I have an outdoor pond. I have to keep a hole in the ice so that my fish will survive the winter. The pond store sold me a small heater for this purpose.
The heater is rated at 1250 watts. It would run pretty much continuously through the winter (it only turns itself off if the temperature rises above 35 degrees F).
To save of electricity costs, the ever friendly pond store wants to sell me a $77 fountain-like device to keep a hole in the ice, and only use the 1250 watt heater when it gets below about 10 degrees F).
I don’t know how to do the cost benefit calculation. If the heater would cost me an extra $15 in electricity a winter, I probably wouldn’t bother with the other device. However, if the heater would cost me an extra $50, then it’s a different story.
(I assume I’d be able to find out my local electric utility’s rate per kilowatt).
Electricity is generally priced in KWh which is Kilo Watt hours. Your heater is a 1.25 KW heater times 24 hours times 30 days is about 900 KWh. Which I might add is a huge amount of electricity. My wife and I use about 350 KWh a month for the both of us. 7 to 15 cents a KWh is about what people pay for electricity in the US so you are looking at spending $63 to $135 a month for the heater. The real question is how much the heater will be on in the second case. You don’t really give much information about that.
Thanks scotth and gazpacho for your quick response.
I guess I could’ve done this myself if I’d have known that the appliance rating was in watts-per-hour. Then, as you both show, the calculation itself is quite simple.
I learned something today!
Well, actually two things… I also learned that I need to buy that $77 thing (rated at only 100 watts).
No, no, no. Electrical service is charged by the kilowatt-hour; there’s no such thing as watts per hour (at least not in this business). You multiply watts times hours to find the total amount of energy used.
An energy flow of 1250 watts is 1.25 kilowatts. Run it for an hour, and it has consumed (hours x kilowatts) of electrical energy, or 1.25kilowatt-hours. Or 30KWH over 24hrs, or 900KWH over 30 days. If one KWH costs between 6 and 8 cents, then the heater eats up between $1.80 and $2.40 per day ($54 to $72 per 30 days.) Expensive, but remember that a guy on a treadmill only puts out around 100 joules per second of energy, so you’d need to pay twelve people running on treadmills to keep your pond hot. Or feed a couple of ponies.
PS
If this sort of calculation is hard to understand, that’s because it’s inside-out! When energy flows from place to place, the flow is measured in Joules per second. But we’ve renamed the joule-per-second. We’ve called it the “watt.” If we start dealing in watts of energy flow, and never mention the energy that’s flowing, then we’ll come up with some weird math:
The fundamental unit of energy is no longer the joule, instead it is the watt-second (watts times seconds.) One joule IS one watt-second. That’s because one watt is really one joule per second, or one joule/sec. To convert energy flow back into energy, multiply the energy flow in joules/sec by the number of seconds the flow has gone on: joules/sec x sec = joules. Since a watt is really just a joule/sec, that’s why we have to MULTIPLY watts by seconds to find out the energy. Beware the common mistake of thinking that kilowatt-hours are kilowatts PER hour. A KWH is really kilowatts TIMES hours.
When I wrote “watts per hour” I was referring ONLY to the concept of what the number stamped on the appliance meant. It means that the appliance consumes x watts per hour. Right?
To put it another way: Power (what’s measured in watts) is energy per time. The utility company doesn’t actually charge for power, they charge for energy. To get energy from power, you need to multiply your power by the amount of time that you’re using the power. Physicists usually multiply a watt by one second, to get their energy units (Joules), but that’s an inconvenient unit for the utility company, so they instead measure by the hour.
I think the calculations by gazpacho are correct. Although, why the device has to run 24/7 is questionable, since the thermostat cuts off once the temperature is 35 deg F. I think the following need consideration too :
1> The density of water is maximum at 4 deg C (39 deg F). That is why though water at the top is nearly at 32 deg F , the bottom is like 39 deg F.
2> The power consumption by the heater will be a function of the exposed area and the mass of water in the pond. I doubt it will be 1250kW continously. It can be calculated if more info was available on the tank itself.
3> Putting a fountain like device will help increase the heat transfer coefficient. So your pond will cool down faster now, even to maintain 10 deg F you may end up spending more power then maintaining the temp. at 35 without the fountain.
4> The fountain itself consumes power, which needs to be considered too.
5> Say you lose power for a few hours (or a kid turns the power off the fountain) and the top of the pond freezes in the meantime, you’r fountain cannot revive it. Whereas, if you were running your heater at 35 deg F, you’d have a better chance of this not happening.
6> I don’t know if this pond has a shelter, you may have to think of the fountain hitting the shelter too, if its the case.
7> Finally, won’t an iron pipe wrapped with some heating tape work to give you a constant air hole on the pool ? What about a small air sparger into the tank ?
I have no experience with fish ponds. The above are just some technical considerations.
2> The power consumption by the heater will be a function of the exposed area and the mass of water in the pond. I doubt it will be 1250kW continously. It can be calculated if more info was available on the tank itself.
The heater is not intended to heat the whole pond. Just enough to keep a hole at the surface. The pond is about 1800 gallons. In a thread last summer I was calculating the amount of water lost by evaporation, and, IIRC the surface area was about 75 square feet.
3> Putting a fountain like device will help increase the heat transfer coefficient. So your pond will cool down faster now, even to maintain 10 deg F you may end up spending more power then maintaining the temp. at 35 without the fountain.
4> The fountain itself consumes power, which needs to be considered too.
The “fountain” is more of a vertical recirculator. It just keeps the water moving enough to keep a hole about a foot in diameter open. The recirculation only goes down about 10 inches (the unit would sit on a “shelf”) so that the water at the bottom stays “warm”. This unit is rated at 100 watts, so it would consume far less electricity.
5> Say you lose power for a few hours (or a kid turns the power off the fountain) and the top of the pond freezes in the meantime, you’r fountain cannot revive it. Whereas, if you were running your heater at 35 deg F, you’d have a better chance of this not happening.
I’d still have my heater handy though. The recirculation unit will only be effective down to about 10 degrees F. If it freezes over, I can set the heater on the ice to melt a hole.
6> I don’t know if this pond has a shelter, you may have to think of the fountain hitting the shelter too, if its the case.
It won’t shoot water up into the air… just keep some water at the surface moving continuously so it doesn’t freeze.
7> Finally, won’t an iron pipe wrapped with some heating tape work to give you a constant air hole on the pool ? What about a small air sparger into the tank ?
Air sparger? Another option is an aerator which generates bubbles. Has the same purpose as the recirculator. I need a bigger hole than heating tape would provide. Even if I wrapped pipe that was in a circular shape, I think the inside of the circle would freeze anyway.