-- AlexanderBleile - 15 Jun 2010

Linear Resistance

Linear resistance of the Nuclotron Cable consists of 4 components:
  • resistance of copper matrix
  • resistance of NbTi filaments
  • resistance of CiNi tube
  • resistance of NiCr wire
The length of the wire $l_w$wounded around cable is bigger than the length of the cable $l$ by factor $c_w$
\[c_w=\frac{l_w}{l}=\frac{\sqrt{t_{twc}^2+\pi(D_{CuNi}+d_w)^2}}{t_{twc}}\]
with cable twist pitch $t_{twc}=47 mm$, outer diameter of CuNi tube $D_{CuNi}=5 mm$ and wire diameter $d_w=0.508 mm$ :

for the NUCLOTRON cable (two layer dipole): $c_w=1.021$

For the length of the NiCr wire wounded around cable:
\[c_{NiCr}=\frac{l_NiCr}{l}=\frac{\sqrt{t_{twNiCr}^2+\pi(D_{CuNi}+2d_w+d_{NiCr})^2}}{t_{twNiCr}}\]
with NiCr wire twist pitch $t_{twNiCr}=0.4 mm$ and NiCr wire diameter $d_{NiCr}=0.2 mm$ :

for the NUCLOTRON cable (two layer dipole): $c_{NiCr}=27.56$

linear resistance of copper matrix

\[rl_{Cu}=\frac{R_{Cu}}{l}=\rho(T,RRT)\frac{l_w}{A_{Cu}l}=\rho(T,RRT)\frac{c_w}{A_{Cu}}\]

\[A_{Cu}=\frac{A\alpha}{\alpha+1}=\frac{{\pi}d_w^2\alpha}{4(\alpha+1)}\]

$\alpha=\frac{A_{Cu}}{A_{NbTi}}$ $A=A_{Cu}+A_{NbTi}=\frac{{\pi}d^2}{4}$
Topic revision: r3 - 2010-06-16, AlexanderBleile
 
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