Superconductors — Question 8

Question 8

A quench can convert stored loop energy into a temperature rise

A loop of inductance L initially carries $I_0$. A quench creates resistance $R_q>0$ and the supplied decay is $I(t)=I_0e^{-R_qt/L}$. Assume all Joule heat remains in a body of constant heat capacity C, with no energy extraction.

1. Find heat deposited up to time t and the final temperature rise.

2. Find the time when half the total heat has been deposited and the maximum initial current for an allowed rise $\Delta T_{\max}$.

Question 8 - Solution

A quench can convert stored loop energy into a temperature rise

(a) Method and calculation

$$Q(t)=\int_0^t R_qI_0^2e^{-2R_qt\prime/L}dt\prime=\tfrac12LI_0^2(1-e^{-2R_qt/L}).$$$$\Delta T_\infty=Q(\infty)/C=LI_0^2/(2C).$$

(b) Interpretation and further calculation

$$1-e^{-2R_qt_{1/2}/L}=1/2\Rightarrow t_{1/2}=\frac{L\ln2}{2R_q}.$$$$I_0\le\sqrt{2C\Delta T_{\max}/L}.$$

Physical check

Resistance determines how quickly heat is released; the total comes from the initial stored energy.