Exercises in the lectures on Exercises in the lectures on “ Superconducting RF - I and - II ” p g Kenji Saito, KEK
Exercise I. Using the Abrikosov’s theory: κ h κ φ λ φ φ c ( hc / 2 ) e = = = 0 = 0 = 0 H , , H 2 c c 2 2 2 2 c c 2 2 2 2 λ λ λ λ π π λξ λξ ξ ξ πξ ξ π λξ 2 2 * e 2 2 2 2 2 2 − 7 2 φ = = × ⋅ hc / 2 e 2.0678 10 Gauss cm 0 -15 15 2 2 × ⋅ =2.0678 10 T m 1) write down ξ , λ λ by H C and H C2 , 2) get the T-dependences of ξ , H C2 , κ , H C RF RF , from the given T-dependences ξ of λ λ and H C : λ (0) ( ) ⎡ ⎡ ⎤ ⎤ 2 = = − λ λ = = H H ( ) ( ) T T H H (0) 1 ( / (0) 1 ( / T T T T ) ) ( ) ( ) T T , ⎣ ⎣ ⎦ ⎦ C C C 4 − 1 ( / T T ) C H RF RF = ⋅ C ,here H is given as H 2 . κ κ C C C C
Exercise II. 1) Get the following formula for the surface resistance Rs for good electric conductor. µω µω µσω µσω 1 1 1 1 = = = R S σ σ σδ 2 2 2) Calculate the δ and R S for a 1300MHz copper cavity, when the σ is given as 1/ σ =1.72E-8 [ Ω m] at 20 O C. 1/ σ 1.72E 8 [ Ω m] at 20 C. 3) If the RRR of the copper material is 40, calculate the Rs at 4.2K. Exercise III. Exercise III By the two fluid model, electric conductivity is given as the bellow: ⎛ ⎞ 2 2 n e n q J E ⎜ n − s s ⎟ = σ σ σ − σ = i E , = i n s ⎝ ⎝ ω ω ⎠ ⎠ v v m m m m e s Put this complex electric conductivity into the formula of surface impedance: Z=R S +iX S, show the surface resistance and admittance for superconductor are: 2 ⋅ 1 1 n n e e 2 2 2 2 3 3 = σ ω µ λ λ = ωµλ λ σ = n R , X and S n L S L n ⋅ 2 v m S n n is the number of unpaired electrons (quatsi particle), then it could be written by Boltzman statistics as: then it could be written by Boltzman statistics as: Δ 2 − e k T σ = n (0) e B n s ⋅ m v Show the formula of surface resistance in case of superconductor as: Show the formula of surface resistance in case of superconductor as: Δ 2 l = λ ξ ⋅ ⋅ − R T f ( , ) A ( , , , T ) f exp( ) S L C k T B
Exercise IV. Get the formulas in lecture note p.65 ⎡ ⎤ ⎛ ⎞ ∂ ω 1 B B e = ∇ ⎜ z ⎟ + εµ ×∇ ⎢ i E ⎥ , ⎛ ⎞ t t z t z ⎝ ⎝ ⎠ ⎠ 2 ⎣ ⎣ ⎦ ⎦ ∂ ω z c 2 2 ⎜ ⎜ εµ − ⎟ ⎟ k k 2 ⎝ ⎠ c ⎡ ⎤ ⎛ ⎞ ∂ ω 1 E E e = ∇ − ×∇ ⎜ z ⎟ ⎢ i B ⎥ ⎛ ⎛ ⎞ ⎞ t t ⎝ ⎝ ⎠ ⎠ z t z 2 ⎣ ⎣ ∂ ∂ ⎦ ⎦ ω ω z z c c 2 εµ − ⎜ ⎟ k 2 ⎝ ⎠ c Exercise V. e c se V. Make design a 1300MHz TM 010 – mode single cell Pill Box cavity 1.What is the diameter of the cell? 2 What is the cell length? 2. What is the cell length?
Exercise VI. Superfish outputs f 0 =1293.77430MHz Ploss=118.1551W RsQ=265 171 Ω RsQ=265.171 Ω Qo=28257.6 (Rsh/Q)=109.24 Ω Hp=1753.44 A/m Hp 1753.44 A/m Ep=0.946176 MV/m Calculate the following cavity RF parameters from above Superfish outputs. Calculate the following cavity RF parameters from above Superfish outputs. Rsh [ Ω ] = Accelerating Voltage V [MV]= RF wave length λ[ λ[ m ] ] = Gradient Eacc = V/L eff [MV/m]= ,defined as L eff = λ /2 eff [ ] , eff Hp/Eacc[Oe/(MV/m)] = , use 1A/m= 4 π 10 -3 Oe Ep/Eacc = E Eacc [MV/m] = Z= [MV/ ] Z Geometrical factor Γ Γ [ Ω ] =
Exercise VII. Calculate the cable correction factors: C in , C r and C t , when measurement results are: when measurement results are: p in =55.5 μ W, p o =50.0mW, p r =10.72 μ W, p t =3.04mW and ’ 39 0 ’ 22 6 ’ 27 9 p o ’ =39.0mW, p in W ’ =22.6mW, p t W ’ =27.9mW W Exercise VIII. Calculate β in ∗ , β , β in , β , β t , P loss [W]’ , Q L , Q in , Q o , Q t , R s [ Ω ], Eacc[MV/m], E p [MV/m], and H p [Oe], when measure results are : h lt f 0 =1303.590529MHz, τ 1/2 =23.6 msec, τ 1/2 23.6 msec, coupling over, p in =3.11mW, p r =192nW, p t =0.142mW. For the cable correction factors, use the results of the exercise VII. RF cavity parameters are given as following: Γ 269 Ω E /E Γ =269 Ω , Ep/Eacc=1.83, Hp/Eacc=45.2 Oe/[MV/m], 1 83 H /E 45 2 O /[MV/ ] and = ⋅ Eacc MV m [ / ] 86.94 P Q [ W ] t t
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