AC recovery cost

Roughly, 10,000 watts(heat) = 3300BTU. Do the math from there.

A heat pump can move heat more efficiently than heat strips can make it.

More precise numbers are not significant since it is only an estimate anyway.

-----------------------SSB-----------------------------------------

All the energy used by a light bulb ends up as heat. So 100W bulb is equivalent to a 100W heater. for practical purposes the same goes for a refrigerator, oven, etc.

These posts have not answered the question.

You are quite correct; all heat input to a building has to be removed, which includes lighting, computers, people, even mobile phone chargers. The power required is determined by the Coefficient of Performance (CoP) of your a/c unit. A reasonable average is 3, so 100watts of heat will take about 33watts of a/c power.

Now this is where the fun starts. The CoP is quoted by the manufacturer and is not unlike the fuel consumption of a vehicle. The manufacturer wants as high a number as possible to put on advertising etc and the vehicle will be designed to perform optimally at the conditions used for standardised testing. Similarly the CoP quoted by manufacturers is set in laboratory conditions and seldom achieved in practice. Domestic units can go as high as 4, or as low as 2. All assuming of course that the unit is well maintained.

In an office environment, with lighting on all day, lighting accounts for a significant a/c load, more than the contribution from people themselves.

BabyGuinness,

On the eastern side of the Pacific, the CoP refers to the heating capability of the heat pump--its ability to move heat from the relatively cold outside to the relatively warmer inside. The energy efficiency ratio (EER) is its ability to move it from the relatively comfortable inside to the relatively hot outside. Although each is pumping heat against a thermal gradient, the air conditioning side has a higher average efficiency than the heating side. The seasonal energy efficiency ratio (SEER) is based on a defined range of temperature differences for a defined set of time intervals. Current energy efficient A/C units can operate with SEER's as high as 14; average older units are as low as 6.

The typical number in many energy conservation manuals is that for every unit of excess energy introduced into a dwelling unit you will spend 1/3 of a unit to remove it. This is the ratio your post suggests. However, the above numbers suggest that the actual expenditure will be lower--on the order of 1/6 or somewhat less.

In very energy-efficient structures, the internal gain from lighting, appliances, equipment, and occupants is enough to supply a significant percentage of the total heating needs and consume a significant percentage of the total cooling needs. Since the typical electrical utility experiences its peak loads during the cooling season, any steps taken to reduce this load will have the greatest environmental benefit. Such steps include optimizing the use of natural lighting, the most efficient artificial lighting available, higher-efficiency appliances and equipment, etc.

In high-rise buildings, the heat produced in the core of the building is often enough to offset the heat loss on its perimeter, so well-designed HVAC systems are simply moving the heat from one area to another (poorly-designed systems are treating the areas as separate heating or cooling loads).

In some super-insulated houses, the internal gain is enough to supply all heating needs down to an outside temperature below freezing. The predominant HVAC load in such structures is for cooling. Few HVAC contractors are prepared to analyze such structures accurately, because their computer models make poor assumptions regarding the heat requirements of such buildings. In homes with large thermal mass, the problem is even worse because the whole heat loss paradigm is based on the assumption that heat loss is directly proportional to the instantaneous temperature difference across the insulating barrier--which is not true when the mass isolates the outside from the inside. The typical calculations and tables of R-values have little meaning in this type of construction; the calculated heating/cooling loads can be off by as much as a factor of 2.

--JMM

Hi Jmueller

Careful there you are mixing up your units. SEER & EER are American terms and confusingly mix BTU/hr with watts in the one ratio. Like US v imperial gallons, this of course is irrelevant as long as you compare like for like. But CoP is dimensionless, comparing watts for watts.

From memory I think all new US aircon has to be above 13 EER. Converting BTU/hr to watts you divide by 3.41, so EER 13 = CoP 3.81. But as I indicated this is an idealised maximum.

Seasonal EER (SEER) is a further adjustment to reflect real working conditions. These figures are not used outside the US. CoP is used for both heating and cooling, and 2 figures should therefore be quoted.

Here in NZ aircon is used for both heating and cooling (i.e. as heat pumps). CoP for cooling is around 3, for heating is around 2, as the compressor losses are created outside. Heating CoP drops to 1 as the temperature approaches zero outside.

With respect to heating effect of lamps, Andy Germany is spot on! Tungsten lamps are about 95-97% inefficient. However this is irrelevant; all energy manifests itself as heat eventually. Light is absorbed by the environment which heats up as a result. (A very small percentage of visible is lost through glazing to the outside, but vene IR is reflected back in).

Rather more than it would take to recover the heat from a low-energy replacement light source (the 100W incandescent light bulb is becoming extinct in the European Economic Community).

• Imagine the concept: heat the building with 100W light bulbs, then cool it using air conditioning! Why not cut out the middle-man?

The less heat you have to remove the better, so replacing as much "hot" lighting with "cool" lighting the better.

The modern "cool" lighting generates more light with far less heat for the same amounts of power and the AC has less work to do in getting rid of of the smaller amounts of heat.....

Even if everything was 100% efficient it would be bad enough, but with all the losses involved as well in getting rid of extra heat, you simply do not want to go there......

Some years ago I remember I was working in our bathroom on a stepladder, where 6 halogen lamps of each around 40-50 watts were doing there job. The air near the (insulated) ceiling was "painfully" hot. I was shocked. Since then I have 4 x 40 watt CFLs and 2 x 3 watt LEDs. The CFLs take a few seconds to get going so the LEDs supply immediately enough light to stop anyone falling over in the dark!!!

Since then, the ceiling area shows no discernible difference in temperature, except when the central heating is on.....

I have only got AC upstairs in our bedrooms, not on the ground floor, it is seldom needed in Germany all over a house!! But as warm air rises and cold falls, the effects downstairs are still felt if and when needed....the house is insulated far far better than most others that I know of here and we notice it working well for us in hot summers and cold winters.

Believe the conversion from watts to BTUH is 3.414 BTUH/watt, but to figure the amount of power needed to compensate for different heat sources, you need to know the efficiency of the source. Believe light bulbs (normal incandescent) are about 30% efficient in producing light, so the rest goes to heat.....for 100W bulb, then, heat is 70W or so, and this produces about 240 BTUH if left on for an hour. Then, you have to also figure the efficiency of the air conditioner in counteracting that heat. A good conversion these days is about 14 BTUH per watt , so it takes about 17 watts of air conditioning power to deal with the 70 watts from the 100 watt light bulb (assuming 100% delivery efficiency--but you should assume maybe 80%, so use about 21 watts of airconditioning power for the 240 BTUH).

Different heat sources have different efficiencies.....hair dryer's intent is to put out heat, BUT some also goes to infrared/red light, so that isn't totally efficient either, but it's probably better than 30%, etc. Pretty much the same for ovens, toasters, etc.....toasters are from 600-1000 watts, ovens can be up to 3000 watts....do the math. Where possible, it's best to exhaust the unwanted heat outside with a fan than try to cool your way down.

A normal 100W bulb uputs out 95% heat........a halogen is about 1/2 % better or so (from memory....!)

Hi h2om.

1. Sum all lighting power.

2. Sum all heating power, like a toaster, hair dryer and other devices with electric resistors

3. Sum all TV's, PC's, HiFi's, etc.

4. Multiply the refrigerator compressor power by three (if it's a good one)

5. Add 50 Watts for every person inside (if they're not running or making exercise)

Finally, add results 1 through 5 and you will get the maximum Thermal Power sources inside. The electrical power is roughly 1/3 the thermal result.

Now, think of the ones which will be working at the same time and re-calculate.

Also, be aware that the thermal insulation of the room and the temperature difference between in and out will be very important.

Cheers,

RGO