Algebra and formulas

[telectrix]

This is 1 out of the book
A machine lifts a mass of 50kg through a distance of 15 metres in 30 seconds what will the power required be?

so that's 110lbs. through 49ft.

answer in ft.lbs. please.

 

[Paulgibson]

This is 1 out of the book
A machine lifts a mass of 50kg through a distance of 15 metres in 30 seconds what will the power required be?

Using power formula P=Force X Distance / Time. So Mass of 50kg X 15 / 30 = 25
Or also apply Newton s apple 1kg=9.81 Newton's about 245.

 

[Paulgibson]

Or a transformer has a 25:1 ratio. If the input voltage is 400v, the output voltage will be?

I'd just like help with the easiest ways to work these out please

Primary/ secondary, input V / output V, input c / output c

400 / 25 = 16. The secondary winding is 16v = a step down transformer.

 

[Paulgibson]

Advice with 202 though I'd say remember the star / delta formula it's self.

In Delta the formula being Vl=Vp so the voltage is balanced. It's the current that's square root of 3 1.732.

In Star it's the other way around.

 

[Jackmcleod89]

Advice with 202 though I'd say remember the star / delta formula it's self.

In Delta the formula being Vl=Vp so the voltage is balanced. It's the current that's square root of 3 1.732.

In Star it's the other way around.

thanks Paul you have helped me out massively getting my head around the, they can be used on the ohms law triangle can't they?

 

[Paulgibson]

Power triangle helps assess the loads and whether ur circuits can handle the loads. P =V/I . A good example is a 700 watt device / 230v supply is 3a so anything over 700 watt you would need a higher amp fuse.

 

[Simonslimline]

Power triangle helps assess the loads and whether ur circuits can handle the loads. P =V/I . A good example is a 700 watt device / 230v supply is 3a so anything over 700 watt you would need a higher amp fuse.

Don't forget that with AC, Power only equal Volts x Amps if the power factor is unity (1). If not then power equals Volts x Amps X power-factor.

 

[telectrix]

Don't forget that with AC, Power only equal Volts x Amps if the power factor is unity (1). If not then power equals Volts x Amps X power-factor.

smarrtarse. now you're getting beyond the Electrical Trainee level. careful.you'll have them all asking wtf is power factor.

 

[Simonslimline]

Advice with 202 though I'd say remember the star / delta formula it's self.

In Delta the formula being Vl=Vp so the voltage is balanced. It's the current that's square root of 3 1.732.

In Star it's the other way around.

Yes a vital one to remember. I had questions on this when i did my 309 science and principles written exam.

STAR - (IL = IP) (VL = VP X 1.732)

DELTA - (VL = VP) (IP = IL / 1.732)

 

[M4rtyj]

I too have got my principles exam next month C&G 2365 202, been doing lots of revision and always stumble on star/delta formulas, just cannot remember which way round they are.

 

[Paulgibson]

Also remember Power factor correction, and SohCahToa for real/reactive and total power. 303 is also more fun.

 

[Paulgibson]

I'd say try writing it down in your own words in a small note book as I find it helps slot for me anyway. Don't mean to try scare you with the 202 it's actually not that bad!

 

[Jackmcleod89]

I appreciate all the feedback and I'm going to write all these in my notebook for my course in ways I understand and also your ways too! Thanks again!

 

[happysteve]

I too have got my principles exam next month C&G 2365 202, been doing lots of revision and always stumble on star/delta formulas, just cannot remember which way round they are.

Alright, how's this:

The first thing to clock is what we mean by line and phase. I think of "line" as the connections (imagine a red 3-phase commando socket), so your line current would be current going to one of the pins on the socket, and your line voltage would be between two of the line connections (L1 to L2, or L2 to L3, or L3 to L1). I think of "phase" as the windings on the supply transformer - the squiggly symbol that looks like an inductor. So phase voltage would be voltage across one of these squiggles, and phase current would be the current through one of these squiggles.

You know that on one of them (star/delta) the current is the same (I[SUB]L[/SUB]=I[SUB]P[/SUB]), and on the other one the voltage is the same (V[SUB]L[/SUB]=V[SUB]P[/SUB]) but which way round? Also, maybe you know that, but can never remember whether it's V[SUB]L[/SUB]=√3 x V[SUB]P[/SUB] or vice-versa.

For the first one of these:

Star is the only one with a neutral. So the voltages must be different, as on delta you can only measure between any two lines and you'll always get the same voltage.

As for which is the bigger number, V[SUB]L[/SUB] (or I[SUB]L[/SUB]) or V[SUB]P[/SUB] (or I[SUB]P[/SUB]), consider this:

On a single phase system, you'd say "the voltage is 230V." This is between a line and neutral, in other words, across one of the squiggly bits on your picture, so it's the phase voltage (V[SUB]P[/SUB]).

On a three phase system, you'd say "the voltage is 400V." This is between two lines, so it's the line voltage (V[SUB]L[/SUB]).

So V[SUB]L[/SUB] must be bigger than V[SUB]P[/SUB]. So V[SUB]L[/SUB] = √3 x V[SUB]P[/SUB].

And on delta, what about the currents? The rule is: same relationship. So in both cases, line (voltage on star, current on delta) is bigger than phase (voltage on star, current on delta. So on delta, I[SUB]L[/SUB] = √3 x I[SUB]P[/SUB].

Hope that's helpful.