1 / 21

Introduction

Introduction. The temperature margin is given by following aspects: Properties of the conductor and operating fields (-> given) heat transfer towards the helium, dominated by the Kapton foil (-> given) helium temperature (-> validation)

hbrockway
Download Presentation

Introduction

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Introduction • The temperature margin is given by following aspects: • Properties of the conductor and operating fields (-> given) • heat transfer towards the helium, dominated by the Kapton foil (-> given) • helium temperature (-> validation) • The helium temperature will vary along the string; therefore the maximum temperature in the system has to be evaluated. • For this discussion following aspects have to be considered: • Basic assumptions • Lattice of SIS300 • Geometric data of the magnets (spec. dipole ) • Heat loads of all individual components and effects • Recooling efficiency • Available processes parameter of the refrigeration cycle

  2. Basic assumptions SIS300 will be supplied in two strings. The total heat load of SIS300 is 3455 W. Therefore the minimum required mass flow rate within one string is 100 g/s.

  3. Lattice of SIS300 mass flow along coil dh,1 mass flow rate along recooler dh,2 ------------------------------------ = -----

  4. Lattice of SIS300 . . Tout, coil,i = f( hin(Tin,I, p,) + Qdipole /mcoil , p) Tout, recooler,i=T2ph+(Tin,i-T2ph)/exp(kA/mrecooler*cp(Tmiddle , p) Tout, dipole,i= f( (hout,coil,I*mcoil + hout,recooler,I*mrecooler)/m , p) Tout, line1,i = f( hout,dipole(Tin,I, p) + Qtube/m , p) Tout, quad, i = f( hout,line(Tin,I, p) + Qquad/m, p) Tout, line2, i = f( hout,quadpole(Tin,I, p) + Qtube/m , p) Tout, lineHX i = f( hout,quadpole(Tin,I, p) + Qtube/m , p) +(Tin,i-T2ph)/exp(kA/mrecooler*cp(Tmiddle , p) . . . . . . . . . . . . .

  5. heat loads in SIS300 Dipole Quadrupole Rest length [m] 7,6 length [m] 1,2 length [m] 4,2 load stat [W] 6,56 load stat [W] 1,69 rest load [W] 2,99 load dyn [W] 46,1 load dyn [W] 7,2 beam pipe[W] 3,03 beam pipe[W] 0,05 Bypass line 0.5 W/m

  6. heat loads in SIS300 => Increasing the heat exchange area outside the dipole => to lower the suction pressure or => to increase the mass flow rate. This needs additional power to recycle the surplus of liquid by a pump.

  7. . . . . m coil = 80 % m m recooler = 20 % m m coil = 20 % m m recooler = 80 % m . . . .

  8. Size: 300*300*800

  9. Temperature profiles of one dipole cell

  10. Temperature profiles of one dipole cell

  11. Two-phase line

  12. It concludes that for the 4 K sub-atmospheric operation one needs bigger heat exchanger tube for the two-phase flow. The tube with 65 mm ID allows us to get the high mass flow rate up to 200 g/s with acceptable pressure drop (50 mbar) and temperature difference (50 ~ 60 mK) over the half ring length (seen in the following two pictures). Two-phase line

  13. Two-phase line

  14. Energy cost aspect: P pump= 2*(Pstat + Pcomp)=2*(110 W +370 W) *230W = 220800W P = Ptotal * (280-230) = 200000W T P h s [K] [Pa] [J/kg] [J/kg K] suction 4,31 1,10E+05 10505 3673 isentrop 11,25 3,00E+05 12013 3673 real process 0,5 19,41 3,00E+05 15029 recooled 4,40 3,00E+05 11328 Difference 3700

  15. l=length, dh= hydraulic diameter, r = density, w= velocity, D p= pressure drop, z = friction factor ; 2300<Re<5*106, .5<Pr<2000,L/d >1 dh= 4 cross section/ wetted surface D pdipole= f(maximal allowed pressure drop along string , number components, safety margin,etc.) D pstring= 0.5bar => D pdipole= 3 mbar

  16. Hole size Tube diameter d d d tube m 0.07 0.07 Dp Pa 300 300 Diameter of ellipse m 0.0833 0.1032 Sensitivity to Pa/m 61695 119333 diameter changes

  17. 100 g/s 200 g/s

  18. 100 g/s 200 g/s

  19. Cooling tube

  20. p_in [Pa] 1,10E+05 T_mid [K] 4,310588 density [kg/m^3] 18,4 visco [Pa s] 1,29E-06 mass flow [kg/s] 0,1 d [m] 0,066 length [m] 8 cross section [m^2] 0,003421 wetted sur [m] 0,207345 hydraulic diameter [m] 0,066 velcosity [m/s] 1,592259 Re [-] 1501248 friction [-] 0,010862 Dp [Pa] 0 friction 0,010811 Pr [-] 1,327879 Nu [-] 2553,918 l [W/m K] 0,009336 a [W/m^2K] 361,2637

More Related