Posted by *Curbie* on April 12, 2009, 4:10 pm

Is there a standard formula used to calculate heat gain from

electrical devices. When considering devices like an oven, OR an

incandescent light, OR a drill press, it wouldn't seem so, but I

really don't know?

Thanks

Curbie

Posted by *RamRod Sword of Baal* on April 12, 2009, 8:06 pm

*> Is there a standard formula used to calculate heat gain from*

*> electrical devices. When considering devices like an oven, OR an*

*> incandescent light, OR a drill press, it wouldn't seem so, but I*

*> really don't know?*

*> Thanks*

*> Curbie*

--------------------

Lights 3.4 BTUs per watt per hour or 3412 BTUs per Kilowatt per hour

Electrical appliances that heat 3.4 BTUs per watt per hour or 3412 BTUs per

Kilowatt per hour x time on

Note fluorescent light wattage should be multiplied by 1.25 then multiplied

by 3.4 BTUs per watt to include the heat from the ballast

IE a toaster is maybe 2 KW but only runs for say 4 minutes, hence 2 Kw for 4

minutes would around 455 BTUs

Electric motors in room

1/4 HP efficiency 64% = 640 BTUs

1/2 HP efficiency 70% = 1,280 BTUs

1 HP efficiency 79% = 2540 BTUs

These numbers are for 1 hour operation. Again you need to know how long the

motor is going to run for, 1/2 an hour would indicate 1/2 the BTUs

Ovens are Kw multiplied 3412 that you need to know how long the actual

element is on for in that hour, as the thermostat switches the element on

and off.

So if you have a 2 KW element and it is on for 1/2 the hour, then it would

be around 3412 BTUs. Time on for the element will change as to how high you

have the thermostat set and how often you open the oven door.

There would be some variation if the kitchen is very cold, compared to a hot

summers day. It is better to over estimate the heat gain than under estimate

it if you are looking at what size cooling equipment that is required.

It is NOT a good idea to wildly overestimate the cooling load, as it can

effect the humidity in the conditioned space.

IE do not install a 5 HP system if the load indicates 3 HP.

In a home I find it better to let the kitchen heat up a bit rather than use

the full load of the cooking appliances in selection of the cooling

equipment, as most of the time these appliances are not running, so it is

easy to oversize the cooling equipment if sizing up the cooling unit for all

day operation

Usually one does not have all the cook top elements on at once plus the

oven, yes it may happen at Christmas, but is it worth a few thousand dollars

extra to cool the kitchen for one day a year. Again you must consider not

over sizing the system for the times the cooking equipment is not running.

A good reputable air conditioning company should be able to advise you.

Mind you the cheapest (and smallest) is not usually the best selection when

getting quotes.

I am assuming here you are looking at a ducted system that does more than

one room, but the heat loads are still accurate for single rooms.

Posted by *Curbie* on April 12, 2009, 9:27 pm

THANKS!

Just what I was looking for, I think I have human, electrical, and the

last is solar heat gain. I won't know exactly what I need there until

I finish some TMY data conversions which should take a few more days,

Know anything about solar heat gain?

Thanks a million.

Curbie

*>Lights 3.4 BTUs per watt per hour or 3412 BTUs per Kilowatt per hour*

*>Electrical appliances that heat 3.4 BTUs per watt per hour or 3412 BTUs per *

*>Kilowatt per hour x time on*

*>Note fluorescent light wattage should be multiplied by 1.25 then multiplied *

*>by 3.4 BTUs per watt to include the heat from the ballast*

*>IE a toaster is maybe 2 KW but only runs for say 4 minutes, hence 2 Kw for 4 *

*>minutes would around 455 BTUs*

*>Electric motors in room*

*>1/4 HP efficiency 64% = 640 BTUs*

*>1/2 HP efficiency 70% = 1,280 BTUs*

*>1 HP efficiency 79% = 2540 BTUs*

*>These numbers are for 1 hour operation. Again you need to know how long the *

*>motor is going to run for, 1/2 an hour would indicate 1/2 the BTUs*

*>Ovens are Kw multiplied 3412 that you need to know how long the actual *

*>element is on for in that hour, as the thermostat switches the element on *

*>and off.*

*>So if you have a 2 KW element and it is on for 1/2 the hour, then it would *

*>be around 3412 BTUs. Time on for the element will change as to how high you *

*>have the thermostat set and how often you open the oven door.*

*>There would be some variation if the kitchen is very cold, compared to a hot *

*>summers day. It is better to over estimate the heat gain than under estimate *

*>it if you are looking at what size cooling equipment that is required.*

*>It is NOT a good idea to wildly overestimate the cooling load, as it can *

*>effect the humidity in the conditioned space.*

*>IE do not install a 5 HP system if the load indicates 3 HP.*

*>In a home I find it better to let the kitchen heat up a bit rather than use *

*>the full load of the cooking appliances in selection of the cooling *

*>equipment, as most of the time these appliances are not running, so it is *

*>easy to oversize the cooling equipment if sizing up the cooling unit for all *

*>day operation*

*>Usually one does not have all the cook top elements on at once plus the *

*>oven, yes it may happen at Christmas, but is it worth a few thousand dollars *

*>extra to cool the kitchen for one day a year. Again you must consider not *

*>over sizing the system for the times the cooking equipment is not running.*

*>A good reputable air conditioning company should be able to advise you.*

*>Mind you the cheapest (and smallest) is not usually the best selection when *

*>getting quotes.*

*>I am assuming here you are looking at a ducted system that does more than *

*>one room, but the heat loads are still accurate for single rooms. *

Posted by *Neon John* on April 12, 2009, 11:01 pm

I stopped reading that BS when he got to the part that made

fluorescent lights generate more heat than electrical input.

The answer to your question is simple. With the one exception of a

heat pump moving heat from the outside, the heat gain of any

electrical device is exactly 1. One watt of heat out for one watt of

electricity in. To convert to BTU, multiply watts by 3.415179 BTU/hr.

John

wrote:

*>THANKS!*

*>Just what I was looking for, I think I have human, electrical, and the*

*>last is solar heat gain. I won't know exactly what I need there until*

*>I finish some TMY data conversions which should take a few more days,*

*>Know anything about solar heat gain?*

*>Thanks a million.*

*>Curbie*

*>>Lights 3.4 BTUs per watt per hour or 3412 BTUs per Kilowatt per hour*

*>>Electrical appliances that heat 3.4 BTUs per watt per hour or 3412 BTUs per *

*>>Kilowatt per hour x time on*

*>>*

*>>Note fluorescent light wattage should be multiplied by 1.25 then multiplied *

*>>by 3.4 BTUs per watt to include the heat from the ballast*

Posted by *Fred F.* on April 13, 2009, 1:33 am

Neon John:

Here is a question (it might be a dumb question), From your answer you

a basically saying that all electricity is converted to heat, is there

anything else that electricity is converted to or is 100% of wattage

input end up as heat?

*>I stopped reading that BS when he got to the part that made*

*>fluorescent lights generate more heat than electrical input.*

*>The answer to your question is simple. With the one exception of a*

*>heat pump moving heat from the outside, the heat gain of any*

*>electrical device is exactly 1. One watt of heat out for one watt of*

*>electricity in. To convert to BTU, multiply watts by 3.415179 BTU/hr.*

*>John*

*>wrote:*

*>>THANKS!*

*>>*

*>>Just what I was looking for, I think I have human, electrical, and the*

*>>last is solar heat gain. I won't know exactly what I need there until*

*>>I finish some TMY data conversions which should take a few more days,*

*>>Know anything about solar heat gain?*

*>>*

*>>Thanks a million.*

*>>*

*>>Curbie*

*>>*

*>>*

*>>>Lights 3.4 BTUs per watt per hour or 3412 BTUs per Kilowatt per hour*

*>>>Electrical appliances that heat 3.4 BTUs per watt per hour or 3412 BTUs per*

*>>>Kilowatt per hour x time on*

*>>>*

*>>>Note fluorescent light wattage should be multiplied by 1.25 then multiplied*

*>>>by 3.4 BTUs per watt to include the heat from the ballast*

> Is there a standard formula used to calculate heat gain from> electrical devices. When considering devices like an oven, OR an> incandescent light, OR a drill press, it wouldn't seem so, but I> really don't know?> Thanks> Curbie