A superconducting magnetic energy storage coil has 10,000 turns of wire, a mean diameter of $10 \mathrm {~m}$, and a length of $2 \mathrm {~m}$. It carries a current of $2000 \mathrm {~A}$.
(a) Calculate the energy storage capacity in MJ.
(b) A device with a larger capacity is required. Calculate the effect of doubling each of the following parameters while leaving the others the same
(i) Length
(ii) Mean diameter
(iii) Current
(iv) Number of turns

Question

Quizplus · Accepted Answer

The Answer is (a) The energy stored in the solenoid (in joules) is given by

E = \frac { 1 } { 2 } r N ^ { 2 } I ^ { 2 } \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 l } { r } ight] }

Using the values from the problem

\begin{array} { l } r = 5 \mathrm {~m} \N = 10,000 \I = 2000 \mathrm {~A} \l = 2 \mathrm {~m}\end{array}

then the energy is calculated to be

E = ( 0.5 ) imes ( 5 \mathrm {~m} ) imes \left( 10 ^ { 4 } ight) ^ { 2 } imes ( 2000 \mathrm {~A} ) ^ { 2 } imes \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 imes 2 \mathrm {~m} } { 5 \mathrm {~m} } ight] } = 3.0 imes 10 ^ { 9 } \mathrm {~J}

or

3000 \mathrm { MJ }

. (b.i) For

l = 4 \mathrm {~m}

the energy is

E = ( 0.5 ) imes ( 5 \mathrm {~m} ) imes \left( 10 ^ { 4 } ight) ^ { 2 } imes ( 2000 \mathrm {~A} ) ^ { 2 } imes \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 imes 4 \mathrm {~m} } { 5 \mathrm {~m} } ight] } = 2.294 imes 10 ^ { 9 } \mathrm {~J}

(b.ii) For

r = 10 \mathrm {~m}

the energy is

E = ( 0.5 ) imes ( 10 \mathrm {~m} ) imes \left( 10 ^ { 4 } ight) ^ { 2 } imes ( 2000 \mathrm {~A} ) ^ { 2 } imes \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 imes 2 \mathrm {~m} } { 10 \mathrm {~m} } ight] } = 7.091 imes 10 ^ { 9 } \mathrm {~J}

(b.iii) For

I = 4000

A the energy is

E = ( 0.5 ) imes ( 5 \mathrm {~m} ) imes \left( 10 ^ { 4 } ight) ^ { 2 } imes ( 4000 \mathrm {~A} ) ^ { 2 } imes \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 imes 2 \mathrm {~m} } { 5 \mathrm {~m} } ight] } = 1.2 imes 10 ^ { 10 } \mathrm {~J}

(b.iv) For

N = 20,000

the energy is

E = ( 0.5 ) imes ( 5 \mathrm {~m} ) imes \left( 2 imes 10 ^ { 4 } ight) ^ { 2 } imes ( 2000 \mathrm {~A} ) ^ { 2 } imes \frac { 3.9 imes 10 ^ { - 5 } \mathrm {~J} / \left( \mathrm { m } \cdot \mathrm { A } ^ { 2 } ight) } { \left[ 9 + \frac { 10 imes 2 \mathrm {~m} } { 5 \mathrm {~m} } ight] } = 1.2 imes 10 ^ { 10 } \mathrm {~J}

The current and number of turns have the greatest effect and the radius is next. The length has a relatively smaller effect.

A Superconducting Magnetic Energy Storage Coil Has 10,000 Turns of Wire