IDENTIFY THE CAPACITY OF MY MAINS SUPPLY CUT-OUT FUSE (PLUS...)

[Carl Haworth]

24kVA TOTAL, 8kVA per phase.

I see. Sorry! I misunderstood what you said. "24 kVA for single phase" doesn't mean 24 per phase. It's the total for the three (3 x 8 kVA).

You suggest leaving 45 A for the 3-phase section (229.63 V per phase, so 230). That is (45 x 230 x 3): 31,050 VA.

DNO said 55 kVA for the 3-phase (230 x 80 x3): 55,200 VA.

The two 3-P appliances proposed need a total of (15 + 11 Amps): 26 A, or 17,940 VA, so I need to allocate only 30 A, not 45 A, to the 3-P (20,700 kVA), still with a margin in hand.

That would leave (55,200-20,700); 34,500 VA for the S-P, meaning 16.67 A per core/group of S-P circuits - a 50 A total allocation to these circuits (34,500/230/3 cores/3 circuit groups); 16.67 A for each group.

Is that right??

If it is, then I would end up with:

3-.P section: 20,700 VA (30 Amps)
S-P section: 34,500 VA (3 x 50 Amps)
TOTAL: 55,200 VA (as stated by DNO)/(180 A. 180/3 = 60 A per phase)

If it this is right, it means that I would have for the S-P circuits 150 A total (3 circuit groups x 50 A each) (34,500 VA) instead of 1 circuit group x 80 A (18,4700 VA).

Looks too good to be true, but the high VA available to the S-P circuits would make it less difficult to divide these into 3 safe groups, and the 3-P allocation seems fine for what it envisaged.

The actual allocations could be varied within the pair of capacities - leaving the possibility of increasing the 3-P load (more powerful boiler and/or V charger?) at the expense of reducing the maximum S-P .load per group of circuits/phase core to bring this closer to present 80 A (or 100 A?) rather than 150 A.

However. Lucien, I stand fully prepared to be corrected - I fear that my result, although carefully checked, may indeed be found by you to be "too good to be true"!

Best wishes,

Carl

 

[Lucien Nunes]

TOTAL: 55,200 VA (as stated by DNO)/(180 A. 180/3 = 60 A per phase)

Actually 55,200 / 230 = 240A not 180A. Hence 80A 3-phase supply as per DNO.

That would leave (55,200-20,700); 34,500 VA for the S-P, meaning 16.67 A per core/group of S-P circuits - a 50 A total allocation to these circuits (34,500/230/3 cores/3 circuit groups); 16.67 A for each group.

By groups, I only meant assigning the loads into three groups for connection to the three phases. Not further subdividing them. (I am refraining from using the word balancing). So with 30A reserved for the 3-phase loads, there would be three groups of loads each of which could have the remaning 50A of each phase available.

If it this is right, it means that I would have for the S-P circuits 150 A total (3 circuit groups x 50 A each) (34,500 VA) instead of 1 circuit group x 80 A (18,4700 VA).

Exactly.

I mentioned avoiding the use of the word 'balancing' because it is not strictly necessary to balance these loads. It is helpful to do so in many cases for two reasons. One is to make the best use of diversity, so that none of the groups has an excessive tendency to exceed its diversified total under unfavourable usage conditions. The other is to attempt to maintain equal voltage drop on all conductors of the supply, to ensure that the voltage seen by 3-phase loads is as nearly symmetrical as possible. That would be of interest in a factory with lots of 3-phase induction motors, for example, which are at their most efficient when the line voltages are equal. If all the baseline single-phase load were allocated to one phase leading to it being fully loaded most of the time, while the others were only loaded intermittently, the voltage drops would often be lopsided and a very small percentage of efficiency thereby wasted. But some types of 3-phase loads, such as boilers and active rectifiers such as are used in car chargers, are not affected by voltage assymmetry in the same way, so that consideration has less impact in your application.

 

[Carl Haworth]

Actually 55,200 / 230 = 240A not 180A. Hence 80A 3-phase supply as per DNO.



By groups, I only meant assigning the loads into three groups for connection to the three phases. Not further subdividing them. (I am refraining from using the word balancing). So with 30A reserved for the 3-phase loads, there would be three groups of loads each of which could have the remaning 50A of each phase available.



Exactly.

I mentioned avoiding the use of the word 'balancing' because it is not strictly necessary to balance these loads. It is helpful to do so in many cases for two reasons. One is to make the best use of diversity, so that none of the groups has an excessive tendency to exceed its diversified total under unfavourable usage conditions. The other is to attempt to maintain equal voltage drop on all conductors of the supply, to ensure that the voltage seen by 3-phase loads is as nearly symmetrical as possible. That would be of interest in a factory with lots of 3-phase induction motors, for example, which are at their most efficient when the line voltages are equal. If all the baseline single-phase load were allocated to one phase leading to it being fully loaded most of the time, while the others were only loaded intermittently, the voltage drops would often be lopsided and a very small percentage of efficiency thereby wasted. But some types of 3-phase loads, such as boilers and active rectifiers such as are used in car chargers, are not affected by voltage assymmetry in the same way, so that consideration has less impact in your application.

 

[Carl Haworth]

Thank you very much, yet again, Lucien, for your response to my wondering if I'd got something wrong in my calculations, given that they showed such a "vast" capacity for the single-phase circuits (albeit divided into three groups).

And thank you also for your further enlightening on diversifying the peak loads possible on all the installation's circuits, so as to apportion these into groups which will each run reliably on c. 50 Amp supply limit (depending on load on 3-Phase). All that we know is that our peak loads currently produce no problems with what may be either a 100 A or an 80 A main fuse.

Before we had the new main fuse fitted, over 20 years ago, the installation was on a 1968 fuse box with a 60 A main fuse. There was no space for extra circuits:-

One fuseway for the front door chime transformer, 3 for lighting circuits (one of which supplied a wall-mounted heater in upstairs bathroom, and nothing else), one for an electric cooker (not used by us), one for the immersion heater/s, and two for ring circuits (each of which covered about half the floor area of both storeys of the house).

We had this replaced by a split consumer unit with twelve breaker slots and a RCCB for numbers 7 to 12.

This now has the following circuits, all with B breakers except 2:-

NOT VIA RCCB
1. Upstairs lighting, with inline RCD for bathroom) and forced ventilation (original circuit for only heater in bathroom - no longer in place).
2. Air conditioner (20D). Supplier of this system stipulated non-RCD-protected circuit.
3. Downstairs lighting (including outside, garage and workshop lights via RCD) plus door chime transformer
4. Downstairs lighting.
5. Downstairs lighting (original circuit)
6. Upstairs lighting (original circuit)

VIA RCCB
7. Ring
8. Ring
9. Immersion heaters (either/or)
10. Ring (original circuit)
11. Immersion heaters (original circuit)
12. Air compressor in my workshop (originally electric cooker circuit. The present appliance needs a circuit of its own with a 32B breaker because of high starting current (3.5 h.p. motor, so 20Dor lower would probably be more appropriate). But spike is largely avoided if compressor is started using the decompressor control.

The electrician who did most of this work advised uprating the main fuse from 60 to 100 A. We followed his advice, as much an anything to put the system into line with then-recent ones. It's conceivable that 60 would have remained OK.

Apportioning these circuits into three groups with roughly similar peak loads looks difficult!

ANOMALIES Which might be overcome when apportioning them into three groups of 4+5+5, say, rather than 4+4+4 (say):-

- One lighting circuit is for lighting plus door chime transformer which, for obvious reasons. it is undesirable to have on a shared circuit.

- If we dump gas completely (the daily standing charge becomes an increasing burden as consumption falls) we would need an extra 32B circuit for the cooker, whose wiring is in place in the ceiling void above the kitchen (with plenty of spare cable to drop down) and , at the other end, ready to connect to the present consumer unit.

Please don't feel obliged to respond to the immediately above. You will have better things to do! It's up to us to get it sorted out appropriately.

ASSYMMETRIC 3-PHASE VOLTAGES

Obviously one should aim to keep these as symmetric as possible, which is relatively straightforward when a complete installation circuit is being designed, but has to use hard-to-estimate-reliably data when the component circuits are established and working!

I suppose that a voltage drop on one phase relative to the other two (or on two relative to the other one) would simply reduce the wattage, and probably only slightly, of a non-induction-motor 3-phase appliance?

What is the effect of asymmetric voltage drop on an induction motor - just less torque, or reductions and increases which cause possibly fluctuating speed as well as loss of output power?

Very best wishes, and many thanks again for your time,

Carl

 

[Carl Haworth]

Gas currently costs just over 10p/kWhr, and is available at this price all day, every day.
There is the odd electricity dual tariff that can match this on the off peak night rate, but the corresponding rate for the majority of the day is at least four times the gas price, and increasing.

DNOs will not fit a main fuse that is larger than the max rating of the meter, and will remove larger ones without notice, or informing the customer, if they come across them in the course of other work. This policy has caused problems for more than one of my customers, where supplies that have been fault free for decades have suddenly started blowing main fuses at short intervals.

A 80A fuse does not happily conduct 80A indefinitely, and then blow if the current is increased to 81A. It will never blow at a little over 100A, and even at 200A will last over six minutes.

Thanks for that!

Electricity cost on my Octopus tariff is now, including daily standing charge, not much more than 2x similarly calculated price of electricity. I'm assuming that heat energy obtained by us from gas is about 75% of total heat energy supplied (we have oldish boiler at about 76%, gas cooker at considerably worse than that, and one gas convector heater which claims 80%).

Your comment about main fuse capacities does make me wonder how technically -wise (rather than just fitting-wise) the two men were who were sent by the DNO to uprate our main fuse all those years ago. I went to the trouble to point out the 40 Amp max marking on the meter (and the original fuse was a 60 Amp). They assured me that the meter was fine for 100 A. I doubt that we draw anything like this (though possibly now more than 60 A at times of peak load). The change was nearly 20 years ago and the meter still works fine.

Should someone more cautious/knowing from DNO spot this, wouldn't it make more sense for us to change the meter (which would have to be a Smart one)? Octopus fit these foc (we all know why foc!), if the fitting of an inadequate fuse was proposed?

 

[Lucien Nunes]

What is the effect of asymmetric voltage drop on an induction motor

Consider a 3-phase motor with three windings, each independently contributing a fraction of the torque needed to drive the rotor. The magnitude and phase angle of current in each winding depends on many factors but under ideal conditions all those factors will be equal and so will be the currents and the torque contributions of the three phases. If the voltage of one phase falls, the tendency for them all to equate will cause a shift in equilibrium that attempts to restore that voltage by transferring workload onto the others and increasing their currents. Because of the non-proportionality of some of the losses this results in higher total heat dissipation, as well as reduced overall torque capability and increased torsional vibration.

In the extreme case of one phase of the supply being completely lost, the motor will not only attempt to drive its mechanical load on the power available from the remaining two, but can also absorb power from them to regenerate the missing phase and attempt to energise any other loads connected to it. This normally results in heavy overcurrent and tripped control gear, although there are cases where a big motor on light load would carry on regardless and special precautions need to be taken to ensure the machine shuts down promptly. Conversely, the phenomenon can also be exploited as a means of converting single-phase supplies to 3-phase, using an 'idler' motor that does not drive a load, in conjunction with capacitors to compensate the reactive power flows.

 

[Carl Haworth]

Consider a 3-phase motor with three windings, each independently contributing a fraction of the torque needed to drive the rotor. The magnitude and phase angle of current in each winding depends on many factors but under ideal conditions all those factors will be equal and so will be the currents and the torque contributions of the three phases. If the voltage of one phase falls, the tendency for them all to equate will cause a shift in equilibrium that attempts to restore that voltage by transferring workload onto the others and increasing their currents. Because of the non-proportionality of some of the losses this results in higher total heat dissipation, as well as reduced overall torque capability and increased torsional vibration.

In the extreme case of one phase of the supply being completely lost, the motor will not only attempt to drive its mechanical load on the power available from the remaining two, but can also absorb power from them to regenerate the missing phase and attempt to energise any other loads connected to it. This normally results in heavy overcurrent and tripped control gear, although there are cases where a big motor on light load would carry on regardless and special precautions need to be taken to ensure the machine shuts down promptly. Conversely, the phenomenon can also be exploited as a means of converting single-phase supplies to 3-phase, using an 'idler' motor that does not drive a load, in conjunction with capacitors to compensate the reactive power flows

Consider a 3-phase motor with three windings, each independently contributing a fraction of the torque needed to drive the rotor. The magnitude and phase angle of current in each winding depends on many factors but under ideal conditions all those factors will be equal and so will be the currents and the torque contributions of the three phases. If the voltage of one phase falls, the tendency for them all to equate will cause a shift in equilibrium that attempts to restore that voltage by transferring workload onto the others and increasing their currents. Because of the non-proportionality of some of the losses this results in higher total heat dissipation, as well as reduced overall torque capability and increased torsional vibration.

In the extreme case of one phase of the supply being completely lost, the motor will not only attempt to drive its mechanical load on the power available from the remaining two, but can also absorb power from them to regenerate the missing phase and attempt to energise any other loads connected to it. This normally results in heavy overcurrent and tripped control gear, although there are cases where a big motor on light load would carry on regardless and special precautions need to be taken to ensure the machine shuts down promptly. Conversely, the phenomenon can also be exploited as a means of converting single-phase supplies to 3-phase, using an 'idler' motor that does not drive a load, in conjunction with capacitors to compensate the reactive power flows.

 

[Carl Haworth]

Dear Lucien,

I'm afraid that I got my figures badly wrong yesterday for the A drawn by the two 3-p appliances. It did look "too good to be true", and I now see that it was.

That was because I said 26A total for the 15 kW boiler and 11 kW charger. Nonsense! 26 is 15 +11. These are the kVA of the two appliances!

It is all too clear that I didn't know how to calculate the current drawn by an appliance of known wattage on UK 3-phase.

How to do this?

Obviously I should not use line to neutral voltage, because this is for S-P

Do I use UK 3-P line to line voltage quoted as 230 x 1.732: 398.36 V?

If so, the proposed 15kW boiler at pf1 draws 37.65A

(Wrong, I suspect!

Why? Because calculator from kW to Amps available via Google gives, for this wattage, and 400V line to line, 21.76A.

Isn't line to line voltage (single phase voltage x 1.732), which gives almost the same value: 398.36V (presumably rounded up by calculator to 400?).

Is the calculator right, and the A is 21.76?

If Yes, can you say how it arrives at this? (No indication given of how calculator works it out. Must be a complex calculation.)

Same calculator gives 15.96A for 11kW charger (also at pf1).

If these values are correct, my 3-P needs 38A (round figures), so say 40A, meaning (at 230V x 3) 27,600VA.

That would leave 40A for S-P, meaning 120A total, so (40 x 230 x 3) VA total: also 27,600VA.

Grand total VA is 55,200, as propsed by DNO (3 x 80 x 230).

Any total amperage at or above 80 should be fine for S-P. As you say, the devil is in the aggregating there to produce 3 reasonably similar and representative loads.

Cheers!

Carl

 

[Lucien Nunes]

You can consider a balanced 3-phase load to be three single-phase loads of 1/3 the power. In the case of electric heating that is often a true representation of what is inside.
So 15+11=26kVA consumes
26000 / 3 / 230 = 38A

You still have 42A available or 126A for the single-phase stuff. Actually is the boiler only 15kVA?

For convenience, one can still calculate this way even for loads that don't have a neutral and operate at 400V connected between the lines. In this case, the calculation ought to be
kVA / sqrt(3) / 400 but the answer for the line current is the same. This is intuitively logical because the line current depends on the power consumed, not how the individual coils are connected within the machine.

The reason for the equivalence is that the loads or coils, when delta-connected at 400 volts between pairs of lines, pass a current of kVA / 3 / 400 i.e. the load can still be thought of as divided into three equal ones. But because of the delta cinnection, the line current is now the vector sum of a pair of coil currents which is higher by a factor of sqrt (3).

In normal 3-phase working, one does not routinely say or think 'per phase'. Circuit current always represents that in one conductor, regardless of the number of conductors. TBH it grates when people unaccustomed to 3-phase systems say 'I've got a pump that takes 5A per phase' or similar, or ask whether that makes 15A total. After all, one does not declare the line and neutral currents separately in a 2-pole device, a 100A DP main switch is not rated at 200A on account of both the line and neutral being good for 100.

But there is nothing sneaky about looking under the hood of a 3-phase device and finding it's just three 230V devices in one box.

 

[Carl Haworth]

You can consider a balanced 3-phase load to be three single-phase loads of 1/3 the power. In the case of electric heating that is often a true representation of what is inside.
So 15+11=26kVA consumes
26000 / 3 / 230 = 38A

You still have 42A available or 126A for the single-phase stuff. Actually is the boiler only 15kVA?

For convenience, one can still calculate this way even for loads that don't have a neutral and operate at 400V connected between the lines. In this case, the calculation ought to be
kVA / sqrt(3) / 400 but the answer for the line current is the same. This is intuitively logical because the line current depends on the power consumed, not how the individual coils are connected within the machine.

The reason for the equivalence is that the loads or coils, when delta-connected at 400 volts between pairs of lines, pass a current of kVA / 3 / 400 i.e. the load can still be thought of as divided into three equal ones. But because of the delta cinnection, the line current is now the vector sum of a pair of coil currents which is higher by a factor of sqrt (3).

In normal 3-phase working, one does not routinely say or think 'per phase'. Circuit current always represents that in one conductor, regardless of the number of conductors. TBH it grates when people unaccustomed to 3-phase systems say 'I've got a pump that takes 5A per phase' or similar, or ask whether that makes 15A total. After all, one does not declare the line and neutral currents separately in a 2-pole device, a 100A DP main switch is not rated at 200A on account of both the line and neutral being good for 100.

But there is nothing sneaky about looking under the hood of a 3-phase device and finding it's just three 230V devices in one box.

Dear Lucien,

Many thanks again!

Ah, it was OK (I later got cold feet, and decided that I had been wrong!).

It's OK only if one knows the kVA of an appliance, or if it is obvious that this must be the same as, or very close to, the kWh rating.

Given that the appliances in question (even the charger?) are predominantly non-inductive, the kVA should be either identical, or very close, to the kWh. The boiler's element, like that of an electric kettle or immersion heater etc. wound using an alloy such as nichrome, with a highly positive coefficient of resistance, will draw a starting current much higher than the operating current. This will very quickly drop as the winding heats up. so the appliance ends up running at a kVA more or less (the winding form, especially if spiral, may well be inductive) identical to the quoted kWh.

So the current (A) calculation is:-

USING kWh AS kVA IF ONLY kW IS GIVEN, AND IF THIS IS OBVIOUSLY VERY CLOSE TO kVA

kVA (of appliance)/3 (phases)/230 (V per phase)

So the two appliances together (15,000VA+11,000VA) work out, as you say. at :-
26,000/3/230 = 37.68A,
38A rounded.

The same calculation can't be done with the 3-P phase to phase line voltage at 398.36V (230 x 1.732) because this does not use (one phase V x 3), so gives a very different result (65.27A!!).

The correct calculation leaves, as you say, (42A x 3 single phases) 128A total for the S-P installation.

The 42 could sensibly be rounded to 45, leaving (35A x 3: 105A total) for the S-P circuits

There can be give on one side and take on the other, or vice-versa.

BOILER: IS 15,000 W SUFFICIENT (Your implied question)?

Our 1997 non-condensing gas boiler generates, in Wh. about (28,000 x 0.76): 21,280 of heat energy. This includes DHW heating. This boiler is probably a bit over-sized, though not to the monstrous extent of some of the combi boilers installed in more in recent years.

So, with DHW being heated directly at 3,000 Wh by one of two S-P immersion heaters rather than by a boiler circuit ,15,000 Wh may be enough for this house, which is much better insulated now than it was in 1997. Coincidentally (and importantly!), it appears that any electrical boiler with an output higher than 15 kW has to be floor-standing. We can't accommodate this - only a wall-hung one.

The heat requirement would have to be worked out on the basis of a proper energy survey of our house.

We could accommodate, an 18,000 Wh boiler:

3-P: (18,000 + 11,000)\ 230\3 = 42A. Say 45A.

That leaves 3 x 35A (105A) for S-P. Should be fine. Given the circuit/splitting/diversity problems it would be sensible not to reduce S-P total below 100A.

THE AMPS-FROM-kWH CALCULATOR THAT I FOUND VIA GOOGLE

This is at

Kilowatts to amps (A) conversion calculator

Kilowatts (kW) to amps (A) conversion calculator.

www.rapidtables.com

fUsing 26 kWh, this calculates (I) 21.77 Amps with voltage entered as 3-phase line to line 398V, and (ii) 37.68 A for 3-phase 230V line to neutral.

No wonder I became unsure!

This calculator looks serious, though calculation methods are not displayed.

What do you make of the current levels that it calculates? They're clearly not right - at least for UK 3-phase. (ii) is closer than (i), which is way out.

Many thanks again, and best wishes,

Carl