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Would one of you electron benders please tell me how I go about computing
cost of operation of an appliance if I know that it operates on 220vac at
2,000 w and 'y' is the cost of electricity/ Kwh. Is there a time factor tied
to the 2,000 (- too many years)? I'm thinking of supplementing the shop's
LPG heat with electric thinking that it would be cheaper than just using the
gas alone. I only heat on an as-needed basis but the last fill up nearly
cost me a good internal organ!

Multiply the watts of power consumed (at any voltage, it does not matter) time the kW rate your utility is charging.

2000 watts (2 kW) at .15 kW = 30¢ per hour

NO! Multiply watts of power by hours to get kilowatt hours of energy!

Learn the difference between power and energy!

Nick

With 7 horses, I know the difference. Do you? :-)

Nick

And your answer to the original question is??????????????

But he does not know how to figure the operating cost of those kWs. He seems to know what he needs, (thus the 2,000 watt figure) just not the cost of operation compared to propane.

Kilowatt-HOURS!!!

Nick

#### Site Timeline

- posted on December 30, 2005, 7:21 pm

- posted on December 30, 2005, 7:27 pm

Multiply the watts of power consumed (at any voltage, it does not matter) time the kW rate your utility is charging.

2000 watts (2 kW) at .15 kW = 30¢ per hour

- posted on December 30, 2005, 8:16 pm

NO! Multiply watts of power by hours to get kilowatt hours of energy!

Learn the difference between power and energy!

Nick

- posted on December 30, 2005, 9:16 pm

snipped-for-privacy@ece.villanova.edu wrote:

Learn the difference between giving advice and being a horse's ass.

Learn the difference between giving advice and being a horse's ass.

- posted on December 30, 2005, 10:56 pm

With 7 horses, I know the difference. Do you? :-)

Nick

- posted on December 30, 2005, 10:14 pm

chill...........

Ed's final answer is correct...............note the units in his reply

30¢ per hour

but his calc needs a slight correction to be dimensionally correct

2000 watts (2 kW) at .15 kW = 30¢ per hour

S/B

2000 watts X 1 kW* /(1000 watts) X 15¢ / *kWatt hr = 30¢ / hr

giving the cost to operate an electrical device in ¢ or $ per hour makes the most sense, that way the user can easily determine the cost since he has a good idea of the total time of usage.

cheers Bob

Ed's final answer is correct...............note the units in his reply

30¢ per hour

but his calc needs a slight correction to be dimensionally correct

2000 watts (2 kW) at .15 kW = 30¢ per hour

S/B

2000 watts X 1 kW

giving the cost to operate an electrical device in ¢ or $ per hour makes the most sense, that way the user can easily determine the cost since he has a good idea of the total time of usage.

cheers Bob

- posted on December 31, 2005, 12:28 am

And your answer to the original question is??????????????

- posted on December 31, 2005, 1:15 am

The original question of "tell me how I go about computing cost of operation
of an appliance" can only be answered by knowing how many BTUs he needs.
Then he can figure out how long an electric heater will have to run.
All that calculating isn't necessary. All he has to do is multiply his gas
usage by his heater's efficiency, then divide that by 3,413 to find out how
many KWs he needs.

wrote:

wrote:

- posted on December 31, 2005, 3:58 am

But he does not know how to figure the operating cost of those kWs. He seems to know what he needs, (thus the 2,000 watt figure) just not the cost of operation compared to propane.

- posted on December 31, 2005, 1:13 pm

Seeing as electric heat is 100% efficient, all he has to do is multiply the
KW times the rate his electric company charges.

By doing it this way, he will be able to arrive at an accurate figure on total operating costs of l.p. vs. electric.

By doing it this way, he will be able to arrive at an accurate figure on total operating costs of l.p. vs. electric.

- posted on December 31, 2005, 1:03 pm

Kilowatt-HOURS!!!

Nick

- posted on December 30, 2005, 7:30 pm

Depending on where you are, electric is usually more expensive than l.p.
Find out what the cost of both are, and then go here to calculate
http://energy.cas.psu.edu/energyselector /

- posted on December 30, 2005, 10:46 pm

C & E wrote:

Watts is an instantaneous figure; no time factor. So w/1000 x y = cost per kWh. E.G. 2000 W heater, cost is $0.10 per kWh. So 2000/1000 x $0.10/kWh 2kW x $.10/kWh = $0.20/h

That probably doesn't tell you much because you won't know how many hours the 2Kw heater will run in a month. Nontheless, heating with electricity is likely to be less costly than heating with LPG. Somebody could tell you the BTU per gallon of LPG and the BTU for a kilowatt of electricity and then you could apply your local cost per gallon of LPG and per kWh of electricity to see which would be cheaper.

Watts is an instantaneous figure; no time factor. So w/1000 x y = cost per kWh. E.G. 2000 W heater, cost is $0.10 per kWh. So 2000/1000 x $0.10/kWh 2kW x $.10/kWh = $0.20/h

That probably doesn't tell you much because you won't know how many hours the 2Kw heater will run in a month. Nontheless, heating with electricity is likely to be less costly than heating with LPG. Somebody could tell you the BTU per gallon of LPG and the BTU for a kilowatt of electricity and then you could apply your local cost per gallon of LPG and per kWh of electricity to see which would be cheaper.

- posted on December 31, 2005, 12:07 am

Compare cost of your gas price to your electric price by btu. for me in
the midwest electric is stlll apx 50% more expensive than Ng, only you
know your costs. Btu comparisons or for Lp are online. In other words
how much is your cost for 100000 btu for each. You likely will see Lp is
best, but it depends on your area.

- posted on December 31, 2005, 1:02 am

m Ransley wrote:

How efficient is a typical "LPG heater"?

I presume they must vent the exhaust outdoors, so what's a reasonable number for the percentage of BTUs in the gas that actually heats the workspace?

Does it vary significantly with the design (and era of manufacture) of the LPG heater?

I expect that electric heat must be near 100% efficient, huh?

My inquiring mind wants to know...

Jeff

How efficient is a typical "LPG heater"?

I presume they must vent the exhaust outdoors, so what's a reasonable number for the percentage of BTUs in the gas that actually heats the workspace?

Does it vary significantly with the design (and era of manufacture) of the LPG heater?

I expect that electric heat must be near 100% efficient, huh?

My inquiring mind wants to know...

Jeff

--

Jeffry Wisnia

(W1BSV + Brass Rat \'57 EE)

Jeffry Wisnia

(W1BSV + Brass Rat \'57 EE)

Click to see the full signature.

- posted on December 31, 2005, 1:18 am

Right OP didnt say what his lpg heater is so how can you figure a
comparison without knowing its efficiency. Im sure its as efficient as
Ng and there are many types and ratings so he could have junk running
40% efficiency or a ventless 99% or furnace at 60-94.5% eficiency
Electric is 100% efficient.

- posted on December 31, 2005, 3:06 am

He is talking to supliment heat thinking electric is cheaper, so cost
comparing the 2 sources is all he needs to know if its worth looking
further

- posted on December 31, 2005, 3:41 am

m Ransley wrote:

efficiency is how they calculate it.

My gas furnace is 80 percent which I assume means that the 80 percent of the BTU's in the gas end up in the house side of the heat exchanger, i.e, useable heat.

But I don't think they ever consider the heat loss from the the vents. How many BTU's are lost through the the two openings of 100 square inches each, one at the floor and one at the ceiling. The flow of air through these vents is continous 24 hours a day. The furnace may put 80 percent of the heat into the house, but how much heat is lost through those vent openings? The greater the difference between inside temperature and outside temperature, the greater the total heat loss. Maybe that is why they don't estimate that loss.

I could imagine that at 0 degrees, as much as 50 percent of the heat could be lost. In other words, at 80 percent AFUE furnace would actually result in only 40 percent of the heat retained in the house compared to Electric heat which would be 100 percent since no vents are used.

efficiency is how they calculate it.

My gas furnace is 80 percent which I assume means that the 80 percent of the BTU's in the gas end up in the house side of the heat exchanger, i.e, useable heat.

But I don't think they ever consider the heat loss from the the vents. How many BTU's are lost through the the two openings of 100 square inches each, one at the floor and one at the ceiling. The flow of air through these vents is continous 24 hours a day. The furnace may put 80 percent of the heat into the house, but how much heat is lost through those vent openings? The greater the difference between inside temperature and outside temperature, the greater the total heat loss. Maybe that is why they don't estimate that loss.

I could imagine that at 0 degrees, as much as 50 percent of the heat could be lost. In other words, at 80 percent AFUE furnace would actually result in only 40 percent of the heat retained in the house compared to Electric heat which would be 100 percent since no vents are used.

- posted on December 31, 2005, 4:03 am

When you say vents do you mean exuast up the chimney, Im sure its
calculated in in Afue rating. The big increase in condensing units is
the powered vent, exterior air, and of course second exchanger.
But an interesting fact is the efficiency difference between an 80% and
95% furnace is not 15 % its an 18.75% increase, time to think again.

- posted on December 31, 2005, 7:19 am

m Ransley wrote:

The vents are for combustion air and ventilation.

18.75 percent increase in efficiency.

If you go the other way, i.e., drop from 95 to 80, it is 95-80 = 15, and 15/95 is .1666, or a drop in efficiency of 16.66percent.

The vents are for combustion air and ventilation.

18.75 percent increase in efficiency.

If you go the other way, i.e., drop from 95 to 80, it is 95-80 = 15, and 15/95 is .1666, or a drop in efficiency of 16.66percent.

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