Recent content by JE93

thanks for you help really appreciate it. Looking back at it all, it shows using the power triangle was a massive help in working it all out. I just need to stop the silly little mistakes haha.

11*10^3/√3=6350.8 6350^2/1.25*10^6 =32.3Ω

sorry, That now gives me 167.7 Ω

Qc=V^2/Xc 1.25=11^2/Xc Xc=11^2/1.25=96.8Ω

could you just use simple ohms law? but using V^2/P so 1.25^2/3.2=0.48kΩ thinking this maybe wrong though

/ you would need 1.25MVAr to cancel out

Sorry must have put it wrong, I worked it out to 3.72MVA using P tan θ 3.2tan11.5 = 0.65

Thanks From re-doing the power triangle for a power factor of 0.98 I now have: Qd= S*Sin θ = 3.2sin11.5 = 0.74MVAr Where would I go from here to calculate the Capacitive Reactance, seem to be getting confused as the equations i know of to calculate Xc you need to know the capacitor value in...

I have re done this now using the power triangle above. I realize now that I used the MVAr rather than the MVA for calculating the p.f. Now i have: s=√(P^2+Q^2 ) = 3.2MVA 2.3MW/3.2MVA=0.72 〖cos〗^(-1) 0.72=43.9° = θ Would I now put the correct power factor into the previous equation, so...

Hi Gneill. I got the power factor from doing the real divided by the reactive power which, would be 1.9mva/3.2mw. looking at a power triangle where Z (impedance) is the hypotenuse and X (reactance) is the opposite side, X=Zsin θ, so would this be: 167.7-j229.2 sin 53.8 presuming the p.f. is...

the full question is: The following phase schematic diagram (FIGURE 4) shows an 11 kV, 50 Hz, 3-phase, short line feeding a load. By calculation or constructing the phasor diagram (use a scale of 1 mm = 2 A) for the load current with VR as reference, determine the capacitive current and •...

Just need a bit of guidance to make sure that I am heading the right way. Question is: Fig 3 shows the block diagram of the control of an electric heating system. The heater is driven from a voltage-controlled power supply, the voltage V1 being derived from a potientiometer. The output...

I understand it now! 2.55 is just 1 unit change so 4 unit changes would be 10.2 drawing the step change in graph form and getting the triangle works out to 10.19 Thanks for the help in understanding this, appreciate it. Presuming you are taking HND Electrical electronic engineering? how are...

Thanks, I have been looking at this equation and trying to get it to work but seem to be miles off. Output Change = Input Change (1-e^-t/T) 4 = 10 (1-e^-t/5) Re-arranged: t = -T ln (Output Change / Input Change) t = -5 ln(4/10) t = 2.55 I must...

Sorry should have been a bit more clear. Looking for help with the question on the original post, for question 4 on TMA 1 . Sorry again for late replying, juggling this as well as work, only seem to get round to doing anything whilst working nights. Thanks