Transcript SEAL AND BEARING FAILURE ON A TWO-STAGE

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
1
Outline

What Happened ?
 In
short: User 'moved / simplified' piping on
the pump, which affected Axial Thrust.

2
Mini-tutorial on calculating Axial thrust for
this pump.
Pump Cross
Section
3
The Problem – What Happened ?

4
Pump design
assumed by the
user to have
both a seal
piping Plan 13
and a Plan 11.
The Problem – What Happened ?
Pump was
simplified to only
use the Plan 11.
 The "Plan 13"
was removed.
[Which actually
was a Balance
Line.]

5
The Problem – What Happened ?

6
Original
Pump
The Problem – What Happened ?


Plan 11, with
Balance Line
Removed:

7
No flow through the
seal = seal failure
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.
8

Consequence
 Seal
is no longer
failing.
 But bearings
are, every 6-9
months.
 Axial Thrust !
9

Axial Thrust:
 Mainly
a function of: Pressure
distribution on the rotor.

10
Also: Momentum force.
Axial Thrust: Pressure Distribution on
Impeller Shrouds

Rule of thumb: 0.75*Pd
(differential pressure) for
the shroud pressure profile



11
(from the wear-ring-labyrinth
to Impeller OD)
Used in this Case Study.
But in reality it is more
complicated...
Axial Thrust: Pressure Distribution on Shroud

Dependent on fluid
dynamics in SideRooms:
 Off-BEP
operation
 Leakage direction and
amount.
 Side-room geometry.
 Rotor to Case
Alignment.

12
For Example:
Example: Effect of leakage on Pressure Distribution
Actual pressure profile is
decreased because:
greater swirl in side-room
due to fluid entering sideroom 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
14
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) ->
2
5 72  .7 7  1
1 


 3 9.9 8lbf

7 22  1 9.6 4 1 5.7 1
15
Calculate Axial Thrust for this Pump
What do we Need to
start ?
 Cross section
 Diameters of
wear ring
labyrinths.
 Pressures




16
Suction
Pressure
= 111 psi
Differential
Pressure
= 340 psi
Flush plan
General idea of
leakage
direction,
labyrinth
clearances,
leakage flow, etc.


We are assuming the simplified
0.75*Pd on Shrouds.
Initially show all pressures on rotor,
and then show the typical
simplification.
17
Sum of All Pressures
on Rotor
Axial Thrust - As designed. (all pressures on rotor)
Item #
I.D. [in]
O.D [in]
=Area [in ]
Pressure
[psig]
1
2
3
4
5
6
7
8
0
6.354
3.234
3.234
6.354
6.354
2.75
0
6.354
12
12
6.354
12
12
6.354
2.75
31.7
81.4
104.9
23.5
81.4
81.4
25.8
5.9
111
239
239
281
409
409
111
0
2
Dir
[-> +] =Force [lbf]
1
1
-1
1
1
-1
-1
-1
3520
19452
-25067
6602
33288
-33288
-2860
0
SUM
1646
Simplified.
Shrouds
Balance.
Axial Thrust - As designed.
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(Simplified)
Item #
I.D. [in]
O.D [in]
=Area [in ]
Pressure
[psig]
1
2
3
4
0
3.234
3.234
2.75
6.354
6.354
6.354
6.354
31.7
23.5
23.5
25.8
111
239
281
111
2
Dir
[-> +] =Force [lbf]
1
-1
1
-1
3520
-5615
6602
-2860
SUM
1646
Simplifi
ed
Further
Axial Thrust - As designed.
19
(Fully simplified)
Item #
I.D. [in]
O.D [in]
=Area [in ]
Pressure
[psig]
1
2
3
0
3.234
3.234
2.75
6.354
6.354
5.9
23.5
23.5
111
128
170
2
Dir
[-> +] =Force [lbf]
1
-1
1
659
-3007
3994
SUM
1646
Axial Thrust – Plan 11 with No Balance Line
20
Axial Thrust – Plan 11 with No Balance Line
Item #
1
2
3
4
21
I.D. [in]
0
3.234
3.234
2.75
O.D [in]
2.75
6.354
6.354
6.354
2
=Area [in ]
5.9
23.5
23.5
25.8
Pressure
[psig]
111
128
170
340
Dir [-> +] =Force [lbf]
1
659
-1
-3007
1
3994
-1
-8762
SUM
-7116
Plan 13 – with no Balance Line

In Series Flow:




22
Hub Seal,
Throat Bush,
Plan 13 Orifice
What is the
pressure behind
the 2nd Stage
Impeller?
Plan 13 – with no Balance Line

Cross section area of each Restriction:
H ub Gap, X-sec Area

4
Throat Gap, X-sec Area

4
Orif ic e, X-s ec Area

4
23

2
2

in

2
2

in
 6 .37 5  6 .35 5  0 .20 0
 2 .90 5  2 .87 5  0 .13 6

( .1 25)
  0 .01 2
2
2
in
2
2
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
q
C A  2 g  1 44
P

q = f low rate [3f /ts] (1f t3/ s=448. 8gpm)
C = orif ic e f low c oef f ic ient [~0.6]
A = cros s s ec tional area orif ic e2[] f t
2 ]s
g = grav it y [ f t/
P = pres sure drop ac ross orif ice [ ps i]
 = weight dens ity f luid[ lbf
f t3/]
340 
 0.012
 2 32.2 144
 0.6
  448.8 5.043 gpm
144
62.43

24
_careful with units_
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.
2
 5 .04 3  6 2.3


4
48
.8


2
P 
q 
2
2 88g
 C A
25
2
0 .20 0
2 88 ( 3 2.2)  ( .6)  

1
44


2
2
 1 .22 1 p si
 Pressure
Drop across Hub = 1.2 psi
 Therefore Pressure Behind 2nd Stg Impeller
is = 338.8 psi
26
Axial Thrust – Plan 13 with No Balance Line
Item #
1
2
3
4
27
I.D. [in]
0
3.234
3.234
2.75
O.D [in]
2.75
6.354
6.354
6.354
Pressure
=Area [in2]
[psig]
Dir [-> +]
5.9
111
1
23.5
128
-1
23.5
170
1
25.8
338.8
-1
=Force
[lbf]
659
-3007
3994
-8731
SUM
-7085
Bearing L10h.

The most simple method for Bearing Life Calculation is "L10"
ISO or AFBMA equation for basic rating life:





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.

28
L10 = (C/P)p
L10h = (1 000 000/ (60*n))*L10
n = RPM
Bearing L10h.
Bearing Lif e C alc ulation
C   3 47 00l
 bf
Basic Load D y namic R at ing f or 5313 Bearing
n   3 56 0
R PM of Pump
Fr   5 0 l bf
Approx imat e R adial Forc e.
Fa   1 64 6l
 bf
Axi al Thrust AS DESI GNED
P   .6 3 Fr  1 .24Fa

Equiv alent dy namic brg load, Ang. C ontact Brgs Dbl R ow.
P  2 07 3l bf
C
P
6
 1 6.7
 C
L1 0h 
 
6 0 n  P 
10
3
L1 0h 2 19 72
L10 li fe i n Hours.
Fa   7 10 0l
 bf
Axi al Thrust WI TH N O B ALANC E LIN E & PLAN 11 o.r 13
P   .6 3 Fr  1 .24Fa

Equiv alent dy namic brg load, Ang. C ontact Brgs Dbl R ow.
P  8 83 5l bf
C
P
29
6
 3 .9
 C
L1 0h 
 
6 0 n  P 
10
3
L1 0h 2 84
L10 li fe i n Hours.
Bearing L10.

30
Alternative, Use C/P
[basic load dynamic rating
/ dynamic load] and
nomograph from the
bearing supplier.
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



31
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.
Questions ?
Thrust
32