1

Seal and Bearing Failure on a Two-Stage Overhung Pump (3x6x13.5 CJA 2 Stage)

John Schmidt, PE

CSS Field Engineering Sulzer Pumps (US), Inc

2

Outline

What Happened ? In short: User 'moved / simplified' piping on

the pump, which affected Axial Thrust. Mini-tutorial on calculating Axial thrust for

this pump.

3

Pump Cross Section

4

The Problem – What Happened ?

Pump design assumed by the user to have both a seal piping Plan 13 and a Plan 11.

5

The Problem – What Happened ?

Pump was simplified to only use the Plan 11.

The "Plan 13" was removed. [Which actually was a Balance Line.]

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The Problem – What Happened ?

Original Pump

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The Problem – What Happened ?

Plan 11, with Balance Line

Removed: No flow through the

seal = seal failure

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The Problem – What Happened ? Since Seal Failure

occurred: Changed the seal

piping from a Plan 11 to a Plan 13.

Plan 13: Flush is restored

through the Seal. But Balance Line

not restored.

9

ConsequenceSeal is no longer

failing.But bearings are,

every 6-9 months.Axial Thrust !

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Axial Thrust:

Mainly a function of: Pressure distribution on the rotor.

Also: Momentum force.

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Axial Thrust: Pressure Distribution on Impeller Shrouds

Rule of thumb: 0.75*Pd (differential pressure) for the shroud pressure profile (from the wear-ring-labyrinth

to Impeller OD) Used in this Case Study.

But in reality it is more complicated...

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Axial Thrust: Pressure Distribution on Shroud

Dependent on fluid dynamics in Side-Rooms:

Off-BEP operation Leakage direction and

amount. Side-room geometry. Rotor to Case

Alignment.

For Example:

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Example: Effect of leakage on Pressure Distribution

Actual pressure profile is decreased because: greater swirl in side-room due to fluid entering side-room with high pre-rotation.

Actual pressure profile is increased because: Less swirl in side-room due to fluid entering hub seal with no pre-rotation.

pressure profile if rotation factor assumed to be 0.5

pressure profile if rotation factor assumed to be 0.5

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Axial Thrust: Momentum Force

Very low. Is typically not included. Thrust due to momentum change. Momentum Force=(Capacity2 x density) / [(Eye Area)x722]

Momentum force in lbf. Capacity in GPM, Density in SG Eye Area in Square Inches. 722 is unit conversion factor. Assumes 90 deg turn of fluid.

For This Pump (at design flow) ->

5722 .77722

1

19.64

1

15.71

39.98 lbf

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Calculate Axial Thrust for this Pump

We are assuming the simplified 0.75*Pd on Shrouds.

Initially show all pressures on rotor, and then show the typical simplification.

What do we Need to start ? Cross section Diameters of

wear ring labyrinths.

Pressures Suction

Pressure = 111 psi

Differential Pressure = 340 psi

Flush plan General idea of

leakage direction, labyrinth clearances, leakage flow, etc.

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Axial Thrust - As designed. (all pressures on rotor)

Item # I.D. [in] O.D [in] =Area [in2]Pressure

[psig] Dir [-> +] =Force [lbf]

1 0 6.354 31.7 111 1 35202 6.354 12 81.4 239 1 194523 3.234 12 104.9 239 -1 -250674 3.234 6.354 23.5 281 1 66025 6.354 12 81.4 409 1 332886 6.354 12 81.4 409 -1 -332887 2.75 6.354 25.8 111 -1 -2860 SUM8 0 2.75 5.9 0 -1 0 1646S

um o

f A

ll P

ress

ures

on

Rot

or

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Axial Thrust - As designed. (Simplified)

Item # I.D. [in] O.D [in] =Area [in2]Pressure

[psig] Dir [-> +] =Force [lbf]

1 0 6.354 31.7 111 1 35202 3.234 6.354 23.5 239 -1 -56153 3.234 6.354 23.5 281 1 6602 SUM4 2.75 6.354 25.8 111 -1 -2860 1646S

impl

ified

.

Shr

ouds

B

alan

ce.

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Axial Thrust - As designed. (Fully simplified)

Item # I.D. [in] O.D [in] =Area [in2]Pressure

[psig] Dir [-> +] =Force [lbf]

1 0 2.75 5.9 111 1 6592 3.234 6.354 23.5 128 -1 -3007 SUM3 3.234 6.354 23.5 170 1 3994 1646S

impl

ified

F

urth

er

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Axial Thrust – Plan 11 with No Balance Line

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Axial Thrust – Plan 11 with No Balance Line

Item # I.D. [in] O.D [in] =Area [in2]Pressure

[psig] Dir [-> +] =Force [lbf]

1 0 2.75 5.9 111 1 6592 3.234 6.354 23.5 128 -1 -30073 3.234 6.354 23.5 170 1 3994 SUM4 2.75 6.354 25.8 340 -1 -8762 -7116

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Plan 13 – with no Balance Line In Series Flow:

Hub Seal, Throat Bush, Plan 13 Orifice

What is the pressure behind the 2nd Stage Impeller?

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Plan 13 – with no Balance Line

Cross section area of each Restriction:

Hub Gap, X-sec Area 46.3752 6.3552 0.200 in2

Throat Gap, X-sec Area 42.9052 2.8752 0.136 in2

Orifice, X-sec Area 4

.125( )2 0.012 in2

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Plan 13 – with no Balance Line The Orifice is greatest restriction (by far)

Find flow rate through the orifice (assume water):

General / Simple Equation for Orifice

0.60.012

144 2 32.2 144

340

62.43

448.8 5.043 gpm

_careful with units_

q = flow rate [ft3/s] (1ft3/s=448.8gpm)C = orifice flow coefficient [~0.6]

A = cross sectional area orifice [ft2]

g = gravity[ft/s2]P = pressure drop across orifice [psi] = weight density fluid[lbf/ft3]

q C A 2 g 144P

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Plan 13 – with no Balance Line

Flow rate through this line is ~5 GPM (or less if we include other restrictions and resistances..)

Given 5 GPM flow what is the pressure drop across the hub seal? Re-Arrange Orifice Eqn, solve for pressure across hub seal gap.

Pq2

288g C2 A2

5.043

448.8

2

62.3

288 32.2( ) .6( )20.200

144

2

1.221 psi

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Pressure Drop across Hub = 1.2 psiTherefore Pressure Behind 2nd Stg Impeller

is = 338.8 psi

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Axial Thrust – Plan 13 with No Balance Line

Item # I.D. [in] O.D [in] =Area [in2]Pressure

[psig] Dir [-> +]=Force

[lbf]

1 0 2.75 5.9 111 1 6592 3.234 6.354 23.5 128 -1 -30073 3.234 6.354 23.5 170 1 3994 SUM4 2.75 6.354 25.8 338.8 -1 -8731 -7085

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Bearing L10h.

The most simple method for Bearing Life Calculation is "L10" ISO or AFBMA equation for basic rating life: L10 = (C/P)p

L10 = basic rating life, millions of revolutions C=Basic dynamic Load Rating (from Bearing Tables) P=Equivalent Dynamic Bearing Load p=Exponent, 3 for ball, 3.333 for roller.

Operating hours at constant speed before onset of fatigue. L10h = (1 000 000/ (60*n))*L10 n = RPM

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Bearing L10h.

L10 life in Hours.L10h 284L10h106

60 nC

P

3

C

P3.9

P 8835lbfEquivalent dynamic brg load, Ang. Contact Brgs Dbl Row. P .63 Fr 1.24Fa

Axial Thrust WITH NO BALANCE LINE & PLAN 11 or 13.Fa 7100lbf

L10 life in Hours.L10h 21972L10h106

60 nC

P

3

C

P16.7

P 2073lbfEquivalent dynamic brg load, Ang. Contact Brgs Dbl Row. P .63 Fr 1.24Fa

Axial Thrust AS DESIGNEDFa 1646lbf

Approximate Radial Force.Fr 50 lbf

RPM of Pumpn 3560

Basic Load Dynamic Rating for 5313 BearingC 34700lbf

Bearing Life Calculation

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Bearing L10.

Alternative, Use C/P [basic load dynamic rating / dynamic load] and nomograph from the bearing supplier.

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Problem Resolution User realized the issue after reading about pump axial

thrust and better understood what they had affected. Solution: Pump User returned the pump to the original

configuration and reliability was improved.

Lessons Learned Continuing Education / pump training should be included

in the maintenance/operation/reliability sections of any plant in order to achieve success.

Modifications can have un-intended consequences. You should contact the OEM as necessary.

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Questions ?

Thrust