I am trying to figure something out and any help is appreciated. I am trying to find the impedance im mohms for a perticular length of cable. It is for a hospital installation where the vendor equipment is restricting the maximum incoming line impedance as (m ohm) 180. I used table 9 to find the ohms to netural per 1000' and got it down to ohms per foot now how do I get it to m ohm's?
Impedance
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ryan_618
Senior Member
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Re: Impedance
Are you talking miliohms? If you are, divide your ohms by 1,000
Edited for english
[ February 01, 2006, 01:58 PM: Message edited by: ryan_618 ]
Are you talking miliohms? If you are, divide your ohms by 1,000
Edited for english
[ February 01, 2006, 01:58 PM: Message edited by: ryan_618 ]
ron
Senior Member
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Re: Impedance
Table 9 will give you a resistance/impedance for 1000'. If you had 1 ohm per 1000', and you ran that conductor for 1000', you would have reached 1 ohm or 1000 milliohms.
There are 1000 milliohms in 1 ohm.
Table 9 will give you a resistance/impedance for 1000'. If you had 1 ohm per 1000', and you ran that conductor for 1000', you would have reached 1 ohm or 1000 milliohms.
There are 1000 milliohms in 1 ohm.
ramsy
NoFixNoPay Electric
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Re: Impedance
When finished, which is the best way to check the impedance?
Z = (VD / I), where VD = (Et - Er) Used by ShureTest St-1THD
or
Z = (Et / It) Book formulas (not tested by me)
[ February 01, 2006, 08:27 PM: Message edited by: ramsy ]
When finished, which is the best way to check the impedance?
Z = (VD / I), where VD = (Et - Er) Used by ShureTest St-1THD
or
Z = (Et / It) Book formulas (not tested by me)
[ February 01, 2006, 08:27 PM: Message edited by: ramsy ]
physis
Senior Member
Re: Impedance
The best guess I can make is that m ohms is meaning miliohms (thousanths of an ohm). Not a term I can remember working with but makes sense non the less.
M ohms (with a capitol M) is megaohms or 1,000,000 ohms. Not an impedance value that would make sense here.
180 miliohms would be 0.18 ohms.
You can use your resistance value per foot that you derived from table 9 (Call that R), multiply that by the run distance (Call that D), times 2 (for single phase).
R x D x 2 = Total Resistance
The Total Resistance of your run has to be .18 ohms or less.
That's if the specs actually mean miliohms or thousansdths of an ohm.
If it's three phase replace the 2 with 1.732 and do the same thing.
The best guess I can make is that m ohms is meaning miliohms (thousanths of an ohm). Not a term I can remember working with but makes sense non the less.
M ohms (with a capitol M) is megaohms or 1,000,000 ohms. Not an impedance value that would make sense here.
180 miliohms would be 0.18 ohms.
You can use your resistance value per foot that you derived from table 9 (Call that R), multiply that by the run distance (Call that D), times 2 (for single phase).
R x D x 2 = Total Resistance
The Total Resistance of your run has to be .18 ohms or less.
That's if the specs actually mean miliohms or thousansdths of an ohm.
If it's three phase replace the 2 with 1.732 and do the same thing.
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Re: Impedance
The easy part of the answer is that the "2" in the single phase merely accounts for the two legs (going out, and coming back).
The square root of three will find its way into any calculation that involves three phase systems. The math is a bit tricky to explain. Let me just say that it is not a simple matter of adding apples to apples. Rather, you are adding one current that has its peaks and valleys occuring at certain points in time to a curren that has its peaks and valleys occurring at different points in time. So you can't arrive at a value of "total current" by adding one current to the other current. You don't get twice the current, you get about 173.2% of the current (i.e., a factor of 1.732, or the square root of three).
The easy part of the answer is that the "2" in the single phase merely accounts for the two legs (going out, and coming back).
The square root of three will find its way into any calculation that involves three phase systems. The math is a bit tricky to explain. Let me just say that it is not a simple matter of adding apples to apples. Rather, you are adding one current that has its peaks and valleys occuring at certain points in time to a curren that has its peaks and valleys occurring at different points in time. So you can't arrive at a value of "total current" by adding one current to the other current. You don't get twice the current, you get about 173.2% of the current (i.e., a factor of 1.732, or the square root of three).
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