The Elimination of erosion phenomena in the main cooling water pumps in thermo power plant (TPP) “Kosova B” : [presentation given October 7, 2009]

Rochester Institute of Technology

RIT Scholar Works

Theses Thesis/Dissertation Collections

2009

The Elimination of erosion phenomena in the main

cooling water pumps in thermo power plant (TPP)

“Kosova B” : [presentation given October 7, 2009]

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Presented by

Abedin KABASHI

Introduction

Problem background

g

Project Description

Project Findings

This Capstone Project will address the erosion phenomena

on the impeller blades of the main cooling water pumps in

Thermal Power Plant (TPP) “Kosova B”. The focus is on

pumps suction line or the line between the cooling tower

basin and the pumps intake

TPP “Kosova B” has two units B1 -- 339 [MW]

B1 339 [MW] B2 -- 339 [MW]

The cooling system is one of the largest and most complex systems in TPP “Kosova B”

The cooling system includes Cooling water

A li

A cooling tower

….. water used for cooling is taken from Decarbonized Water Treatment Plant

Inhibitors against scaling or deposit formation (3DT 149)

Inhibitors against corrosion (N-73190) Inhibitors against corrosion (N 73190)

Biological inhibitors (a mixture of N-3434 and NaOCL)

Cooling water is used for:

Cooling water is used for:

Condenser

Turbine oil coolers

Generator hydrogen coolers Exciter air coolers

The primary task of the

cooling tower to reject heat

into the atmosphere

The technical data and features of

th

li

t

t TPP “K

into the atmosphere

Reinforced Concrete Shell

the cooling tower at TPP “Kosova

B”

The type of the cooling tower: “Natural draught cooling tower”

Fill

Drift eliminator

Natural draught cooling tower The height of the cooling tower basin: 132 [m]

Th di t f th li

Diagonal Columns

Cooled Water

Warm water

Tower basin

The diameter of the cooling tower basin is 103 [m]

Discharge head Bearing with

bb i id

Thrust Bearing Shaft seal

The technical data and features of the main

cooling water pumps at TPP “Kosova B”

head rubber inside

Pumps manufacturer CCM SULZER

The pump water capacity is 4.84 [m3/s]

Pump shaft Column _assembly

The power of the driver motor is 1.3 [MW]

The supply voltage of driven motor is 6.3 [KV]

Th t th h h i 148 [A]

Bowls Bearing with

rubber inside

The current through phases is 148 [A]

The number of rotations is 495 [rpm].

Single stage mixed pumps

Impeller rubber inside

Single stage mixed pumps.

Bronze intermediate bearings with rubber inside.

Bronze impeller. p

The surfaces of the

impeller blades

affected from the

The broken blade due to

the slimming of the

blade thickness in the

The erosion phenomenon has existed since the very beginning of the TPP

“Kosova B” operation (since 1986)

affected from the

erosion phenomena

blade thickness in the

affected place from the

erosion

Kosova B operation (since 1986), but has never been analyzed before

Cavitation occurs

when the pump cannot get enough liquid and the resulting reduction

in pressure causes p

Necessary Resources for conducting the Capstone Project

The archives in the library at TPP “Kosova B” The archives in the library at TPP Kosova B

The library in the Mechanical Faculty at the University of Prishtina

Th lib i th M h i l F lt t th U i it f Ti

The library in the Mechanical Faculty at the University of Tirana

Documents related to this problem from the library of the CCM SULZER

Recourses form Master Program at AUK

Context an Trends

Introduction to Project Management

Asset Management

First step

First step – Collection of information pertaining to the pumps suction line

Second step

Second step -- Collection of reference books and web sites that have to do with Fluid Mechanics, Water Quality , Mixed Flow Pumps and Cooling Towers.

Thi d t

Thi d t Th t f th t l it

Third step

Third step -- The measurement of the water velocity

Fourth step

Fourth step -- Water flow definition and defining the friction factor

Fifth step

Fifth step -- The analyze of the devices efficiency (pumps plus driver motors)

Sixth step

Sixth step – Calculation of head losses through suction line, and

Last step

The open channel, at a length of 10.827 meters, is a common channel for both units, B1 & B2, which is

di id d i id i l h l

SCREEN

divided into two identical channels through a dividing concrete wall

Open channel

Pipe miter bends

The dimensions and features of the pipe

line are as follows

L = 42 [m] is the length of the first part of the

L1, d Detail C

Detail C

L₁= 42 [m] – is the length of the first part of the pipeline

L₂= 42 [m] – is the length of the second part of the pipeline

L2, d

p p

L₃= 42 [m] - the length of the third part of the pipeline

L₃, d

L₄, d

amber

L₄= 42 [m] - the length of the fourth part of the pipeline

d = Ø 2176 [mm] = Ø 2.176 [m] – is the pipe

i di t

Suction Ch

a

Pipe miter

bends _{The end of}

inner diameter

δ = 22 [mm] – the wall thickness of the round pipe,

The end of pipe ]

[ 72 . 3 4

14 .

3 2 2

m d

The suction chamber is

common chamber for the

two cooling water pumps

that are placed in parallel

Real pump capacity

Q_p = η_p x Q = 17427.84 [m/h] = 4.84 [m/s]

Where:

η_p=0.87

Q = 20032 [m3/h] -- is the ideal pump’s capacity

η_p = 0.87 – is the pump efficiency

ηη

=

=

ηη

P

82

.

0

2

1

=

=

=

EL MECH p

p

P

P

η

η

67

.

0

82

.

0

82

.

0

2

1

⋅

=

⋅

=

=

tot

η

η

η

-- The total efficiency of a device (pump & driver motor)

-- The total efficiency of devices (pumps & driver motors)

Where

P_MECH = Q_p x p_p = 1084160 [W] = 1084.16 [KW] – The mechanical power of pump

η

p1 =

η

p2

p_p=2.24 [bar] = 2.24 x 105 _[N/m2_{] – is the pump discharge pressure}

P_EL= I x U x cosφ x 3^(0.5) = 1327.50 [KW] – the Electrical power of driver motor

I = 145 [A] -- current

U = 6.3 [KV] – supply voltage

The water flow may be:

Laminar Re < 2000

i i l

Transitional 2000 < Re > 4000

Turbulent Re > 4000

ρ

⋅

d

⋅

V

μ

ρ

⋅

d

⋅

V

=

Re

ρ = 994.08 [kg/m3] – is the water density for the cooled water temperature at 35 [°C] -- the Reynolds number for water

flow through the pipeline

d = Ø 2.176 [m] – is the pipe inner diameter,

μ = 0.00072 [kg/m s] – is the dynamic viscosity of water at 35 [°C]

For only one pump in operation

Re_pipe(I) = (ρ x d x V_(I) / μ )= 3.905 x 106 _{> 4000}

For two pumps in parallel operation For two pumps in parallel operation

For only one pump in operation

Re_channel(I)= Re_pipe(I) / 4 = 0.976 x 106 ₄₀₀₀

d

δ_WALL … the flow is

turbulent

λ _δ/D

106_>4000

For two pumps in parallel operation

δ/D

Re_channel(II)= Re_pipe(II)/ 4 = 1.32 x 106

> 4000

• Friction factor

where:

δ_WALL = 22 [mm] – is the pipe Rechannel= 0.976 x 106

6

WALL [ ] p p

wall thickness

d = 2176 [mm] –is the pipe inner diameter

Rechannel = 1.32 x 106

Re_pipe= 3.905 x 106 Re_pipe = 5.29 x 106

W t V l it V [ / ]

The measurement of the water velocity in the open channel is done by the INKOS Institute

Water Velocity V [m/s] Point 1 Point 2 Point 3

0.588 0.569 0.716 0 566 0 586 1 13

Noise [dB] 96 97 0.566 0.586 1.13

0.565 0.684 1.15 99

1 2 ) ( 2 ) ( Q

V_{I ⎜}⎛ _{I ⎟}⎞

For only one pump in operation

First Row of Screens

(Strainers

2 _{0 1}

) ( 2 ) ( 1 ) ( 1 m g A Q h g V h

H_{L str I} I I =

⋅ ⋅ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⋅ + = − −

For two pumps in parallel operation

V

V

H_{L str II} II II =

⋅ ⋅ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − − ⋅ + = − −

h = 2 895 [m] is the water depth before entering

2 II

V

h₁ h₂

h₁ = 2.895 [m] – is the water depth before entering to first row of screens or strainers

h₂ = 2.745 [m] – is the water depth after first row of screens or strainers

⎤ ⎡ +

+V V m

V ⎥⎦ ⎤ ⎢⎣ ⎡ = ⋅ = s m Q Q V V II I II

I 0.46

) ( ) ( ) ( )

( -- The water velocity before first row of screens

⎥⎦ ⎤ ⎢⎣ ⎡ = + + = − − − s m V V V

VII I II III 0.613 3

2 2

2 )

( -- The water velocity before first row of screens for two

(

)

5

.

957

[

]

) ( 1

1

0

A

A

A

m

A

=

−

+

=

Where

-- The clear surface of screens or strainers

PLATES ] [ 01 . 9 2 2

1 W D m

A = ⋅ ⋅ =

Where

W = 1.58 [m] -- the width of useful screens or strainers

-- The total surface of submerged screens in water

D = 2.85 [m] – the height of useful screens or strainers.

] [ 877 . 1 ) ( 2 2 )

( A A m

A_{w H+V} = ⋅ _wh + _wv = -- is the total occupied surface

f th h i t l d ti l

M_Wh= 34 WIRES

δWIRE= 4 [mm]

from the horizontal and vertical wires ] [ 5297 . 0 2 m N L

A_wh = δ_W ⋅ _h ⋅ _wh = -- is the occupied surface from the horizontal wires

SCREEN (STRAINER) Wh [mm] ] [ 409 . 0 2 m N L

A_wv = δ_W ⋅ _v ⋅ _wv = _{-- the occupied surface from the} vertical wires ] [ 176 1 ) 08 0 75 0 6 08 0 85 2 ( 2 2

A Th i d f _{MWv= 44}

[mm] ] [ 176 . 1 ) 08 . 0 75 . 0 6 08 . 0 85 . 2 ( 2 2 m

A_plate = ⋅ ⋅ + ⋅ ⋅ = -- The occupied surface

Second Row of

Screens (Strainers)

H_{L str I} I I _⎟⎟ ⋅ =

⎠ ⎞ ⎜⎜ ⎝ ⎛ − − + = − −

For only one pump in operation

2 2⋅g ⎜_⎝A₀₋₂⎟_⎠ ⋅g

] [ 107 . 0 2 1 2 2 ) ( 3 2 ) ( 2 1 2 ) ( 2 m A Q h V h

H_{L str II} II II _⎟⎟ ⋅ =

⎠ ⎞ ⎜⎜ ⎝ ⎛ − − + = − −

For two pumps in parallel operation

V

2

2 _{0 2}

) (

g A

g ⎜_⎝ ⎟_⎠ ⋅

⋅ ₋

Where

h₂ = 2.745 [m] – is the water depth before entering to

th d f th t d th b f th

h₂ h₃

the second row of screens or the water depth before the first row of the screens or strainers

h₃ = 2.745 [m] – is the water depth after the second row of screens or strainers

⎥⎦ ⎤ ⎢⎣ ⎡ = ⋅ = − − s m V A A

V _{I (I)} 0.304

1 1 0 2 ) ( 2 1 ⎤ ⎡

-- the water velocity between the first and second row of screens or strainers for only one pump in operation

⎥⎦ ⎤ ⎢⎣ ⎡ = ⋅ = − − s m V A A

V _{II (II)} 0.41

1 1 0 2 ) ( 2

1 -- the water velocity between the first and second row of screens

(

)

) ( 2

2

0 A A A m

A ₋ = − _{w H+V} + _plate =

Where

-- The clear surface of the second row of screens or strainers

] [ 564 . 8 2 2

2 W D m

A = ⋅ ⋅ = Where

-- The total surface of submerged screens in water

] [ 734 . 1 ) ( 2 2 )

( A A m

A_{w H+V} = ⋅ _wh + _wv = -- The total occupied surface from the horizontal and vertical wires

] [ 489 . 0 2 m N L

A_wh =δ_W ⋅ _h ⋅ _wh =

] [ 378 . 0 2 m N L

A_wv =δ_W ⋅ _v⋅ _wv =

-- is the occupied surface from the horizontal wires

-- the occupied surface from the ti l i

WIRES

δWIRE= 4

vertical wires ] [ 14 . 1 ) 08 . 0 75 . 0 6 08 . 0 63 . 2 ( 2 2 m

A_plate = ⋅ ⋅ + ⋅ ⋅ = -- The occupied surface

of screens from the plates

MWh= 34 [mm]

[mm]

…..the sum of head losses through

the first and second row of

H

= H

+ H

screens (strainers)

For only one pump in operation

]

[

256

.

0

) ( 2 )

( 1 )

(

H

H

m

H

=

+

=

y p p p

]

[

216

.

0

) ( 2 )

( 1 )

(

H

H

m

H

=

+

=

Open channel L1= 42 [m]

L₂= 42 [m] L₃= 42 [m] L₄= 42 [m]

…. are caused due to friction between

water and pipe wall

L₁, d

4 [ ]

d = Ø 2176 [mm]

For only one pump in operation

] [ 137 . 0 2 2 2 2 ) ( 2 ) ( m d L A g Q g V d L

H_{LF I} p I ⎟=

⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

=λ λ

] [ 247 . 0 2 2 2 2 ) ( 2 ) ( m d L A g Q g V d L

H_{LF II} p II ⎟=

⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

=λ λ

For two pumps in parallel operation

Wh _L

2, d

L₃, d

L₄, d

Where

λ=0.0395 – friction factor ( Moody chart)

L = L₁+ L₂+L₃+L₄=87.58 [m] – the total pipeline length

Q Q 4 84 [ 3_{/ ] th it f l} 3,

n

Chamber

Q_(I)=Q_p=4.84 [m3_{/s] – the capacity of only one pump}

Q_(II)=(Q_p1+Q_p2) η_tot=(4.84+4.84) 0.67=6.486[m3/s] – the capacity of two pumps in parallel operation

A = 3.14 x d2_{/4 [m}2_{] – The pipeline cross section area}

Suctio

n

d = Ø 2.176 [mm] – the pipe inner diameter

ζ_IN Open channel

….. are losses through the pipe miter bends are classified as “minor losses”. Minor losses include energy losses resulting from the rapid changes in the direction or the

it d f th t l it i th i li

ζ₁

ζ₂

ζ_IN= 1

ζ₁= 0.20

ζ₂= 0.20

ζ₃= 0.10

ζ₄= 0.35

For only one pump in operation

] [ 248 . 0 2 2 2 ) ( ) ( m g A Q

H _{LM I} p I ⋅ =

⋅ ⋅

= ζ

magnitude of the water velocity in the pipeline

ζ₄ 0.35

ζ₅ = 0.024

ζ_EX= 1

For two pumps in parallel operation

Where g ] [ 446 . 0 2 2 2 ) ( ) ( m g A Q

H _{LM II} p II ⋅ =

⋅ ⋅

= ζ

Mitered Bend Angle ζ

θ= 5o _0.016

ζ₃ er 874 . 2 ) ( + ₁+ ₂+ ₃+ ₄+ = =

Σζ ζ_IN ζ ζ ζ ζ ζ_EX -- the sum of coefficients due to pipe miter bends

θ= 15o _0.05

θ= 30o _0.10

θ= 45o _0.20

θ 60o _{0 35}

θ ζ4 ζ_{5 tion Chamb}

e

θ θ= 60

o _0.35

θ= 90o _0.80 ζEX

ζ₅

Suc

….. the sum of head losses due to friction

and due to pipe miter bends

H

= H

+ H

Where

H_LF [m] -- are the head losses due to the friction in the pipeline

H_LM [m] -- are the head losses due to the minor losses (pipe miter bends) in the pipeline

For only one pump in operation

H

H

+ H

0 137 + 0 248 0 385 [ ]

H_LM [m] -- are the head losses due to the minor losses (pipe miter bends) in the pipeline

H

= H

+ H

= 0.137 + 0.248 = 0.385 [m]

For two pumps in parallel operation

…. are caused due to sudden

contraction duct

For only one pump in operation

] [ 18 . 0 2 2 ) ( ) ( m g V

H_{L CONTRACTION I CONTRACTION} out I =

⋅ ⋅ =

− ζ

For two pumps in parallel operation

2 g ] [ 35 . 0 2 2 ) ( ) ( m V

H_{L−CONTRACTION II} =ζ_CONTRACTION ⋅ out II = [ ] 2 ) ( g N CONTRACTIO II N CONTRACTIO

L ζ _⋅

Where

71 0

OUT

A

A_IN = 3.26 [m2_]

AOUT/AIN 0.01 0.10 0.20 0.40 0.50 0.60 0.70 0.80

42 . 0 = N CONTRACTIO

ζ _for 0.22

26 . 3 71 . 0 = = IN OUT A A 2 N CONTRACTIO

ζ 0.05 0.49 0.42 0.33 0.30 0.25 0.20 0.15

[ ]

] [ 93 . 2 ) ( ) ( s m A A V V OUT IN I IN I

OUT = ⋅ = -- The water velocity at the duct outlet for only one pump in operation

] [ 07 . 4 ) ( ) ( s m A A V V OUT IN II IN II

OUT = ⋅ =

-- The water velocity at the duct outlet for two pumps in parallel operation ⎤ ⎡ L ⎥⎦ ⎤ ⎢⎣ ⎡ = − ⋅ = ₋ s m B L V

V_{IN (I) Out pipe (I)} (1 ) 0.64 -- The water velocity at the duct inlet only one pump in parallel

⎥⎦ ⎤ ⎢⎣ ⎡ = − ⋅ = ₋ s m B L V

V_{IN(II) Out pipe(II)} (1 ) 0.89 -- The water velocity at the duct inlet for two pumps in parallel

i ⎦ ⎣ operation ⎥⎦ ⎤ ⎢⎣ ⎡ = − s m

V_{Out pipe (I)} 1.3 -- The water velocity at the outlet pipe or the water velocity at the

suction chamber for only one pump in operation

⎥⎦ ⎤ ⎢⎣ ⎡ = − s m

VOut pipe (II) 1.7 -- The water velocity at the outlet pipe or the water velocity at the

suction chamber for two pumps in parallel operation

L 9600 [ ] 9 6 [ ] i h di f h i l h f h L = 9600 [mm] = 9.6 [m] – is the distance from the pipe outlet to the centers of the

…. are all losses caused due to screens or strainers that are placed in the open

channel, friction between water and pipe wall, pipe miter bends and sudden

contraction duct at pumps intake

H

= H

+ H

+ H

co t act o duct at pu ps ta e

Where Where

H_L-CHANNEL[m] – Head losses through the open channel H_L-PIPE [m] – Head losses through the pipeline

For only one pump in operation

H_L-INTAKE [m] – Head losses through the pumps intake

H

= H

+ H

+ H

= 0.256 + 0.385 + 0.18 = 0.821 [m]

For two pumps in parallel operation

…. is the head that is present at

the suction side of the pump

NPSH_A– should be calculated

NPSH_R = 6.857 [m] -- from the pumps manufacturer

For only one pump in operation For only one pump in operation

For two pumps in parallel operation

] [ 956 . 11

) ( )

( H H H H m

NPSHA I = A + ST − L I − VP =

] [ 51 . 11

) ( )

( H H H H m

NPSH_{A II} = _A+ _ST − _{L II} − _VP =

Where

] [ 75 .

9 m

g p H atm

A= ρ_⋅ = -- the head due to the atmospheric pressure

NPSH_R= 6.857 [m]

] [ 95 .

0 bar

p_atm = -- the atmospheric for the altitude of 531.80[m] above sea level

Q = 4.84 [m3_/s]

H

…. the vertical distance between the

surface of water in the cooling tower

basin and the centerline of the pumps

H_ST = 3600 [mm] = 3.60 [m] – the hydrostatic head

basin and the centerline of the pumps

…… the absolute pressure at

which water will change from

liquid to steam at a specific

] [ 574 .

0 m

g p

H VP

VP =

⋅ =

ρ -- the water vapor pressure head

temperature

[ ]

m N bar

--- According to the pumps manufacturer CCM SULZER and ANSI standard HI 9.6.1:

)

5

.

2

3

.

1

(

−

−

=

NPSH

NPSH

For only one pump in operation

74 . 1 857 . 6 956 . 11 ) ( ) ( = = I R I A NPSH NPSH

-- the pump will operate cavitation free

For two pumps in parallel operation

265 . 1 19 9 628 . 11 ) ( )

(II _{= =} A

NPSH NPSH

-- the pumps in parallel

19 . 9 ) (II R NPSH ] [ 19 9 2 NPSH NPSH Where

Th i d N t

operation will operate with cavitation ] [ 19 . 9

2 _{( )}

)

( NPSH m

NPSH_{R II} = ⋅ _{R I} ⋅η_tot = -- The required Net Positive Suction Head for two pumps in parallel operation

NPSHA/NPSHR= 1.3

67

.

0

=

Two pumps in parallel operation will operate with the presence of cavitation.

Hydrostatic head is the only parameter that can be changed

If VP II L ST A II

A

H

H

H

H

NPSH

=

+

−

−

] [ 02 4 3 1 ' m H H H NPSH

H + + _{Th difi d h d i h d}

) ( )

(II 1.3 R II

A NPSH

NPSH = ⋅ Then

] [ 02 . 4 3 .

1 NPSH _{( )} H H _{( )} H m

H_ST = ⋅ _{R II} − _A+ _{L II} + _VP = _{-- The modified hydrostatic head}

] [ 947 . 11 ) ( ' ' )

( H H H H m

NPSH_{A II} = _A+ _ST − _{L II} − _VP = _{-- The modified available Net Positive Suction Head}

' NPSH 30 . 1 19 . 9 947 . 11 ) ( ) ( _{= =} II R II A NPSH NPSH ] [ 42 ] [ 42 . 0 ' cm m H H

H_ST = _ST − _ST = =

Δ -- the difference between the modified hydrostatic head --- Free cavitation zone

] [ 3 . 31 ) ( 2 )

( H H cm

H_{ST REAL} =Δ _ST − _{L str II} =

Δ ₋

Where

ST ST

ST

to the existed hydrostatic head

-- The water level in the cooling tower basin that should be increased

Where ] [ 107 . 0 ) ( 2 m

Budget

22 [m3] x 500 [€/m3] =

11000

[€]

Where

22 [m 3] -- is the amount for the reinforced concrete …

500 [€/m3] – is the total price of the reinforced concrete per cubic meter

Events

Each square meter of the cooling tower basin foundation will be loaded for 19 [%] more than it is.

The efficiency of the cooling tower will drop 2 [%] from the air inlet side

Commissioning phase:

The efficiency of the cooling tower will drop 2 [%] from the air inlet side.

The total amount of water that should be added for 31 centimeters in the cooling tower basin is 2620 [m3]

Commissioning phase:

Calibration of new the water level in the cooling tower basin

The measurement of the noise level of the pumps with a sound level meter (should be < 90 [dB]

The outcomes will be to:

Extend the working life of the pumps

Increase the safe operation of the pumps

Increase reliance on the cooling system