Cost of Power

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jaymiller

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Question: What does it cost per year (at $8.60 cents per kWh) for the power loss of two 10 AWG circuit conductors that have a total resistance of 0.30 ohms with a current flow of 24A? Please explain your answer/show work.
 
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How close do yu want the answer to be?

Leap year or regular year?

Calendar year or astronomical year?

$130.32 the year I used. :eek:hmy:

For an ACCURATE answer, you need to provide the hour by hour ambient temperatures AND the wire insulation type, spacing, wire orientation, etc..... also, or are you controlling the 0.3 ohms by adjusting the wire temperature or just making a constant assumption?

Apparently also an assumption of daytime and night power rates the same, many place not.

Tell yer teacher to provide a better test question if this is an engineering class, if just for electricians, just give the 3% loss number :lol:
 
Mike Holt's textbook is my teacher!! haha.

I'm trying to understand how the textbook turns $8.60 into $0.0086? The textbook answer is $130, I'm sure rounded. Can you explain why? That's where I'm having trouble understanding.
 
jaymiller said:
Question: What does it cost per year (at $8.60 cents per kWh) for the power loss of two 10 AWG circuit conductors that have a total resistance of 0.30 ohms with a current flow of 24A? Please explain your answer/show work.
Click to expand...

(24)^2 x 0.30 x 24 x 365 x 0.086/1000 = $130.18 per year
watt x hr/day x day/year x $/kwh / 1000 w/kw = $/year
 
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jaymiller said:
Question: What does it cost per year (at $8.60 cents per kWh) for the power loss of two 10 AWG circuit conductors that have a total resistance of 0.30 ohms with a current flow of 24A? Please explain your answer/show work.
Click to expand...

P = (I^2)(R) = (24.0A^2)(0.3 ohms) = 172.8W.

If you assume steady state loading, E in kW-hr/year/circuit = (P)(hours/year)/(W/kW-hr) = (172.8W)(365 days/year)(24 hr/day)/(1000W/kW) = 1513.7 kW-hr/year/circuit

$ = (# of circuits)(E)(cost/kW-hr) = (2 circuits)(1513.7 kW-hr/circuit)($0.086/kW-hr) = $260.36

This assumes that the question as stated means 0.3 ohms per circuit, not the total for the two circuits.
 
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ggunn said:
P = (I^2)(R) = (24.0A^2)(0.3 ohms) = 172.8W.

If you assume steady state loading, E in kW-hr/year/circuit = (P)(hours/year)/(W/kW-hr) = (172.8W)(365 days/year)(24 hr/day)/(1000W/kW) = 1513.7 kW-hr/year/circuit

$ = (# of circuits)(E)(cost/kW-hr) = (2 circuits)(1513.7 kW-hr/circuit)($0.086/kW-hr) = $260.36

This assumes that the question as stated means 0.3 ohms per circuit, not the total for the two circuits.
Click to expand...

Rereading the question, it looks like it means one circuit with two conductors with a total R of 0.3 ohms, hence half what I said, i.e., $130.18.
 
jaymiller said:
Mike Holt's textbook is my teacher!! haha.

I'm trying to understand how the textbook turns $8.60 into $0.0086? The textbook answer is $130, I'm sure rounded. Can you explain why? That's where I'm having trouble understanding.
Click to expand...

Apparently you have a grasp of what to do here, but apparently the confusion is $8.60 is eight dollars, sixty cents not 8.6 cents

Poorly written and full of conflict if it actually says "$8.60 cents"
 
kwired said:
Apparently you have a grasp of what to do here, but apparently the confusion is $8.60 is eight dollars, sixty cents not 8.6 cents

Poorly written and full of conflict if it actually says "$8.60 cents"
Click to expand...

If electricity is $8.60/kWh I'd look for a new POCO.
 
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