Question

solve part c

Hìnt: The following identities will be useful: LWI) PVI sin e sin 2ot (1-59) cos a cos B = [cos (a - B) + cos (a + B)] cos (a - B) - cos a cos B + sin a sin B 1-21. A linear machine shown in Figure P1-15 has a magnetic flux density of 0.5 T di- rected into the page, a resistance of 0.25 N, a bar length I=- 1.0 m, and a battery voltage of 100 V. (a) What is the initial force on the bar at starting? What is the initial current flow? (b) What is the no-load steady-state speed of the bar? (c) If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady-state speed? What is the efficiency of the machine under these circumstances? Facp - Find ilB Gndy 90 UB- TB The 0.25 2 B = 0.5 T eids UB-iR FIGURE P1-15 c TB'= 100 V the X. ing' potkendei efficircy Pin= Vb* pad Pin The linear machine in Problem 1-21.

Answer 1

solution Given Data Resistance R=0.25 Islom voltage = loov 0.25 tuo B=0.51 loov starting & what is @ what is the initial Force on the bar at the Initial Current flow ? Initial curent (18) = = 100) 40017 B=0.5 Vg =1000 Initial Force F. (LXB) 400 (LX0-5) ? tuo t o = 400% you F = 200N @nohat is the No-Load Steady state speed of the bor? VB - VBA Ve ar Be v= No - Load steady state speed of the bar Ve 100 V = 100 do 0.5X1 100 = 200m/s @ If the bar is loaded with a force of 254 opposite to the clirection of Motion what is the new steady state speed? what is the efficiency of the Machine under these Grumstances. V=VB VB = Pind - Venir VB - 100 VB = 100- iR B

i Q5 0.5X1 Vo=100-(0.25)50 (ve. 87.55] V = 87,5 0.5X1 V = 84.5 90 po 17 loitat |= 175 m/s 100 g ex of = Pout xioo Pino (axi) - Ponte Vexi = Pout Posting Vgxixe Pin = 87.5x50 100x5out - = 4375 = 5000 + n = 5000 . 4345x100 boas on 5000 875 2 = 4875 If you have any doubts ok ne pleage Rate my answe a ita fc7H50