nce 1 2 d  by oved Distance moved by load
Iutput
2 
=

-
d
Velocity ratio of wheel and axle
Ratio
work
Iutput
Velocity Output work Dista  m 2  load Distance moved by load
=
=
πD/
Efficiency =
mov

by
Distance

moved
Mechanical Advantage × 100% Output work = πD ∗ Distance ed  by load load
2
Iutput
work

=
2
 1
π.

4
= Output
Velocity Ratio work Efficiency = Iutput work Dista d nce - d moved by load πD D


2
Efficiency = 5 = πD ∗ = =
Iutput
Efficiency = Output work πD  1 D - d 2   πd d
work
2D

=
=
π.
d
=
4
2 D
MA = 5 work πd d =  1 πD d d = 
= Load Iutput
-
d
2D πd

Effort Load× Distance moved by load Velocity ratio of wheel and axle = D
= D
=
Load× Distance moved by load Effort × Distance moved by effort D 2 15 d
Load
1
= MA = d d = - d = 1
D


2
1
d
6
mov
Distance
moved
load

Ef Load×ancefort × Dist Effort 75 ed by effort - A wheel and axle consist of a larger wheel (the wheel) and a smaller cylindrical rod (the axle) that are connected.
by
 πd
− πd
of
=

= = 1  1 d 2  2  
D− πdd
 π
15 
Effort × Distance moved by effort Distance moved by load = 2 of - When force is applied to the wheel, it rotates the axle, allowing for the transfer of force over a distance.
Load
25
150
= Load
=
1 

1


=
2
1
d
= of
6 = − πd
dπd


=
Load × Distance moved by load Effort × Distance moved by effort 1 = 4 Differential Wheel and Axle π  1 − d 2   
75
x
Effect

=
2


2
 1
2
=
π
− d
d

Ef Load Distancefort Dista mov 3 moved nceed by effort load = Differential wheel and axle is shown in figure below. It consists of wheel and two axles. Wheel and axles are
by
25
1 

= × = 2 Load  1 150   1 2 
2
15
= 15 − d
d
75 π
= =

d −
d 
π
Effort Distanc 4 moved e by = effort Advantage = 1 keyed to the same shaft, which is mounted on ball bearings in order to reduce the effect of frictional resistance to
=
x =
Effect ×


Mechanical
2 1
= Mechanical Advant 1 Velocity Ratio π  d − 2  6 8 =  2 
3
100 d 
minimum. The effort string is wound round the wheel.
=age =



15 
2
75 15
1
15
Dist moved Ratio = = Another string is wound round the larger axle; after passing round the pulley (load attached to it) is wound round
4ance Velocity
effort
d −
by 1
d 
π
×
=
= Velocity Ratio =Advantageanical = 2   Distance moved by Load
=
Mech
100
6 8
Velocity
Ratio =
8 =
=
Distance moved by Ratcity Loadio Load the smaller axle. String wound on smaller axle is in the opposite direction of string wound on larger axle. It should
Velo
moved
Distance

Load
by

Load

moved
Distance
2

by
15
15 150
= Velocity Ratio = 2π l moved by effort MA = Effort = Velocity Ratio = be ensured that the string wound on “wheel” and “smaller.
Distance

Load
by
Load
= moved e

by
Load
Distance
Distanc
moved

MA =
8
=
8
x
Load axle is the same direction
Distance
Effort p Load moved by Load = Velocity Ratio = Distance moved by Load = πD
MA = l 1200 = 150 = 15 π  1 - d 2 
πD
2π Effort
d

=


88
x 150x 
πD
 1
1200 = 300 D = Diameter of wheel to which effort is applied
2
= d
-
d
π

p
=
Distance
= Velocity Ratio = 1200 moved by effort = 2ππ d = Diameter of larger load axle  1 2 
1 




300
2
8 1
= Distance moved by Load p − p 2  1 π d - d  π  d - d  
1
150x8  2 
300
= Velocity Ratio = Distance moved by Mechanical Advantage π  d -  d   2  = πD/ 2
2ππ
effort
=
8 1
1
Mechanical Advantage moved by Load Velocity p Ratio 2 = πD/ π  d - d  
Distance
2
p −
= Mechanical Pitch of screw = πD/ 2 = πD ∗ 2
Ratio
Advantage
Velocity
F ×
/
2 Velocity Ratio nce of screw 4 = πD ∗ 2 π.  1 - d 2   170
Circumfere
d


∗ d
π.
= F × = / Pitch of screw = 5 = πD  1 - d 2 2   CITS : WCS - Electrical - Exercise 17
4
2 5 Circumfere nce of screw Load  2D π.  1  - d 2    = 1 2D 2
d
4

=
Load 5 MA = Effort = d 1 d- 2 2D d - d
MA = =
Effort Load d 1 - d 2 D = 15
=
MA =
Effort 75 D = 15 d 6
=
75 = 25 d D 6 = 15 Load 150
=
=
25 75 Load d 150 6 = Effect = x
= = =
25 3 Ef Loadxfect 150
=
3 4 = = 75 15 15
= Effect x = × =
4 3 = 75 × 15 = 15 100 6 8
= 100 75 6 15 8 15
4 Distance moved by effort = × = 15
=
15
= Velocity Ratio = Distance moved Velocity Ratio = Distance moved by Load 100 6 8 8
effort
by
Distance mov ed by Load effort 8 15 150 15
moved
by
Distance
= Velocity Ratio = 2π l 150 15 =
8
=
Distance
2π l moved by Load p = x 8
=
p 2π l x 150 8 = 15
= x 8 150x8
p Distance moved by effort 2ππ
150x8
Velocity
=
= Velocity Ratio = Distance moved by effort Ratio = Distance moved by Load = p − 8 p 2 150x8 8
2ππ
1
=
Distance
by
moved
Distance mov ed by Load effort 1 p 2 2ππ
p −
= Velocity Ratio = = 8
Distance moved by Load p − p 2
1
Pitch of screw
= F × / Pitch of screw = F × / Circumfere nce of screw
2
2 Circumfere nce of o screw f screw
Pitch
= F × / Circumfere nce of screw
2