Recent content by Joe85

Thank you gneill. You’ve been a real gent.

So having quickly whacked the figures into Symbolab Calc on my phone, the PSO = 377627.86 Watts or 378KW. So the answer was correct but the method wrong, hence what you were saying about the entire source being lost?

Thanks gneill, I'm eternally grateful thus far, but why can't we just carry out the subtraction for VS-VR since we know what both VS and VR are? And then insert this into the power absorbed equation?

Sorry it was PS from post 5. Or 378KW.

There’s so little in the learning materials to go on here. isn’t A1+jA2 the reverse voltage gain. So (A1+jA2)2/ ( VS2/P) would equal the power absorbed?

Doh, sorry Gneill. That's what i get for copy and pasting! PS= Re{VSIS*} I didn't get as PSO (which i assume will be the result of PR - PS? S & R being Sending and receiving?) And would PR = Re{VRIS*} ?

VS=VR(A1+jA2)+IR(B1+jB2) IS=VR(C1+jC2)+IR(D1+jD2) Given the parameter values in TABLE C and an open-circuit received voltage measured as 88.9 kV, calculate the values of VS and IS and hence the power (PSO) absorbed from the supply by the transmission line on open circuit. VR= 88.9×103 Table...

Thank you, Sir.

Equations: Pr : Pm : Pc = 1 : 1-S : S Total Power into rotor: Pr=I'22R'2/S Mechanical Power Output: Pm=I'22R'2(1-S)/S Power loss in rotor: Pc=I'22R'2So I'm a little skeptical of my answer here. Seems a little too easy, which normally means I've missed to mark by about 100 miles. Attempt...

Thank you. Your assistance and patience is very much appreciated.

Just before you replied i edited this in above, honest :D Edit: Ok so i believe by using the following equation i can isolate for n2: n∝V/Φ and Φ∝If So n∝E/If and n=(E/If)*K Which gives: n2=( n1E2If1) )/( E1If2) :confused:

Right and the only equation that seems of any use to be able to resolve for flux is: E1/E2=Φn1/Φn2. I feel like a complete idiot but I cannot see how to isolate Φ. Edit: Ok so i believe by using the following equation i can isolate for n2: n∝V/Φ and Φ∝If So n∝E/If and n=(E/If)*K Which gives...

Thanks again, So i should begin by trying to work out the flux before the resistance inserted and then after it's inserted to be able to know what the change in speed is?

Thank you cnh. And the speed will just be the same as above: n2 = (n1xe2)/e1 = 606RPM?

Ok so: 21.365 x 2.273 = 48.558A 500/270 = 1.852A. 48.558/1.852A= 26.22A?