Calculating the time to heat up a room 4

The equation you used is for Joule heating; it does not include anything to do with transient behavior. Your R values are tied in with a power equations Q = h*A*delta_T It's like filling a tank that has a small hole at the bottom. Your equation deals with how much water is required to fill the tank. The power equation deals with the leak.

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If i understand your explanation correctly, does that mean that the heat loss of the room that i have calculated, based on Q=U*A*delta_T, is actually the heat that is losing in the system where my 1000 W heater is how much heat is added into the system?

In this case, the total heat transfer in the system will be the delta_Q = Q_Joule Heating - Q_power equation.

But to determine how much time it will reach to the desired temperature, the Joule heating equation still applies, right?

I guess what i am trying to say is that to find the time for the heat to reach to the desired temperature, the following equation is derived:

Q [w] = m [kg/s] * cp [J/kgK] * delta_T [ºC] ====> rearrange this equation

t = (cp * Volume of the Room * density of air * delta_T)/Q where Q is the total heat transfer of the system

heating up a room is one thing, while heating up room air is another thing.

if you want to heat up a room, not only room air, you need to heat up walls, floor, ceiling to the same temperature. mass of building elements counts much.

that is non-stationary process, while equations mentioned apply to stationary state, and seemingly do not include time as variable.

the calculation is quite complex, but you do not need it at all when you have both heater and room defined... you just need to measure time.
if you still want to calculate it, you can refer to some heating load design standard. the simplest recommendation i know about is to add 30% provision to value rectified from heat loss calculation to have "acceptable" warm up time. the more complex recommendations include tables which give values according to many parameters - mass of building elements, desired warm up time, lowest outdoor temperature and similar.

Heat-loss calculation is for to keep up the room at design temperature. It is necessary to add %7-30 for start heating, depends on non-heating time. You need specific heat of the materials in room and room construction, for accuracy in calculation.

You need to take a course in heat transfer.....this is not something you can learn on a web forum.

thread403-232882

It's about, excess power. As noted above, if everything is cold soaked you need to consider that also.

Thank you for all the suggestion to this problem. I know it's not a easy problem to solve and I just want to get a brief idea of where the logic of solving this type of problem.

MintJulep, thanks for the link. From that thread, that equation is probably the same with the one i posted. Thank you for confirming that.

"From that thread, that equation is probably the same with the one i posted"

No, it's not, MJ clearly stipulated:

"Qx is the quasi-instantaneous excess capacity of your cooling system."
"Qx is rather more difficult. You could get a reasonable feel for the system's cooling capacity from the compressor's performance curves. The excess capacity is the difference between that and the cooling load at the same conditions"



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I apologize for the lack of knowledge for quasi-instantaneous excess capacity. I have never heard of that term before.

Basically if i understand it correctly, in my case, the excess capacity of heating is the difference between the heating capacity of the heater and the heating load.

Can you elaborate what you mean by "same conditions"? same humility of air? same pressure? same temperature? etc.

The excess capacity is a function of temperature, humidity, pressure, etc., since the heat loss is a function of same.

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i don't know where that formula comes from, but it does not seem to be correct to the best of my knowledge.

as mentioned before, the mass of room objects and surrounding counts much, but is totally ignored in the mentioned formula.

that is even more influential in cooling load calculations which is referred in formula originally. all cooling loads are calculated with time delay to take masses into account. in stationary heating state that is not so important as sun moving over the building does not change load so much as in cooling, but in warm-up heating it counts.

even if you ignore room heating as a concept and focus on room air heating only, the equation still does not apply as every degree you are coming closer to your desired temperature, rate of heat exchange drops accordingly.

an equation should include time variable if time is to be calculated.

@Drazen:

It came from my brain, and is clearly identified as a first order approximation.

You are absolutely correct about the effect of the mass of things in the space other than air, and about the need for time.

"Quasi-instantaneous excess capacity" is a time-dependent function, or more correctly it is a condition-dependent function, with conditions varying with time. It could be defined with a differential equation, or you can perform a calculation every second, or minute or hour or whatever as appropriate for the system of interest.

I understand the point about the effect of the mass of things in the room. It does make a change regarding to the heat exchanging rate. However, just want to consider the empty space first because i assume it's easier to be calculated. Of course, if there are stuffs which need to be heated up in the room in a cold environment, the amount of time to reach to the temperature will change or possibly rise.

What I am still confused about is the relationship between the heat that is generated by the heater versus the heat loss/gain. As IRstuff has mentioned, it seems to be like an example of filling up a box with a small leakage. The amount of heat loss in the room is pretty much like a "negative Q". And the heater is the amount of heat "+Q". However, I have a feeling that it cannot be just the energy in and energy out situation.

As Drazen suggest, a calculation of a standard heating load design is performed. The heat loss that is calculated is probably loosing heat since the system is being heated.

@MintJulep, thanks for explaining the term of "Quasi-instantaneous excess capacity". I will look into the differential equation of this term.

Are you an ME at all? Even if you didn't take thermo in college, the concept presented here is barely at the level of Calculus I.

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I see, Mint, that could be something like framework for further refining.

Currently I use new thermal calculation software and there is one feature which is new for me "heat up time". Earlier, heat up was either taken as rule-of-thumb addition or recommendations are given on bases of masses and fixed "usual" heat-up time, which was 2 hours most of time.

I am not familiar with algorithm, though. It is very likely I will keep using rule-of-thumb until learning how it works.

Mechwill,

You are on the right track with Q = m Cp delta T, however, everyone that has commented is also correct in that what you are looking for is not a simple thing to calculate.
Let me first offer a correction to your equation so it reads Q=M*c_p*⌂T where Q and M are not time dependent. Then you have q = m*c_p*⌂T in which q and m are time dependent and q = Q/t and m = M/t. Then when you are asking how long it takes to heat a room, you are looking for a value for t which can be found with t = M*c_p*⌂T/q. Where M is in kg, c_p is kJ/kg*K, ⌂T is in °Celsius or Kelvin and q is kJ/t.
Note here that kW is a unit of power which is energy per unit of time; 1 kW = 3600 kJ/hr.

As others have mentioned earlier, while you are adding heat to the room with your heater, you are also loosing heat through the walls, windows, doors and infiltration. So q is really q_net = q_in - q_out. Where q_in is your 1kW heater + heat from lights + body heat (if your sitting in the chamber) + heat from a stove and other appliances + heat from the sun (assuming it is up). And q_out is found roughly with q=U*A*⌂T; where U is 1/R-value of your walls, windows, doors and roof; A is the area of the walls, windows, doors and roof; ⌂T is the temperature difference between the inside of the chamber and outside air.

Doing some conversions and using typical values for the constants in the equation above I came up with:

t = V*⌂T/(49.6*q_net)

Where:
t = minutes
V = Volume of space in m³
⌂T = change in temperature in °Celsius or Kelvin
q_net = q_in - q_out = q_heater - U*A*(T_outside - T_{inside,average}) in kW

If you ignore the heat loss through the walls, doors and such, then add 30-50% to the results of this equation, you should be in the ballpark.

Stay warm.

@dbill74

Thank you for the detail explanation to confirm with the approach of solving this problem. I was just confused of the relationship between the heater and heat loss. I know the concept of it, but I don't have the confidence in my own concept.

@IRstuff

Sorry if I have upset you at any point or ask too many questions. I did take a course in thermodynamic in college and I am a mechanical technician not an engineer. Just want to make sure that my understanding is correct. One thing about calculus is that I know how to work with the equation and understand the equation but I don't know how to apply the calculus with real life situation accurately. Once again, I apologize for upsetting you at any point.

Thanks for all you guys' help. I greatly appreciate all the helps.