volt drop calcs and 230 v

[jackhammerJIM]

why do we have to do the calcs to 230v when the supply never is 230v

measured voltage at source is 248 volts at the site i am on at the moment ,

can somebody shed some logic on the idea before i start engineering an 18 volt drop into my cable sizes

( a joke of course but why not ? ) :ihih:

 

[7029 dave]

With that kind of voltage it should be no supply, nothing will work just too low!!

 

[Flanders]

With that kind of voltage it should be no supply, nothing will work just too low!!

7just realised Typo should of been 249 volts 251 volts:81:

 

[7029 dave]

yes lol coz I had job last year nothing working but had polarity, measured 155 v on the screws passed over to DNO joint had gone down in street.

 

[Jabbajaws]

yes lol coz I had job last year nothing working but had polarity, measured 155 v on the screws passed over to DNO joint had gone down in street.

Now this makes more sense. Nothing working which l didn't think it would....

 

[Knobhead]

Never mind the IET and EU harmonisation.

The specification for a DNO local transformer is for up to 1000KVA, 11/.433KV 4.5% reactance. 1500KVA the reactance rises to 5%

433V not 400V

 

[cmac70]

You could always do the good old way a taste to see what flavour voltage it is 230 or 400.

 

[Knobhead]

You could always do the good old way a taste to see what flavour voltage it is 230 or 400.

Which planet are you on?

230 = 400/√3, to all intense and purpose the same.

Which flavour of vanilla?

If you really want to be pedantic 230.94 = 400/√3

 

[Silly Sausage]

If you really want to be pedantic 230.94 = 400/√3

If you want to be really pedantic 230*√3 = 398.37

 

[Engineer54]

No idea what you're all arguing about you'll never going to be seeing 230 Volts or 400 volts in the UK or Europe for literally 10's of years, if ever!! lol!! It's all just fantasy world stuff!!

 

[Knobhead]

It could be done using the voltage regulators on the 33KV OLTC’s. The knock on would be disastrous. The LV network can’t cope with the current loading as it is now.

I tried lowering the voltage on an 80MVA system, I soon put it back to where it was.

 

[Knobhead]

If you want to be really pedantic 230*√3 = 398.37

 

[westmac]

What do the IET expect to happen where this occurs? Im a little confused here. If l listen to my maths teachers from sparky college, he said that a voltage drop can cause a circuit to overload.

Is that whats happening here, at the transformer perhaps? Ive no idea personally and would love to know the answer as ive never seen 223V anywhere before. Always been 236-253V...

I think your maths teacher needs to think about this "overload" comment, it's simply incorrect. A drop in the voltage will cause the load to decrease.

 

[i=p/u]

This guy above is correct , as I've been put right before and explained too .. Burning forget.

but I'd say its a misunderstood equation , higher voltage = lower current ..

 

[Knobhead]

I think your maths teacher needs to think about this "overload" comment, it's simply incorrect. A drop in the voltage will cause the load to decrease.

Time you learnt about workload and efficiency. Not every load is resistive, in fact most aren’t.
Take a motor, input power = output power – efficiency.

1KW at the shaft needs 1KW feed in to the motor and a bit more for the losses.

1KW = 1000W no mater how you look at it.

 

[westmac]

Time you learnt about workload and efficiency. Not every load is resistive, in fact most aren’t.
Take a motor, input power = output power – efficiency.

1KW at the shaft needs 1KW feed in to the motor and a bit more for the losses.

1KW = 1000W no mater how you look at it.

This is the same as another post in which this came up. Items on electrical equipment are given a power rating according to a nominal voltage, they do this as voltages are never constant.

With no voltage applied a motors windings are a theoretical linear load, once voltage is applied the load becomes inductive and so the impedance opposes the flow of current and and limits it. Even with no rotor, a motors windings will limit the flow of current.

Once a rotor is introduced, the windings induce a voltage (and so a current) within it and as the sine wave of the supply voltage fluctuates at the given frequency, the rotor tries to keep up. As the load on the motor is increased, the slip of the rotor increases also and with this there is a greater difference in speed between the supply frequency and rotor, creating a larger current in the rotor.

With these points made, how would a lower voltage on a given size of motor result in a greater current? What happens if a 240V voltage is placed across the coil of a 24V relay? The relay will have a stated holding current at its nominal voltage (24V) but applying 240V won't reduce this current ten fold to keep the coil power consumption constant.

Although a motors load may indeed be constant at 1kW, this does not mean the voltage and current will flex to meet this fixed load. A motor may be capable of providing the torque to meet a load of 1kW at a given geared speed negating the slip. This very same motor may be incapable of generating the torque at a few volts less as it would simply stall if the voltage was not high enough to generate a high enough current to drive the load before losses.

If a motor is loaded at 100% of its rating, it must be supplied with its rated nominal voltage or higher to drive the load. A lower voltage will not just simply result in a constant power output through an increased current.

I think the time for me to learn has passed and it is you who needs more time to learn.

 

[Lucien Nunes]

With these points made, how would a lower voltage on a given size of motor result in a greater current? What happens if a 240V voltage is placed across the coil of a 24V relay? The relay will have a stated holding current at its nominal voltage (24V) but applying 240V won't reduce this current ten fold to keep the coil power consumption constant.

This is a non-sequitur because a motor is not the same as a relay coil.

A relay coil has constant impedance at a given frequency because it is a simple RC series network, therefore its current will be approximately proportional to the voltage (ignoring magnetic non-linearities). Note that the electrical power absorbed at a given voltage is determined mainly by the characteristics of the coil.

A typical AC motor is a much more complex animal. Its stalled-rotor impedance is relatively low, therefore the electrical power absorbed at a given voltage is determined not so much by the characteristics of the motor but those of the load. I.e. it does not look like a constant impedance from the electrical side.

For synchronous and induction motors, within sensible limits that do not cause the motor to stall, the variation of speed with voltage is zero or small. For any conventional load torque-speed curve the variation of torque with voltage is thus also zero or small, hence also the output power. But as the voltage changes, so do the load angle and power factor. It is necessary to know the characteristics of the motor to determine precisely how the current and power factor will vary with voltage at constant torque, however the in-phase component of the current must vary inversely with the voltage and can be calculated from the input power as Tony points out above.

A more clearcut case occurs with a power-factor-corrected switched-mode power supply that operates at constant efficiency and constant output voltage. That most certainly does have negative dynamic resistance, and will consume current inversely proportional to the voltage, within the range of voltages it will accept.

That's about as clear an explanation as I can give without pages of algebra!

 

[Knobhead]

This is the same as another post in which this came up. Items on electrical equipment are given a power rating according to a nominal voltage, they do this as voltages are never constant. True

With no voltage applied a motors windings are a theoretical linear load, once voltage is applied the load becomes inductive and so the impedance opposes the flow of current and and limits it. Even with no rotor, a motors windings will limit the flow of current. Try it, keep a fire extinguisher handy

Once a rotor is introduced, the windings induce a voltage (and so a current) within it and as the sine wave of the supply voltage fluctuates at the given frequency, the rotor tries to keep up. As the load on the motor is increased, the slip of the rotor increases also and with this there is a greater difference in speed between the supply frequency and rotor, creating a larger current in the rotor. You need a slipring motor to prove the rotor currents. You may be surprised by the readings for slip current.


With these points made, how would a lower voltage on a given size of motor result in a greater current? What happens if a 240V voltage is placed across the coil of a 24V relay? The relay will have a stated holding current at its nominal voltage (24V) but applying 240V won't reduce this current ten fold to keep the coil power consumption constant.

Although a motors load may indeed be constant at 1kW, this does not mean the voltage and current will flex to meet this fixed load. A motor may be capable of providing the torque to meet a load of 1kW at a given geared speed negating the slip. This very same motor may be incapable of generating the torque at a few volts less as it would simply stall if the voltage was not high enough to generate a high enough current to drive the load before losses.
You will never escape:
Power in = Power out + losses


If a motor is loaded at 100% of its rating, it must be supplied with its rated nominal voltage or higher to drive the load. A lower voltage will not just simply result in a constant power output through an increased current.
Want to rethink that?
I think the time for me to learn has passed and it is you who needs more time to learn.

I’m willing to learn at any time. Lead on.

 

[i=p/u]

The question is , does lower voltage = higher current. Won't an item pull what it needs.

 

[Lucien Nunes]

The question is, do you understand enough electrical theory to work this out from first principles?

It depends on the 'item'. A heating element connected to the mains will pass a current that increases with rising voltage. A laptop power supply will pass a current that decreases with rising voltage. If you understand how something works you can calculate how its current will vary.

 

[i=p/u]

Higher voltage = less current is the principle I know.. But it obv isn't the the case. Ty I shall dig my heels in read on...

 

[Knobhead]

The question is, do you understand enough electrical theory to work this out from first principles?

It depends on the 'item'. A heating element connected to the mains will pass a current that increases with rising voltage. A laptop power supply will pass a current that decreases with rising voltage. If you understand how something works you can calculate how its current will vary.

First principals just aren’t taught anymore.

Lucien, you and I are from a different age!

 

[Lucien Nunes]

I=P/U, your username only applies when dP/dU is zero. While I'm at it, here's one for you to think about...

Assume a heating element has constant resistance, so P=V²/R and I=V/R. i.e the current increases proportionally with voltage. Fit 100 of those same heating elements in thermostatically controlled heaters in a building, let the temperatures stabilise and now try varying the supply voltage, all other factors remaining equal. What happens to the supply current?
Now, on which basis do you calculate the worst-case voltage drop?

BTW Tony I am not sure I like the way you phrased that. What age? Bronze?

 

[ejgajg]

Depends On The load of the transformer, cable lenght and supplied voltage of medium voltage. All conform EN 50160 regulation

 

[i=p/u]

You learn ohms law and power triangle but they don't talk about different loads ... Are they really an issue. Just theory which isn't relevant???