Warm-up time for a hollow cylinder

In summary: The application involves sending information via an infrared module (TFDU4101, 3.3V powered). The circuit also includes a Teensy 4.0 and a 12V to 5V converter to power the Teensy.The circuit powers the Teensy 4.0, which then sends the data.
  • #1
defrag
8
0
Hello all ,

I need to estimate the time it would take for the inside of a hollow cylinder to reach a limit temperature (detailed data below), in order to adjust the wall thickness and choose an insulating material to delay heating as much as possible. Being a total beginner in thermodynamics, I don't understand how to proceed, what modes of heat transmission to use.

The situation is as follows:

I have a hollow cylinder with internal diameter D1, external diameter D2 and height h. The cylinder is immersed in oil at a constant temperature T2 and has an initial internal temperature T1. The oil is stationary relative to the cylinder.
I therefore tried to make an initial estimate of the time it would take to heat up by means of a heat flow calculation, but I find a result that I find totally inconsistent.

Data :

D1 = 35mm ; D2 = 40mm ; h = 60mm
T1 = 40°C ; T2 = 90°C

I've done the calculation for walls covered with aerogel (it's not realistic, but it's a first estimate), thermal conductivity λ=0.03W.K^(-1).m^(-1), thickness e = 7mm.

The surface area of my cylinder is therefore :

S=2πrh=2π×0.02×0.06=0.00753 m2

For the flux, I get :

ϕ=((λ.ΔT)/e)*S=((0.03×(90-40))/0.07)*0.00753=1.61567 W

To obtain the heating time, I divide the flow by the volume of air in the cylinder multiplied by the heat capacity of the air (1.256 kJ.m-3).

Volume of air in the cylinder (removing the volume taken up by the aerogel):

V=π*(r^2)*h=π*(0.02-0.007×2)^2×0.06=6.7858e-6

So the time I find is :

t=(cp×V)/ϕ=(1256×6.7858e-6)/1.61567=0.005 s

I don't think this is the right way to do it, should heat transfer be calculated in all modes (conduction, conduction, radiation)?

An additional problem that I haven't yet taken into account, but which is very important, is that the cylinder contains an electronic circuit that also heats up. By insulating the cylinder, the temperature stabilizes at around 40°C.

Thank you very much for your help
 
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  • #2
defrag said:
For the flux, I get :

ϕ=((λ.ΔT)/e)*S=((0.03×(90-40))/0.07)*0.00753=1.61567 W
That looks right.
defrag said:
To obtain the heating time, I divide the flow by the volume of air in the cylinder multiplied by the heat capacity of the air (1.256 kJ.m-3).

Volume of air in the cylinder (rem

So the time I find is :

t=(cp×V)/ϕ=(1256×6.7858e-6)/1.61567=0.005 s
That looks wrong; did you flip-flop the density and specific heat of air? In any case, the errors basically cancel out....though you also seem to have dropped the decimal point. Also, that's seconds per degree. Or maybe I'm not understanding where that equation came from....
defrag said:
I don't think this is the right way to do it, should heat transfer be calculated in all modes (conduction, conduction, radiation)?
Not for a problem like this, for a first approximation. The outer surface will be almost exactly the oil temperature due to high conductivity. The inner surface will not exactly match the inside air temp, but it's a small container and will equalize fairly quickly. The bottom line here is that the contents will heat up quickly.
defrag said:
An additional problem that I haven't yet taken into account, but which is very important, is that the cylinder contains an electronic circuit that also heats up. By insulating the cylinder, the temperature stabilizes at around 40°C.
What does that mean? How much heat does the circuit produce (how much electricity does it use?)?

What, exactly, is the application here?
 
  • #3
Thanks for your reply.

I realize that I first miscalculated the volume by multiplying the insulation thickness by 2, even though we're talking about radius here and not diameter.
Instead, we have :
V=π*(r^2)*h=π*(0.02-0.007)^2×0.06=3.1855e-5 m3

russ_watters said:
That looks wrong; did you flip-flop the density and specific heat of air? In any case, the errors basically cancel out....though you also seem to have dropped the decimal point. Also, that's seconds per degree. Or maybe I'm not understanding where that equation came from....
What do you mean here? Wouldn't you have to take the heat capacity to get the time it would take to heat up the cylinder? What formula would you use?
Incidentally, you're right, the unit is s.K-1, not s.

russ_watters said:
What does that mean? How much heat does the circuit produce (how much electricity does it use?)?

What, exactly, is the application here?
The application involves sending information via an infrared module (TFDU4101, 3.3V powered). The circuit also includes a Teensy 4.0 and a 12V to 5V converter to power the Teensy.
 
  • #4
I'm trying to get a mental image of what you mean.
Does it look like this?

1690044074924.png


Then, you want to find how the air temperature inside the barrel (material unknown) changes with time depending on how much insulation (material unknown, initial estimate is aerogel) you put in. Is that it?
How much power does the circuit generate by the way?
 
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  • #5
Juanda said:
Does it look like this?
Hi Juanda,

This diagram represents the situation well, thank you very much for your involvement, it would have been up to me to do it in my first post.
Juanda said:
Then, you want to find how the air temperature inside the barrel (material unknown) changes with time depending on how much insulation (material unknown, initial estimate is aerogel) you put in. Is that it?
Yes, that's it.

I don't know what material or combination of materials I'm going to use.
Perhaps the outer material (in blue) will be aluminum, to limit radiant heating, and the inner material an insulator. But for the calculation I'm only considering the insulation, aerogel in this case.
Juanda said:
How much power does the circuit generate by the way?
The circuit includes a Teensy 4.0 which consumes around 100mA with a 5V supply voltage, so 500mW, an infrared module consuming around 10mA for 3.3V, so 33mW, and an MCP2122 microcontroller with a maximum dissipation of 800mW.
So I guess not more than 1,5W.
 
  • #6
All right. If the problem were stationary it would be much easier but you're interested in how it evolves with time.
I'll try to solve it using MANY simplifications.
  1. The relevant heat exchange only happens with the walls of the cylinder. Ground and top could be considered but they only make the problem more tedious and I'm just trying to post an example of how I would approach a problem like this.
  2. Heat flux in cylinders has a different equation than the one you showed because as the radius increases, the surface area increases too. However, since the inner and outer radii are pretty similar, I think it's OK to use the heat flux equation for "flat surfaces".
    $$\dot{Q}_{in}=k \frac{dT}{dx}$$
  3. I will consider the temperature distribution in the cylinder to be linear. This will be definitely not the case but the problem is quite harder if we don't take that. As a first shot, it should still provide some numbers to get an idea of how the system will behave. If the temperature in the cylinder is linear then
    $$\dot{Q}_{in}=k \frac{dT}{dx}=k \frac{\Delta T}{\Delta x}$$
  4. Only the insulation layer will be studied. I feel like the aluminum will get as hot as the oil pretty quickly anyways so I'll skip it. The problem could be solved considering the aluminum too but it'd require adding an annoying intermediate point.
  5. The insulator in contact with the oil (assuming aluminum walls don't exist) will have the same temperature. The insulator in contact with the air inside will share the same temperature as well.
  6. I'll assume the mass of air in the cylinder is constant and has a uniform temperature distribution.
  7. I'll ignore the mass of the circuit because nothing was mentioned about it. I'll simply assume the temperature of the circuit is the same as the air surrounding it.

So the simplification looks like this.
1690132855889.png


The surface for the heat to travel is
$$S = Jh = 2 \pi \frac{r_1+r_i}{2}h$$

The change in energy for the mass of air will be a combination of the heat input from the oil and the heat generated in the circuit.
$$\frac{dU}{dT}=\dot{Q}_{in}+\dot{Q}_g$$

If we treat the air as an ideal gas, its internal energy will be a function of its temperature.
$$U=mC_vT_{air} \rightarrow \frac{dU}{dT}= mC_v \frac{dT}{dt}$$

The mass of air can be found with the volume inside the cylinder and the atmospheric conditions before warming it.
$$m_{air} = \rho V = \rho \pi r_i^2h$$

The input heat will be.
$$\dot{Q}_{in} = kS\frac{T_{oil}-T_{air}}{r_1-r_i}$$

The heat generated has been declared as
$$ \dot{Q}_{g} = 1.5W$$

Finally, it is possible to solve the differential equation. (T is the temperature of the air in the container)
$$mC_v \frac{dT}{dt}=kS\frac{T_{oil}-T}{r_1-r_i}+1.5$$

Be careful with the units! I wrote 1.5 because I'd use the international system.

I cannot solve it now but the solution should look like this.
1690133837662.png


Where the stabilization temperature will actually be slightly higher than the oil temperature due to the circuit generating heat. Basically, the oil will end up refrigerating the air inside.

There are a gazillion simplifications I had to do to solve the problem but heat transfer problems are pretty hard. Especially if they involve changes in time. Hopefully, this approach will be realistic enough for the situation you're facing.
You should be able to input your values to get a numerical answer to your problem.
 
Last edited:

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