tanks lecture 3

8/10/2019 Tanks Lecture 3

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Lecture 3

January 23, 2006


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Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 2

In this lecture

Modeling of tanks Time period of tanks


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Modeling of tanks

As seen in Lecture 1 liquid may be replaced byimpulsive and convective mass for calculation ofhydrodynamic forces

See next slide for a quick review


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Modeling of tanks

Rigid

mc

Kc/2Kc/2

mihi

(hi*)

hc

(hc*)

mi= Impulsive liquid mass

mc= Convective liquid mass

Kc= Convective spring stiffness

hi= Location of impulsive mass(without considering overturnig

caused by base pressure)

hc= Location of convective mass(without considering overturning

caused by base pressure)

hi*= Location of impulsive mass

(including base pressure effect onoverturning)

hc*= Location of convective mass

(including base pressure effect onoverturning)

Mechanical analogue

orspring mass model of tank

Graphs and expression for these parameters are given in lecture 1.


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Approximation in modeling

Sometimes, summation of miand mcmay not beequal to total liquid mass, m

This difference may be about 2 to 3 %

Difference arises due to approximations in the

derivation of these expressions More about it, later

If this difference is of concern, then

First, obtain mcfrom the graph or expression

Obtain mi = m mc


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Tanks of other shapes

For tank shapes such as Intze, funnel, etc. : Consider equivalent circular tank of same

volume, with diameter equal to diameter at thetop level of liquid


7/70Sudhir K. Jain, IIT Kanpur E-Course on Seismic Design of Tanks/ January 2006 Lecture 3 / Slide 7


Example:An Intze container has volume of 1000 m3. Diameter ofcontainer at top level of liquid is 16 m. Find dimensions ofequivalent circular container for computation of

hydrodynamic forces.

Equivalent circular container will have diameter of 16 m

and volume of 1000 m3.Height of liquid, h can be obtained

as :/4 x 162x h = 1000

h = 1000 x 4/(x 162) = 4.97 m




Thus, for equivalent circular container,h/D = 4.97/16 = 0.31

All the parameters (such as mi, mcetc.) are to beobtained using h/D = 0.31

16 m

Intze containervolume = 1000 m3

16 m 4.97 m

Equivalent circular container



Effect of obstructions inside tank

Container may have structural elements inside For example: central shaft, columns supporting

the roof slab, and baffle walls

These elements cause obstruction to lateral

motion of liquid

This will affect impulsive and convective masses




Effect of these obstructions on impulsive andconvective mass is not well studied

A good research topic !

It is clear that these elements will reduceconvective (or sloshing) mass

More liquid will act as impulsive mass




In the absence of detailed analysis, followingapproximation may be adopted:

Consider a circular or a rectangular container ofsame height and without any internal elements

Equate the volume of this container to netvolume of original container

This will give diameter or lateral dimensions of container

Use this container to obtain h/D or h/L




Example: A circular cylindrical container has internaldiameter of 12 m and liquid height of 4 m. At the centerof the tank there is a circular shaft of outer diameter of2 m. Find the dimensions of equivalent circularcylindrical tank.

12 m

4 m

Hollow shaft of2 m diameter

12 m

Elevation

Plan


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Solution:Net volume of container = /4x(12222)x4 = 439.8 m3

Equivalent cylinder will have liquid height of 4 m and itsvolume has to be 439.8 m3.

Let D be the diameter of equivalent circular cylinder, then/4xD2x4 = 439.8 m3

D = 11.83 m

Thus, for equivalent circular tank, h = 4 m, D = 11.83mand h/D = 4/11.83 = 0.34.

This h/D shall be used to find parameters of mechanicalmodel of tank


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Effect of wall flexibility

Parameters mi, mcetc. are obtained assumingtank wall to be rigid

An assumption in the original work of Housner(1963a)

Housner, G. W., 1963a, Dynamic analysis of fluids incontainers subjected to acceleration, Nuclear Reactors andEarthquakes, Report No. TID 7024, U. S. Atomic EnergyCommission, Washington D.C.

RC tank walls are quite rigid

Steel tank walls may be flexible

Particularly, in case of tall steel tanks


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Wall flexibility affects impulsive pressure distribution It does not substantially affect convective pressuredistribution

Refer Veletsos, Haroun and Housner (1984) Veletsos, A. S., 1984, Seismic response and design of liquid

storage tanks, Guidelines for the seismic design of oil and gaspipeline systems, Technical Council on Lifeline Earthquake1Engineering, ASCE, N.Y., 255-370, 443-461.

Haroun, M. A. and Housner, G. W., 1984, Seismic design of liquidstorage tanks, Journal of Technical Councils of ASCE, Vol. 107,TC1, 191-207.

Effect of wall flexibility on impulsive pressure dependson Aspect ratio of tank Ratio of wall thickness to diameter

See next slide


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Effect of wall flexibility on impulsive pressure distribution

From Veletsos (1984)

h

z Rigidtank

tw/ D = 0.0005

tw/ D = 0.0005

Impulsive pressure on wall

twis wall thickness

h/D = 0.5


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If wall flexibility is included, then mechanicalmodel of tank becomes more complicated

Moreover, its inclusion does not change seismicforces appreciably

Thus, mechanical model based on rigid wallassumption is considered adequate for design.


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All international codes use rigid wall model forRC as well as steel tanks Only exception is NZSEE recommendation

(Priestley et al., 1986) Priestley, M J N, et al., 1986, Seismic design of storage

tanks, Recommendations of a study group of the NewZealand National Society for Earthquake Engineering.

American Petroleum Institute (API) standards,which are exclusively for steel tanks, also use

mechanical model based on rigid wallassumption API 650, 1998, Welded storage tanks for oil storage,

American Petroleum Institute, Washington D. C., USA.


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Effect of higher modes

miand mc described in Lecture 1, correspondto first impulsive and convective modes For most tanks ( 0.15 < h/D < 1.5) the first

impulsive and convective modes togetheraccount for 85 to 98% of total liquid mass

Hence, higher modes are not included

This is also one of the reasons for summation ofmiand mcbeing not equal to total liquid mass

For more information refer Veletsos (1984) andMalhotra (2000)

Malhotra, P. K., Wenk, T. and Wieland, M., 2000, Simpleprocedure for seismic analysis of liquid-storage tanks,Structural Engineering International, 197-201.


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Modeling of ground supported tanks

Step 1: Obtain various parameters of mechanical model

These include, mi, mc, Kc, hi, hc, hi*and hc

*

Step 2:

Calculate mass of tank wall (mw), mass of roof(mt) and mass of base slab (mb)of container

This completes modeling of ground supported

tanks


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Modeling of elevated tanks

Elevated tank consists of container and staging

Elevated tank

Container

Staging

Wall

Roof slab

Floor slab


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Liquid is replaced by impulsive and convectivemasses, miand mc All other parameters such as hi, hc, etc, shall be

obtained as described earlier

Lateral stiffness, Ks, of staging must beconsidered

This makes it a two-degree-of-freedom model

Also called two mass idealization


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hi

mc

mi hc

hs

Spring mass model Two degree of freedom system

OR

Two mass idealization of elevated tanks

Ks

mi + ms

mc

KcKc/2 Kc/2


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msis structural mass, which comprises of : Mass of container, and

One-third mass of staging

Mass of container includes

Mass of roof slab

Mass of wall

Mass of floor slab and beams


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Two Degree of Freedom System

2-DoF system requires solution of a 2 2 eigenvalue problem to obtain

Two natural time periods

Corresponding mode shapes

See any standard text book on structuraldynamics on how to solve 2-DoF system

For most elevated tanks, the two natural time

periods (T1and T2) are well separated. T1 generally may exceed 2.5 times T2.


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Ks

mi+ m

s

mc

Kc

Two degree of freedom system

Ks

mi + ms

mc

Kc

Two uncoupled

single degree of freedom systems

when T1 2.5 T2


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Priestley et al. (1986) suggested that thisapproximation is reasonable if ratio of two timeperiods exceeds 2.5

Important to note that this approximation is

done only for the purpose of calculating timeperiods

This significantly simplifies time period calculation

Otherwise, one can obtain time periods of 2-DoFsystem as per procedure of structural dynamics.


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Steps in modeling of elevated tanks Step 1:

Obtain parameters of mechanical analogue

These include mi, mc, Kc, hi, hc, hi*and hc

*

Other tank shapes and obstructions inside the container shallbe handled as described earlier

Step 2:

Calculate mass of container and mass of staging

Step 3:

Obtain stiffness of staging


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Recall, in IS 1893:1984, convective mass is notconsidered

It assumes entire liquid will act as impulsive mass

Hence, elevated tank is modeled as single

degree of freedom ( SDoF) system

As against this, now, elevated tank is modeledas 2-DoF system

This 2-DoF system can be treated as twouncoupled SDoF systems


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Models of elevated tanks

Ks

mi + ms

mc

Kc

Ks

m +ms m = Total l iquid mass

As per the Guideline As per IS 1893:1984


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Example: An elevated tank with circular cylindricalcontainer has internal diameter of 11.3 m and waterheight is 3 m. Container mass is 180 t and staging massis 100 t. Lateral stiffness of staging is 20,000 kN/m.Model the tank using the Guideline and IS 1893:1984

Solution:

Internal diameter, D = 11.3 m, Water height, h = 3 m.

Container is circular cylinder,

Volume of water = /4 x D2x h= /4 x 11.32x 3 = 300.9 m3.

mass of water, m = 300.9 t.


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h/D = 3/11.3 = 0.265From Figure 2 of the Guideline, for h/D = 0.265:

mi/m = 0.31, mc/m = 0.65 and Kch/mg = 0.47

mi = 0.31 x m = 0.31 x 300.9 = 93.3 tmc= 0.65 x m = 0.65 x 300.9 = 195.6 t

Kc= 0.47 x mg/h

= 0.47 x 300.9 x 9.81/3 = 462.5 kN/m


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Mass of container = 180 tMass of staging = 100 t

Structural mass of tank, ms

= mass of container +1/3rdmass of staging= 180 +1/3 x 100

= 213.3 t

Lateral stiffness of staging,Ks= 20,000 kN/m


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Ks

mi + ms

mc

Kc

Ks

m+ ms

Model of tank as perthe Guideline Model of tank as perIS 1893:1984

mi= 93.3 t, ms= 213.3 t, mc= 195.6 t,Ks= 20,000 kN/m, Kc= 462.5 kN/m

m = 300.9 t, ms= 213.3 t,Ks= 20,000 kN/m


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Time period

What is time period ? For a single degree of freedom system, time

period (T ) is given by

K

MT 2

M is mass and K is stiffness

T is in seconds M should be in kg; K should be in Newton per

meter (N/m)

Else, M can be in Tonnes and K in kN/m


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Time period

Mathematical model of tank comprises ofimpulsive and convective components

Hence, time periods of impulsive and convectivemode are to be obtained


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Time period of impulsive mode

Procedure to obtain time period of impulsivemode (Ti) will be described for following threecases:

Ground supported circular tanks

Ground supported rectangular tanks Elevated tanks


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Ti for ground-supported circular tanks

Ground supported circular tanks Time period of impulsive mode, Tiis given by:

Et/D

h

CT ii

2i

0.067(h/D)0.3h/D0.46h/D

1

C

= Mass density of liquidE = Youngs modulus of tank material

t = Wall thickness

h = Height of liquid

D = Diameter of tank


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Tifor ground-supported circular tanks

Ci can also be obtained from Figure 5 of theGuidelines

0

2

4

6

8

10

0 0.5 1 1.5 2

Ci

Cc

h/D

C


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This formula is taken from Eurocode 8 Eurocode 8, 1998, Design provisions for earthquake

resistance of structures, Part 1- General rules and Part 4 Silos, tanks and pipelines, European Committee forStandardization, Brussels.

If wall thickness varies with height, thenthickness at 1/3rdheight from bottom shall beused

Some steel tanks may have step variation of wallthickness with height


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This formula is derived based on assumptionthat wall mass is quite small compared to liquidmass

More information on time period of circular

tanks may be seen in Veletsos (1984) andNachtigall et al. (2003)

Nachtigall, I., Gebbeken, N. and Urrutia-Galicia, J. L., 2003, On theanalysis of vertical circular cylindrical tanks under earthquakeexcitation at its base, Engineering Structures, Vol. 25, 201-213.


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It is important to note that wall flexibility isconsidered in this formula

For tanks with rigid wall, time period will be zero

This should not be confused with rigid wall

assumption in the derivation of miand mc Wall flexibility is neglected only in the evaluation

of impulsive and convective masses

However, wall flexibility is included while

calculating time period


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This formula is applicable to tanks with fixedbase condition

i.e., tank wall is rigidly connected or fixed to thebase slab

In some circular tanks, wall and base haveflexible connections


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Ground supported tanks with flexible base aredescribed in ACI 350.3 and AWWA D-110 ACI 350.3, 2001, Seismic design of liquid containing

concrete structures, American Concrete Institute,Farmington Hill, MI, USA.

AWWA D-110, 1995, Wire- and strand-wound circular,prestressed concrete water tanks, American Water WorksAssociation, Colorado, USA.

In these tanks, there is a flexible pad between

wall and base Refer Figure 6 of the Guideline


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Such tanks are perhaps not used in India

Types of connections between tank wall and base slab


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Impulsive mode time period of groundsupported tanks with fixed base is generallyvery low

These tanks are quite rigid

Tiwill usually be less than 0.4 seconds

In this short period range, spectral acceleration,Sa/g has constant value

See next slide


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Impulsive mode time period of ground supportedtanks likely to remain in this range

Sa/g


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Example: A ground supported steel tank has waterheight, h = 25 m, internal diameter, D = 15 m andwall thickness, t=15 mm. Find time period ofimpulsive mode.

Solution:h = 25 m, D = 15 m, t = 15 mm.

For water, mass density, = 1 t/m3.

For steel, Youngs modulus, E = 2x108kN/m2.

h/D = 25/15 = 1.67. From Figure 5, Ci = 5.3


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Et/D

hCT ii Time period of impulsive mode,

8i

2x100.015/15

1.0255.3T

= 0.30 sec

Important to note that, even for such a slender tank of

steel, time period is low. For RC tanks and other short tanks, time period will be

further less.


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In view of this, no point in putting too muchemphasis on evaluation of impulsive mode timeperiod for ground supported tanks

Recognizing this point, API standards have

suggested a constant value of spectralacceleration for ground supported circular steeltanks

Thus, users of API standards need not findimpulsive time period of ground supported tanks


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Tifor ground-supported rectangulartanks

Tifor ground-supported rectangular tanks Procedure to find time period of impulsive mode

is described in Clause no. 4.3.1.2 of theGuidelines

This will not be repeated here

Time period is likely to be very low and Sa/g willremain constant

As described earlier Hence, not much emphasis on time period

evaluation

f l d k


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Ti for Elevated tanks

For elevated tanks, flexibility of staging is important Time period of impulsive mode, Tiis given by:

s

sii

K

mm2T

mi= Impulsive mass of liquid

ms= Mass of container and one-third mass of staging

Ks= Lateral stiffness of staging= Horizontal deflection of center of gravity of tank when a

horizontal force equal to (mi+ ms)g is applied at the

center of gravity of tank

ORg

2T


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These two formulae are one and the same Expressed in terms of different quantities

Center of gravity of tank refers to combinedmass center of empty container plus impulsive

mass of liquid


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Example: An elevated tank stores 250 t of water. Ratio ofwater height to internal diameter of container is 0.5.Container mass is 150 t and staging mass is 90 t.Lateral stiffness of staging is 20,000 kN/m. Find timeperiod of impulsive mode

Solution: h/D = 0.5, Hence from Figure 2a of theGuideline, mi/m = 0.54;

mi= 0.54 x 250 = 135 t

Structural mass of tank, ms

= mass of container + 1/3rdmass of staging

= 150 + 90/3 = 180 t


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Time period of impulsive modes

si

iK

mm2T

00020

180135

2 ,Ti

= 0.79 sec.

L t l tiff f t i K


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Lateral stiffness of staging, Ks

Lateral stiffness of staging, Ksis force requiredto be applied at CG of tank to cause acorresponding unit horizontal deflection

CGP Ks= P/


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For frame type staging, lateral stiffness shall beobtained by suitably modeling columns andbraces More information can be seen in Sameer and

Jain (1992, 1994) Sameer, S. U., and Jain, S. K., 1992, Approximate

methods for determination of time period of water tankstaging, The Indian Concrete Journal, Vol. 66, No. 12,691-698.

Sameer, S. U., and Jain, S. K., 1994, Lateral load analysis

of frame staging for elevated water tanks, Journal ofStructural Engineering, ASCE, Vol.120, No.5, 1375-1393.

Some commonly used frame type stagingconfigurations are shown in next slide


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4 columns 6 columns 8 columns

9 columns 12 columns

Plan view of frame staging configurations


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24 columns 52 columns


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Explanatory handbook, SP:22 has consideredbraces as rigid beams SP:22 1982, Explanatory Handbook on Codes for

Earthquake Engineering, Bureau of Indian Standards, NewDelhi

This is unrealistic modeling Leads to lower time period

Hence, higher base shear coefficient

This is another limitation of IS 1893:1984

Using a standard structural analysis software,staging can be modeled and analyzed toestimate lateral stiffness


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Shaft type staging can be treated as a verticalcantilever fixed at base and free at top

If flexural behavior is dominant, then

Its stiffness will be Ks= 3EI/L3

This will be a good approximation if height todiameter ratio is greater than two

Otherwise, shear deformations of shaft wouldaffect the stiffness and should be included.

Time period of convective mode


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Convective mass is mcand stiffness is Kc Time period of convective mode is:

c

c

c K

m

T

2


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mcand Kcfor circular and rectangular tanks canbe obtained from graphs or expressions

These are described in Lecture 1

Refer Figures 2 and 3 of the Guidelines


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For further simplification, expressions for mcand Kcare substituted in the formula for Tc Then one gets,

)/68.3(68.3

2

DhtanhCc

D/gCT cc

L/gCT cc

For circular tanks:

For rectangular tanks:

))/(16.3(16.3

2

LhtanhCc


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Fig. 5 For circular tanks

0

2

4

6

8

10

0 0.5 1 1.5 2

Ci

Cc

C

h/D


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2

4

6

8

0

0 0.5 1 1.5 2

Fig. 7 For rectangular tanks

h/L

Cc


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Example: For a circular tank of internal diameter, 12 mand liquid height of 4 m. Calculate time period ofconvective mode.

Solution: h = 4 m, D = 12 m,

h/D = 4/12 = 0.33

From Figure 5 of the Guidelines, Cc= 3.6

D/gCTcc

Time period of convective mode,

12/9.813.6Tc = 3.98 sec

At the end of Lecture 3


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At the end of Lecture 3

Based on mechanical models, time period forimpulsive and convective modes can beobtained for ground supported and elevatedtanks

For ground supported tanks, impulsive modetime period is likely to be very less

Convective mode time period can be very large

tanks lecture 3

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