Top-6 best power plant operation and maintenance books

 1.An Introduction to Thermal Power Plant Engineering and Operation: For Power Plant Professionals Paperback – 27 October 2018

Rs 610 

2-Practical Boiler Operation Engineering & Power Paperback – 25 July 2015

Viva Questions & answers for preparation of BOE exam & interview

3-Industrial Maintenance Process: Mechanical Maintenance Paperback – 9 March 2020

 

                              

4-Industrial Maintenance Process (Volume II): Electrical, Instrumentation & Control Paperback – 10 August 2020

                                                              

5-Practical Approach to Power Plant Operation and Maintenance: Based on Professional Field Experience Paperback – 28 February 2019

                                        

                                                                 

6-Boiler Operation Engineer Exam Question and Answer Paperback – 1 January 2017

                                          

Steam Turbine SOPs

15-Emergencies in power plant operation

How do you carryout performance Guarantee (PG) test of power plant equipments??

Challenging situations & troubleshooting during Boiler light up & start up

 

1-Unavailability of start-up equipments/Competent Man power etc:

As we aware after long shutdown or after maintenance activities, we face following common problems

• Not closing of work permits,
• Not normalizing equipment like Boiler fans, Fuel feeding system equipments from electrical department
• No trial of equipments prior to Boiler light up.

Solution:

• Preparation of check lists for pre light up of Boiler
• Equipments trial & interlock testing
• Ensuring competent persons for light up activities

Factors considered for Boiler design

Boiler calculations for Boiler operation engineer (BOE) exam

2-Boiler light up failure:

The main reasons for Boiler light up failure are;

• Improper selection of fuel (coal, bagasse) : High moisture content fuel may lead to late light up or complete failure
• Foreign materials in fuel may lead to clinker
• Improper fuel: air mixture can lead to clinker formation
• Following wrong SOP of light up
• Not achieving 3T’s of combustion (Temperature. Time & Turbulence)
• Failure of fuel feeding system
• Choke up of fuel feeding system

Solution:

• Following SOP
• Selecting good quality fuel
• Deputing competent person for Boiler light up

3-Drum level fluctuation:

The reasons for drum level fluctuations are;

• Not following boiler pressure raise curve (Fast raising of Boiler pressure)
• Sudden opening or closing of start-up vent control valve
• Improper controlling of blow down valve
• Improper Boiler loading
• Uncontrolled & uneven firing rate

Solutions:

• Following Boiler start up SOP
• Maintaining good control over Start up vent control valve & IBD valves
• Deputing competent Boiler Engineer for operation

4-Air Pre Heater (Tubes jamming):

Reasons for APH tubes jamming;

• Not bypassing the APH during initial start-up (Some Boilers have 100% APH bypass arrangement & some Boilers have 25-30%).If APH not bypassed during Boiler light up, cold FD air will cause condensation of flue gas passing through APH tubes. Initially APH inlet flue gas quantity & temperature is less and FD air quantity will be more. This causes condensation of flue gas in tubes resulting into tubes jamming.
• Maintaining lower flue gas temperature at APH inlet
• Leakages in APH tubes: APH tubes leakage may also lead o jamming
• Operating Boiler at more excess air

QnA on AFBC Boilers

Solutions:

Bypassing the APH during light up & start ups.Generally APH is taken online when APH outllet flue gas temperature crosses 140 Deg C

5-Draft fluctuation & Boiler back fire:

Initially there will not be any control systems working in Auto mode, so controlling the draft & combustion is totally based on communication with field people

Probable reasons are;

• Unbalanced draft
• Poor communication of field people & control room engineers
• Uncontrolled fuel feeding

Solutions:

Maintaining proper communication of field & control room engineer is good remedy to control back fire & draft fluctuation

6-Incomplete & Secondary combustion:

Incomplete combustion is mainly due to poor air fuel mixture leading to more unburnt & secondary combustion at Super heaters, APH & ESP.

Incomplete combustion is also due to not achieving 3T’s of combustion (Lower bed temperature, over feeding, low excess air)

Power plant safety Questions & Answers

Solutions:

• Maintaining 3Ts during boiler light up & start up
• Maintaining balanced draft to avoid secondary combustion due to carryover of fuel
• Ensure OFA nozzles are open & dampers opening is optimized as per requirement

7-Flue gas duct explosions:

This is mainly due to more negative draft, carry-over of fuel particles & secondary combustion. Secondary combustion at the lack of O2 lead to formation of CO, this CO when exposed to rich oxygen leads to explosion.

How CO formation leads to explosions??

During Light up & start-ups unburnt fuel particles get carryover & accumulate at corners of ducts (APH, ESP etc).As we know incomplete combustion leads to formation of Co instead of Co2.This CO formed by unburnt particles when comes in contact with oxygen start to burn (combustion), due to low volume at ducts corners, ESP , this combustion leads to sudden explosion.

Solution:

• During light ups & start up maintain balanced draft, or draft slightly positive (up to + 2 to 4 mmwc) in furnace.
• Maintain sufficient turbulence for combustion
• Ensure all flue gas & air ducts are clean & free from obstacles
• Get feedback from field people on furnace draft, combustion & chimney condition etc.

Note: More black smoke at stack outlet shows unburnt carryover & incomplete combustion or insufficient air

8-No water flow to Boiler:

This condition exists;

• On tripping of pumps,
• Instruments malfunctions
• Faulty flow meter
• Stuck up of control valve etc
• BFP suction strainer choke up

Solution:

• Ensure pre start up checks have been followed properly
• Keep standby equipments healthy & should be available readily

9-Stuck up of Main Steam Stop valve (MSSV):

• Main steam valves stuck up due to long shutdown & no frequent operation
• Unequal pressure at inlet & outlet of MSSV
• Problems related to actuator (low torque set)
• Not operating MSSV bypass valve before opening MSSV

Solution:

• Carryout preventive maintenance (PM) of MSSV periodically
• If PM not conducted, follow valve operation schedule during long shutdowns
• Ensure actuator open & close torques have been set properly
• Before opening the Main valve, do open the bypass line valve (1” or 1.5”), which helps for pressure equalization at inlet & out let side of main valve.

Note:

Functions of Main steam line bypass line:

• Pressure equalizer
• Initial line charging

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10-Main steam line hammering

Steam line hammering is the phenomenon when steam & water mix at high pressure. Water hammer is a pressure surge or wave caused when a fluid (usually a liquid but sometimes also a gas) in motion is forced to stop or change direction suddenly (momentum change).

As soon as steam leaves the boiler, it starts losing heat. As a result, steam stats condensing inside the pipe work. The rate of condensate formation is high particularly during the start ups when the system is cold. As a result of the condensation, the droplets of water are formed. These droplets of condensate get built up along the length of steam pipework forming a solid slug. When this slug encounters any obstacle such as a bend, it will be brought to a halt abruptly. All the kinetic energy of the condensate slug will get converted into pressure energy which has to be absorbed by the pipe work. This gives rise to the phenomenon of water hammer.

Understanding the term water hammer:

After condensate is formed, the flow inside the pipe has two components, steam and the condensate. The flow velocity of steam is much higher than that of the condensate. During such dual phase flow, the heavy condensate which flows at the bottom of the pipe is pulled by high speed steam. This results in formation of water slug which is much denser than steam travelling with the velocity of steam. When this slug is stopped by any abruption like a bend or equipment, the kinetic energy of the slug will be suddenly converted into pressure energy which will create a shock wave in the entire pipework. The pipework will keep on vibrating until this energy is dissipated in the structure.

Solution:

• Charge main steam line by opening bypass valves only, if line is cold give sufficient time to warm the line
• Keep open all drains of main steam line & ensure condensate is draining from those line
• Ensure operator opens the valve very slowly (if MOV is not provided)
• Ensure all steam traps are working properly
• Ensure all steam lines are covered with insulation to avoid steam condensation
• Ensure NON return valves of attemperator (desuper-heating line) water line is working properly

11- Boiler & steam line uneven thermal expansion:

This condition exists when;

• Boiler light up & start-ups are not done as per OEM recommendation
• Quick light up & start ups
• Improper refractory during carrying out maintenance activities
• Obstacles or foreign material left in Boiler during maintenance
• Damage of steam line supports or stuck up
• Overloading the Boiler or steam line

Solution:

Following post shutdown Boiler maintenance checks

• Ensuring proper refractory during shutdown
• Ensuring no any pressure parts are welded with external parts of boilers like platform, columns, beams etc. which do not undergo thermal expansion
• Conduct preventive maintenance steam line supports & spring hangers
• Strictly following SOPs for Boiler light up & start-ups as per time curve given by manufacturer

12-Not achieving the Boiler rated parameters (Pressure, temperature & load)

This is condition is due to;

• Improper combustion
• Selecting improper fuel (high moisture. High ash content fuel)
• Problems related to combustion system
• Improper air & fuel distribution
• Choked fuel feeding system
• Choked air nozzles
• Leakages in & out of the Boilers
• Not operating the Boilers as per SOP

Solution:

• Check issues related to combustion system like proper opening of air & flue gas duct dampers, direction of rotation of FD, PA & ID fans
• Check & ensure field & DCS instruments are showing correct readings
• Check & ensure proper air & fuel distribution in furnace/bed
• Arrest leakages.

10-Tips to reduce LOI

13-Frequent jamming of fuel feeding system

• This is due to wet fuel
• Operating the fuel feeding at lower speed
• Not providing sufficient air pressure
• Foreign materials in fuel

Solution:

• Ensure correct moisture fuel is being used for Boiler light up
• Ensure proper air pressure for fuel distribution

14-Pressure parts failure

During initial start-up of Boilers there is more chances of failure of super heater coils & water wall tubes.

This is due to;

• Less steam flow through super heater coils: This happens if pressure rising is done without opening the start-up vent valve, or super header drains.
• Low water level
• High firing rate
• High drum level (>100%) can cause water hammering in super heater coil & eventually failure

Solution:

• Follow Boiler light up & start up SOP
• Ensure enough steam is flowing through super heater coils coils during boiler start up
• Control firing rates
• Ensure proper water level & cross check DCS level with local gauges

Why does Boilers explode???

15-Clinker formation:

• Reasons for clinker formation;
• Improper air fuel mixture
• Not achieving 3TS during light up
• Foreign materials in coal
• Foreign materials in bed material or poor quality bed material

Solution:

• Using low moisture coal
• Maintaining proper air fuel mixture
• Achieving proper combustion
• Using coal with no foreign materials
• Avoiding frequent startups and stops
• Use coal with high ash fusion temperature

Other General problems associated during Boiler light up & start-ups are

• Fuel handling chutes jam due to wet coal
• Coal crusher jamming
• Low water level in DM storage tanks
• Power supply constraints for running WTP & other BOPs
• Ash handling chutes & hoppers jam

 

 Questions & Answers on Spent wash Boilers

Troubleshooting guide for Boilers

 

 Viva Questions & Answers on Boiler Safety Valves

 

 

 

 

 

Power plant maintenance calculations

 

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

Power plant Safety QnA

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

 Based on above all data, we can calculate the tensile strength of the belt

TS = Initial tension of belt X 5.4 /Belt splice efficiency

For 3-ply belt efficiency will be 75%

 So, TS = 37.87 X 5.4/0.75 =272.67 N/mm

 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

 N1 = 1475 RPM

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/sec² 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 3.142² 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.

 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 10⁶/60N……………..e = 3 for ball bearing & 10/3 for roller bearings

 L10 = (175/22)³ X 10⁶ /(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 10²

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 for welding in Amps= Welding electrode size in mm X 40 +/- 20

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

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

 We have Rockwell hardness number = BHN / 10.81 = 100 / 10.81 = 9.25 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

 We have weight of material = Volume of material in M3 X density in kg/m3

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

 W = Volume X Density

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

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

 W = Volume X Density

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

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

 W = Volume X Density

 Volume of cylinder = 2∏RH

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 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)

 Divide equation by 44

1 + 3.63 = 3 +1.63

 From above result it is clear that 3.63 kg of Oxygen is required to burn 1 kg of Propane to achieve 100% combustion.

So for burning 19 kg of commercial LPG, need 19 X 3.63 = 68.97 Kg of oxygen

 Volume of oxygen cylinder in cylinder = 6.9 M3 compressed at 140-150 kg/cm2

Convert 6.9 to kg by dividing oxygen density, we get weight of O2 in cylinder = 9.1 kg

 So total O2 cylinders required = 68.97 / 9.1 =7.58 Nos

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

 Convert volume to kg by multiplying the density of the gas

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)

 Divide equation by 52

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

 Normal operating vibrations = 2400 / √N = 2400 / √3915 = 38.96 microns

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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