Recent content by Guineafowl

Ok, so a cosphi of 0.65 gives an angle of 49.5 degrees: $$Z=36.4\Omega <49.5° = 23.6 +j27.7\Omega$$ $$C=\frac 1 {2\pi f 27.7} = 115 \mu F$$

VFD manual should be linked in the first post. Yes, I thought lifting the car slower might reduce the loading of the system. Or, running at 60Hz might improve the power factor, according to the motor plate. All doable with a twiddle of a pot. It seems very inefficient. Lifting a 2-tonne car at...

As a first attempt, looking at the motor plate (connected in star): $$Z=\frac V I = \frac {400} {11} = 36.4\Omega$$ Find capacitor with Xc of that value at 50Hz: $$ C = \frac 1 {2\pi f X_c} = 87\mu F$$

1. Done - it has automatic screw lubricators which have been cleaned out and topped up. 2. Yes, certainly something to try. I assume 3 equal values of capacitance - start at 5uF? As I understand it, the motor is custom-designed to be used for minutes in the hour (pf not a priority), but must...

Motor plate, for interest:

This one is a 3ph 400V in and out model. Can’t be changed. Cheap enough for a punt on this problem; if no go, I can use it for something else. It’s not worth buying a 1ph 230V in, 3ph 230V out model for this one car - it would have to be specced a lot higher than the 3.5kW, given the lift...

Power supply to the 7.5kW rotary is 1ph, 240Vac. Supply voltage drop at full load is within 5%. The rotary outputs (roughly) 415Vac 3ph to the screw-type car lift, capacity 2.7 tonne. The lift motor is a 3.5kW, 50Hz, 4-pole, asynchronous induction motor with a power factor of just 0.65. Phase...

I’ve also seen it written: $$\int fg’ \,dx = fg - \int f’g \,dx$$ Maybe you could preface with ‘f and g are functions of x’ and save all the (x)s.

Thanks, that all makes sense. All I will say for the ##uv## version is, it’s briefer, and for the integral bits, there’s a pleasing symmetry to: $$\int u\frac{dv}{dx}…\int v\frac{du}{dx}$$ All those ##(x)## bits look a bit messy to me:smile:

Another thing - I didn’t know you could evaluate the ##uv## part of IBP between the limits, as you have done in the square brackets. I haven’t seen that in any guide to integration, internet or book-based.

I thought the comment about initial terms vanishing was a maths/physics in joke, but there it is made real. Your method is a lot quicker and simpler than mine, which in turn was simpler than the integral calculator, which used u-sub in a very long-winded fashion. If you’d been my maths...

Now, at my stage, the way I know to solve a definite integral is do exactly this. In the real world, are you saying this isn’t always possible? If so, how is it done without dropping the bounds?

Stripping away the ##\lambda##, researching examples of the type: $$\int_0^\infty u^n e^{-u} du$$ I found that this is the gamma function for n+1: ##\Gamma (n+1)=n!## Obviously, if ##\lambda <0## in the original equation then we will no longer have ##e^{-\lambda u}## but ##e^{+\lambda u}## and...

Right, I tried to knock your question into bits, and accidentally found something called the Gamma function. First, the origin of the ##\lambda ^{n+1}## term. How did it get underneath, and why n+1? If I substitute ##u=\lambda x, x=\frac{u}{\lambda}, dx=\frac {du}{\lambda}##, I get...

I see. I assumed the wrong sign must have crept in during the difficult integration bit, not the easy final bit. Putting my water wings back on for the top line, quoted in post #12: $$-(2ax^2e^{-2x/a})-(2a^2xe^{-2x/a}+a^3e^{-2x/a})$$ $$-2ax^2e^{-2x/a}-2a^2xe^{-2x/a}-a^3e^{-2x/a}$$...