The Mother of all Story Problems

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A/A Fuel GTX

A/A Fuel GTX

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WI & AZ
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Electrician
Ok, I'm sure this is a first. I have a client that has rental properties. There is a Landlord/Tenant dispute revolving around a garage door opener and the kwh it would consume over the course of a year. The tenant has paid for the electricity to to operate the overhead door opener in a garage that is fed off of his meter. The door is operated once per day, one up cycle and one down cycle. The duration of each cycle is 13 seconds so we have a 1/2 hp, 120v load operating for 26 seconds per day. How many kwh would that translate into over the course of a year? If possible, I need the formula so I can put the calculations and answers in writing:cool:
 
A/A Fuel GTX said:
Ok, I'm sure this is a first. I have a client that has rental properties. There is a Landlord/Tenant dispute revolving around a garage door opener and the kwh it would consume over the course of a year. The tenant has paid for the electricity to to operate the overhead door opener in a garage that is fed off of his meter. The door is operated once per day, one up cycle and one down cycle. The duration of each cycle is 13 seconds so we have a 1/2 hp, 120v load operating for 26 seconds per day. How many kwh would that translate into over the course of a year? If possible, I need the formula so I can put the calculations and answers in writing:cool:
Click to expand...

1hp = 0.746 kw

Assume that the motor draws its nameplate horsepower for the full time.

1/2 Hp = 0.746 kw/2 = 1.492 kw

Time of operation = 26 seconds x 365 days per year / 3600 secs/hour = 2.64 hours.

kwh's used per year = 1.492 kw x 2.64 hours = 3.94 kwh

Is there a 40 Watt bulb that stays on for a couple of minutes mounted on the opener? Need to add that power in also.
 
OK 1/2 HP is 372 Watts
26 Seconds is .43333 Minutes
.43333 Minutes is
.00722 Hours
So you have 2.6866 Watt hours per day
365 days would be 980.61 Watthours per year or 0.98 kwh per year
 
rcwilson said:
1hp = 0.746 kw

Assume that the motor draws its nameplate horsepower for the full time.

1/2 Hp = 0.746 kw/2 = 1.492 kw

Time of operation = 26 seconds x 365 days per year / 3600 secs/hour = 2.64 hours.

kwh's used per year = 1.492 kw x 2.64 hours = 3.94 kwh

Is there a 40 Watt bulb that stays on for a couple of minutes mounted on the opener? Need to add that power in also.
Click to expand...

What about the power necessary to run the electronics 24/7? May seem trivial, but there's 8765.812771 hours in a year.
 
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Yes Bob.....I forgot to mention the 40w bulb that stays on for 2 minutes per cycle so it would be a 40w bulb on for 4 minutes a day...can you recalculate?
 
A/A Fuel GTX said:
The door is operated once per day, one up cycle and one down cycle.
Click to expand...

Once per day?

Trying to imagine the possibilities;
- door stays open all day, gets shut at night.
- used as an exit or entry means once per day, then closed.
:-?
 
celtic said:
Once per day?

Trying to imagine the possibilities;
- door stays open all day, gets shut at night.
- used as an exit or entry means once per day, then closed.
:-?
Click to expand...
The garage is used for storage so the landlord goes in via the overhead door, does his thing and then closes the door once per day.
 
Don't forget "line loss" -

P=IR2​

For both the wire (if #14, we're talking a fight for every penny) @ 3.07 ohms per 1000', per Table 8 chapter 9, and is there anything else potentially energized (like a light in the push buttons)?
 
4 minutes/60 Minutes*40 W= 2.666 Watthours per day
So total with bulb and motor= 5.3526 watt hours per day

1953.7 Watthours per yar (1.9537 kWH per year)
 
A/A Fuel GTX said:
Ok, I'm sure this is a first. I have a client that has rental properties. There is a Landlord/Tenant dispute revolving around a garage door opener and the kwh it would consume over the course of a year. The tenant has paid for the electricity to to operate the overhead door opener in a garage that is fed off of his meter. The door is operated once per day, one up cycle and one down cycle. The duration of each cycle is 13 seconds so we have a 1/2 hp, 120v load operating for 26 seconds per day. How many kwh would that translate into over the course of a year? If possible, I need the formula so I can put the calculations and answers in writing:cool:
Click to expand...

My advice is stay out of it. Anyone who will argue over the amount of energy used by a single garage door opener in a year is someone you don't want to do business with. They're looking for a fight, and obviously don't care whether they find a good one or not.
 
Per Table 430.248, a 1/2 HP, 115 V motor draws 9.8 amps.

9.8 times 115 is 1,127 watts, or 1.127 KW.

1.127 KW times 26 seconds, then divided by 3600 (to convert seconds to hours), and finally multiplied by 365 days per year, gives me 2.97 kilowatt-hours per year.

To that we add 40 watts times 240 seconds, then divided by 3600 (to convert seconds to hours), and finally multiplied by 365 days per year, gives me 0.97 kilowatt-hours per year.

Total is therefore 3.94 kilowatt-hours per year. Assuming a rate of 10 cents per KWH, tell the landlord to credit the tenant a half a buck, and call it a day. :cool:
 
wirenut1980 said:
I have to laugh at this...1 kw-hr per year at a whopping 11 or 12 cents/per kw-hr. Whoa! Don't break the bank on that one!
Click to expand...

Thats what I was thinking also - the cost is probably very trivial - even with the 40 watt lamp, the motor, and the electronics on 24x7.

Also, don't forget: the starting power of the motor will be a lot higher than the running power. And the motor won't use the full 1/2 HP to lower the door. Still, I think were probably talking peanuts.

Steve
 
rcwilson said:
1hp = 0.746 kw

Assume that the motor draws its nameplate horsepower for the full time.

1/2 Hp = 0.746 kw/2 = 1.492 kw

Time of operation = 26 seconds x 365 days per year / 3600 secs/hour = 2.64 hours.

kwh's used per year = 1.492 kw x 2.64 hours = 3.94 kwh

Is there a 40 Watt bulb that stays on for a couple of minutes mounted on the opener? Need to add that power in also.
Click to expand...

My bad. I multiplied by 2 instead of dividing. That's what I get for being quick and incorrect. Here's the correct math.

1/2 Hp = 0.746 kw/2 = 0.373kw

Time of operation = 26 seconds x 365 days per year / 3600 secs/hour = 2.64 hours.

kwh's used per year = 0.373 kw x 2.64 hours = 0.985 kwh

Add in 40W on 4 minute per day: 40 watts = 0.040 kw

(0.040 kw x 4 minute x 365 days) / 60 minutes/hour = 0.973 kwh per year

0.985 + 0.973 = 1.96 kwh per year total for the motor and light. (Assuming the motor draws 1/2 HP)

If the electronics and control power transformer draw power continuously , it can't be much - say 5 watts. 5 W x 8760 hours per year /1000 = 43.8 kwhours
 
My power conversion function on my calculator shows 1 HP = 746W as others have stated.

But my handy-dandy Square-D motor data calculator shows a 1/2 HP motor at 120V has a full load current of 9.8A. So 9.8A * 120V = 1.176kW. The difference is probably peak vs. running.

26 seconds = 0.007222 hr
4 minutes = 0.06667 hr
1.176kW * 0.007222 hr * 365 days = 3.1kWh
0.04kW * 0.06667 hr * 365 days = 0.973kWh
For a grand total of 4.073kWh per year. Or about 40 cents.

But as minimal as this is it is generally against the law to have the tenant meter pay for common area devices. Like in an apartment building if there's a common hallway or stair the lights in there have to be feed from a landlord panel rather than the nearest tenant panel. I don't remember where I read this law but if this is an applicable law in your area then any fine or penalty would be more then 40 cents.
 
Mr. Bill said:
.....But as minimal as this is it is generally against the law to have the tenant meter pay for common area devices. Like in an apartment building if there's a common hallway or stair the lights in there have to be feed from a landlord panel rather than the nearest tenant panel. I don't remember where I read this law but if this is an applicable law in your area then any fine or penalty would be more then 40 cents.
Click to expand...

Maybe not 'the law', but 210.25(B) pretty much sys so.
 
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