NR > 4000 turbulent
For channel flow
NR < 500 laminar
NR > 2000 turbulent
11
Another number for channel flow!
Froude Number [NF] (gravity versus inertial forces)
h
F
gy
v
N=
Where yh is referred to as the hydraulic depth and given as
yh = A/T
where A is
20 36
Since 1.333 is closer to 1.38 as compared to 1.4286.
Therefore, tC = 24 and tD = 36
The top gear requires direct connection between input shaft and output shaft.
3.8.3 Power Transmitted by Simple Spur Gear
When power is bring transmitted by a spur g
Devices
Figure 3.30 : Compound Gear Train
Let N1, N2, N3, . . . be speed in rpm of gears 1, 2, 3, . . . , etc. and t1, t2, t3, . . . ,
etc. be
number of teeth of respective gears 1, 2, 3, . . . , etc.
Gear Ratio 1 1 2 1 3
42424
NNNNN
NNNNN
24
13
tt
tt
The
V
2
cos
60 2
n
N tm
PF
where t is number of teeth and m is module.
Figure 3.32
Example 3.2
An open flat belt drive is required to transmit 20 kW. The diameter of one of the
pulleys is 150 cm having speed equal to 300 rpm. The minimum angle of contact
may
intermediate gears fill the space between input and output gears and have effect on
the
sense of rotation of output gear.
SAQ 8
(a) There are six gears meshing externally and input gear is rotating in
clockwise sense. Determine sense of rotation of the ou
due to the belt slip. The coefficient of the friction is 0.3. Determine
(a) power transmitted,
(b) power lost in friction, and
(c) efficiency of the drive.
Solution
Data given :
Diameter of driver pulley (d1) = 1.2 m
Diameter of driven pulley (d2) = 0.5 m
TT
T
Maximum power transmitted (Pmax) = (T1 T2) V
Velocity of belt (V) 1 1 1.2 25
22
d
V = 15 m/s
P V max (1200 623.6) (1200 623.6) 15
8646 W or 8.646 kW
111
Power Transmission
Devices
Example 3.6
A shaft carries pulley of 100 cm diameter which rotates a
12
Uniform steady flow and Mannings Equation
When discharge remains the same and depth does not change
then we have uniform steady flow.
In this condition
The surface of water is parallel to the bed of the channel
Or S = Sw
Where S is the slope of the ch
2
1
B
21
Line of A
contact
98
Theory of Machines Crossed-Helical Gears or Spiral Gears
They can be used for any two shafts at any angle as shown in Figure 3.23 by a
suitable choice of helix angle. These gears are used to drive feed mechanisms on
machine t
lap and coefficient of friction. If coefficient of friction is same on both the pulleys
smaller angle of lap will be used in the formula. If coefficient of friction is
different, the
minimum value of product of coefficient of friction and angle of lap wil
or, b = 45.4 mm
The effect of the centrifugal tension increases the width of the belt required.
Example 3.3
An open belt drive is required to transmit 15 kW from a motor running at 740 rpm.
The diameter of the motor pulley is 30 cm. The driven pulley run
thickness equal to that in (a). The density of the belt material
is 1.0 gm/cm3.
Solution
Given that Power transmitted (p) = 20 kW
Diameter of pulley (d) = 150 cm = 1.5 m
Speed of the belt (N) = 300 rpm
Angle of lap () 170 2.387 radian o 170
180
Coefficien
the rope is limited to 800 N. Determine :
(a) number of ropes and rope diameter, and
(b) length of each rope.
Solution
Given data :
Diameter of driving pulley (d1) = 100 cm = 1 m
Speed of the driving pulley (N1) = 500 rpm
Speed of the driven pulley (N2) =
- cross section should remain unchanged referred to as a
prismatic channel
Varied steady flow when depth changes but discharge
remains the same
(how can this happen?)
Varied unsteady flow when both depth and discharge change
along a channel length of inte
3.10 KEY WORDS
Spur Gears : They have straight teeth with teeth layout is
parallel to the axis of shaft.
Helical Gears : They have curved or straight teeth and its
inclination with shaft axis is called helix angle.
113
Power Transmission
Devices
Herringbo
Where Q is in m3/s
For uniform flow, Q is referred to as Normal discharge
16
The above equation can also be re-arranged such that
1/ 2
2/3
S
nQ
AR =
The left hand term is simply based on channel geometry.
17
Problem 14.2
Determine normal discharge for a
0.25
1.5 4 (1 0.5 0.0625) 8.72 m
2
.
3.9 SUMMARY
The power transmission devices are belt drive, chain drive and gear drive. The belt
drive
is used when distance between the shaft axes is large and there is no effect of slip
on
power transmission. Chain dr
gear engagement, it may use sliding mesh arrangement, constant mesh
arrangement or synchromesh arrangement. Discussion of these arrangements is
beyond the scope of this course. We shall restrict ourselves to the gear train. It can
be explained better with