I have a job where there is a 40 m long 2 phase sub-main power supply to a building, installed about 1965. The cable configuration is:

1 x single core 16 sq.mm cable with 2 layers of insulation

1 x 2 core 16 sq.mm screened cable with 1 phase, 1 neutral and the screen is earth.

Run underground, enclosed in what looks like alkathene at the supply end into the DB.

I'd like to know how much current I can allow for. The single phase loads are unbalanced. In GenCalc I've entered the values shown and selected 230v, 3 phase, unbalanced. Is the correct result 35.1A for each phase? Thank you.

## 2 phase current calculation

### Re: 2 phase current calculation

I would treat this as a single-phase calculation.

Adding the extra phase will result in N-current being less than the sum of phase currents, and therefore less contribution to volt-drop from the N;

but to determine how much less you have to do a separate calculation for the N.

Which is difficult since that the load profile of each phase in any multi-phase system supplying single-phase loads is varying constantly, so the value for N-current can only ever be a guestimate.

Even Gencalc will be making assumptions about the relative loads in each phase for it's unbalanced load calculations;

and by ticking "three phase" those assumptions cannot be truly relevant for a 2-phase system .

In fact they will err on the wrong side; because they will assume that there is a 3rd phase that will be improving the overall load balance and further reducing N-current compared to reality.

So taking the N/S as a single-phase standalone calculation; and your chosen limit of 2.5 % V-drop

Table 42 provides a VD value of 2.43 (@ conductor temp 75 C); for a 3-phase circuit.

adjust for single phase gives 2.81 mV / A/ m.

40 m x 2.81 = 112.4 mV / A

2.5 % = 5.75 V / 5750 mV

5750 / 112.4 = 51.15 A

Using Gencalc for same set-up; gives 58.8 A

Note that Gencalc uses a VD value; based on lower conductor temp (underground for most of the route)

That's probably valid; though the Tables don't go down that far .

For a 3-phase 400 V unbalanced load; Gencalc gives 57.7 A

I think the problem is that you have entered it as 3-phase in combination with 230 V;

which has confused the Gencalc "brain" and resulted in total permitted v-drop being 3.30 V .

This has de-railed the entire calculation, because in your example Gencalc is working on 230 V between phases.

I seldom use Gencalc or similar; as for a single-length calculation like this it's very easy to do it yourself - which gives you a good feel for how changing any variable affects the outcome.

For more complex calculations; i haven't yet found one that can deal with multiple stages.

For example; your choice of 2.5 % for this submain is entirely arbitrary - and probably not valid.

It's long been normal practice to apply a rule-of thumb of 2.5 % for mains, and 2.5 % for the rest.

That's mostly because before 2010 ESRs, we were required to work on basis of 2.5 % from main switch.

Now it's 5% from POS.

But when there's a submain involved there are three calculations to do.

First for the mains - which may be contributing more or less than 2.5 %.

But, assuming for now that 2.5 % is valid; then if you use the other 2.5 % for this submain, there is nothing left for the final subcircuits.

Chances are those subcircuits are using at least 2 %.

Eg a single-phase socket circuit rated at 20A and with 15 m route length of 2.5 mm2 TPS:

15 m x 10 A x 18.02 mv =2703 mV (1.17%).

So reality dictates that allowing 2.5 % for the submain is unwise (at best);

as a rule of thumb should be max 1.25 % with another 1.25% for final subcircuits; and 2.5 % for mains.

(Outside of a domestic installation; most existing mains are fairly well loaded up).

Which would give 31.1 A for this submain (by Gencalc; single phase).

Enough for only a few subcircuits of typical loadings & cabling.

The fact that there are 2 phases suggests there's more than just a few subcircuits in the end of this;

So chances are good that there's already excessive volt drop in this system

And if you want it any closer; you have to actually calculate all three stages.

One option is to keep plugging in arbitrary values into Gencalc; until you find a split that looks acceptable.

But actually easier to start with total allowable 5% (11,500 V)

deduct calculated V-drop for mains

eg 20m x 50 A x 1.11 (25 mm2 3-phase) = 1540 mV

leaving only 11500 - 1540 = 9960 mV available at DB1.

Step 2; deduct for worst-case final subcircuit

9960 - 7251 = 2709 mV available for the submain (2.7%)

That would allow for up to 63 A in the submain (single phase);

We can't just wing this stuff.

We have to look at the entire situation; not just one part of it.

And our CoC amounts to a declaration that max volt drop cannot be exceeded no matter what the end-user does.

So before you re-do the "how much current can i draw from this existing submain" exercise;

you need to research the size, loading, and approx route length of the mains,

and also look at the existing final subciruits for cable sizes, route lengths, and approx loads.

There may well be no capacity left for whatever you're planning.

Disclaimer: I've amended these figures several times in order for it to be a realistic scenario;

also there may be a typo or other error in them.

The point isn't to provide an accurate figure; just to show that looking at a submain in isolation is very likely to lead us into non-compliance

Adding the extra phase will result in N-current being less than the sum of phase currents, and therefore less contribution to volt-drop from the N;

but to determine how much less you have to do a separate calculation for the N.

Which is difficult since that the load profile of each phase in any multi-phase system supplying single-phase loads is varying constantly, so the value for N-current can only ever be a guestimate.

Even Gencalc will be making assumptions about the relative loads in each phase for it's unbalanced load calculations;

and by ticking "three phase" those assumptions cannot be truly relevant for a 2-phase system .

In fact they will err on the wrong side; because they will assume that there is a 3rd phase that will be improving the overall load balance and further reducing N-current compared to reality.

So taking the N/S as a single-phase standalone calculation; and your chosen limit of 2.5 % V-drop

Table 42 provides a VD value of 2.43 (@ conductor temp 75 C); for a 3-phase circuit.

adjust for single phase gives 2.81 mV / A/ m.

40 m x 2.81 = 112.4 mV / A

2.5 % = 5.75 V / 5750 mV

5750 / 112.4 = 51.15 A

Using Gencalc for same set-up; gives 58.8 A

Note that Gencalc uses a VD value; based on lower conductor temp (underground for most of the route)

That's probably valid; though the Tables don't go down that far .

For a 3-phase 400 V unbalanced load; Gencalc gives 57.7 A

I think the problem is that you have entered it as 3-phase in combination with 230 V;

which has confused the Gencalc "brain" and resulted in total permitted v-drop being 3.30 V .

This has de-railed the entire calculation, because in your example Gencalc is working on 230 V between phases.

I seldom use Gencalc or similar; as for a single-length calculation like this it's very easy to do it yourself - which gives you a good feel for how changing any variable affects the outcome.

For more complex calculations; i haven't yet found one that can deal with multiple stages.

For example; your choice of 2.5 % for this submain is entirely arbitrary - and probably not valid.

It's long been normal practice to apply a rule-of thumb of 2.5 % for mains, and 2.5 % for the rest.

That's mostly because before 2010 ESRs, we were required to work on basis of 2.5 % from main switch.

Now it's 5% from POS.

But when there's a submain involved there are three calculations to do.

First for the mains - which may be contributing more or less than 2.5 %.

But, assuming for now that 2.5 % is valid; then if you use the other 2.5 % for this submain, there is nothing left for the final subcircuits.

Chances are those subcircuits are using at least 2 %.

Eg a single-phase socket circuit rated at 20A and with 15 m route length of 2.5 mm2 TPS:

15 m x 10 A x 18.02 mv =2703 mV (1.17%).

So reality dictates that allowing 2.5 % for the submain is unwise (at best);

as a rule of thumb should be max 1.25 % with another 1.25% for final subcircuits; and 2.5 % for mains.

(Outside of a domestic installation; most existing mains are fairly well loaded up).

Which would give 31.1 A for this submain (by Gencalc; single phase).

Enough for only a few subcircuits of typical loadings & cabling.

The fact that there are 2 phases suggests there's more than just a few subcircuits in the end of this;

So chances are good that there's already excessive volt drop in this system

And if you want it any closer; you have to actually calculate all three stages.

One option is to keep plugging in arbitrary values into Gencalc; until you find a split that looks acceptable.

But actually easier to start with total allowable 5% (11,500 V)

deduct calculated V-drop for mains

eg 20m x 50 A x 1.11 (25 mm2 3-phase) = 1540 mV

leaving only 11500 - 1540 = 9960 mV available at DB1.

Step 2; deduct for worst-case final subcircuit

9960 - 7251 = 2709 mV available for the submain (2.7%)

That would allow for up to 63 A in the submain (single phase);

We can't just wing this stuff.

We have to look at the entire situation; not just one part of it.

And our CoC amounts to a declaration that max volt drop cannot be exceeded no matter what the end-user does.

So before you re-do the "how much current can i draw from this existing submain" exercise;

you need to research the size, loading, and approx route length of the mains,

and also look at the existing final subciruits for cable sizes, route lengths, and approx loads.

There may well be no capacity left for whatever you're planning.

Disclaimer: I've amended these figures several times in order for it to be a realistic scenario;

also there may be a typo or other error in them.

The point isn't to provide an accurate figure; just to show that looking at a submain in isolation is very likely to lead us into non-compliance

**Rating:**16.67%

- Mazdaman
**Posts:**16**Joined:**Tue Apr 14, 2020 12:00 pm**Location:**Canterbury-
**Has thanked:**35 times -
**Been thanked:**5 times

### Re: 2 phase current calculation

Thank you very much AlecK for the considerable time you have spent helping me. Sorry I neglected to include additional information such as the main supply and loads.

The property is a small church.

Main supply is 25mm NS, approximately 30m long, mostly buried to the meter box then 0.6m of 16mm's to the DB.

The loads on the distribution board in the church building are fairly balanced, being 3 fan heaters and 6 infra red heaters and a little lighting. About 22-25A per phase.

The sub main loads in a separate hall at the end of the 40m long, buried 16mm, 2 phase cable are 4 infra reds, 1 domestic oven, 1 fan heater and 1.5kW Zip water heater, say a total of 60-70A when all running.

Both buildings are not usually in use at the same time however I'll allow for that in recalculating.

I'll take current and voltage measurements with all loads running next week when in the area.

Thanks again!

The property is a small church.

Main supply is 25mm NS, approximately 30m long, mostly buried to the meter box then 0.6m of 16mm's to the DB.

The loads on the distribution board in the church building are fairly balanced, being 3 fan heaters and 6 infra red heaters and a little lighting. About 22-25A per phase.

The sub main loads in a separate hall at the end of the 40m long, buried 16mm, 2 phase cable are 4 infra reds, 1 domestic oven, 1 fan heater and 1.5kW Zip water heater, say a total of 60-70A when all running.

Both buildings are not usually in use at the same time however I'll allow for that in recalculating.

I'll take current and voltage measurements with all loads running next week when in the area.

Thanks again!