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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Various IBR forms, acts & regulations for a power plant engineers

Sl No.	IBR Form No.	Details of forms
1	I	Memorandum for Boiler inspection or registration
2	II	Issued by inspection authority during new construction
3	II A	Issued by inspection authority during new construction for foreign code boilers
4	III	Certificate of manufacture issued by constructor
5	III A	Steam pipe line test certificate  issued by manufacturer
6	III B	Tubes  test certificate issued by manufacturer
7	III C	Test certificates for  fittings & mountings (Valves,bend,elbow,tee etc)
8	III F	Test certificates of casting & forging issued by manufacturer
9	V	Boiler use provisional order issued by Boiler inspector authority
10	VI	Certificate issued by  directorate of steam boiler for boiler use
11	VII	Inspecting authority (directorate of steam boiler) certificate for constructional Boiler
12	XI	Certificate for the use of an Economiser
14	XII	Welder qualification record
15	XIII	IBR welder certificate
16	XVI-F	Certificate for approval of tube maker
17	XVI-G	Certificate for approval of pipe maker
18	XVI-H	Certificate for approval of material testing laboratory
19	XV-I	Certificate for approval for RLA organisation
20	Form No.C	Certificate for 2nd class Boiler attendant issued by Directorate of steam Boilers under rule-41
21	Form No.B	Certificate for 1st class Boiler attendant issued by Directorate of steam Boilers under rule-41
22	Form No.B	Certificate for BOE  issued by Directorate of steam Boilers under rule-40 Boiler operation engineer-2011

Power plant consruction materials & welding electrodes used

Various sections and their significance of IBR act 1923.

Sl. No.	Section No.	Used for
1.	Section-7	New boiler registration
2.	Section-8	Renewal of boiler on form VI
3.	Section-9	Provisional order on form V
4.	Section-12	Boiler alteration, modification
5.	Section-23	Penalties on illegal use of boilers

Power plant maintenance Questions & Answers

Various regulations and their significance of IBR regulation 1950.

Sl. No.	Regulations	Used for
1.	376	Preparation for boiler inspection
2.	379	Boiler hydraulic test
3.	390	Boiler open inspection
4.	391A	Boiler RLA test or NDT tests

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