1-A V
pulley & Belt system is used for fuel feeding system. One of the pulley of
size 6” is mounted on motor shaft & other 8” on gear box shaft. Then
calculate the input speed to gear box if motor rated speed is 1450 RPM is
transmitted through V belt to 8” pulley.
We have,
Speed of
motor N1 = 1450 RPM
Size of
drive pulley (fitted on motor shaft) D1 = 6”
Size of Non
end drive pulley (fitted on Gear box input shaft) D2 = 8”
Speed of
pulley fitted on Gear box shaft N2 =?
We have the
formula for calculating the speed of non-drive end side
N1 / N2 = D2
/D1
N2 = N1 X D1
/ D2
N2 = 1450 X
6 / 8
N2 = 1087.5
RPM
Speed of Non
end drive end pulley N2 = 1087.5 RPM
Note: If drive end pulley is smaller than
non-end drive pulley, then non end drive pulley speed is always more than drive
end pulley & Vice Versa
2-A gear box input
& output speeds are 1475 RPM & 100 RPM respectively, then calculate the
gear box reduction ratio
We have,
Gear box
input speed N1 = 1475 RPM
Gear box
output speed N2 = 100 RPM
Gear box
reduction ratio R = N1 / N2 = 1475 / 100 =14.75
Ratio is
14.75:1
That is gear box output shaft will rotate by 1 rotation on 14.75 revolution of input shaft
3-A V
belt is fitted on pulley of size 6” at drive end side & 6” at non drive end
side, both pulleys rotating at speed 1450 RPM, then calculate the speed of the
belt in meter/sec
Given that,
D1=D2= 6”
=0.15 meter
N1=N2=1450
RPM
Speed of the
belt V = ∏ X D X N / 60
V = 3.142 X
0.15 X 1450 / 60
V = 11.38
meter/sec
4-A
concrete mixing drum need to rotate at 25 RPM, an engineer has planned to use
Simplex chain drive system. He has following materials
1-Motor 3.75 KW &
1475 RPM
2-Simplex chain pitch
1”
3-A reduction gear box
of ratio R = 25:1
4-Chain sprocket of
size 10”
Calculate the other
side chain sprocket size to maintain mixer speed around N3 = 25 RPM
Power plant safety Questions & Answers
We have
input speed to gear box N1 = 1475 RPM
Output speed
of gear box N2 = N1 / R = 1475 / 25 = 59 RPM
We have 10“chain
sprocket & we need 25 RPM. If we fit 10” sprocket at DE, then we should
require higher size sprocket at NDE.
And if we
select smaller sprocket at DE, then we can get reduced speed at NDE
Case-1:
Let us
decide 10” sprocket will be fitted on mixer machine shaft, now calculate the
size of the sprocket to be fitted on gear box output shaft
N2 / N3 =
D3/D2..Assume D3 = 10” & D2 to be found out)
D2 = 10 X 25
/ 59 = 4.23”
Select
standard size either 4” or 5”. If selected 4” you will get slightly less RPM
than 25 & if you select 5” you will get slight higher speed than 25.
Case-2:
Let us
decide 10” sprocket will be fitted on gear box output shaft, now calculate the
size of the sprocket to be fitted on concrete mixer machine shaft
N2 / N3 =
D3/D2..Assume D2 = 10” & D3 to be found out)
D3 = 59 X 10
/ 25 = 23.6”
Select
standard size 24” sprocket.
Case-2 is costlier as compared to
case-1, so better to select case-2
5- A belt conveyor of length 300 meter need to install VGTU
system calculate the minimum length of take up.
Minimum length of take up = Conveyor length
(centre to Center) X 1.5%
= 300
X 1.5/100 = 4.5 meters
6-A coal
conveyor belt of width 800 mm & thickness 10 mm is coiled on 100 mm steel
pipe. No.of turns of belt are 75. Calculate the length of the belt
Given that,
Belt width W
= 800 mm = 0.8 meter
Thickness t
= 10 mm = 0.01 m
Belt coil ID
= D1 = 100 mm = 0.1 m
Belt coil OD = Inner diameter of belt + 2 X
No.of coils X Thickness
Belt OD = (2 X 75 X 0.01) = 1.5
meter
Length of belt L= (D1 + (D2-D1)/2) X π X N
L = (0.1 + (1.5-0.1)/2) X 3.142 X 75
L = 188.52 meters.
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Width of the
belt W= 1600 mm
Length of
the belt conveyor L= 200 m
Conveyor
speed V= 0.95 m/sec
Operating
power of the conveyor P = 18 KW =18000 W
We have,
Belt start
up tension T = P X 3.2 / (V X W) = 18000 X 3.2 / (0.95 X 1600) = 37.87 N/mm
TS = Initial
tension of belt X 5.4 /Belt splice efficiency
For 3-ply
belt efficiency will be 75%
D1 = 6” =
150 mm
R1 = 75 mm
D2 = 10’ =
300 mm
R2 = 150 mm
Centre
distance C =1500 mm
For open
belt drive the direction of rotation of both DE & NDE pulleys is same.
Length of
belt = Lo = ∏ (R1+R2) + 2C + (R2-R1)2/2C
Lo = 3.142 X (75 +
150) +2 X 1500 + (150-75)2/(1500)
Lo =3710.7mm
Convert it into inches Lo =
3708.82/25.4 = 146”
If pulley section is B, then V belt
size is B-146
For cross belt drive direction of rotation of both DE & NDE pulleys is opposite
& length of belt required is larger than open belt drive.
Length of
belt Lc = ∏ (R1+R2) + 2C + (R2+R1)2/C
Lo = 3.142 X (75 +
150) +2 X 1500 + (150+75)2/(1500)
Lo =3740.72 mm
P1 = 7.5 KW
Gear box
efficiency ηg = 95%
Reduction
ratio, R = 34:1
N2 = N1/R =
1475 /34 =43.38 RPM
P2 = P1 X
95/100
P2 =7.5 X
95/100 = 7.12 KW
Torque
developed on output shaft = P2 X 60 / (2∏N2)
T = 7.12 X 60 /(2 X 3.142 X 43.38) = 1.56 KNm
10-What will be the size required
for a hub to be fit on a 100 mm shaft?
Hub OD = 2 X
Shaft size = 2 X 100 = 200 mm
Hub length L
= 1.5 X D = 1.5 X 100 = 150 mm
For deep
groove ball bearing
Shaft size =18
X 5 = 90 mm..…(Last two digits of bearing number X 5)
12-A bearing vibration shows 50
microns (peak to peak), calculate the bearing vibrations in mm/sec when shaft
rotates at 1500 RPM
Vibrations
in mm/sec =(Vibration in displacement peak - peak (microns) X 2 X ∏ X N)/(60 X
2 X 10000)
Vibrations
in mm/sec = 50 X 2 X 3.142 X 1500 / (60 X 2 X 1000) =0.392 mm/sec
Vibrations
in mm/sec =(Vibration in displacement peak - peak (microns) X 4 X ∏2
X N)/(60 X 2 X 1000)
Vibrations
in mm/sec2 = 50 X 2 X 3.1422 X 1500 / (60 X 2 X 100000) =0.123 mm/sec2
14- A 6315 2Z bearing has OD 160 mm and width 37 mm, what amount
of grease is required for first lubrication?
We have,
Bearing lubrication quantity in grams =
Bearing OD (D) X Width (B) X 0.05
= 160 X 37 X 0.05 = 296 grams.
N = 1500 RPM
C = 175 KN, P = 22 KN
Bearing life L10 = (C/P)e X 106/60N……………..e = 3
for ball bearing & 10/3 for roller bearings
L10 = (175/22)3 X 106
/(60 X 1500)
L10 = 5593 hours
Module M =
PCD/No.of teeth
M = 125 / 44
= 2.84
Select
standard value of module from chart
R = 175:1
So, R = N1 /
N2
N2 =
1475/175 = 8.4 RPM
We have
Nylon rope size D= 1”
Nylon rope
safe working capacity = D X D = 1 X 1 = 1 MT
19-Calculate the safe working
load of 1” steel wire rope?
Steel wire
rope safe working capacity = 8 X D X D = 8 X 1 X 1 =8 MT
20-Calculate the safe working
load of Chain block having Load link diameter 10 mm
We have Safe
working load of chain block = 80 X 0.4 X D2
D is the
link diameter in mm
SWC = 80 X 0.4 X 102
SWC = 3200
kg = 3.2 MT
Current
required = 3.15 X 40 +/-20 =146/106 Amps
23-Calculate the safe working load
of bolt of grade 4.8
A bolt of
grade 4.8 implies that,
Tensile
strength = 4 X 100 = 400 N/mm2
Efficiency
or with standing load = 0.9 X 100 =90%
This means
that, a bolt of metric grade 4.8 with stands a load 40.7 kg/mm2 of (400/9.81)
& fails at 90% load (37 kg)
Tensile
strength = 80 X 10 = 800 Mpa or 800 N/mm2
W = 6 X 1.25
X (8/1000) X 7800 ….Take all values in meter & Density of mild steel is
7800 kg/m3
W = 468 Kgs
Volume of
circle plate = ∏ X D2/4 X Thickness = 3.142 X (0.12/4) X
20/1000 =0.0001571 kg/m3
W = 0.0001571
X 7800 = 1.22 Kg
Volume of round
rod = ∏ X D2/4 X Thickness = 3.142 X (0.052/4) X 5 =0.0098
kg/m3
W = 0.0098 X
7800 = 76.44 Kg…. Density of SS is also
around 7800 kg/m3
We have OD
of cylinder D1 = ID + 2 X thkickness = 1 + 2 X (12/1000) = 1.024 meter
R1 = D1/2 =
1.024/2 = 0.512 meter
R2 = ½ = 0.5
meter
Volume = 2∏R1H-=
2∏R2H-2∏R1H = (2X3.142X0.512X3)-(2X3.142X0.5X3) =0.22 M3
W = 0.22 X
7800 = 1716 Kgs
Commercial LPG
(C3H8) has 19 kg weight that is 19 kg of propane
Combustion
equation of propane
C3H8 + 5O2 =
3CO2 + 4 H2O
44 + 160 =
132 + 72 (Molecular weight of C = 12, O = 16, H = 1)
1 + 3.63 = 3
+1.63
So for
burning 19 kg of commercial LPG, need 19 X 3.63 = 68.97 Kg of oxygen
Convert 6.9
to kg by dividing oxygen density, we get weight of O2 in cylinder = 9.1 kg
30-Calculate the number of Oxygen
cylinders required to consume 1 no.of dilute acetylene cylinder for gas cutting
operation
DA (C2H2) cylinder
has 8 m3 of acetylene
8 X 0.899 =
7.192 kg
Combustion
equation of propane
2C2H2 + 5O2
= 4CO2 + 2H2O
52 + 160 =
176 + 36 (Molecular weight of C = 12, O = 16, H = 1)
1 + 3.07 = 3.38
+0.69
So for
burning 7.192 kg of DA, need 7.192 X 3.07 = 22.07 Kg of oxygen
So total O2
cylinders required = 22.07 / 9.1 =2.42 Nos
31-A steam Turbine has rated speed
3915 RPM,calculate its thrust bearing’s operating , alarm & trip vibration
levels
Alarm level
vibration = 4500 / √N = 4500 / √3915 = 71.93 microns
Trip level
of vibration = 6600 / √N = 6600 / √3915 = 105.6 microns
32-How do you judge the bearing temperatures by hand physical touch?
Generally in power plants or in any other industries, bearing temperature are judged by infra red temperature sensors or by hand touch during field rounds.
Following table gives the actual range of temperature based on hand touch.
Assumption: Actual bearing temperature is 10 deg C more than its plummer block top surface temperature.
Plummer block temperature
1-40-45 deg C -Hand touch with standing time > 1 minute
2-45-50 deg C-Hand touch with standing time < 12 Sec
3-50-55 deg C-Hand touch with standing time < 8 sec
4-55-60 deg C-Hand touch with standing time < 5 sec
5-60-70 deg C-Hand touch with standing time < 2 sec
6-> 70 deg C -Hand touch with standing time < 1 sec
