Heat transfer questions

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hello.

I'm kind of conducting a very simple study at work on heat transfer to try to compare 2 different heating systems of enclosed spaces but I'm having a bit of a struggle to understand the meaning of some of the values I'm getting and I'm seeking some help to see if the values makes sense in the context!

So, the idea is to use 2 formulas (Q = m*C*ΔT and Qloss = A * ΔT / Rt) to calculate energy needed to heat up a place and to calculate the heat losses so that, in the end of the day I can look to the values and say something like:

- with system A1 I need X1 amount of energy and it will cost me Y1€ to run it over a given period of time.
- with system A2 I need X2 amount of energy and it will cost me Y2€ to run it over the same period of time.

I've been working on this using a spreadsheet of course.
So, to start I need to give some context.
I'm running these formulas to calculate energy needed to heat up a room which I'm going to assume the following properties:

A is area = 101.49m² (ignoring the floor)
V is volume = 97.89m³ (ignoring the floor)

Then, the room wall is composed of 2 walls of plasterboard and an air gap. Dimensions are as follows:
dp is thickness of plasterboard = 0.01m (x2 walls of it)
da is thickness of air = 0.04m

To calculate energy needed to heat this space up, I have considered two different temperature targets to account to colder and hotter time frames.
My target temperature is Tin, which is the temperature I want to keep constant inside the enclosed space.
Tout1/2 is the ambient temperature which I am considering two fo them, so this will give me 2 different ΔT, ΔT1 and ΔT2.
So:
Tin = 17ºC
Tout1
= 3.8ºC
Tout2
= 10.4ºC
ΔT1
= 13.2ºC
ΔT2
= 6.6ºC

So, I get the following results:
1712487614003.png

What I'm not sure if I'm doing correctly is for the heat losses. The formula I'm using gives me Watts [W] or Joules per second [J/s] which is a rate. So, to calculate the heat losses over a day, I should simply multiply the results of Qloss by 86400 seconds which is the number of seconds of a day and finally add that value to the heat needed to initially heat up the space?
 

MrAl

Joined Jun 17, 2014
11,693
Hello.

I'm kind of conducting a very simple study at work on heat transfer to try to compare 2 different heating systems of enclosed spaces but I'm having a bit of a struggle to understand the meaning of some of the values I'm getting and I'm seeking some help to see if the values makes sense in the context!

So, the idea is to use 2 formulas (Q = m*C*ΔT and Qloss = A * ΔT / Rt) to calculate energy needed to heat up a place and to calculate the heat losses so that, in the end of the day I can look to the values and say something like:

- with system A1 I need X1 amount of energy and it will cost me Y1€ to run it over a given period of time.
- with system A2 I need X2 amount of energy and it will cost me Y2€ to run it over the same period of time.

I've been working on this using a spreadsheet of course.
So, to start I need to give some context.
I'm running these formulas to calculate energy needed to heat up a room which I'm going to assume the following properties:

A is area = 101.49m² (ignoring the floor)
V is volume = 97.89m³ (ignoring the floor)

Then, the room wall is composed of 2 walls of plasterboard and an air gap. Dimensions are as follows:
dp is thickness of plasterboard = 0.01m (x2 walls of it)
da is thickness of air = 0.04m

To calculate energy needed to heat this space up, I have considered two different temperature targets to account to colder and hotter time frames.
My target temperature is Tin, which is the temperature I want to keep constant inside the enclosed space.
Tout1/2 is the ambient temperature which I am considering two fo them, so this will give me 2 different ΔT, ΔT1 and ΔT2.
So:
Tin = 17ºC
Tout1
= 3.8ºC
Tout2
= 10.4ºC
ΔT1
= 13.2ºC
ΔT2
= 6.6ºC

So, I get the following results:
View attachment 319334

What I'm not sure if I'm doing correctly is for the heat losses. The formula I'm using gives me Watts [W] or Joules per second [J/s] which is a rate. So, to calculate the heat losses over a day, I should simply multiply the results of Qloss by 86400 seconds which is the number of seconds of a day and finally add that value to the heat needed to initially heat up the space?
Hi,

Are you saying simply that you are getting a result in Watts, and you need to know the total energy usage over one day?
Often the energy is calculated in Watt Hours or Kilowatt Hours. If you have a 100 watt light bulb lit for 20 hours that's 2000 watt hours or 2 kw hours.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hi,

Are you saying simply that you are getting a result in Watts, and you need to know the total energy usage over one day?
Often the energy is calculated in Watt Hours or Kilowatt Hours. If you have a 100 watt light bulb lit for 20 hours that's 2000 watt hours or 2 kw hours.
What I need to know is that if that second formula I'm using to calculate heat loss, gives me J/s, I can't simply add it to the other result I get from the first formula I use to calculate the energy needed for the initial heat. One is in Joules and the other comes in J/s. One is an asolute value and the other is a rate.

So, on one side, I have heat needed to inittialy heat up a given volume.
Q = m * C * ΔT and the units here is Joules
Qloss = K*A*ΔT/d
or Qloss = A*ΔT/Rt comes in J/s

So, my question is if I want to know the totall heat losses over a day I multiply the last formula result by the number of seconds in a day and then I can add that value to the one calculated by the previous formula?
 

MrAl

Joined Jun 17, 2014
11,693
What I need to know is that if that second formula I'm using to calculate heat loss, gives me J/s, I can't simply add it to the other result I get from the first formula I use to calculate the energy needed for the initial heat. One is in Joules and the other comes in J/s. One is an asolute value and the other is a rate.

So, on one side, I have heat needed to inittialy heat up a given volume.
Q = m * C * ΔT and the units here is Joules
Qloss = K*A*ΔT/d
or Qloss = A*ΔT/Rt comes in J/s

So, my question is if I want to know the totall heat losses over a day I multiply the last formula result by the number of seconds in a day and then I can add that value to the one calculated by the previous formula?
Hello again,

If you are getting two results for the same basic units then you should only have to go over the units for each one and see what went wrong.
If you get say 1.234 in feet per second and another 0.567 in feet and they are both supposed to be in the same units, then one is wrong and that should be found by going over each step to find out what was entered. Some quantity was not entered right or it was not assigned the right units to begin with.
The final units should be watt hours or similar. In home energy calculations usually kilowatt hours are used because that's usually how the bill comes out from the power company, in units of kilowatt hours.

You should be able to find it like this, but if not you would have to show each step and each result so myself or someone else could look it over.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Actually it's not a matter of wrong units. Or better, is it but it was not a mistake of mine. It was rather an unerstanding problem from me.
The formulas I'm working with are:
Q = m*C*ΔT [J]
Qloss = A * ΔT / Rt [W]

But then I need to be able to sum the two above to get total Energy needed, for instance, over a day.

So, what I did was to use the 1st formula to calculate the initial energy needed, in [J]
Then, I converted that value to [kWh] by dividing it by 3.600.000[J].

Then I calculated the rate of the heat loss in [W] or in [J/s] and converted it to [kWh] over a day by multiplying by 24h and dividing by 1000, so that I could add these two values to be able to get the quantity of energy needed over a day to heat up and mantain the space at the given temperature!

So, for instance, for a given space with 97.89m² of area I got a total of 2.425.248,14 J or 2,4252481 MJ.
Converting to kWh I get 0.6737 kWh.

Then, I calculated the Heat Loss rate over a day and I got a total of 1196.11 W. Converting to [kWh] I got 28.707 [kWh].

Now, that I have both values in [kWh] I can add them to get the total enegy needed over a day, if I did everything correctly.
I get 29.38 kWh.

Considering I have a heating device rated at 3500W, I used the following formula to calculate how mugh time I need to heat up the space initially:
W = Q / T <=> T = Q / W <=> T = 2.425.248,14 / 3500 = ~693s

I did this spreadsheet and I split the values in 2, to account for lower and higher temperature differences and at the end, I get the same 693s but split in two of 7m:46s and 3m:48s.

1713276239496.png

Does this makes sense?
 
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MrAl

Joined Jun 17, 2014
11,693
Hello again,

In line 10, what is "Heat Capacity [J]" and where did that come from?
In line 30, what is "Heat Loss [W or J/s] and where did that come from?
 

MrAl

Joined Jun 17, 2014
11,693
Hello again,

Oh I see what you did there, you were just reiterating the item and the units, that was not a minus sign that was just an 'indicator' like a colon.
"Heat Loss: 12345 Watts" => "12345 - Heat Loss (W)".

Ok I think I see what confused you now. In the original post you had used the idea of "per day" by using a division sign, which made the time factor look a little stranger.
In the second post with the info, you used the better description of "kWhr over one day". That's a better way of describing it.

So it looks much better now.

I did get slightly different numbers than you did so that's why I initially thought you were subtracting a small amount from the calculation and calling it something. I'm not sure why I got different numbers than you did, but they were only slightly different so maybe it was roundoff errors to blame, but you could check that if you like. I started from the first line and worked down, then another calculation starting with a later line and worked down, and got different numbers than you did, but again they were just slightly different.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hello again,

In line 10, what is "Heat Capacity [J]" and where did that come from?
In line 30, what is "Heat Loss [W or J/s] and where did that come from?
In line 10,Heat Capacity is the name I gave to te formula and the [J] is the SI Units.
In line 30 is the Heat Loss through the insulationg materials. That forumla is used to calculate total thermal resistance for a composite wall. Its SI units is W (Watt) or J/s.
Hello again,

Oh I see what you did there, you were just reiterating the item and the units, that was not a minus sign that was just an 'indicator' like a colon.
"Heat Loss: 12345 Watts" => "12345 - Heat Loss (W)".

Ok I think I see what confused you now. In the original post you had used the idea of "per day" by using a division sign, which made the time factor look a little stranger.
In the second post with the info, you used the better description of "kWhr over one day". That's a better way of describing it.

So it looks much better now.

I did get slightly different numbers than you did so that's why I initially thought you were subtracting a small amount from the calculation and calling it something. I'm not sure why I got different numbers than you did, but they were only slightly different so maybe it was roundoff errors to blame, but you could check that if you like. I started from the first line and worked down, then another calculation starting with a later line and worked down, and got different numbers than you did, but again they were just slightly different.
Yes, someting like <Symbol assigned to the value - informal description - [SI unitt]

But I still think that the time I'm getting to heat up that space is yet too short!
As you see in the spreadsheet, I have 2 different temperature differences. Those are to account for the colder and hotter parts of the day.
At line 7, you see 2 values. 3.8ºC for MTN and 10.4ºC for MTX. MTN is average minimum temperature for January from 1981 up to 2010. MTX is the average of maximum temperatures of the same time frame.
An for instance, for the MTN, I get only 10mins and 18sec to heat that space from 3.8ºC up to 17ºC ( or by other words raise the temperature by 13.2ºC), using only a 2500W heating device! I fin it too fast heating the space up. I expected longer time.
 

MrAl

Joined Jun 17, 2014
11,693
In line 10,Heat Capacity is the name I gave to te formula and the [J] is the SI Units.
In line 30 is the Heat Loss through the insulationg materials. That forumla is used to calculate total thermal resistance for a composite wall. Its SI units is W (Watt) or J/s.


Yes, someting like <Symbol assigned to the value - informal description - [SI unitt]

But I still think that the time I'm getting to heat up that space is yet too short!
As you see in the spreadsheet, I have 2 different temperature differences. Those are to account for the colder and hotter parts of the day.
At line 7, you see 2 values. 3.8ºC for MTN and 10.4ºC for MTX. MTN is average minimum temperature for January from 1981 up to 2010. MTX is the average of maximum temperatures of the same time frame.
An for instance, for the MTN, I get only 10mins and 18sec to heat that space from 3.8ºC up to 17ºC ( or by other words raise the temperature by 13.2ºC), using only a 2500W heating device! I fin it too fast heating the space up. I expected longer time.
Hi,

Oh I thought you only had a problem with some units or something like that.

Now looking again, I would ask, when you calculate the time to heat up initially, did you figure in the heat losses as well as the volume and specific heat capacity?

If you take a block of aluminum and find that, when fully and completely insulated, it takes 10 minutes to heat up to 10 degrees C above ambient, when you remove the insulation it is going to take longer to heat up to that temperature. In fact, once you remove the insulation, it may never be able to reach that temperature with a limited amount of heating power. If it heats up with 1000 watts in 10 minutes with the insulation, without the insulation if may take 20 minutes or it may require 1500 watts to reach the right temperature.
1000 watts with zero loss is 1000 watts, 1500 watts with 500 watts loss is still just 1000 watts.

You could check that out.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hi,

Oh I thought you only had a problem with some units or something like that.

Now looking again, I would ask, when you calculate the time to heat up initially, did you figure in the heat losses as well as the volume and specific heat capacity?
I'm not sure I understand the question!
But when I calculated the time to heat upt the space initially, I only considered that volume, air heat capacity, mass of air and the temperature difference. I used the values in Joules from B10 and C10 and divided this sum by 3600000 J. That returned 0.6737kWh. So, this is not accounting for any losses!

If you take a block of aluminum and find that, when fully and completely insulated, it takes 10 minutes to heat up to 10 degrees C above ambient, when you remove the insulation it is going to take longer to heat up to that temperature. In fact, once you remove the insulation, it may never be able to reach that temperature with a limited amount of heating power. If it heats up with 1000 watts in 10 minutes with the insulation, without the insulation if may take 20 minutes or it may require 1500 watts to reach the right temperature.
1000 watts with zero loss is 1000 watts, 1500 watts with 500 watts loss is still just 1000 watts.

You could check that out.
Yes, I understand what you mean. But this initial calculation doesn't take any time into account. While when calculating losses, I calculate the rate of loss, meaning Wh or kWh. So, I have 28.71kWh of heat losses over a day and I need 0.6737kWh to heat up the space by 13.2ºC.

Now, that calculation I did for time to heat up the space, I only took into account theenergy needed to initially heat up the space!
 

MrAl

Joined Jun 17, 2014
11,693
I'm not sure I understand the question!
But when I calculated the time to heat upt the space initially, I only considered that volume, air heat capacity, mass of air and the temperature difference. I used the values in Joules from B10 and C10 and divided this sum by 3600000 J. That returned 0.6737kWh. So, this is not accounting for any losses!



Yes, I understand what you mean. But this initial calculation doesn't take any time into account. While when calculating losses, I calculate the rate of loss, meaning Wh or kWh. So, I have 28.71kWh of heat losses over a day and I need 0.6737kWh to heat up the space by 13.2ºC.

Now, that calculation I did for time to heat up the space, I only took into account theenergy needed to initially heat up the space!
Ok see your last sentence.
How can you calculate the time (10 minutes, or 7m and 3m), if you only took into account the energy needed to initially heat up the space but without considering the losses at the same time?

In other words, I do not think you actually calculated the time to heat up the space initially because you left something out. The specific heat capacity alone (with the mass and volume) is not enough the losses are also relevant. The SHC and M and V account for the heating with ideal insulation. With non-ideal insulation the main heater power will not all go toward heating up the mass.

You can also think about what would happen if you left the heater running forever never turning it off and never allowing the thermostat to turn it off. If the losses were irrelevant, the temperature would rise forever. Obviously, the temperature would stop rising once equilibrium is reached, where the losses equaled the applied power.

Consider these next two examples...

[1]
You do the same thing as before but now the room is insulated with very efficient insulation, so efficient that it lets absolutely NO heat out at all, ever, no matter how high the temperature gets inside. If you do not turn off the space heater (I guess it is 2500 watts) the temperature inside the room will rise indefinitely, theoretically approaching infinite temperature. That's with absolutely perfect insulation as mentioned.
Measuring the time for a rise in temperature delta_T we may get 10 minutes depending on the volume of the room, and the specific heat capacity and the mass, ONLY.

[2]
You do the same thing as before but now the room has very thin copper walls, so thin that they let ALL of the heat out, and the temperature outside is at 10 K. Using a space heater of 2500 watts, the temperature inside the room never goes up even 1 degree.
Measuring the time for a rise in temperature delta_T, we get no answer because the temperature never rises.
Using a little bit of insulation we may see the temperature rise by delta_T but with just a little insulation it will take a while because we lose some of the power heating (2500 watts) through the walls. If we lost 2400 watts we'd be left with just 100 watts, which will mean it will take longer to heat up the room. If we lost just 100 watts, we'd be left with 2400 watts which may not be that bad, but it will take longer than if we had the full 2500 watts.

Conclusions:
When there is room insulation the heat loss is less than with no insulation so the temperature rises.
When there is no insulation the temperature rises more slowly because some of that heating power (2500 watts) is lost through the side walls even during that shorter time that it is just starting to heat up.
The amount lost is subtracted from the generator (the space heater).

One electrical equivalent would be to have a constant current charging a capacitor in parallel with a resistor. Some of that current is lost through the resistor, and the higher the voltage gets across the capacitor the more current is lost.
It reaches equilibrium when all the current goes through the resistor, but for our real life problem it will never be allowed to go that high as the current would be switched off for a time once the capacitor voltage got up to some desired voltage.

In a lot of these problems, we don't bother to calculate the time to heat initially because it's considered a short time relative to the time it will be running. An example is a transistor heat sink.
It is interesting though that if you place the transistor in a water bath that is relatively large and you over power the transistor, you'd have to know how long it takes for the water to get up to a certain temperature so the transistor temperature does not go too high or the water boils away and leaves the transistor with no heat sink.
 
Last edited:

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Ok see your last sentence.
I know, I'm considering perfect insulation, otherwise I would be running in circles. Explained below.

How can you calculate the time (10 minutes, or 7m and 3m), if you only took into account the energy needed to initially heat up the space but without considering the losses at the same time?
Well, to calculate losses, I would need a time frame because the units are W or J/s. And what would be this time frame if I wanted to include the losses in the energy needed to heat up the space initially? The 7 mins + 3 mins? But if I include the losses for 10 mins, then, the time to heat up the place is no longer the 10 mins. So, I'm not sure how to include the losses in the inital heat up!
 

MrAl

Joined Jun 17, 2014
11,693
I know, I'm considering perfect insulation, otherwise I would be running in circles. Explained below.



Well, to calculate losses, I would need a time frame because the units are W or J/s. And what would be this time frame if I wanted to include the losses in the energy needed to heat up the space initially? The 7 mins + 3 mins? But if I include the losses for 10 mins, then, the time to heat up the place is no longer the 10 mins. So, I'm not sure how to include the losses in the inital heat up!
Think of the room as a giant heat sink. The temperature will rise T1 degrees per watt, no time involved there and that is because the loss is considered a constant, after the room heats up. Once you have that figure, you should be able to work it like a heat sink where you know the loss in watts.

For a heatsink, we would have that figure in degrees C per watt. If that is say 2 degrees per watt, then if we apply 20 watts the temperature will rise to 40 degrees C above ambient.
If we only knew the 40 degree rise, then if we apply 1 watt we'd get a temperature rise of 2 degrees, so we'd know the thermal resistance was 2 degrees per watt, so we would know that we would need at least 20 watts.
What did we do there. We took the degrees and divided by the watts, and we got the thermal resistance of the heat sink. If we applied 3 watts the temperature would have gone up by 6 degrees, and 6 divide by 3 is again 2deg/watt.

I think you already calculated the loss (watts) for a given temperature rise, didn't you?
If this doesn't help I'll read everything over again :)

Oh, then there is the number of air changes in the room per hour or whatever, "drafts". That of course represents another loss.

I should have added that if your loss is 1000 watts then you are heating the thermal "mass" with only 2500-1000=1500 watts. I think that will get you to your time.
This would be modeled as a current source charging a capacitor, and the capacitor has a resistor across it that acts to discharge the cap or just conduct some current. Once the cap gets up to a certain voltage, you can divide the voltage by the current to get the resistance of that resistor, provided the voltage does not change anymore. Once you have that, you can create a time equation knowing the capacitance and the resistance. The analogy would then be:
V=Temperature
I=watts
R=still resistance
C=specific heat capacity

[Note: if we consider the voltage to rise indefinitely then we have to use the time equation i*R-i*R*e^(-t/(C*R)) so we would like to avoid that situation. ]

The current through the cap is:
iC=I-V/R
therefore the time equation is:
dv=i*dt/C
where
dv is the change in temperature,
i is the applied watts,
dt is the time,
C is the specific heat capacity of the mass.
Note we can't do this without the R.

This should work but there are some caveats such as we have not considered air leaks in the room.
Also, if anyone else wants to check this over I invite them to do so.

A question I thought of while looking at this was, is there a roof over the ceiling? If so, there may be an attic, and that would add to the insulation to some degree. Of course it also depends on the insulation in the attic.
That reminds me, does this room have any insulation other than the solid construction materials and the air spaces. If not, that might be a little unusual.

I should also mention that once R is determined, we can use the exponential form to calculate the time to a given temperature. That analysis can be found with the analogous electric circuit.
 
Last edited:

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Think of the room as a giant heat sink. The temperature will rise T1 degrees per watt, no time involved there and that is because the loss is considered a constant, after the room heats up. Once you have that figure, you should be able to work it like a heat sink where you know the loss in watts.
Yes, I understand. My reluctancy about this is that Watt, by definition involves time, as Watt is the rate of energy usage per unit of time. Let's then see if I can move on with this.

I think you already calculated the loss (watts) for a given temperature rise, didn't you?
Yes, I calculated the losses as around 800W for colder temperatures (ΔT is higher) and around 400W wen temperatures are higher (ΔT is lower).

So, from what I understand, Knwing the losses, instead of considering the heating device as rated at 2500W (or whatever) I shoul subract the losses to the rating of the device and go from there. Is that it?

Regarding the analogy with capacitors and dis/charging resistors, tbh, I get a bit lost with them, so, I'm reading what you write about them but with due respect, I'm not thinking too much about them!
 

MrAl

Joined Jun 17, 2014
11,693
Yes, I understand. My reluctancy about this is that Watt, by definition involves time, as Watt is the rate of energy usage per unit of time. Let's then see if I can move on with this.



Yes, I calculated the losses as around 800W for colder temperatures (ΔT is higher) and around 400W wen temperatures are higher (ΔT is lower).

So, from what I understand, Knwing the losses, instead of considering the heating device as rated at 2500W (or whatever) I shoul subract the losses to the rating of the device and go from there. Is that it?

Regarding the analogy with capacitors and dis/charging resistors, tbh, I get a bit lost with them, so, I'm reading what you write about them but with due respect, I'm not thinking too much about them!
Hello again,

Ok I'll see if I can draw something up that illustrates all this. Part of what I have told you so far is in common use for dealing with heat sinks. Heat sinks are specified in units of T/W usually in degrees C over Watts, so a typical heat sink rating might be 1/2 degree C per watt. That means that you can take the watts (known from the circuit behavior) and multiply that rating by that power, and that gives you the temperature rise. So with that heat sink if we had to dissipate 40 watts we could expect a temperature rise of 0.5*40=20 degrees C. Usually it's the other way around though, we have to find a heat sink that can dissipate the power we have while keeping the temperature rise down to whatever we need, such as that 20 degrees C. This means usually we would take the 20 degrees C and that 40 watts, and then calculate the requirement of the heat sink which would come out to 1/2 degree C per watt, then we would go an buy a heatsink that could do at least that good.
An important point here is that with the 20 degrees C, that is the final temperature, after all exponential terms have settled out, and that means, in theory, at a time very distant into the future. In practice though, it doesn't take that long because it gets so very close to the final temperature that we call that the final temperature anyway. So it may be just 5 minutes for example, but we do not calculate that because we don't consider that important in most cases.
If we did have to calculate that, then we have to bring in values like the specific heat capacity of aluminum (the most common heatsink metal).

I'll see if I can get a better illustration of how this works sometime today.
In actual practice with room heating I don't think they worry about the time to heat up to a given temperature anyway, because that's just a small part of the total heating costs over days or weeks or months. This means that it may just be a waste of time to try to calculate the time to heat up to a given temperature and just go with the static quantities. Doing that, you would just multiply the quantity you've calculated by the number of days (or hours or whatever) and figure out the total operating costs from that. Since that would be a large part of the total cost to operate, the time to heat up would be very insignificant. That's something to think about also, is it really necessary to calculate the time to heat up to a given temperature.

Oh BTW, do you do any circuit analysis or study any circuits? I can't remember if you do or not. I ask because electrical circuit analogies work very well to illustrate the way other systems work, including heating and mechanical systems.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Even before reading your last post, I would like to add to my previous reply the following, because I missed it and I just read it and wante to reply:

A question I thought of while looking at this was, is there a roof over the ceiling? If so, there may be an attic, and that would add to the insulation to some degree. Of course it also depends on the insulation in the attic.
That reminds me, does this room have any insulation other than the solid construction materials and the air spaces. If not, that might be a little unusual.
Yes, the area considered in my calcs include a ceilling. However, I decided to ingore/exlcude the floor. So, the area includes 4 vertical walls (2x side walls plus front and back) and the ceilling. You can see in my post #5 spreadsheet line 19 that the area includes 3 parts which are the ones I just described!

Will reply here to your last post after I carefully read it and write the reply. Will take a few minutes! :)

Edited;

Ok, after reading your reply, yes, it makes sense to ignore the time it takes to heat up initially. Absolutely.
I think I don't need illustration but if you are commited I'll be happy to see them. Maybe you will add something I've beem missing the whole time, even regarding any other subject other than the time to initially heat up the space. I'm now more convinced to ignore that calculation.

Ok, to try to make things a bit simpler, let's recap with values to see if I can work this out once and for all.
I have calculated the energy needed to initially heat up the space.
Considering the 2 deltas, I got 1.630MJ and 800kJ.

Then I calculated the losses.
I got 800W and 400W for the 2 deltas.

So, if I got your explanation right, I subtracted the 800W + 400W (roughly) to the heating device rated power (2500W) to get around 1300W.

Is this the correct way? Or at least the way you tried to explain to me?

About if I do any circuit analysis, yes, I did in my grauation but I never practiced it too much, so in more complex stuff I may struggle to do circuit analysis, but yes, I have the very basic knwoledge of Ohms law and that shoolar basic stuff!
 
Last edited:

MrAl

Joined Jun 17, 2014
11,693
Even before reading your last post, I would like to add to my previous reply the following, because I missed it and I just read it and wante to reply:



Yes, the area considered in my calcs include a ceilling. However, I decided to ingore/exlcude the floor. So, the area includes 4 vertical walls (2x side walls plus front and back) and the ceilling. You can see in my post #5 spreadsheet line 19 that the area includes 3 parts which are the ones I just described!

Will reply here to your last post after I carefully read it and write the reply. Will take a few minutes! :)

Edited;

Ok, after reading your reply, yes, it makes sense to ignore the time it takes to heat up initially. Absolutely.
I think I don't need illustration but if you are commited I'll be happy to see them. Maybe you will add something I've beem missing the whole time, even regarding any other subject other than the time to initially heat up the space. I'm now more convinced to ignore that calculation.

Ok, to try to make things a bit simpler, let's recap with values to see if I can work this out once and for all.
I have calculated the energy needed to initially heat up the space.
Considering the 2 deltas, I got 1.630MJ and 800kJ.

Then I calculated the losses.
I got 800W and 400W for the 2 deltas.

So, if I got your explanation right, I subtracted the 800W + 400W (roughly) to the heating device rated power (2500W) to get around 1300W.

Is this the correct way? Or at least the way you tried to explain to me?
Hi,

That looks right. It makes sense that if we lose 100 watts and we are using a power source of 500 watts, then that 100 watts does not do anything to raise the temperature we only have 400 watts left to raise the temperature.
If we had two separate losses then we would have to add them up to get the total loss, and that would mean we have less power to raise the temperature. If we had 100 watts loss and 50 watts loss the total loss would have to be 150 watts and with that same 500 watt power source that would mean only 500-150=350 watts left to raise the temperature.

I suppose the loss could be looked at as a parallel resistance R2 in parallel with a resistance R1. We want to calculate the power in R1 so that we know what power rating to buy. If R1+R2 is using 500 watts and R2 is using 100 watts, then the rise in temperature of R1 will be due to only 400 watts because that's all it sees. So the entire 500 watts does not go to raising the temperature of R1 because R2 is dissipating some of that 500 watts. That in turn means R1 does not get as hot, and if we wanted to use that to heat something we'd certainly have to take into account the loss in R2 from the total power.

I'll have to read your most recent post next.
Oh I see now that was a reply to someone else's post, I was wondering what that all was :)

I think I will create a circuit to illustrate these ideas, and a plot.
 

hrs

Joined Jun 13, 2014
414
There are a few things I'm not sure what or why they are like you used them.
Like, I assume Cp is the specific heat capacity! What would be the htc?
Also, why you have Q_in expressed in W/m²K?
htc is the heat transfer coefficient. It is 1/Rt.
Q_in should be W. I think the auto-complete feature of excel was trying to help me again.
 
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