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Dump Heating Temperature (revision 2_ part 1)

Dump Heating Temperature (revision 2_ part 1). Ang Lee March 16, 2010. FEA Model. Material Property. Kxx (W/ mK ) Cp(J/ kgK ) density (kg/m^3) Copper 383 385 8900 Steel 44 470 7800 Concrete 0.91 961 2300. Geometry and Loading. Given by Fernanda G/Larry A./Rob R as shown Fig 1.

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Dump Heating Temperature (revision 2_ part 1)

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  1. Dump Heating Temperature(revision 2_ part 1) Ang Lee March 16, 2010

  2. FEA Model

  3. Material Property Kxx(W/mK) Cp(J/kgK) density (kg/m^3) Copper 383 385 8900 Steel 44 470 7800 Concrete 0.91 961 2300

  4. Geometry and Loading • Given by Fernanda G/Larry A./Rob R as shown Fig 1. • 9hz@10 usec@400 Mev@35 mA1.26 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

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

  6. Temperature Result

  7. Temperature Result

  8. Temperature after 30 days50% contact

  9. Temperature (K) Result after 30 days

  10. 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).

  11. Extra case study

  12. Extra case study

  13. Structural Calculation(Working in progress, expect to be done for next meeting , in ~2 weeks ? Hopefully) • For structural calculation: • Buckling calculation: to find critical collapeingpressure 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). • 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). • The thermal stress due to temperature rise. • The combine stress due to the thermal and structure load (vacuum). • The deformation if it is required/interested .

  14. StructuralCalculation • 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 psi Su(ultimult)=30.6 ksi (? Depending upon the what kind copper it is ). The collapeingpressure 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 as σ=p*r/t=15*4/0.148=405 psi << 2/3*Sy=3220 psi

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

  16. Stress due 15 psi for the 6 sided tube

  17. Stress due 15 psi for the 6 sided tube

  18. Stress plot along section AA

  19. 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

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