How long can I heat up a container?

[mechwill]

hi guys,

I have a little cabin in my backyard. Recently, i have dug out a 1000 Watt air heater and installed in the cabin.

I like to find out how long i need to run the heater to warm up to the room temperature of 21 ºC from the ambient of 12 ºC. Either the heater needs to be bigger or let it run longer to heat up the room

I initially start the calculation with this equation Q = m Cp delta T. Re-arrange the equation and it becomes m = Q / (Cp * delta T). Once i have find the mass flow rate, i know the volume of the room and the density of the air. I can calculate the time right?

However, one of my main concern is about the insulation and the infiltration which changes the time to heat up the room right? but that equation does not state any U value or R value. so... did i do something wrong or use the wrong equation?

Thanks

 

[Danger]

Welcome, Mechwill.
Not being educated, I would take this to be a trial-and-error problem. While formulae are a great idea, they don't always reflect reality. I see a couple of problems. One is that you claim to know the density of the air. Have you accounted for the fact that it will change as a function of the temperature rising? Secondly, and to me more importantly, is that since heat rises you will be filling up the top of the room with warm air first, which is of no use to you. While I recommend that you wait for someone knowledgeable to respond, my first thought is that you might increase your efficiency by installing a ceiling fan to blow the warm air back down.

 

[mechwill]

Welcome, Mechwill.
Not being educated, I would take this to be a trial-and-error problem. While formulae are a great idea, they don't always reflect reality. I see a couple of problems. One is that you claim to know the density of the air. Have you accounted for the fact that it will change as a function of the temperature rising? Secondly, and to me more importantly, is that since heat rises you will be filling up the top of the room with warm air first, which is of no use to you. While I recommend that you wait for someone knowledgeable to respond, my first thought is that you might increase your efficiency by installing a ceiling fan to blow the warm air back down.

I haven't considered that option until you've mentioned it. That way you will circulate the hot and cold air in the room.

The reason why i am asking this is because that i want to know how much electricity i am paying to have that heater running. Maybe if i know the time to heat up the room, i can size down the heater and add a fan on top of it to save some money.

Thanks for the suggestion

 

[Danger]

There's no need to down-size the heater, especially since you already have one that you don't have to pay for. Just lower the duty cycle so it doesn't run more than necessary. That's what thermostats are for. It should have one built in, but if it doesn't you can nab one for 5 or 10 bucks at a hardware store. It won't use any more energy than a smaller unit running more frequently.
You could always supplement it with a good old-fashioned potbelly stove or build a nice little fireplace.

 

[etudiant]

Your heater is conveniently sized to use 1 kwhr per hr.
You therefore can find out very quickly how much you pay to run your heater, simply multiply your electric $ cost/kw hr shown on your electric bill by the number of hours the heater runs.
Separately, depending on where you live and the size of your cabin, a 1000w heater is unlikely to provide much in the way of space heat. If it has a fan built in. it may be enough for a warm spot near your workbench.

 

[Danger]

a 1000w heater is unlikely to provide much in the way of space heat. If it has a fan built in. it may be enough for a warm spot near your workbench.

That's more a function of the insulation than anything else. If it's very efficient, 1KW should be fine. (I suppose that I should qualify my comments by pointing out that I'm Canadian. Once our outdoor temperature hits anything over 5°, we're out on the lawn in our T-shirts slugging beer and ogling the neighbourhood girls in their bikinis...)

 

[Travis_King]

You know the volume of air in the box, and the density won't change all that much from 12 C to 21 C, so you know the mass of the air in the box (to the accuracy you'd need). At 10 C air is more dense than at 20 C, so use that. It's 1.247 (kg/m3)

You know the specific heat capacity of the air in the box, since it's pretty much the same at those temps (at typical room conditions, Cp=1.012)

Your delta T is obviously just 21 C - 12 C (or more correctly 294.15 K - 285.15 K) = 9 degrees C.

This gives you an estimate of the heat requirement requirement. Since you know your power in wattage, you just take the heat requirement and divide by your power available and you'll get how long it will take to heat up. Then multiply that by your electricity costs.

Then, of course, you'll need to do some estimation on how much it energy it will take to maintain that temperature.

 

[mechwill]

You know the volume of air in the box, and the density won't change all that much from 12 C to 21 C, so you know the mass of the air in the box (to the accuracy you'd need). At 10 C air is more dense than at 20 C, so use that. It's 1.247 (kg/m3)

You know the specific heat capacity of the air in the box, since it's pretty much the same at those temps (at typical room conditions, Cp=1.012)

Your delta T is obviously just 21 C - 12 C (or more correctly 294.15 K - 285.15 K) = 9 degrees C.

This gives you an estimate of the heat requirement requirement. Since you know your power in wattage, you just take the heat requirement and divide by your power available and you'll get how long it will take to heat up. Then multiply that by your electricity costs.

Then, of course, you'll need to do some estimation on how much it energy it will take to maintain that temperature.

I think in order to maintain at the same level of temperature. Heat loss of the building needs to be performed. The calculation starts off with calculating the heat loss through the walls, roof, floor, windows, door, infiltration. The total amount of the heat that is loss from the room is known. Since I have 1 kW heater, which is essentially the heat source, the net heat is known. Finally, I can use the equation that you suggest to estimate the amount of time that is required to heat up the room.

Not sure if i am missing any step here. Thanks for the clarification and the step by step explanation

 

[mechwill]

That's more a function of the insulation than anything else. If it's very efficient, 1KW should be fine. (I suppose that I should qualify my comments by pointing out that I'm Canadian. Once our outdoor temperature hits anything over 5°, we're out on the lawn in our T-shirts slugging beer and ogling the neighbourhood girls in their bikinis...)

Now i see what you do with your spare time. haha