Aluminum tubular ampacities and temperature ratings what do they mean?

[bdn2004]

We have an switchyard with open bus where we are wanting connect a 15kV, 30kVA transformer through an air switch. Most everything in the yard is connected with aluminum alloy tubular bus. Most of it is 1-1/4", with a wall thickness of 0.191.

Are there standards on how this is to be installed? And the current ratings are different with different temperature rises. For example, for this bus, it's rated: 585A for a 10deg C rise, 823A for a 20deg C rise, and 1385A for a 60 deg C rise. Can someone explain why the big differences in ampacity...?

 

[rterickson]

Bus conductors (wire or standard shapes) are rated according to temperature rise above ambient with said ambient being a round number, typically 20C, 30C, or 40C.

So in your example, assuming that your bus is rated for 40C ambient, you could run at 585 amps without exceeding 50C, up to 1385 amps without exceeding 70C.

If it routinely gets hotter than the rated ambient in your location (or is significantly cooler) you can calculate correction factors in determinining your local ratings.

 

[bdn2004]

terickston,

I'm still a bit confused.

I meant to say 30MVA, not kVA, with a FLA of 1150A. This is the Midwest. Ambient conditions range between -4degF (-20deg C) and 104degF (40degC). And the 40 deg C is the rated ambient temperature of the bus.

So what you are saying is...when it's 40degC outside (104 deg F) if 1385A were flowing in this tubular bus on that day, the actual temperature of the bus would not exceed 70 deg C. If the termination on the transformer is rated 90 deg C this will be ok for this installation? But the transformer is rated with a 55 deg C temperature rise...is that the number this should be designed for? In this case per the spec sheet at 50degC the rating is 1140A...and therefore this bus is too small?

 

[rterickson]

Sorry, math mistake:

1385A at 60C rise over 40C ambient would be operating at 100C.

 

[rterickson]

The most common temperature rating for aluminum bus, bus fittings, and disconnect switches is 30C over 40C ambient, or 70C operating temperature.

The 1385A @ 60C rise (100C operating) sounds like a short-term emergency type rating.

You will need to look at every component in your system, and determine how hot you are willing to operate.

But yes, 1-1/4" schedule 80 sounds a little small, I would use at least 2" bus for 1140A on our system.

 

[jghrist]

See

IEEE STANDARD
605-2008 - IEEE Guide for Bus Design in Air Insulated Substations

Description: A proper design of the substation bus ensures a safe and reliable operation of the substation and the power system. Two different types of buses are used in substations, the rigid bus and the strain (cable). This guide provides information on the different bus arrangements used in substations stating the advantages and disadvantages of each. Also it provides information as related to each bus type and construction. Once the bus type is selected, this guide provides the calculation tools for each bus type. Based on these calculations, the engineer can specify the bus size, forces acting on the bus structure, number of mounting structures required, and hardware requirements.

 

[7anoter4]

I think ,nevertheless aluminum pipe of 1-1/4" schedule 80 may be loaded with 1140 A at 40oC ambiant air 2ft/sec wind velocity for only 30oC temperature rise [70oC bus temperature].
following calculation as follows[limited for these conditions]:
Heat produced by current flow:
wcurrent=ResAL*skin*Irated^2
wcurrent=103.24 w/m
neglecting proximity factor ResAL=1/38*1/409.96*(1+0.00403*(70-20))=7.71E-05 ohm/m
skin effect =1.03 [See:

http://www.copperinfo.co.uk/busbars/pub22-copper-for-busbars/sec3.htm

Aext=pi()*Dext*1=pi()*31.75/1000*1=0.099746 m^2
DeltaT=70-(40)=30oC
Heat produced=103.24 w/m
Evacuated Heat:
wconvection=kconvection*Aext*DeltaT
vel=velocity of wind=2 ft/sec=0.6096 m/sec
niuair=cinematic viscozity of air at 40oC =17E-6 m^2/sec
lambda=thermal conductivity of dry air at 40oC=0.027 W/m/Oc
kconvection=Nu*lambda/Dext
Nu [Nussfeld no.]=0.21*Re^0.6*Pr^0.4
Pr=0.71 for dry air
Re=vel*Dext/niuair
Re=0.6096*31.75/1000/17*10^6
Re=6536.76
Nu =0.21*6536.76^0.6*0.71^0.4=35.64
kconvection=35.64*0.027/31.75*1000=30.3
wconv=30.3*0.099746*30=90.7 w/m
wrad=5.7/10^8*epsbus*((Tbus+273)^4-(Tair+273)^4)*Aext
epsbus=0.5
wrad=5.7/10^8*0.5*((70+273)^4-(40+273)^4)*0.099746=12.06
Total evacuated HEAT=wconv+wrad=90.7+12.06=102.8 w [0.4% ERROR]

 

[7anoter4]

Correction:
Nu=Nusselt Number
Re=Reynolds Number
Pr=Prandtl Number