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Calculating Cooling Times 7
- Thread starter mmelville
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There's pretty much only two equations that are ever used in heat transfer,
Q = htc*area*deltaT
Q = emiss*sigma*area*(T_1^4-T_2^4)
You calculate the latent heat, the heat transfer coefficient (htc), emissivity, and determine what kind of lumped model you're going to use to model where the heat energy resides in the material.
TTFN
FAQ731-376
Q = htc*area*deltaT
Q = emiss*sigma*area*(T_1^4-T_2^4)
You calculate the latent heat, the heat transfer coefficient (htc), emissivity, and determine what kind of lumped model you're going to use to model where the heat energy resides in the material.
TTFN
FAQ731-376
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As a simple solution you could assume that the pipe wall acts as a lumped mass with a negligible temperature gradient across the wall thickness as it cools. You then solve V.p.Cp.dT/dt = h.A.(T-Ta) to get the temperature T as an exponential of time t. Values of the heat transfer coefficient, h, you'd have to calculate based upon the air speed, and loss of heat to the outer ambient.
ex-corus (semi-detached)
ex-corus (semi-detached)
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Mmelville,
You’ve already got valid inputs from the posts above.
The solution of the Newton’s cooling equation (the one reportes in corus’ post) will lead to an exponential time-decaying profile of the temperature.
T(t) = Tamb + (Ti - Tamb)*exp[-A*h*t/(m*cp)]
Where:
T(t) = wall temperature at time t
Tamb = ambient temperature
Ti = starting wall temperature at time t = 0 s (as noted by corus with assumption of constant temperature through the pipe wall)
A = surface area
h = convection coefficient
m = pipe mass
cp = pipe material specific heat
t = time
If you want to go any further take a look at the link below (besides the Newton’s equation this paper takes into account a more sophisticated model, which also considers radiation heat transfer ).
You’ve already got valid inputs from the posts above.
The solution of the Newton’s cooling equation (the one reportes in corus’ post) will lead to an exponential time-decaying profile of the temperature.
T(t) = Tamb + (Ti - Tamb)*exp[-A*h*t/(m*cp)]
Where:
T(t) = wall temperature at time t
Tamb = ambient temperature
Ti = starting wall temperature at time t = 0 s (as noted by corus with assumption of constant temperature through the pipe wall)
A = surface area
h = convection coefficient
m = pipe mass
cp = pipe material specific heat
t = time
If you want to go any further take a look at the link below (besides the Newton’s equation this paper takes into account a more sophisticated model, which also considers radiation heat transfer ).
- Thread starter
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This problem recurs from time to time and I believe no one has gotten it right. It turns out to involve solving a very difficult partial differential equation so the best thing to do is solve it numerically. The last time I saw this problem I used EXCEL and broke up the pipe into discrete sections of delta x length and deltat=deltax/V from the known flow velocity V moving the gas along the pipe iteratively,exchanging heat and repeating the process until "equilibrium" is obtained.With EXCEL, you could write a MACRO to do this so that only one row could be updated with each increment of time, delta t.Using a counter you determine the time t=delta t*N
If the OP would furnish the geometry of the pipe and the air mass flow rate I might take a crack at it in my spare time.
If the OP would furnish the geometry of the pipe and the air mass flow rate I might take a crack at it in my spare time.
- Thread starter
- #7
Spare time!! Whats that?
Unfortunately, I work for an automotive supplier and cannot really give out the 3d data.
I really just want to "ballpark" the cooling time required to achieve a given temperature. The equation ione gave got me close, i just need an approximate "h" value.
Until I get samples of the parts I am designing around, I cannot perform experiments and must rely on calculations to get a rough idea of the cooling time.
Unfortunately, I work for an automotive supplier and cannot really give out the 3d data.
I really just want to "ballpark" the cooling time required to achieve a given temperature. The equation ione gave got me close, i just need an approximate "h" value.
Until I get samples of the parts I am designing around, I cannot perform experiments and must rely on calculations to get a rough idea of the cooling time.
hc, convection coefficient depends on flow velocity, diameter and more weakly on gas temperature; hr, the equivalent radiation coefficient depends on emissivity and temperature of pipe and gas; so unless you can "ballpark" the geometry, ( e.g. a pipe with diameter of x opening up to a diameter of y over a distance z) we can't ballpark the h.
mmelville,
The equation given in my previous post was just the solution of the differential equation suggested by corus.
If you are looking for a correlation to calculate h in regime of forced convection (assuming the pipe is cooled just from the outside) you can use one of the correlations in the attached paper (from 9.13 to 9.18).
I also suggest you to give a shot to the calculator at the link below.
The equation given in my previous post was just the solution of the differential equation suggested by corus.
If you are looking for a correlation to calculate h in regime of forced convection (assuming the pipe is cooled just from the outside) you can use one of the correlations in the attached paper (from 9.13 to 9.18).
I also suggest you to give a shot to the calculator at the link below.
If you check the internet or heat transfer books you'll find heat transfer convective coefficients (h's) values and formulae for forced air. You will need to determine the inside velocity under the 750 cfm air flow and probably other parameters such as bulk viscocity and air density.
Following up on my objection to using lumped parameters without proof, I think you could make that determination by sending a slug of cool air through the pipe and if the drop in temperature is small next to the overall temperature difference, then you could write the 1st order linear differential eqaution
rho*A*c*dT/dt=hi*l*(Tc-T)+ho*l*(T0-T)
whose solution is
T=T(0)exp(-t/tau)+hi/(h1+h0)*Tc+h0/(hi+h0)*To
T(0) initial temperature pipe
tau= time constant= rho*A*c/(hi+ho)l
h1,ho inside and outside film coefficients
rho = density coolant
c specific heat
l = perimeter
A cross sectional area of pipe
To test for the slug of air, assume that the pipe remains at the initial temperature. T(0) for the time it takes the slug of cools air to pass through the pipe.
Then the delta temperature of the coolant is
delta Tc=to* hi*l*(T(0)-Tc)/(rho*A*c)
to= transit time thru pipe
If the temperature change is small, then all subsequent changes will be even smaller, thus making the case for the lumped model.
rho*A*c*dT/dt=hi*l*(Tc-T)+ho*l*(T0-T)
whose solution is
T=T(0)exp(-t/tau)+hi/(h1+h0)*Tc+h0/(hi+h0)*To
T(0) initial temperature pipe
tau= time constant= rho*A*c/(hi+ho)l
h1,ho inside and outside film coefficients
rho = density coolant
c specific heat
l = perimeter
A cross sectional area of pipe
To test for the slug of air, assume that the pipe remains at the initial temperature. T(0) for the time it takes the slug of cools air to pass through the pipe.
Then the delta temperature of the coolant is
delta Tc=to* hi*l*(T(0)-Tc)/(rho*A*c)
to= transit time thru pipe
If the temperature change is small, then all subsequent changes will be even smaller, thus making the case for the lumped model.
To check whether the heat transfer lumped model is acceptable, it is necessary to evaluate the Biot number:
Bi = (h* V/A)/k
Where:
h = convective heat transfer coefficient
V = volume of the mass to be cooled
A = surface of the mass to be cooled
k = thermal conductivity of the mass
For low Biot number (Bi<0.1) it is acceptable to adopt lumped model and Newton’s cooling equation works fairly well. A low Biot number implies that the thermal resistance between mass/cooling fluid (related to convection) exceeds the thermal resistance through the pipe wall (related to conduction) and so the pipe wall can be considered constant.
In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable.
Bi = (h* V/A)/k
Where:
h = convective heat transfer coefficient
V = volume of the mass to be cooled
A = surface of the mass to be cooled
k = thermal conductivity of the mass
For low Biot number (Bi<0.1) it is acceptable to adopt lumped model and Newton’s cooling equation works fairly well. A low Biot number implies that the thermal resistance between mass/cooling fluid (related to convection) exceeds the thermal resistance through the pipe wall (related to conduction) and so the pipe wall can be considered constant.
In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable.
- Thread starter
- #13
"To check whether the heat transfer lumped model is acceptable, it is necessary to evaluate the Biot number:
Bi = (h* V/A)/k..................."
Not that easy.
That is true of radial conduction part of the problem. I had assumed that but it does not answer the flow problem here which must address the longitudinal gradient of gas temperature so that one could assume an "average" gas temperature over the length of pipe, reducing it to the total lumped model. Without that, you have no validity in doing it as a lumped model.
Bi = (h* V/A)/k..................."
Not that easy.
That is true of radial conduction part of the problem. I had assumed that but it does not answer the flow problem here which must address the longitudinal gradient of gas temperature so that one could assume an "average" gas temperature over the length of pipe, reducing it to the total lumped model. Without that, you have no validity in doing it as a lumped model.
If the cooling air moves orthogonally to the pipe (outer diameter 40 mm), the fluid temperature should not be that affected, and evaluation of Biot number should give a reliable answer.
So while I still consider evaluation of Biot number a valid approach to define whether a lumped model is applicable or not for the transient analysis, I am now prone to say that the lumped model (Newton cooling law) is not applicable here (or at least it should give misleading results).
Despite of my previous statement (“In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable”), here we are dealing with forced convection and so assuming the lumped model valid for Bi < 0.1:
h = Bi * k * A/V = 0.1 * 0.2 / 0.004 = 5 W/(m^2 * K)
Being:
k = 0.2 W/(m*K) plastic thermal conductivity (it could be also greater but the order of magnitude should be right)
V/A = s = 4 mm (pipe wall thickness)
The value of h, computed as shown above, undoubtedly reminds values more typical of natural convection than forced convection.
So while I still consider evaluation of Biot number a valid approach to define whether a lumped model is applicable or not for the transient analysis, I am now prone to say that the lumped model (Newton cooling law) is not applicable here (or at least it should give misleading results).
Despite of my previous statement (“In the case of a plastic pipe the thermal conductivity is low but the wall thickness is low as well, so the lumped model should be applicable”), here we are dealing with forced convection and so assuming the lumped model valid for Bi < 0.1:
h = Bi * k * A/V = 0.1 * 0.2 / 0.004 = 5 W/(m^2 * K)
Being:
k = 0.2 W/(m*K) plastic thermal conductivity (it could be also greater but the order of magnitude should be right)
V/A = s = 4 mm (pipe wall thickness)
The value of h, computed as shown above, undoubtedly reminds values more typical of natural convection than forced convection.
ione, I'm not sure of your limit of 0.1 for the Biot modulus as I have seen elsewhere that 0.2 should be a limit, and hence double your value of h. In either case the lumped mass model is only intended to give an approximate answer that can be derived fairly easily if no other methods were avaliable.
If it were possible, I'd suggest using finite difference methods or finite elements if the OP were familiar with those. For this 1D case there might be free software available that could do the job.
ex-corus (semi-detached)
If it were possible, I'd suggest using finite difference methods or finite elements if the OP were familiar with those. For this 1D case there might be free software available that could do the job.
ex-corus (semi-detached)
corus,
My source concerning Bi<0.1 is “Introduction to thermodynamics and heat transfer” by Yunus A. Cengel (Italian edition), but the same statement is easily found with a Google search (“Technology of thermoforming” )
I agree that lumped model could give an approximate solution, and the approximation is as rougher as the more prevailing is the convection heat transfer at the interface pipe/fluid over conduction through the wall pipe (increasing Biot number). For a metal pipe the lumped model would have worked fine.
My source concerning Bi<0.1 is “Introduction to thermodynamics and heat transfer” by Yunus A. Cengel (Italian edition), but the same statement is easily found with a Google search (“Technology of thermoforming” )
I agree that lumped model could give an approximate solution, and the approximation is as rougher as the more prevailing is the convection heat transfer at the interface pipe/fluid over conduction through the wall pipe (increasing Biot number). For a metal pipe the lumped model would have worked fine.
My source "Temperature Measurement By Bela G. Liptak"
says "If the Biot modulus os less than 0.2 no significant gradient is expected". So the answer is you takes your pick what would be acceptable.
ex-corus (semi-detached)
says "If the Biot modulus os less than 0.2 no significant gradient is expected". So the answer is you takes your pick what would be acceptable.
ex-corus (semi-detached)
OK, you guys need to get a room ;-)
There are few things in nature that are discontinuous; near as I can tell, the Biot number is a guide, no more. I mean, what happens if the Biot number is 0.11, or 0.21, for Corus? You can still do the calculation, but the accuracy simply gets worse. It's not like it goes completely wonkers, is it?
TTFN
FAQ731-376
There are few things in nature that are discontinuous; near as I can tell, the Biot number is a guide, no more. I mean, what happens if the Biot number is 0.11, or 0.21, for Corus? You can still do the calculation, but the accuracy simply gets worse. It's not like it goes completely wonkers, is it?
TTFN
FAQ731-376
"agree that lumped model could give an approximate solution, and the approximation is as rougher as the more prevailing is the convection heat transfer at the interface pipe/fluid over conduction through the wall pipe (increasing Biot number). For a metal pipe the lumped model would have worked fine."
It will NOT work fine for a long pipe like 200 feet.
If you think so , then what is the "average" fluid temperature over that distance. You have a 2 dimensional heat equation and you have only gotten the radial part approximately right, not the axial.
So, unless that you can prove that there is an "average" temperature over the transient time, you have no valid solution.
Try doing the classic problem of how long it takes for hot water to reach a faucet 200 feet away from the the source. What will you take for the average temperature of the water.
To solve that accurately or even reasonably accurately, you can't, in general, lump the model.
It will NOT work fine for a long pipe like 200 feet.
If you think so , then what is the "average" fluid temperature over that distance. You have a 2 dimensional heat equation and you have only gotten the radial part approximately right, not the axial.
So, unless that you can prove that there is an "average" temperature over the transient time, you have no valid solution.
Try doing the classic problem of how long it takes for hot water to reach a faucet 200 feet away from the the source. What will you take for the average temperature of the water.
To solve that accurately or even reasonably accurately, you can't, in general, lump the model.
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