208 Voltage too low for machine

[HEXx99]

I ran 208 v power to a rottler em69p cnc machine. it calls for 208/240 v 3 phase. 60amp rottler em69 installation manual
everything is wired using 3 #6 hot and #8 ground
to 6/4 SO cord drop to the machine . inside the machine again the wiring is 3 #6 and #8. I checked voltages and got 208-209v across two etc.
when the machine tech. came he said the voltage is too low that its calling for 210V and his readings are only 205v. I have to go back tomorrow to check it out but im not sure where to start. thanks in advance for any advice .

 

[marconi]

1. In the manual (page 2-2) it says something about the high leg, if it exists, being connected to L3:

This machine requires between 208 and 240 Volts AC, Three Phase, 50/60 Hz power supply. For voltages above 240 or below 208 VAC, a 30kva transformer will be required. Measure the incoming voltage between L1 and L2, L2 and L3, and L1 and L3. Current requirements for this machine is 60 amps. Measure the incoming AC voltage at least twice during installation. 1. L1to L2 VAC, L2 to L3 VAC, L1 to L3 ______ 2. L1to L2 VAC, L2 to L3 VAC, L1 to L3 ______ ______Measure each leg of the incoming supply to ground. Sometimes you may find a “high” leg to ground. When this happens make sure the high leg is running to L3. L1 to ground __________VAC L2 to ground __________VAC L3 to ground ________VAC

2. Have you done a cable volt drop calculation to size the supply conductors? The ampacity of the cables is not the only consideration on a long run.

3. Is the supply transformer for the site rated with sufficient kVA? This machine adds a further 30kVA. There must be some margin so that the transformer is not operating at full load when voltage regulation will be poorer.

4. What is the power factor of the site before the machine is added? What is the power factor of the new machine? Does the site need power factor correction equipment installed to free up kVA for the new machine and/or reduce overall site load on supply transformer?

 

[pc1966]

I would add to @marconi questions to ask what is the voltage elsewhere on your site?

Do you have your own transformer, and if so has it any tap adjustment to get a few percent high voltage?

 

[HEXx99]

1. In the manual (page 2-2) it says something about the high leg, if it exists, being connected to L3:

This machine requires between 208 and 240 Volts AC, Three Phase, 50/60 Hz power supply. For voltages above 240 or below 208 VAC, a 30kva transformer will be required. Measure the incoming voltage between L1 and L2, L2 and L3, and L1 and L3. Current requirements for this machine is 60 amps. Measure the incoming AC voltage at least twice during installation. 1. L1to L2 VAC, L2 to L3 VAC, L1 to L3 ______ 2. L1to L2 VAC, L2 to L3 VAC, L1 to L3 ______ ______Measure each leg of the incoming supply to ground. Sometimes you may find a “high” leg to ground. When this happens make sure the high leg is running to L3. L1 to ground __________VAC L2 to ground __________VAC L3 to ground ________VAC

2. Have you done a cable volt drop calculation to size the supply conductors? The ampacity of the cables is not the only consideration on a long run.

3. Is the supply transformer for the site rated with sufficient kVA? This machine adds a further 30kVA. There must be some margin so that the transformer is not operating at full load when voltage regulation will be poorer.

4. What is the power factor of the site before the machine is added? What is the power factor of the new machine? Does the site need power factor correction equipment installed to free up kVA for the new machine and/or reduce overall site load on supply transformer?

Thank you for answering. I apologize in advance if my answers seem insufficient this isnt my field of expertise ive just been around enough to have some knowledge of the basics . Im actually heading over there today so I would have more info later, but for now i know i have to check the power at the main lug of the subpanel and beyond and basically show the customer this is the voltage you are getting from the city . I would then suggest a buckboost transformer to bring the voltage up to the ideal which is 230 or at least 210 . which is operational.
Can you explain number 2 up there about the supply conductors? Additionally would using a higher gauge wire effect this at all?

 

[HEXx99]

I would add to @marconi questions to ask what is the voltage elsewhere on your site?

Do you have your own transformer, and if so has it any tap adjustment to get a few percent high voltage?

ty for answering . This i am going to check on right now as well

 

[marconi]

Reference the cable drop question. Wires/cables have resistance which is as it suggests an impediment to the easy flow of electricity along them. The greater the resistance of the supply cables the more effort is required to push the required current through them. The amount of effort is indicated by the voltage difference between the beginning of the cable and the end of it where it connects to the equipment. This difference in voltage is called 'voltage drop'.

Generally speaking, longer cables have more resistance end-to-end. Thinner cables have more resistance end-to-end. Warmer cables have more resistance end-to-end. So, long, thin, warm cables will cause the most voltage drop when large currents flow through them

To minimise the voltage drop along the cable so that the voltage at the end is as close as possible to the supply voltage the cables have to be sized in cross-sectional area so that for the load current and length of cable run and ambient temperature they have sufficiently low resistance that the voltage drop V, which is Current I x Resistance R, does not cause the voltage at the machine to drop below its minimum requirement.

For long cable runs this often means a much thicker cable is required than is strictly necessary to carry the load current. End of lecture

This is a straight forward calculation for a competent electrician to do. I will not do it for you because I am not familiar with AWG cable sizes, etcetera. What is needs is the voltage at the incoming supply breaker, the length of cable between this point and the equipment, the current in the cable and the ambient temperature, and the manufacturer's resistances/conductivities for the standard AWG cable sizes and types.. There are probably on-line calculators that will do it for you as there in the UK for our cable sizes and types.

Others can expand especially American colleagues such as Megawatt.

PS: It is actually a little more complicated but not impossible to do because one needs to know the resistance of the supply between the main breaker and the supply transformer. But just looking at the cable on site between main breaker and equipment and measuring the incoming line voltages are the first steps.

PS2: So yes it may be the problem is too much voltage drop along the present equipment supply cable which can be remedied by a thicker cable (or maybe a parallel cable of the same size).

PS3: It says in the manual that the supply must be adequate and of good quality otherwise the machine may not work or will work unreliably or not perform as well as it could.

 

[HEXx99]

Reference the cable drop question. Wires/cables have resistance which is as it suggests an impediment to the easy flow of electricity along them. The greater the resistance of the supply cables the more effort is required to push the required current through them. The amount of effort is indicated by the voltage difference between the beginning of the cable and the end of it where it connects to the equipment. This difference in voltage is called 'voltage drop'.

Generally speaking, longer cables have more resistance end-to-end. Thinner cables have more resistance end-to-end. Warmer cables have more resistance end-to-end. So, long, thin, warm cables will cause the most voltage drop when large currents flow through them

To minimise the voltage drop along the cable so that the voltage at the end is as close as possible to the supply voltage the cables have to be sized in cross-sectional area so that for the load current and length of cable run and ambient temperature they have sufficiently low resistance that the voltage drop V, which is Current I x Resistance R, does not cause the voltage at the machine to drop below its minimum requirement.

For long cable runs this often means a much thicker cable is required than is strictly necessary to carry the load current. End of lecture

This is a straight forward calculation for a competent electrician to do. I will not do it for you because I am not familiar with AWG cable sizes, etcetera. What is needs is the voltage at the incoming supply breaker, the length of cable between this point and the equipment, the current in the cable and the ambient temperature, and the manufacturer's resistances/conductivities for the standard AWG cable sizes and types.. There are probably on-line calculators that will do it for you as there in the UK for our cable sizes and types.

Others can expand especially American colleagues such as Megawatt.

PS: It is actually a little more complicated but not impossible to do because one needs to know the resistance of the supply between the main breaker and the supply transformer. But just looking at the cable on site between main breaker and equipment and measuring the incoming line voltages are the first steps.

PS2: So yes it may be the problem is too much voltage drop along the present equipment supply cable which can be remedied by a thicker cable (or maybe a parallel cable of the same size).

PS3: It says in the manual that the supply must be adequate and of good quality otherwise the machine may not work or will work unreliably or not perform as well as it could.

alright i tested the voltage at the subpanel which has a breaker instead of a main lug . at that breaker the voltage is 205v and at the machine the voltage is the same . The cable i ran isnt long the whole run is 40 ft #6 in conduit and 16 ft 6/4 SO cord
there is a transformer that is feeding the subpanel ..its literally suspended above ground i will send picture . my new question is if this transformer might be adjustable. and maybe give me a higher voltage output

 

[HEXx99]

ty for answering . This i am going to check on right now as well

there is a transformer supplying the power to the panel im not familiar with transformer modifications but im going to call someone that does know about them to get a better understanding. or if it can be done with this particular transformer.

 

[HEXx99]

alright i tested the voltage at the subpanel which has a breaker instead of a main lug . at that breaker the voltage is 205v and at the machine the voltage is the same . The cable i ran isnt long the whole run is 40 ft #6 in conduit and 16 ft 6/4 SO cord
there is a transformer that is feeding the subpanel ..its literally suspended above ground i will send picture . my new question is if this transformer might be adjustable. and maybe give me a higher voltage output

also this is the transformer info

 

[marconi]

also this is the transformer info

The transformer you show has the facility to change 'taps' on the incoming primary side which has the effect of raising or lowering the voltage on the secondary or load side. This could be the solution to your voltage problem.

You are wise to ask for advice from an electrician who knows about this type of transformer. You need to confirm ownership of this transformer too. If the power company own it then they are responsible for its connection including the taps and you will have to ask them to change them. Otherwise you need a competent electrician familiar with this transformer and how to isolate it before changing the taps.

You should carry out voltage measurements before tap change and after tap change and record them before re-applying power to the site and its equipment. And then measure the site voltage at the main breaker again when all loads except the Rottler are re-energised and then all loads including the Rottler are energised. Finally check voltage at input terminals to Rottler to confirm they are in specification.

Please keep us informed as we will be interested to know the outcome.

 

[marconi]

Just one extra thing to measure using a clamp ammeter - the line and neutral currents at the main breaker with only site load and then with site load and Rottler machine and record them. Please tell us what the four readings are please. What we want to see is the line currents about equal and the neutral current no more than a few amps.

 

[marconi]

also this is the transformer info

The information on this plate enables us to calculate the voltage drop at the output terminals of the transformer as the load current increases. The transformer is a 75kVA with 3 phase Y output of 208V/240V.

The rated secondary phase current is 75000/(3 x 208) = 25000/208 = 120A

The % Impedance figure of 5.54 tells us the percentage of the rated winding voltage which must be applied to it with the the other winding short circuited for the rated currents to flow in each winding. Thus if the primary is short circuited one would have to apply a voltage of 208 x 5.54/100 = 11.5V across the secondary for 120A to flow in the secondary and the transformed down current in the primary winding.

The effective impedance then of each of the secondary phase windings Zs is 11.5/120 = 0.096 Ohms

The volt drop for a particular phase current is Iphase x Zs. This is why I asked you to measure the phase currents.

At full load of 75kVA, or 120A per phase in the secondary windings the voltage drop is 11.5V, which means the terminal voltage is the secondary induced emf (208V) - 11.5V = 196.5V.

If we just consider just the 30kVA load of the Rottler machine fed by the transformer and assume it is balanced 10kVA per phase, the volt drop just due to the supply transformer will be (10/25) x 120 x Zs = 4.6V ...The secondary voltage will be 208 - 4.6 = 203.4V which is out of specification.

There are of course further voltage drops due to all the cabling and switchgear between the transformer secondary terminals and the terminals of the Rottler machine and also the currents for the rest of the site load.

In summary then, primary winding tap changing may be sufficient alone to achieve voltages within specification for the Rottler but be prepared that to supply the site load and the Rottler may require a new higher Kva site transformer or a dedicated transformer for the Rottler - 30kVA is specified. Interestingly the manual states a 60A per phase supply is required which is 208 x 60 x 3 = 37kVA. Whether 30 or 37kVA this is half of the rated capacity of the present 75kVA supply transformer. A review of the site load and power factor may be necessary to see if they can contribute to improve the site voltage using the present transformer.

 

[marconi]

alright i tested the voltage at the subpanel which has a breaker instead of a main lug . at that breaker the voltage is 205v and at the machine the voltage is the same . The cable i ran isnt long the whole run is 40 ft #6 in conduit and 16 ft 6/4 SO cord
there is a transformer that is feeding the subpanel ..its literally suspended above ground i will send picture . my new question is if this transformer might be adjustable. and maybe give me a higher voltage output

I used an online US cable voltage drop calculator:

Voltage Drop Calculator - https://www.calculator.net/voltage-drop-calculator.html?necmaterial=copper&necwiresize=4&necconduit=steel&necpf=0.9&material=copper&wiresize=0.4066&resistance=1.2&resistanceunit=okm&voltage=208&phase=ac3&noofconductor=1&distance=50&distanceunit=feet&amperes=60&x=43&y=23&ctype=nec

For 50ft of 6AWG for the Rottler cable run the voltage drop between start and end at 60A is 2.44V. You can see now how the voltage drops in the transformer (4.66V) and cabling (2.44V) are adding up to cause the problem of low voltage at the Rottler. 208- 4.66- 2.44 = 201V

You may have to run another triple of the same 6AWG (or thicker but not thinner)cables parallel (electrically and physically) to reduce this to 1.22V. Or replace by one thicker complete cable run sized to achieve 'good enough' voltage drop at 60A.



See:

Voltage Drop Calculator​

This is a calculator for the estimation of the voltage drop of an electrical circuit. The "NEC data" tab calculates based on the resistance and reactance data from the National Electrical Code (NEC). The "Estimated resistance" tab calculates based on the resistance data estimated from the wire size. Click the "Other" tab to use customized resistance or impedance data, such as data from other standards or wire manufacturers.

Result​

Voltage drop: 2.44
Voltage drop percentage: 1.17%
Voltage at the end: 205.56

The calculation result above is based on alternating current resistance and reactance data of 3-phase, 60 Hz, 75°C from National Electrical Code (NEC). The actual voltage drop can vary depending on the condition of the wire, the temperature, the connector, the frequency etc.

NEC data Estimated resistance Other Wire material Copper Aluminum Wire size 14 AWG 12 AWG 10 AWG 8 AWG 6 AWG 4 AWG 3 AWG 2 AWG 1 AWG 1/0 AWG 2/0 AWG 3/0 AWG 4/0 AWG 250 kcmil 300 kcmil 350 kcmil 400 kcmil 500 kcmil 600 kcmil 750 kcmil 1000 kcmil Material of conduit PVC Aluminum Steel Power factor (PF) Voltage Phase DC AC single phase AC 3-phase Number of conductors single set of conductors 2 conductors per phase in parallel 3 conductors per phase in parallel 4 conductors per phase in parallel 5 conductors per phase in parallel 6 conductors per phase in parallel 7 conductors per phase in parallel 8 conductors per phase in parallel 9 conductors per phase in parallel Distance (one-way) feet meters miles kilometers Load current Amps

 

[HEXx99]

I used an online US cable voltage drop calculator:

Voltage Drop Calculator - https://www.calculator.net/voltage-drop-calculator.html?necmaterial=copper&necwiresize=4&necconduit=steel&necpf=0.9&material=copper&wiresize=0.4066&resistance=1.2&resistanceunit=okm&voltage=208&phase=ac3&noofconductor=1&distance=50&distanceunit=feet&amperes=60&x=43&y=23&ctype=nec

For 50ft of 6AWG for the Rottler cable run the voltage drop between start and end at 60A is 2.44V. You can see now how the voltage drops in the transformer (4.66V) and cabling (2.44V) are adding up to cause the problem of low voltage at the Rottler. 208- 4.66- 2.44 = 201V

You may have to run another triple of the same 6AWG (or thicker but not thinner)cables parallel (electrically and physically) to reduce this to 1.22V. Or replace by one thicker complete cable run sized to achieve 'good enough' voltage drop at 60A.



See:

Voltage Drop Calculator​

This is a calculator for the estimation of the voltage drop of an electrical circuit. The "NEC data" tab calculates based on the resistance and reactance data from the National Electrical Code (NEC). The "Estimated resistance" tab calculates based on the resistance data estimated from the wire size. Click the "Other" tab to use customized resistance or impedance data, such as data from other standards or wire manufacturers.

Result​

Voltage drop: 2.44
Voltage drop percentage: 1.17%
Voltage at the end: 205.56

The calculation result above is based on alternating current resistance and reactance data of 3-phase, 60 Hz, 75°C from National Electrical Code (NEC). The actual voltage drop can vary depending on the condition of the wire, the temperature, the connector, the frequency etc.

NEC data Estimated resistance Other Wire material Copper Aluminum Wire size 14 AWG 12 AWG 10 AWG 8 AWG 6 AWG 4 AWG 3 AWG 2 AWG 1 AWG 1/0 AWG 2/0 AWG 3/0 AWG 4/0 AWG 250 kcmil 300 kcmil 350 kcmil 400 kcmil 500 kcmil 600 kcmil 750 kcmil 1000 kcmil Material of conduit PVC Aluminum Steel Power factor (PF) Voltage Phase DC AC single phase AC 3-phase Number of conductors single set of conductors 2 conductors per phase in parallel 3 conductors per phase in parallel 4 conductors per phase in parallel 5 conductors per phase in parallel 6 conductors per phase in parallel 7 conductors per phase in parallel 8 conductors per phase in parallel 9 conductors per phase in parallel Distance (one-way) feet meters miles kilometers Load current Amps

UPDATED EDIT**
ah i just saw your post about the transformer .

 

[HEXx99]

Just one extra thing to measure using a clamp ammeter - the line and neutral currents at the main breaker with only site load and then with site load and Rottler machine and record them. Please tell us what the four readings are please. What we want to see is the line currents about equal and the neutral current no more than a few amps.

The technician will not come out to start the machine until we confirm the voltage he wants at the machine lol

 

[HEXx99]

The information on this plate enables us to calculate the voltage drop at the output terminals of the transformer as the load current increases. The transformer is a 75kVA with 3 phase Y output of 208V/240V.

The rated secondary phase current is 75000/(3 x 208) = 25000/208 = 120A

The % Impedance figure of 5.54 tells us the percentage of the rated winding voltage which must be applied to it with the the other winding short circuited for the rated currents to flow in each winding. Thus if the primary is short circuited one would have to apply a voltage of 208 x 5.54/100 = 11.5V across the secondary for 120A to flow in the secondary and the transformed down current in the primary winding.

The effective impedance then of each of the secondary phase windings Zs is 11.5/120 = 0.096 Ohms

The volt drop for a particular phase current is Iphase x Zs. This is why I asked you to measure the phase currents.

At full load of 75kVA, or 120A per phase in the secondary windings the voltage drop is 11.5V, which means the terminal voltage is the secondary induced emf (208V) - 11.5V = 196.5V.

If we just consider just the 30kVA load of the Rottler machine fed by the transformer and assume it is balanced 10kVA per phase, the volt drop just due to the supply transformer will be (10/25) x 120 x Zs = 4.6V ...The secondary voltage will be 208 - 4.6 = 203.4V which is out of specification.

There are of course further voltage drops due to all the cabling and switchgear between the transformer secondary terminals and the terminals of the Rottler machine and also the currents for the rest of the site load.

In summary then, primary winding tap changing may be sufficient alone to achieve voltages within specification for the Rottler but be prepared that to supply the site load and the Rottler may require a new higher Kva site transformer or a dedicated transformer for the Rottler - 30kVA is specified. Interestingly the manual states a 60A per phase supply is required which is 208 x 60 x 3 = 37kVA. Whether 30 or 37kVA this is half of the rated capacity of the present 75kVA supply transformer. A review of the site load and power factor may be necessary to see if they can contribute to improve the site voltage using the present transformer.

This transformer is running power to at least two panels the machines these panels are powering are not ran constantly. they run maybe one of them about 8 times a day and maybe 10 min at the mist each time. the cnc wouldnt be running all day either . but there are also lights getting power from these panels . Im now thinking a 30 kva transformer 3 phase just for this machine is the best bet . unfortunately these are not cheap. unless i just find 240v andpower it from that, right? but then it looks like i made a mistake of supplying inefficient voltage to the machine . which is true, its not enough , but their own manual says 208 v ..im going to look for 240v

 

[marconi]

This transformer is running power to at least two panels the machines these panels are powering are not ran constantly. they run maybe one of them about 8 times a day and maybe 10 min at the mist each time. the cnc wouldnt be running all day either . but there are also lights getting power from these panels . Im now thinking a 30 kva transformer 3 phase just for this machine is the best bet . unfortunately these are not cheap. unless i just find 240v andpower it from that, right? but then it looks like i made a mistake of supplying inefficient voltage to the machine . which is true, its not enough , but their own manual says 208 v ..im going to look for 240v

Do not beat yourself up too much. The rottler electrical spec is far from clear and certainly not definitive in the way the acceptable voltage is written.

Please measure some phase currents and voltages at main breaker for different states of site load to establish what margins exist to supply the demanding rottler load. Would site accept load shedding to allow rottler to power up?

 

[marconi]

HEXx99. I have been reviewing everything you have said and the information in the Rottler manual on the electrical requirements. I have made a mistake because I thought the 205V you mentioned as a low voltage and the 208V mentioned by Rottler were phase voltage. They are actually line voltage. The Rottler requires a 3 phase 3 wire 60Hz supply with a line voltage in the range 208-240V. In terms of phase voltages, this is a range of 208/rt3 to 240/rt3 = 120 to 139V.

The supply transformer line voltage of 205V is a phase voltage of 118V. Tap changing out to be able to bump this up a few volts. At 125V phase voltage the line voltage would be 216V, or 8V above the minimum.

The cable calculation redone using 10kVA per phase at 120V is a current of 10000/120 = 83A.

the #6AWG cable run will drop between its ends 3.9V or a machine phase voltage of 116V or 201 line voltage. Clearly #6AWG is too thin. A double run of #6AWG would halve the volt drop to circa 2 V, thus a machine line voltage of (120-2) x rt3 = 204V - still not good enough.

#2AWG gives a volt drop of 1.71V; (120-1.7) x rt3 = 204V line voltage at the machine.

And this is before we consider the transformer volt drop and volt drop due to site loads. It will be very difficult to achieve the required machine voltage using the current transformer.

I think you are at the stage of placing a transformer 480V/240V - both line voltages - close to the machine to keep the machine's supply cable voltage drop as small possible without having to use a very thick cable and yet stay well within the 208 - 240V specification.

 

[marconi]

The 75kVA transformer with a secondary phase induced voltage of 120V, will have a full load phase current of 25000/120 = 208A.

5.54 percent impedance means 120 x 5.54/100 = 6.5V applied to secondary with primary short-circuited would circulate full load phase current. This Zs is 6.5/208 = 0.031 Ohms.

The Rottler machine when drawing 83A per phase would create a secondary terminal phase voltage of 120 - (83 x 0.031) = 117V.

117V is a line voltage of 202V - unsatisfactory for the Rottler even before any other loads connected to transformer.

The Rottler requires its own transformer. Hope this helps. Happy to discuss.

Regards

Marconi

 

[marconi]

Here is an 8 minute video on the Rottler 5 axis machine cutter - very impressive.