the arc s periphery, and so it must equal(2rR, & n. Representing the temperature gradient Ra by(roughly)(- we obtain Ra 加2aE2=2tRk2-C R E=2 k(T-7)1 R and since o=aT-T) E=2 k R This important result indicates that the arc field, and hence its voltage, is inversely proportionally radius: the dissipation must increase if the arc is constrained more tightly, which improves its cooling. But note that Ra itself is not yet known, since it is only R, the constrictor diameter that is prescribed The total arc current is 1=[2mm(oEr. Once again, using o=0,we obtain I= TR--E (5) and substituting(4)here, 1=z2(C a Ra =x√2ak(z-7)R (6) Note also that, multiplying(4 )and (6) together we obtain EI=4mk (T-T which is another way to express the heat balance. Its main message is that the arc centerline temperature Tc increases linearly with arc power per unit length, El 16.522, Space Propulsion Lecture 11-12 Prof. Manuel Mar Page 3 of 18the arc’s periphery, and so it must equal 2πRa ( ) k ∂T ∂r ⎛ ⎝ ⎞ ⎠ r = Ra . Representing the temperature gradient Ra by (roughly) − ∂T ∂r ⎛ ⎝ ⎞ ⎠ Ra ≅ 2 Tc − Te Ra , we obtain π / Ra 2 / 1 2 σ cE2 = 2π /Rakc 2 Tc − Te Ra or E = 2 2kc (Tc − Te ) σ c 1 Ra (3) and since σc = a Tc − Te ( ), E = 2 2 kc a 1 Ra (4) This important result indicates that the arc field, and hence its voltage, is inversely proportionally to its radius: the dissipation must increase if the arc is constrained more tightly, which improves its cooling. But note that Ra itself is not yet known, since it is only R, the constrictor diameter that is prescribed. The total arc current is I = 2πr(σE) . Once again, using o R ∫ dr σ ≅ 1 2 σ c , we obtain I = πRa 2 σ c 2 E (5) and substituting (4) here, I = πRa 2 a T( ) c − Te 2 2 2 ka a 1 Ra I = π 2akc Tc − Te ( )Ra (6) Note also that, multiplying (4) and (6) together we obtain EI = 4πkc Tc − Te ( ) (7) which is another way to express the heat balance. Its main message is that the arc centerline temperature Tc increases linearly with arc power per unit length, EI. 16.522, Space Propulsion Lecture 11-12 Prof. Manuel Martinez-Sanchez Page 3 of 18