cooling unit

[GHIZMO]

Hello everyone, someone can tell me how to calculate (very aptly!) the cooling time of a piece out of an oven and left in the air calmly knowing: initial and final temperature of the material piece (al) and mass of the piece, air temp. (Do I also need the surface of the piece?). later, in case the weather was too long I would need to know how to always calculate the cooling time by investing the piece from the air (at a known temp) with a speed to define
thanks in advance

 

[Paolo Molniya]

from "transmission of heat" bonacina-cavallini-mattarolo cleup publisher

"...detail t1 the temperature of the surface and t2 the temperature, appropriately defined, of the fluid that lambishes, the relationship that expresses the thermal flow established by newton in 1701 is : q=a * α * (t1-t2) with α (alpha) convection coefficient. . "

in natural convention for vertical flat or cylindrical walls:

Rolling system: nu = 0.59 * ra^(1/4) with 10^4 < 10^9

turbulent regime: nu = 0.138 * gr^0.36 *(pr0.175 - 0.55) with ra>10^9

where:

nusselt= alfa*l/ lambda [ alfa coefficiente convezione, L dimensione caratteristica, lambda conduttività termica fluido]ra = number of rayleigh = gr*pr

(l^3*rho^2*beta*g*deltat)/(mu^2)
[rho densità fluido, beta coefficiente isobaro espansione termica, g acc. gravità, deltaT differenza temperature, mu viscosità dinamica]prandtl= c*mu/lambda
[c calore specifico]it is not so simple (I hope I have not made big mistakes) you should look for the value of the coefficient on some manual, because calculating it is very difficult and not always get correct values due to the empirical formulas....

 

[GHIZMO]

first of all thank you! for the immediate answer.
that it was not a simple matter I had more than intuition in fact I had specified: ... very rough!
but it is not possible to reach a value to spannons by applying a formula that considers data more "quotidian".
My application, in summary, is this: I have a component in the weight of cira 8kg coming out of an oven to 140 °C, if I leave it in calm air to 23° how long do I have to wait so that it can be manipulated down to a temperature of at least 30°? and if this time was too high and had it invested from a laminating flow to 0.5 m/s - 23°c, and if the flow was instead at 15°.
That's all.

 

[paulpaul]

Hi.
This is a problem of heat transfer in a transitional regime: In the first line I would apply the method of concentrated capacity, that is, considering all the solid at the same temperature, instant by instant, otherwise the calculations become too complex. This method is all accurate when the conductive strength in the solid is much less than the convection resistance between solid and environment, i.e. when the number of biot bi = hl/k is small, indicatingly less than 0.1. h is the convective exchange coefficient that we do not know for now, and k is the conductivity of aluminum. l =v/a is the ratio between volume and surface of the solid.

said this, using this method you can use the formula which is reported for example by the work and which is derived by integrating the equation of energy in a transitional form

de/dt =
-ha(t-tamb)

by making calculations you get the time you need:

♪ [(rho*V*c)/(h*A)]*n[(Ti-Tamb)/(Tf-Tamb)]rho is the density of the aluminum, and its specific heat, the initial tf, the final one you want to impose.

remains h, we said, whose calculation as it says paolo is difficult and imprecise in the case of external flow and natural convection, since in the texts there are formulas (complexes) only for relatively simple geometries (spheres, cylinders, flat walls, etc.).
I would consider h = 5...20 w/m^2k for natural convection, with lowest values for very calm air.
an extra order of magnitude for forced convection.

Bye!

 

[maxopus]

Thermal camera?
Sometimes the measurement can be more immediate than the calculation.

 

[Fulvio Romano]

Thermal camera?
Sometimes the measurement can be more immediate than the calculation.

temperature measurement is certainly more immediate...but the heaters do not measure temperature:frown:

the camera measures the irradiation from an object, and this, as stefan-boltzmann teaches, is linked to the temperature by a variable called emissivity.

emissivity depends on the material, color, surface finish, temperature itself.. .
therefore also with a measure that seems direct, all depends on variables that if randomly give wrong information.

 

[PierArg]

temperature measurement is certainly more immediate...but the heaters do not measure temperature:frown:

the camera measures the irradiation from an object, and this, as stefan-boltzmann teaches, is linked to the temperature by a variable called emissivity.

emissivity depends on the material, color, surface finish, temperature itself.. .
therefore also with a measure that seems direct, all depends on variables that if randomly give wrong information.

I was convinced that the heaters, measuring the irradiation, still managed to get back to the field of temperature.

 

[paulpaul]

other punctualization as rightly detected: I have considered as q out only the heat transmitted by convection. In fact there would also be the uncorrecting quota if the t of the piece is high, but in this case formula of the cooling time would not be so simply obtainable by hand.

 

[maxopus]

temperature measurement is certainly more immediate...but the heaters do not measure temperature:frown:

the camera measures the irradiation from an object, and this, as stefan-boltzmann teaches, is linked to the temperature by a variable called emissivity.

emissivity depends on the material, color, surface finish, temperature itself.. .
therefore also with a measure that seems direct, all depends on variables that if randomly give wrong information.

interesting your consideration.
It deserves an insight, to understand how to interpret the data of the above tools.

 

[Fulvio Romano]

I was convinced that the heaters, measuring the irradiation, still managed to get back to the field of temperature.

try to point a thermocamera on a painted black satin cube and on a polished steel ball, at the same temperature. you see them of different colors.

 

[maxopus]

try to point a thermocamera on a painted black satin cube and on a polished steel ball, at the same temperature. you see them of different colors.

Perhaps there are some presets?
What kind of response do we have?

 

[Fulvio Romano]

Perhaps there are some presets?
What kind of response do we have?

typically in the heaters there is a list of standard emissivity to choose from. according to the material, choose the one that comes closer.

if the body is glowing means that it is beginning to emit radiation in the visible. typically a camera can see an infrared field that does not arrive in the visible, so all components above the sensor cutting frequency are simply ignored.

nothing to do with emissivity.

 

[Paolo Molniya]

hi I tried to make two contiguous with the following data:
== sync, corrected by elderman ==
tamb=23°C
tf=30°c
specific heat pure aluminum 896 j/kg*k
aluminium density 2707 kg/m^3

the air values are taken from table for a temperature of 127°c:
cp=1014 j/kg
thermal conductivity air 0.03365 w/m*k
cinematic viscosity air 25.9 10^-6 m^2/s
air density 0.8826 kg/m^3
coefficient beta= 1/ t (media) =~ 0.00333 1/k

the shape of the body is very important for calculations, so I hypothesized for simplicity that it was a cube of 8 kg aluminum and therefore with total area of 0.1244 m^2 and side (or characteristic size l) of about 144 mm.
at this point I have resumed the formulas that I posted at the beginning for laminating regime in natural convention (for simplicity) and I go to calculate alpha of convection alone so:

alpha = 0.59 * (lambda_aria^(3/4))/(l^1/4) *g^(1/4))

(ii)

time(in seconds)= (8kg*896 j/kg*k)/(area_contact* alpha)

from my calculations I get a single convection alpha of 7.66 w/k*m^2 and a time of about 2 hours. (if you need to place all calculations)
then obviously to simplify I did not take into account:
-property of air considered for average temperature value;
-body shape;
- conductive thermal exchange, on the face in support;
-radiative thermal exchange if nearby there are other objects with different temperatures.

 

[Paolo Molniya]

from my calculations I get a single convection alpha of 7.66 w/k*m^2 and a time of about 2 hours.

dog world, I forgot to multiply for the logarithm in the formula of time is to multiply by ln(((ti-tamb)/(tf-tamb))) and therefore comes a time of about 6 hours that is entirely inverosimile....so to you the conclusions recalling what I said before

 

[GHIZMO]

Please, thank you again... in both cases the weather is improposable, I think it will be necessary to ventilate and that the air will also have to be treated by a chiller, but of course now arises the problem of establishing speed and air temperature as the target of cooling is about 15min max. definitely the shape of the piece is relevant (I apologize for not having mentioned it) that is a tubular frame welded with spess. max around 3 mm.

Hi.

 

[paulpaul]

because of the logarithm, if you are aware of 50 °c as the final temperature (instead of 30 °c) you could almost halve time.
considering a coefficient of exchange in forced convection of 180 w/m2k also leaving unaltered air temperature (23 °c), to have your block at 30 °c you would employ only 15 '.
If you have a frame, I'd say you can't do anything but improve, because you're improving the exchange coefficients and the exchange area itself.
you could also hypothesize the scope you need, always with a certain approximation.

Goes to fashion the heat exchange on this forum lately!