How do you calculate the efficiency of pumps???

 

How do you calculate the efficiency of a pump?

Efficiency of the pump is the ratio of output power to the input power

Pump efficiency = Output power X 100 / Input power

Centrifugal pumps efficiency varies from 40% to 95% well.

Pump efficiency is equal to the power of the water/liquid produced by the pump divided by the pump’s shaft power input.

A pump’s output power is determined by how much water and how much pressure it delivers 

On what factors pumps efficiency depends on?

Pumps efficiency depends on

• In put power
• Output power
• Type of liquid
• Fluid temperature
• Flow
• Suction & discharge head
• Losses & leakages
• Fluid viscosity
• Pump operating load
• Pumps internal frictions

What do you mean by volumetric efficiency of a pump?

It is the ratio of actual flow delivered by the pump to the theoretical flow

How do you calculate the Mechanical or hydraulic efficiency of the pump?

Mechanical/Hyd.efficiency = Pump output power X 100 / Pump input power or pump shaft power

Mechanical efficiency is also calculated as

 = Theoretical torque required to drive the pump X 100 / Actual torque provided to drive the pump

Calculations:

A centrifugal pump delivers 0.2 m3/sec flow at total head 27 m, calculate its hydraulic power. Assume density of water 995 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Pump hydraulic power = 0.2 X 27 X 995 X 9.81 / 1000

Pump hydraulic power = 52.7 KW

A centrifugal pump of capacity 0.05 m3/sec flow has to lift the water from 3.5 m deep well & has to discharge the water at 45 meter height. Calculate the pump efficiency if its shaft power is 30 KW

Assume water density 998 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Total head = Discharge head + Suction lift

Total head H = 45 + 3.5 = 48.5 meter

Pump hydraulic power = 0.05 X 48.5 X 998 X 9.81 / 1000

Ph = 23.74 KW

Pump efficiency =Pump hydraulic power X 100 / Pump shaft power

Pump efficiency =23.74 X 100 / 30

Pump efficiency =79.13%

A centrifugal pump of delivering  0.65 m3/sec flow at 75 meter discharge head, pump has positive suction head around 2.7 meter from overhead tank.Pump efficiency is 59%, calculate the motor input power if efficiency power if efficiency of motor is 92.5%.

Assume water density 1000 kg/m3

Pump hydraulic power = Pump flow in m3/sec X pump total head in meter X Fluid density in kg/m3 X 9.81 m/sec2 / 1000

Total head = Discharge head -Suction head

Total head H = 75 -2.7 = 72.3 meter

Pump hydraulic power = 0.65 X 72.3 X 1000 X 9.81 / 1000

Ph = 461 KW

Pump efficiency =Pump hydraulic power X 100 / Pump shaft power = Ph X 100 / Ps

59 =461 X 100 / Ps

Ps = 781.35 KW

Motor efficiency = Pump shaft power X 100 / Motor input power =Ps X 100 / Pm

92.5 = 781.35 X 100 / Pm

Pm = 844.7 KW

A centrifugal pump of efficiency 63% develops hydraulic power 27.5 KW, calculate the motor efficiency if motor input power is 45 KW

Pump shaft power, Ps = Pump hydraulic power Ph / Efficiency of pump

Ps = 27.5 / 0.63 =43.65 KW

Motor efficiency = Pump shaft power X 100 / Motor input power

Motor efficiency= 43.65 X 100 / 45

Motor efficiency = 97%

 

 

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50-Interview questions & answers on centrifugal pumps

 

1. What are the centrifugal pumps?

Centrifugal pumps are the mechanical devices which pump or transport various fluids by converting their rotational kinetic energy into hydrodynamic energy.

2. Why the name centrifugal pump?

A centrifugal pump uses centrifugal force

3. Where the centrifugal pumps find applications in power plants? 

Read reference books for power plant O&M

• Boiler feed water pump 
• Auxiliary & main cooling water pumps 
• Raw water transfer pumps  
• Condensate extraction pump,  
• Deaerator & feed water tank make up pumps 
• Firefighting water pumps 
• UF & RO feed water pumps 
• MGF feed pump 
• Degassed water transfer pumps 
• Sometimes lube & control oil pumps 

4. How do you specify the centrifugal pumps? 

Centrifugal pumps are specified as bellow 

• Flow in M3/Hr or M3/sec 
• Head or discharge pressure in meter or bar or kg/cm2 
• Shutoff head 

5.What are the various parts of centrifugal pumps? 

Centrifugal pumps have following parts 

• Pump casing or diffuser 
• Impeller 
• Wear ring 
• Shaft 
• Lantern ring 
• Stuffing box 
• Inlet vertex 
• Mechanical seal or gland packing 
• Shaft sleeve 
• Bearings 

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6.What are the energy conversions take place in centrifugal pumps 

In centrifugal pumps hydraulic energy is being converted into kinetic energy  

7.What types of reducers are used at pump suction & discharge ends? 

Suction side: Eccentric type & Discharge side: Concentric 

8.What are the two main types of centrifugal pumps?

Axial flow & Radial flow

9.What is the function of impeller in centrifugal pumps?

It converts kinetic energy of pump into hydrodynamic energy by rotary motion

10.What is the function of pump casing?

Casing converts velocity head from impeller into pressure head & also guides the flow to the discharge end.

11. What are the types of pump casing?

Volute & diffusers are two different types of pump casing

12. What do you mean by volute?

A volute is a spiral-like geometry with an increasing through-flow area, reducing the velocity of the fluid and increasing the static pressure

13. What are the different types of volutes?

Single volute & Double volute

15. Write down the working principle of centrifugal pumps

In centrifugal pumps, fluid enters the impeller through inlet eye & exists along the circumference between the vanes of impeller. This impeller is connected to shaft & in turn to motor, this rotary motion of the impeller converts kinetic energy of the fluid into hydrodynamic energy.

16.What are the types of impellers?

Open impeller: As its name suggests, an open impeller has vanes that are open on both sides without any protective shroud. These are structurally weak.

These are used for low flow & low head applications. Generally used for pump solids or sludge. These require much NPSH.

Semi open impeller: Semi-open impellers have a back-wall shroud that adds mechanical strength to the vanes.

Closed impeller: Are very robust & require low NPSH

Impellers are also classified as single suction & Double suction

17.What are the rotary & stationary parts of the pumps?

Rotary parts:

• Shaft
• Impeller
• Shaft sleeve
• Bearings

Stationary Parts

• Pump casing
• Gland packing or mechanical seal
• Lantern ring

18.Why eccentric reducers are used at pump suction side? 

To avoid air locking & cavitation eccentric reducers are used at suction side 

19. What do you mean by the NPSH in pumps? 

It is the net positive head required at pump suction to avoid cavitation 

20. What do you understand by the term cavitation? 

Cavitation is the formation & collapsing of vapor bubbles at pump’s suction 

21. How the cavitation does affect the pump’s life? 

• Cavitation causes 
• Vibrations in pump 
• Damage of impellers 
• Heavy noise 

22. What are the factors considered for centrifugal pumps design? 

• Flow required 
• NPSH available & NPSH required 
• Total head 
• Pump efficiency 
• Fluid used 

23. What are the materials used for pump casing? 

Generally cast steel or cast iron are used for single stage centrifugal pumps 

24. What are the materials used for Impellers? 

Impellers are made up of cast iron, gun metal & stain less steel 

25. What is the function of wear ring? 

As the name indicates it protects the wear & tear of impeller 

26. What do you mean by static suction head in pump?

Therefore, the static suction head is the vertical distance from the center line of the pump to the free level of the liquid to be pumped.

27. What do you mean by static suction head in pump?

Static discharge head is the vertical distance between the pump centerline and the point of free discharge or the surface of the liquid in the discharge tank.

28. What do you mean by total static head?

Total static head is the vertical distance between the free level of the source of supply and the point of free discharge or the free surface of the discharge liquid.

29. What do you mean by total head?

It is total dynamic discharge head plus total dynamic suction head

Note: If source water level is below the pump center line, then

Total head = Discharge head Suction lift

If source Water level is above the pump suction line, then

Total head = Discharge head-Suction head

30. What are the problems associated with centrifugal pumps? 

Following are the common problems associated with pumps 

• Low discharge pressure 
• Low delivery 
• Cavitation 
• High vibrations 
• Pump seize 
• Over load 
• More suction lift 
• Air locking & No priming 

 31. What are the reasons for no delivery or no discharge in centrifugal pumps?

•  Probable reasons are
• Air lock in pump suction
• Suction valve closed
• Low tank level
• 32. What are the reasons for low delivery?
• Suction valve partially opened
• Reverse rotation of pump
• Low speed of pump
• Suction strainer is chocked

 33. What are the reasons for over load of pump?

•  More flow
• High speed
• Reverse rotation of pump
• Pump discharge kept open to atmosphere
• Internal friction in impeller & wear ring or impeller & casing
• More tightened gland packing
• No lubricant in bearing or bearing seized

 34. What are the potential reasons for pump vibrations?

•  Overloading of pump
• Reverse rotation of pump
• Impeller rubbing inside the casing
• Misalignment
• Damaged bearing
• Shaft run out
• Shaft imbalance

 35. Too much noise coming from pump inside, what does this mean?

•  Air lock in pump
• Overloading of pump
• Pump discharge line is less than actual required
• Cavitation
• No lubricant in bearings

 36. What are the common mistakes done during pump installation?

• Choosing poor foundation
• Note: Pump foundation weight should be 3 to 4 times the pump weight
• Lesser size suction pipe line
• Lesser size discharge pipe line
• Interchanging concentric & eccentric reducers

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37. What are the safety protections & interlocks given for a centrifugal pumps?

• Over load
• Low load
• High bearing vibrations
• High bearing temperature
• High suction DP
• Source water level low

38. How do you increase the head & flow of pump by modifying impeller size?

By increasing the impeller diameter head & flow can be increased

By increasing the impeller width flow can be increased

 39. What are the reasons for reduction of pump efficiency?

• Operating the pump at lower capacity
• Operating the pump at higher load
• Throttling the discharge valve
• Increase in impeller & wearing clearance
• Lower suction head
• High suction lift

Calculation part

 40. How do you calculate NPSHA?

 NPSHA is Net positive suction available

NPSHA = Atmospheric pressure + static head - vapor pressure - pressure loss in the suction piping - pressure loss due to the suction strainer.

 41. A centrifugal pump of rated capacity 75 M3/Hr & total head 35 meter is supplying water to fill a tank in 2 hours, calculate the total power consumption. Consider pump & motor efficiency 50% & 85% respectively

 Power consumption = Pump flow in m3/sec X Pump total head in meter X fluid density X g / (1000 X Pump eff. X Motor eff)

Power consumption = (75/3600) X 35 m X 1000 kg/m3 X 9.81m/s2 / (1000 X 0.5 X 0.85)

Power consumption = 16.83 KWH

Power consumption in 2 hours = 16.83 X 2 = 33.66 KW

 42. A centrifugal pump having hydraulic power 22 KWH, discharge & suction head 55m & 12m respectively

Calculate the pump flow in m3/hr, assume density of water 990 kg/m3

Pump flow = Pump hydraulic power X 1000 / (Pump total head X density of fluid kg/m3 X 9.81 m/s2)

Pump flow = 22 X 1000 /( (55-12) X 990 X 9.81)

Pump flow = 0.052 m3/sec

Pump flow in M3/hr = 0.052 X 3600 = 189.6 M3/hr

 43. A centrifugal pump having hydraulic power 15KWH & pump efficiency 65% calculate the pump shaft power

 Pump shaft power = Pump hydraulic power / Pump efficiency = 15 / 0.65 = 23 KW

 44. A centrifugal pump produces flow 20M3/hr (Q1) flow at rated speed 1500 RPM (N1) , then calculate the flow of pump at 1000 RPM(N2)

 We have pump affinity law

 Q1/Q2 = N1/N2

20 / Q2 = 1500 / 1000

Q2 = 13.33 M3/hr

45. A centrifugal pump consumes power of 25KW (P1) at speed of 1500 RPM (N1), after reducing certain RPM its power consumption reduces by 5 KW (P2), calculate that speed

 We have pump affinity law

 P1/P2 = (N1/N2)3

25 / 5 = (1500 / N2)3

N2 = 877.2 RPM

 46. A centrifugal pump produces 150 m (H1) head at 3000 RPM (N1), calculate the head produced if its speed reduced to 50%

We have pump affinity law

H1 / H2 = (N1/N2)2

N2 = N1 X 50% = 3000 X 0.5 = 1500 RPM

150 / H2 = (3000 / 1500)2

H2 = 37.5 meter

47. A centrifugal pump having impeller diameter 250 mm produces flow 250 M3/hr, calculate the diameter of impeller to produce flow 300 M3/hr

We have

Q1 / Q2 = D1 / D2

250 / 300 = 0.250 / D2

D2 = 0.35 m = 350 mm

48. A centrifugal pump having impeller diameter 300 mm produces 250 m head & what could be the diameter if we want to reduce the head by 30m

Reduced head = 250 – 30 = 220 m

We have

H1 / H2 = (D1/D2)2

250 / 220 = (300 / D2)2

D2 = 281.4 mm

49. A centrifugal pump having impeller diameter 150 mm (D1) consumes 15 kw (P1), what is the size of impeller if we want reduce power by 4 KW

P2 = P1-4 = 15-4 = 11 KW

We have

P1 / P2 = (D1 / D2)3

15 / 11 = (150 / D2)3

D2 = 135.2mm

 

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How do you calculate the efficiency of pumps

Interview questions and answers on Boiler fans

 

1-What do you mean by fan & and what are the various fans used in power plant Boilers?

Fan is a Mechanical device , used to move volume air, vapour or flue gas at specific static pressure

Fans used are;

1-Induced draft fan (ID fan)

2-Forced draft fan (FD fan)

3-Secondary air fan (SA fan)

4-Cinder reinjection fan (CR fan)

2-What type of fans used in Boiler?

Generally centrifugal radial, forward & backward curved fans

3-What are the function of Boiler fans in Boiler

SL No.	Fan type	Functions
1	ID fan	1-To suck & expel the flue gas from Boiler 2-To maintain negative & balanced draft in furnace
2	FD fan	1-To provide combustion air for Boiler 2-To maintain balanced draft in Boiler 3-To maintain fluidisation on bed
3	SA fan/PA fan	1-To carry fuel into the furnace or bed 2-To assist fuel in suspension burning (OFA) 3-To spread the fuel in case of Biomass Boilers
4	CR fan	To re-inject unburnt fuel from Bank, Economiser & APH back into the furnace

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4-Which type of fan is more efficient & why?

Centrifugal backward curved is more efficient as they do not over load & speed controlling is very easy. Fan operates beyond the maximum efficiency point at lower in put power

5-Differentiate between forward & backward curved centrifugal fans

Sl No.	Backward curved	Forward curved
1	Less number of blades	More number of blades
2	For this type of impeller, flow is produced in a radial direction because the impeller develops static pressure across the longer length of blade.  On the front side of the blade a positive pressure is generated pushing the air outwards and on the reverse side of the blade a negative pressure is generated	For this type of impeller, flow from the impeller is produced in a tangential direction.  The forward curve of the blade imparts kinetic energy to the air requiring a scroll housing to convert the kinetic energy into static pressure
3	Does not over load the motor at higher flow rates	Overloads the motor at higher flow rate
4	High pressure	Low pressure
5	Lower & medium flow rate	Higher flow
6	Higher efficiency	Lower efficiency

 6-What are the various parts of Boiler fans (centrifugal type)?

• Parts of Boiler fans
• Casing
• Inlet cone
• Evase
• Inlet box
• Impeller blade
• Impeller shroud
• Impeller back plate
• Shaft & coupling
• Bearings

7-What are the material of composition (MOC) of Fan’s various parts

SL No.	Particular Fan part	MOC
1	Casing	Mild steel IS 2062
2	Inlet cone & inlet box	Mild steel IS 2062
3	Evase	Mild steel IS 2062
4	Impeller blade	SAILMA 350
5	Impeller shroud	SAILMA 350
6	Impeller back plate	SAILMA 350
7	Shaft	EN-8

 8-What is thickness of fans casing

Casing of ID fans is 8mm & that of FD, SA & PA fans is 6mm

9-What are factors considered for Boiler fans design?

Following factors are considered while designing the Boiler fans

• Required flow
• Inlet & outlet static pressure
• Total pressure
• Operating & design temperature
• Density of gas/air
• Dust load
• Relative humidity
• Nose level
• Fan speed

 10-What type of bearings used for Boiler fans?

Generally, spherical roller bearings for fans having speed up to 1500 TPM

11-What are the various instruments used for fans?

Draft sensor (draft transmitter)

Flow meter (Aerofoil)

Temperature sensors

12-What are the protections given for fans?

Start permissive,

Inlet Damper close

Bearing temperature normal

Bearing vibration normal

Protections:

High bearing vibrations trip > 7mm/sec

High bearing temperature trip>90 deg C (for journal & rolling contact bearings)

High/low static pressure

Over load

13- Explain the terms static pressure, total pressure and velocity pressure.

Static Pressure is

Resistance to flow

Equal in all directions

Can be Positive or Negative

Independent of air velocity

Measured by pressure tap perpendicular to airflow

Used for fan selection

 

Total Pressure is

 

A fluid in motion will exert a Total Pressure on an object in its path.

Total Pressure measured by pressure tap pointed directly into the air stream.

Used to find velocity pressure.

Velocity Pressure

Cannot be measured directly.

A Pitot tube uses both Static pressure and Total pressure taps.

Used for measuring CFM in a system.

Total Pressure = Static Pressure + Velocity Pressure

17- What is the differential static pressure?

 It is the difference between the static pressure at fan inlet and out let

18- What are the flow control methods employed in fans?

 Controlling inlet guide vanes

Discharge damper control method

 V belt position change on pulley (If fan is stepped pulley driven type)

 Incorporating Variable Frequency Drives (VFD) to fan motors to control the speed

19-Calculate the differential static pressure of ID fan, whose inlet static pressure is -300 mmwc & outlet 0 mmwc

We have,

Differential static pressure = Outlet static pressure-Inlet static pressure = 0-(-300) = 300 mmwc

20- Calculate the differential static pressure of FD fan, whose inlet static pressure is -10 mmwc & outlet 250 mmwc

Differential static pressure = Out let static pressure-Inlet static pressure = 250-10= 240 mmwc

21-What size of FD air duct is required to get flow 70 m3 /sec. at 11 m/sec. speed?

We have,

 Volumetric flow = Area of duct (M²) X Velocity of air (m/sec.)

Area of duct = 70 / 11 = 6.36 M²

 

22- State the relation between fan’s flow, static pressure, and speed and power consumption.

Following are the some affinity laws related to fans:

A. Fan speed is directly proportional to its flow.

(N1/N2) = (Q1/Q2)

Where N1 & N2 are speed RPM & Q1 & Q2 are flow in M3/hr

B. Static pressure is directly proportional to square of the speed.

(SP1/SP2) = (N1/N2)²

Where, SP1 & SP2 are static pressure at two different speed/RPM

C. Power consumption is directly proportional to cube of speed.

(P1/P2) = (N1/N2)³

Where, P1&P2 are power consumption at speed N1 & N2

23-Calculate the power consumed by an ID fan if its speed reduced from 900 RPM to 750 RPM, take 120 KW power consumption at 900 RPM

As per affinity law

(P1/P2) = (N1/N2)³

(120/P2) = (900/750)³

P2 = 69.44 KW

24-Calculate the anticipated static pressure when FD fan RPM is increased from 1100 RPM to 1175 RPM, static pressure at 1100 RPM is 220 mmwc

As per affinity law

(SP1/SP2) = (N1/N2)²

(220/SP2) = (1100/1175)²

SP2 = 268.3 mmwc

25-Boiler ID fan of flow capacity 55 m3/sec having static inlet pressure & efficiency are  is 300mmwc & 78% respectively, calculate the rated power required to run the fan if motor of efficiency 96% is used to drive the fan.

We have,

Motor rated power = Flow (M3/sec) X Static differential pressure (mmwc) / (102 X Fan static efficiency X motor efficiency)

Motor rated power P = 55 X 300 / (102 X 0.78 X 0.96)

Motor rated power P =216.03 KW

26-A Boiler FD fan of capacity 24m3/sec having total pressure 265 mmwc & Mechanical efficiency 82%, then calculate the fan shaft input power

Fan shaft power = Flow (M3/sec) X Total differential pressure (mmwc) / (102 X Fan mechanical efficiency)

Fan shaft power = 24 X 265 / (102 X 0.82) =76.04 KW

27-Calculate the rated capacity of Boiler FD fans based on following data.

SL No.	Particular	Value
1	Boiler capacity	100 TPH
2	Coal consumption	17 TPH
3	Coal to air ratio	1:6
4	Fan inlet static pressure	5 mmwc
5	Draft loss in ducts from Fan to APH	15 mmwc
6	Draft loss in APH	80 mmwc
	Draft loss aerofoil flow meter	25 mmwc
7	Draft loss in ducts from APH to plenum	20 mmwc
8	Draft loss in Grate	50 mmwc
9	Static efficiency of the fan	75%
10	Motor efficiency	95%
11	Air temperature	28 deg C

 

Solution,

Total air required for burning the coal in Boiler = 17 X 6 =102 TPH

Take 20% extra margin on air flow , then air flow = 102 X 120% = 122.4 = 123 TPH

Convert air flow from TPH to m3/sec

Density of air at temperature 28 deg C = 273 X 1.293 / (28+273) = 1.17 kg/m3

Where, 1.293 is density of air at 0 deg C

Air flow = 123000 kg/hr /1.17 =29.2 m3/sec

Based on above flow we can select, two fans of capacity 29.2/2 = 14.6 or 15 m3/sec

Now calculate the total static pressure required to over come the resistance in air flow path

Pressure/draft loss drop = Draft loss in APH inlet ducts + Draft loss in flow element + Draft loss in APH + Draft loss in APH to grate ducts + Draft loss in grate

= 15 + 80 +25 + 20 + 50 = 190 mmwc

Take 20% extra margin on draft

Discharge static pressure = 190 X 120% = 228 = 230 mmwc

Now calculate the motor rated power, P = Flow (M3/sec) X Static differential pressure (mmwc) / (102 X Fan static efficiency X motor efficiency)

P = 15 X (230-5) / (102 X 0.75 X 0.95) = 46.43 KW

Take 15% extra margin on motor capacity

Then, compensated rated motor power = 46.43 X 115% =53.4 KW

Select nearest standard size motor i.e 55 KW

Therefore for boiler of capacity 100 TPH, FD fans capacity 55 KW X 2 are required

28-Why does ID fan trip during Boiler operations?

Reasons are

1-Over load

2-High bearing vibrations

3-High bearing temperature

4-Electrical& Instrumentation related issues like short circuit, mal functions etc

5-Low drum level

6-High Boiler pressure & temperature (if interlocks are provided)

7-Low pressure (more negative pressure) in furnace

29-Why does FD fans trip during Boiler operations?

Reasons are

1-Over load

2-High bearing vibrations

3-High bearing temperature

4-Electrical& Instrumentation related issues like short circuit, mal functions etc

5-High drum level

6-High Boiler pressure & temperature (if interlocks are provided)

7-Low pressure (more negative pressure) in furnace

8-Upon tripping of ID fan

30-What will happen if Boiler ID fan trips (one fan out of two)?

1-Same stream FD fan trips

2-Same stream SA fan trips

3-If pressure drops fuel feeding system trips

4-Boiler pressure becomes low & eventually boiler load reduces

31-What will happen if Boiler FD fan trips (one fan out of two)?

1-Same stream SA fan trips

2-If pressure drops fuel feeding system trips

3-Boiler pressure becomes low & eventually boiler load reduces

32-What are the maintenance activities that you will carryout on Boiler fans

• Fans cleaning
• Casing thickness checking
• Impeller inspection, includes Shroud plate, blade and back plate thickness checking
• Wear liner inspection and hard facing
• Inlet cone inspection
• Drive end non drive end bearings overhauling and grease replacement
• Shaft run out checking
• Coupling inspection
• Shaft ultrasonic continuity test
• Suction and discharge dampers and pneumatic cylinders overhauling
• Motors servicing
• Realignment
• Manhole doors and inspection doors proper sealing
• Fan trial and dynamic balancing at site

33-Write down the start of steps of Boiler fans

Start-up checks

• Ensure all the maintenance activities are carried out on fans and equipments are normalized
• Ensure all the instruments like bearing temperature sensor, vibration sensor etc are fitted & showing readings in DCS
• Ensure none of the interlocks are bypassed
• Ensure Fans suction & discharge dampers are in closed condition
• Ensure coupling & bearing cooling fan guards are fitted properly
• Ensure local emergency push button is healthy & in good condition
• Give clearance to DCS or electrical person to start at 10% initially
• Check abnormal sound from fan casing, bearing & couplings, if found normal open discharge damper slowly & subsequently suction damper

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