Help a mechanical engineer who is playing an electrical engineer

[Girswald]

Hello, as the title states, I am a mechanical engineer that has been thrust into an electrical engineers position for a short time while we look for another to replace ours.

My question/issue is that I am testing thermal rise in a copper bus system by loading current and shorting out the opposite end of the bus system, it is 3 phase bus and I have a 3 phase current generator that can output 4000 amps at 24 volts. Testing a simple bus run is pretty straight forward, I short one end and place thermocouples along the bus in the manner prescribed by UL to measure the temperature rise...simple stuff even for a mechanical engineer! However, I have been thrown for a big loop in that I now have a main horizontal bus system that is rated at 3200 amps and connected to it is a vertical bus system that is rated at 800 amps. The joint between these two bus systems is where I need to check temperature rise. I short the horizontal bus system, like always (now the tough part) and put an 800 amp load on the vertical section. That should give me 2400 amps in the horizontal and 800 in the vertical. Problem is, how do I load the vertical bus? A solid short only puts around 250 amps of load. I added lengths of wire to the end of the vertical bus to add resistance, 3 pieces per phase of 350 MCM @ 12 feet lengths. That number was given to me by the guy that retired. When I started the generator, I actually got LESS resistance in the vertical down to 120 amps. That just doesn't seem right...

What if I add a steel bar instead of the copper bar that is used to short the vertical, steel has +/- 10 times the resistance as copper, so that should just act as a 19,200 watt heater? All I would need to do is calculate the required size of steel, I think.

Any other ideas? Remember, like most companies, I can't spend any money on this...I'm sure you all know how that goes.

Thanks

 

[ATSman]

Bus Short Location

Bus Short Location

I am not sure why you are shorting both the Horiz and Vertical buses.

To prove that the 800A vertical bus will carry the rated current then only
short the end of the vertical bus and run current up to 800A.

To prove the rating on the horiz bus then do a separate test by shorting
the end of the horiz bus and run up to its rated current.

 

[Girswald]

I am not sure why you are shorting both the Horiz and Vertical buses.

To prove that the 800A vertical bus will carry the rated current then only
short the end of the vertical bus and run current up to 800A.

To prove the rating on the horiz bus then do a separate test by shorting
the end of the horiz bus and run up to its rated current.

What I am trying to test is that the connector between the horizontal bus and the vertical bus does not go over temperature when the SYSTEM is fully loaded. The device is rated for 3200 amps but only 800 amps can be pulled off by the vertical bus system. There can be multiple 800 amp vertical systems attached, but worst case is only one (2400 in horizontal and 800 in vertical). If either is not loaded, heat will be pulled out by the unloaded system.

 

[winnie]

From your description, you had insufficient current on the section of vertical bus, but then proceeded to add resistance to it. Added resistance will _decrease_ current flow.

You would need to add resistance to the shunt on the main bus so that more current will flow on the vertical bus through its short.

-Jon

 

[ggunn]

From your description, you had insufficient current on the section of vertical bus, but then proceeded to add resistance to it. Added resistance will _decrease_ current flow.

For a given voltage, yes, but if he is feeding the system a fixed current the current will remain the same and the I^2R heat will increase when he adds series resistance.

 

[winnie]

ggunn: yes, I agree that adding resistance to the load on a constant current supply will increase heating.

What I did not clarify is that the OP has a parallel circuit connected to the test current source: the short at the end of the main bus, and the short at the end of the tapped (vertical) bus. Adding resistance to _one_ of the parallel paths will cause less current to flow on that path. The resistance will change the proportion of current following one path or the other.

-Jon

 

[Sahib]

Girswald:
It is a matter of current division between two parallel resistances. Let the resistance of 3200A horizontal bus bar be R1 and R2 of 800A vertical bus bar. If you feed 4000A into this parallel resistances, current I1 flowing through R1 is

I1=4000*R2/(R1+R2)-------------(1)

and I2 flowing through R2 is

I2=4000*R1/(R1+R2)--------------(2)

Now you want I1 to be 3200A and I2 to be 800A.

Substitute those values in equations(1) and (2) and solve for R1 and R2. If the actual resistances (measure them!) of the two bus bars are r1 and r2 respectively, additional parallel resistances A1 and A2 to be connected to r1 and r2 are given by

(1/r1)+(1/A1)=1/R1----------------(3)

(1/r2)+(1/A2)=1/R2----------------(4)

From equations (3) and (4), you can calculate the additional resistances A1 and A2 for loading the horizontal and vertical bus bars at their ratings.

 

[Girswald]

Thanks for the info. I am going to measure the resistance of the two bus systems and try out the equations.

 

[Smart $]

Thanks for the info. I am going to measure the resistance of the two bus systems and try out the equations.

Umm... you will likely have to account for temperature rise. Measure volts across tap junction at desired current level. Using ohms law should give you impedance at that current level and temperature rise.

 

[gar]

150910-0910 EDT

To add to what winnie has said.

At the tap point the voltage to both stubs is the same. Thus, the ratio of the two stub resistances is Rratio = Ihoriz/Ivert = Rvert/Rhoriz. Rvert also includes any joint resistance between the vertical bus and the horizontial bus. I assume the horizontial is solid thru the place of this tap. The resistances of your shorts at the bus ends are part of the bus stub resistances.

For the particular tap you were referencing. To get the desired currents you need to increase the resistance at the end of the horizontial bus as winnie said. If your vertical tap was closer to the source end of the current source, then added resistance might be needed on the vertical bus.

For the currents you are seeing the current source may not be able to supply enough voltage to get the desired load currents.

Also you should check the voltage drop across tap point joint.

.

 

[Girswald]

Umm... you will likely have to account for temperature rise. Measure volts across tap junction at desired current level. Using ohms law should give you impedance at that current level and temperature rise.

Yea...couldn't measure the resistance of the systems separately, the joint that is being measured is not accessible once the system is installed into the structure. That gives me no place to start.

I'm going to add resistance to the end of the horizontal bus and just short the vertical next.
The guy that left told one of the lab workers that it would take 10 500 MCM cables per phase to create the resistance I need. Not only is that a lot of money, but with the horizontal bus system just under 8 feet in the air, it will put a heck of a strain on the bus. In the past, the lab techs have clamped heavy steel bars around each phase to add resistance as well.

How do I measure the voltage drop with a 3 phase shorted system? Measuring from the transformer tap to ground on each phase gives me around 24 volts on all 3 phases.

If anybody knows an engineer that we might contract to look over our current generator, I would like to chat. I am in North Carolina.

Thanks

 

[winnie]

Voltage must always be measured between two points.

When you measured from transformer tap to ground, you just measured the output voltage of the (presumably grounded) transformer.

When you want to measure 'voltage drop' you need to select two points 'around' the region where you want to measure the 'drop'. To measure the voltage drop across the tap junction you would place probes on a single phase of the bus, one 'upstream' and one 'downstream' of the junction.

I do not know if you need to make this sort of voltage drop measurement; you presumably have the specifications for what you are testing! From your original post it looks like you are simply required to measure temperature rise.

If your testing requirements are to measure temperature rise under full rated current, then all the discussion about measuring voltage, resistance, voltage drop, etc. is to provide you with information to easily and accurately adjust the current flow to meet your test requirements.

Use the equation that gar posted: you simply need to determine the resistance of the vertical bus and the section of horizontal bus between the junction and the end. The ratio of these two resistances needs to match the inverse of the desired current flow distribution. You need to add resistance (or other impedance) to the horizontal bus to put more current on the section of vertical bus.

You can simply try this out by trial and error, but as you note the necessary conductors are bulky and expensive.

You should have access to datasheets for the bus assembly, which should provide you with resistance values that you can use to estimate the resistance you need to add.

When copper cables to add resistance you bump into the problem that you need enough length of cable to add significant resistance, but also need enough total mass of material to safely carry the current. The '10 500 MCM' cables per phase is presumably set to safely carry the current, but that huge amount of copper in parallel means you have very low resistance per unit length...you would need quite a bit of length to get the required resistance.

Using very conservative values, 500 MCM wire is rated to carry 320A, so you would need 10 cables to carry the full 3200A. This rating is based on NEC usage for installed building wire, with additional limits that apply when you bundle cables together in a conduit.

The note about 'clamped heavy steel bars around each phase' raises an important point. At the low voltage and high current levels used for these tests, _magnetic effects_ are very significant. If you actually clamp steel to a bus bar, you will add conductive material and _lower_ resistance. But because this is _AC_ the magnetic effects of the steel will add _inductance_. This will act to impede the flow of _AC_ without adding resistance.

There are other options for adjusting the current flow between bus sections, but they would require other tools, eg additional drive transformers or non-copper resistance conductors.

-Jon

 

[steve66]

Can you install 4 vertical sections, and short each vertical section? Then set the generator for 3200 amps.

The 1st vertical section will have a little more current than the next, and so on. But you could either tweak the resistances to be a little less on each farther vertical bus, or it might be close enough to pass the test.

Just say for example you have 3200 amps total on the horiz bus, and again, just as an example, 820 amps on the 1st vertical, and it passes the test, then I think you can call it a day.

 

[Girswald]

Can you install 4 vertical sections, and short each vertical section? Then set the generator for 3200 amps.

The 1st vertical section will have a little more current than the next, and so on. But you could either tweak the resistances to be a little less on each farther vertical bus, or it might be close enough to pass the test.

Just say for example you have 3200 amps total on the horiz bus, and again, just as an example, 820 amps on the 1st vertical, and it passes the test, then I think you can call it a day.

No, the test criteria are established by UL. Although your idea might have merit as the only result of this test is to determine if the new joint can handle the same current and not go over temperature as the previous version. I'll bring that up to my manager...

 

[gar]

150910-1954 EDT

Girswald:

You need to think outside the box somewhat.

We need more information, but I will make some assumptions to show a possible direction. Note: a constant current source can not simultaneously be a constant voltage source.

You mention that when both the horizontal and vertical stubs are shorted at their ends that the short stub current is 250 A. What was the horizontal stub current? Assume the 2400 A you mentioned, then the vertical stub is 2400/250 = 9.6 times the resistance of the horizontal.

Assume the cross-sectional area of the horizontal bus is the same over its entire length, then knowing the distances on each side of the tap point you can estimate each resistance if you measure the total resistance of the horizontal bus. From this you can estimate the vertical bus resistance.

Supose the voltage input to the horizontal bus is 5 V to produce 3200 A with the vertical bus open, then the horizontal bus resistance is about Rht = 5/3200 = 0.0016 ohms. Next assume the tap point is at 50%, then the horizontal stub resistance is 0.0008 ohms. In turn the vertical stub is 9.6 *0.0008 = 0.0075 ohms.

To get the desired current distribution ratio of 800 to 3200 requires the horizontal stub path resistance to be 0.0075*800/3200 = 0.0019 ohms. This means 0.0019-0.0008 = 0.0011 ohms of added resistance is required at the end of the horizontal bus.

0.0011 ohms at 3200 A is about 11300 W. That is not a lot power to dissipate. We need some real numbers and then look for a way to make a suitable adjustable resistor.

Check my nath and procedure.

.

 

[Sahib]

Yea...couldn't measure the resistance of the systems separately, the joint that is being measured is not accessible once the system is installed into the structure. That gives me no place to start.

I'm going to add resistance to the end of the horizontal bus and just short the vertical next.
The guy that left told one of the lab workers that it would take 10 500 MCM cables per phase to create the resistance I need. Not only is that a lot of money, but with the horizontal bus system just under 8 feet in the air, it will put a heck of a strain on the bus.

Well, Girswald, I would like to suggest another method, though indirect, then. Your aim is to measure the temperature rise of the tap joint of the two bus bars. The joint passes the total current irrespective of current sharing between the two bus bars. So measure the temperature rise of the joint for various currents through the joint after temperature equilibrium is established and plot the values on a graph; it will be a straight line. So you may find the temperature rise of the joint for a current of 4000A from the graph by extending the straight line.

 

[gar]

150911-0856 EDT

Sahib:

Your approach may not show the temperature rise that results from the joint resistance between the horizontal bus and the vertical bus. 800 A needs to be flowing thru the actual joint to to the vertical bus to indicate if the joint has a higher than should be resistance. And the 3200 A needs to be flowing to the end of the horizontal bus because the resulting elevated temperature may change the joint resistance.

A test with the horizontal bus unloaded and 800 A flowing to the vertical bus could provide some useful information, but that is not what is being required for the specified UL test. Generating the required currents at the joint is the only option unless UL will allow something else.

.

 

[Girswald]

150911-0856 EDT

Sahib:

Your approach may not show the temperature rise that results from the joint resistance between the horizontal bus and the vertical bus. 800 A needs to be flowing thru the actual joint to to the vertical bus to indicate if the joint has a higher than should be resistance. And the 3200 A needs to be flowing to the end of the horizontal bus because the resulting elevated temperature may change the joint resistance.

A test with the horizontal bus unloaded and 800 A flowing to the vertical bus could provide some useful information, but that is not what is being required for the specified UL test. Generating the required currents at the joint is the only option unless UL will allow something else.

.

You are correct, UL specifies the set-up. One of the things overlooked by using just electrical theory is that the mass of copper on the horizontal bus run will actually pull heat away from the joint. That's a good way to keep temperatures down, but copper is very expensive, so it isn't a very economical way to control temperatures.

Heck, spraying flat black paint on the busbar will take heat out, and a tin or silver plated bar will hold the heat in.

Everything you'd expect to happen in a thermal test can be thrown out the window. The concepts work, but in practice its not that easy!

We did get the correct amperages in one phase of the system this morning utilizing cable and steel wraps to induce resistance. Now, if our current source would behave correctly...three identical transformers all wired the same and the outputs are vastly different. A few percent I could understand, but we're not even close.

Oh well, I'll keep trying. Got no choice.

 

[gar]

150911-1354 EDT

Girswald:

That your not getting equal current on each phase means we need to know more about your source.

First, is your source supposed to be an actual constant current source, but adjustable? At this current level I doubt that it is. More likely it is an adjustable voltage source.

If it is an adjustable voltage source, then are each of the three output voltages the same? Are voltages measured line-to-line, or line-to-neutral?

What are the values of these voltages when you produce your output of 4000 A or so?

Is the system a wye or delta?

.

 

[Sahib]

UL specifies the set-up.

It is hard to believe UL specifies full current i.e 4000A should actually be passed through the bus bar joint to test it.

One of the things overlooked by using just electrical theory is that the mass of copper on the horizontal bus run will actually pull heat away from the joint. That's a good way to keep temperatures down, but copper is very expensive, so it isn't a very economical way to control temperatures.

You should wait for temperature equilibrium and then take measurement of the temperature.

We did get the correct amperages in one phase of the system this morning utilizing cable and steel wraps to induce resistance. Now, if our current source would behave correctly...three identical transformers all wired the same and the outputs are vastly different. A few percent I could understand, but we're not even close.

Check no load output voltages of the transformers: they should be same.