Confusing problem

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GeorgeW

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I recently ,(actually today) took my Journeymans test in New Hampshire, and came across a problem that has me at a loss:
Given a hot water heater, 208v 3 phase, compute ampacity draw.
It's three phase. How do you do this problem?
 
iwire said:
Watts divided by volts divided by 1.73 = amps
Click to expand...
Another way to say this is : W / V / 1.73

Dennis Alwon said:
Or--- watts/360.

1.73= square root 3

208 X 1.73 = 360
Click to expand...
Another way to say this is : W/ V x 1.73

Two different formulas. Who's right?
 
080415-2058 EST USA

Here is a way to visualize and help you understand the basics.

Consider a Y connected resistive load for your heater. For a balanced load each of the three resistors dissipates 1/3 of the power. Beacuse of the Y connection the current in the supply line is equal in magnitude and in phase with the current in its connected resistor.This should be intuitively obvious.

The "leg to leg" to "leg to nuetral" ratio can be determined by a vector diagram where "leg to leg" = 2 * "leg to nuetral" * sin 60 = 2 * 0.866 * "leg to nuetral" = 1.732 * "leg to nuetral". Therefore, for your 208 supply the "leg to nuetral" = 120.09, but 120 is a number you should have already known, and thus you would know that ratio or could calculate it.

From the above it should be obvious that you divide total watts by 3 to get the dissipation in one resistor of the Y load, then divide that wattage by 120 to get the current is one supply line. Writing an equation for this we have I * V / 1.732 = W/3. It should be noted that the exact value of 2 * sin 60 = sq-root of 3. But 1.732 is quite close and it is the year of George Washington' birthday to help you remember. Now you see the final equation becomes I = W/(1.732*V) = 0.577*W/V. Where I is line current, V is leg to leg voltage, and W is total power of the heater.

If you can remember this train of thought you might in the future be able to derive the the equation yourself.

It is not true that the phase angle of the current in one element of a delta load is in phase with the line current. So if the question had been posed as a delta load that might have been a tricker question because it might have thrown you off. You might not have considered changing the load to a Y to force the line current phase angle to equal load current phase angle.

.
 
jghrist said:
No, put parentheses in to say: (W/V)/1.73 = W/(V?1.73) = W/360


Same thing.
Click to expand...
W/V/1.732 is appropriate without parentheses. The one that needs parentheses is W/V*1.732, which should be W/(V*1.732). If you don't believe me, try entering the formula however you like in a spreadsheet. That will tell you real quick which is the proper formatting.

They're all based on W = V ? I ? 1.732 ? PF
 
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jghrist said:
No, put parentheses in to say: (W/V)/1.73 = W/(V?1.73) = W/360


Same thing.
Click to expand...
Ok. I plugged in some numbers. It works out the same....
 
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Smart $ said:
W/V/1.732 is appropriate without parentheses. The one that needs parentheses is W/V*1.732, which should be W/(V*1.732). If you don't believe me, try entering the formula however you like in a spreadsheet. That will tell you real quick which is the proper formatting.

They're all based on W = V ? I ? 1.732 ? PF
Click to expand...

You're correct. For some reason I saw this as W/ (V/1.73) which is incorrect.
 
Smart $ said:
W/V/1.732 is appropriate without parentheses. The one that needs parentheses is W/V*1.732, which should be W/(V*1.732). If you don't believe me, try entering the formula however you like in a spreadsheet. That will tell you real quick which is the proper formatting.
Click to expand...

The reason that works is because of the law of mathematics. The calculator will divide W/V first then divide by 1.732.

The formula would actually look like this W/V / 1.732. This is the same as
(W/V) x (1/1.732).
 
Smart $ said:
W/V/1.732 is appropriate without parentheses. The one that needs parentheses is W/V*1.732, which should be W/(V*1.732).
Click to expand...
This is correct, because of the standard convention that was adopted in my youth (circa mid-1300's :D ). When you have a combination of multiplication and division, you do the operations from left to right. Also, anything inside parentheses is performed first, before combining what is inside with what is outside.

The same convention says that when you have a combination of addition and subtraction, you do the operations from left to right.

The final aspect of this same convention is that when you have a combination of multiplication, division, addition, and subtraction, you first do the multiplication and division operations from left to right, then you do the addition and subtraction operations from left to right.

All calculators (other than the RPN calculators you commonly see manufactured by HP) are programmed with these conventions.
 
jerm said:
From 4th grade: "Please My Dear Aunt Sally"

Parenthesis, then Multiplication and Division, then Addition and Subtraction.
Click to expand...


You forgot "EXCUSE" for exponents
Parenthesis, then Exponents, then Multiplication and Division, then Addition and Subtraction.
 
mikeames said:
You forgot "EXCUSE" for exponents
Parenthesis, then Exponents, then Multiplication and Division, then Addition and Subtraction.
Click to expand...

Doh. It's been a long time since fourth grade. :grin:
 
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