**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**

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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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**7-A belt conveyor of width 1600
mm & length 200 meter running at speed 0.95 m/sec.What would be the belt
tension during initial start-up if operating power of the belt conveyor is 18KW**

** **Given that,

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%

**8-A V belt drive system, having
two pulleys DE 6” & NDE 10”, the centre distance between two pulleys is 1.5
meter. Calculate the length V belt required for open belt drives & cross
belt drives**

** **Given that,

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**

**9-A chain
conveyors gear box input speed & power are 1475 RPM & 7.5 KW. The
efficiency of the gear box is 95% & has reduction ratio 34:1, then
calculate the torque developed on output shaft**

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

**11-What is the shaft size on which a deep groove ball bearing of
size 6318 C3 fits**

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

**13- A bearing vibration shows 50
microns (peak to peak), calculate the bearing vibrations in mm/sec ^{2}
when shaft rotates at 1500 RPM**

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 3.142^{2} X 1500 / (60 X 2 X 100000) =0.123 mm/sec^{2}

**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.

**15-Calculate
the ball bearing life if it is rotating
at 1500 RPM having dynamic load & dynamic equipment load 175 KN & 22 KN
respectively**

** **We have,

N = 1500 RPM

C = 175 KN, P = 22 KN

Bearing life L10 = (C/P)^{e} X 10^{6}/60N……………..e = 3
for ball bearing & 10/3 for roller bearings

L10 = (175/22)^{3} X 10^{6
}/(60 X 1500)

L10 = 5593 hours

**16- A 44 teeth spur gear has 125 mm PCD, calculate its module.**

Module M =
PCD/No.of teeth

M = 125 / 44
= 2.84

Select
standard value of module from chart

**17-A planetary type reduction
gear box has input speed 1475 RPM & reduction ratio 175:1, calculate its
output speed**

** **We have N1 =
1475 RPM, N2 =?

R = 175:1

So, R = N1 /
N2

N2 =
1475/175 = 8.4 RPM

**18-Calculate the safe working
load of 1” Nylon rope?**

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

** **Where 80 is
the grade of chain block steel material

D is the
link diameter in mm

SWC = 80 X 0.4 X 10^{2}

SWC = 3200
kg = 3.2 MT

**21-What is the welding current
required for welding 6 mm MS plate by 3.15 mm welding electrode?**

Current
required = 3.15 X 40 +/-20 =146/106 Amps

**22-How do you convert Brinnel
Hardness 100 BHN to Rockwell hardness number HRC**

**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)

**24-Calculate the tensile strength
of A4-80 grade Austenitic stain less steel bolt**

Tensile
strength = 80 X 10 = 800 Mpa or 800 N/mm2

**25-Calculate the weight of 8mm MS
steel of size 6m X 1.25m required for fabrication work**

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

**26-Calculate the weight of 20 mm
MS plate having diameter 100 mm**

Volume of
circle plate = ∏ X D^{2}/4 X Thickness = 3.142 X (0.1^{2}/4) X
20/1000 =0.0001571 kg/m3

W = 0.0001571
X 7800 = 1.22 Kg

**27-Calculate the weight of 50 mm SS
solid round bar having length 5 meter**

Volume of round
rod = ∏ X D^{2}/4 X Thickness = 3.142 X (0.05^{2}/4) X 5 =0.0098
kg/m3

W = 0.0098 X
7800 = 76.44 Kg…. Density of SS is also
around 7800 kg/m3

**28-Calculate the weight of 3 meter
hollow cylinder of internal diameter 1 meter & thickness 12 mm used for
storing HCL.**

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

**29-Calculate the number of Oxygen
cylinders required to consume 1 no.of industrial LPG cylinder for gas cutting
operation**

Commercial LPG
(C3H8) has 19 kg weight that is 29 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 (C3H8) 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

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

**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

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Pls share a method and question and answer sesson on turibe and gearbox shaft alignment

ReplyDeleteThanks & Regards

Gaurav Kumar