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Dump Heating Temperature(revision 2_ part 1)

Ang LeeMarch 16, 2010

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FEA Model

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Material Property

Kxx(W/mK) Cp(J/kgK) density (kg/m^3)

Copper 383 385 8900Steel 44 470 7800Concrete 0.91 961 2300

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Geometry and Loading

• Given by Fernanda G/Larry A./Rob R as shown Fig 1.• 9hz@10 usec@400 Mev@35 mA1.26 KW• 15 Hz@4usec@400 Mev@35mA-->0.84KW• Beam spot =24 cmx4 cmx1” as well as 12 cmx4 cm x 1” (*

Provided by L. Allen & F. Garcia)• Top plate is 1 inch thick copper. • The copper tube is 0.25” thick.• Surrounding Steel is ~36” in diameter & 108” in length and 33”

thick concrete.• The boundary condition for the outer surface of the concrete

is assuming to be natural convection with hc=~5 W*m^2*K

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Slice from earlier result (March 2nd,2010) indicated that 25% contact front and 25% contact at the back is almost same as the No or little contact case

*Table 1 Summary of the ResultMaximum temperature (K) after 1 min (900 pulses)

(copper melting point ~1273 K)

Contact area (%)

100 % 50% (down stream)

<~1% at the very front

25% front and 25% back

50% At the center

10% at the center

5% at the center

1% at the center

0% no contact at all

T (K) for the beam spot of 24 cmx4 cm x 1 inch

374 396 414 414 375 376 386 403 414

T (K) For the beam spot of 12cmx4 cm x 1 inch

420 440 470 470 420 420 429 454 470

* Note: The bold phase temperature are probably most likely case if the proper care being taken to insure a contact around the beam spot.

3/2/2010, A. Lee

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Temperature Result

0 5 10 15 20 25 30 35300

350

400

450

500

550

600

Hot Spot Temperature vs timefor 9 Hz, 10 usec,1.26 kw

24 cmx4 cmx1 in_100% contact 24 cm x 4 cm x 1 in_50% contact 12 cmx4 cmx 1in _100% contact 12 cmx4 cmx 1 in_50% contact

time (days)

Tem

pera

ture

(K)

50% contact area

100 % contact area

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Temperature Result

0 5 10 15 20 25 30 35300

320

340

360

380

400

420

440

460

480

Hot spot Temperature vs time for 15 Hz,4usec,0.84 kw

24 cmx4 cmx1 in_100% contact 24 cm x 4 cm x 1 in_50% contact 12 cmx4 cmx 1in _100% contact 12 cmx4 cmx 1 in_50% contact

time (days)

Tem

pera

ture

(K)

50% contact area

100 % contact area

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Temperature after 30 days50% contact

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Temperature (K) Result after 30 days

9Hz@10usec@400Mev@35mA=1.26KW

15 Hz@4usec@400 MeV@35mA=0.84kw

Contact area 100% 50% 100% 50%

Beam area= 24 cmx4 cm x

1 inch412 (K) 539 (K) 374 (K) 459 (K)

Beam area=12 cmx4 cm x 1

inch422 (K) 554 (K) 381 (K) 469 ( K)

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Extra case study

• Message from Fernanda:“Below I am sticking with my understanding of power, therefore I am calling KW what you call KJ.Anyway, for Operational scenarios, your question1) current: 9 Hz @ 10 usec @ 400MeV @ 35 mA = ~2.0E13 pps equivalent of 1.260 KW.(DONE)2) Near future: 15Hz @ 4 usec @ 400MeV @ 35 mA = 1.3 E13 pps equivalent of 0.840 KW.- (DONE)These are the numbers that we should use in the computation for "operational" scenarios.Now, every day there is a 1 min or 3 min of 35 mA @ 15 Hz @ 20 usec @ 400MeV sent to the dump at least 3 times a day in conjunction with the operational scenario. - extra caseI am not sure how you will handle this complexity”.

• So, I did some extra simulation to see what happen with above case. By assuming 35 ma@15 Hz@20 usec- gives 4.2 KW (Fernanda, can you double check it ?) and run it for 3 min x 3 times consecutively (worst case).

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Extra case study

0 5 10 15 20 25 30 35300

350

400

450

500

550

600

650

700

750

Hot Spot Temperature vs time1.26 KW+4.2 KW(3 min;3 times/day consecutively

case 1_1.26 KW_100% contact case 2_1.26KW_50% contact

case 3, 1.26KW +4.2 KW (3 times*3 min,consecutively)_100% contact case 4, 1.26 KW+4.2 KW (3 times/day* 3 min, consecutively),50% contact

time (days)

Tem

pera

ture

(K)

50% contact

100% contact

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Extra case study

0 5 10 15 20 25 30 35300

320

340

360

380

400

420

440

460

480

Hot spot temperature vs timeFor 1.26KW+4.2 KW (3min,3 times/day;30 min apart)

1.26KW+4.2 KW(3 min,3 times/day; 30 min apart)_50% 1.26 kw+4.2Kw(3min, 3times/day;30 min apart)_100%

time (days)

Tem

pera

ture

(K)

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Structural Calculation(Working in progress, expect to be done for next meeting , in

~2 weeks ? Hopefully)

• For structural calculation:1) Buckling calculation: to find critical collapeing pressure Pcr for a given structure (E

and t) and compare it with CGA (SF=2)or ASME (SF=3) code for the safety factor (SF).

2) Stress calculation : if the structure passes the buckling SF requirement, then the working stress needs to be checked under the operating load (p=15 psi).

3) The thermal stress due to temperature rise.4) The combine stress due to the thermal and structure load (vacuum).5) The deformation if it is required/interested .

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Structural Calculation

• Buckling calculation for the cylindrical section:a) 8” copper pipe with the wall thickness:

t=(8.625”-8.329’)/2=0.148”And E=16.7 mpsi, Sy (yield)=4830 psiSu(ultimult)=30.6 ksi (? Depending upon the what kind copper it is ).

The collapeing pressure Pcr can be found for “ thin tube under uniform lateral external pressure “

Pcr=(1/4 )*E/(1-u^2)*(t/r)^3=232 psi

such that SF=Pcr/Pop=232/15=15.4 15.4>>SF= 2 (CGA) and 15.4>>SF=3 ( ASME)

And the hoop stress due 15 psi vacuum load can be found asσ=p*r/t=15*4/0.148=405 psi << 2/3*Sy=3220 psi

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Buckling for the 6 sided copper tubesame material; SF=91 for p=15 psi external

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Stress due 15 psi for the 6 sided tube

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Stress due 15 psi for the 6 sided tube

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Stress plot along section AA

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Structural calculation

• Still working in progress.• Need to look at the thermal stress due to the

temperature .• Need to look at the thermal+vacuum load

combined case.• And deflection if interested