Q vs Delta T for Forced Convection 3

[Maxham1]

I have an engineering colleague who seems to recall that he was taught that the amount of heat removed from a body (lumped mass) varies with delta T as follows:

1. Delta T - for natural convection

2. Delta T^2 - for forced convection

3. Delta T^4 - for radiation

I familiar with 1 and 3 above, since 1 is Newton's law of cooling and 3 is the radiation equation, but I'm not sure of number 2. Can we say that for a simple lumped mass body, that the amount of heat removed is proportional to the velocity squared?

Thanks

 

[Maxham1]

Sorry, I made a mistake at the end of my question above. I meant to ask:

Can we say that for a simple lumped mass body, that the amount of heat removed is proportional to Delta T squared (not velocity squared)?

And if so, where does this relation come from?

Thanks!

 

[IRstuff]

Actually #3 above is incorrect. It's proportional to T1[sup]4[/sup]-T2[sup]4[/sup], where T1 and T2 are absolute temperature in Kelvins.

I'm not sure about #2, since forced convection is usually treated as an extension of natural convection with the same deltaT, but different h.

At a sufficiently small deltaT, you can likewise treat #3 as a linear difference.

TTFN

 

[electricpete]

Regarding #1 - Natural convection varies in the neighborhood dT^0.25 to dT^0.33

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[electricpete]

Regarding #2 - I'm trying to make sense out of it. It seems like for forced convection there will be established a heat transfer coefficient relatively independent of temperature, except for effect of temperature on fluid properties. Once heat transfer coefficient h is known, heat flux is proportional to h*dT^1

h will depend on velocity, fluid properties, and geometry.

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[jrhols]

I think what your friend remembered was that forced convection is associated with the square of the surface
area to volume ratio, rather than just the surface area.
In forced convection, the delta T is much less important
than the Reynolds number for turbulance as the mass and
flow characteristics determine the air transfer coefficients. It's exacting work, as you can see if you
look at fin cooling coefficients for natural convection.
There isn't much difference in heat for different materials, it mostly depends on the temperature of air if the surface is heated on some end or side. But it involves hyperbolic tangents and dimensionless analysis, so unless the problem is plastics, I'm sure it would not be a valid assumption to use the square of delta T.

 

[25362]

For all modes of heat transfer Q = hA(T[sub]1[/sub]-T[sub]2[/sub]). The question is what values are given to h in each case, as function of T[sub]1[/sub], T[sub]2[/sub], [Δ]T, or T[sub]avg[/sub].

In heat transfer, the geometrical configuration of the hot surface as well as the fluid chatacteristics (air, water, oil, etc.) have a strong influence on the coefficient h.

Thus no hard-and-fast rules can be applied. Having said that the recollections of Maxham1's colleague are in the ballpark as follows.

Assuming one speaks of a steady state process in which both the lumped body and its surroundings stay at the same temperatures, both in space and time, conclusions appear to be as follows:

1. Natural (laminar flow) convection in still air at atmospheric pressure:

h [ε] [Δ]T[sup]0.25[/sup]/T[sub]avg[/sub][sup]0.18[/sup], thus

Q [ε] [Δ]T[sup]1.25[/sup]/T[sub]avg[/sub][sup]0.18[/sup]

where [ε] means proportional to

When T[sub]1[/sub] >>> T[sub]2[/sub]

Q [ε] [Δ]T[sup]1.07[/sup]​

2. Forced convection in air at atmospheric pressure and a given mass velocity of air:

h [ε] T[sub]avg[/sub], thus

Q [ε] [Δ]T[×] T[sub]avg[/sub]

For T[sub]1[/sub] >>> T[sub]2[/sub]

Q [ε] T[sub]1[/sub][sup]2[/sup]​

3. Radiation from a given surface to a given surface with given emissivities:

h [ε] (T[sub]1[/sub]+T[sub]2[/sub])(T[sub]1[/sub][sup]2[/sup]+T[sub]2[/sub][sup]2[/sup]), thus

Q [ε] (T[sub]1[/sub][sup]4[/sup]-T[sub]2[/sub][sup]4[/sup])

When T[sub]1[/sub] >>> T[sub]2[/sub]

Q [ε] T[sub]1[/sub][sup]4[/sup]​

 

[Maxham1]

Thanks for the responses. I've learned something from all of you, plus knocked out some cobwebs in the process.

 

[jrhols]

I should not omit this fact: the square of temperature does appear in conductivity equations, in the steady state equation.

For an isotropic homogenous body filling region R
the triple integral (3D) of the scalar laplacian R times temperature change squared plus conductivity times the triple integral of the surface S times temperature squared (both quantities differetiated with respect to themselves)
is equal to zero. Also known as Green's identity.

 

[electricpete]

jrhols - now you have got my curiosity up. What you said doesn't make sense to me.

You are no doubt referring to some version of the condution equation alpha * del^2*T = dT/dt in absense of heat generation.

Green's as I remember allows us to change the form to convert into an integral equation.

But you say that T^2 is involved really makes me struggle to find any physical meaning.

First of all since we have T as continuous function rather than lumped model, there is no delta-T, so we must be talking about T as you have said. But the original equation deals only with temperature differences and rates of change which can be solved in terms of relative temperature scales (for instance Celsius and Fahrenheit). But when you square a temperature you really need to know an absolute temperature. I can see no reason why absolute temperature is relevant to heat condution.

Bottom line it doesn't sound right to me, but maybe I am mistaken. Can you clarify?

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[25362]

The second partial derivative of T with respect to x in the fundamental Fourier equation Q=-kA[∂]T/[∂]x is zero: [∂][sup]2[/sup]T/[∂]x[sup]2[/sup]=0, also when taken over the three dimensions x, y, z.

By definition, this is simply the result of assuming a constant heat flow Q in steady state heat transfer from one side of a wall to the other.

1. In steady state heat conduction, without internal heat generation, heat flow normal to the surface A, is

Q=-kA[Δ]T/[Δ]x​

2. When considering k (thermal conductivity) as a linear function of T, such as in k=k[sub]o[/sub](1+bT), one can envisage a quadratic dependence on temperature.

q=-(k[sub]o[/sub]A/[Δ]x)[(T[sub]2[/sub]-T[sub]1[/sub]) + [½]b(T[sub]2[/sub][sup]2[/sup]-T[sub]1[/sub][sup]2[/sup])]​

However b may adopt positive, negative or zero values.

3. The value of b is positive for gases, pure aluminum, insulating materials, and some stainless steels. With other materials, especially solids, either b[≈]0, or is negative. Water, as usual, is anomalous in the sense that k rises up to ~270[sup]o[/sup]F, and then decreases at higher temperatures.

4. For all cases where T[sub]2[/sub] >>> T[sub]1[/sub], we"ll call T[sub]2[/sub]=T

When b is positive,

Q [ε] mT+nT[sup]2[/sup]​

a parabola

when b = 0,

Q [ε] mT​

when b < 0,

Q [&epsilon;] mT-nT[sup]2[/sup]​

an inverted parabola.

m and n are constants.

 

[sailoday28]

Maxham1 (Mechanical) Jan 26, 2005
ORIGINAL QUESTION RELATES TO "LUMPED MASS"
Can we say that for a simple lumped mass body, that the amount of heat removed is proportional to Delta T squared (not velocity squared)?

25362 (Chemical) Jan 29, 2005
STATES
The second partial derivative of T with respect to x in the fundamental Fourier equation Q=-kA?T/?x is zero: ?2T/?x2=0

I interpret a lumped mass to have an equivalent uniform temperature. No temp gradients.

For example, to calculate the time to heat up a soldering iron with a surrounding convective atmosphere. The iron would haved a "lumped" temp dependent solely on time and not spacial coordinates. mc(dT/dt)=I^2R-hA(T-Ta)
where m=mass of soldering iron
c=specific heat of iron
T=Lumped temp of soldering iron
I^2R= wattage of electrical input
h= convect coef,
A=surface area
Ts= temp of surroundings
Clearly, the above diff eq can also incorporate net radiation to the surroundings.

 

[25362]

I think sailoday28 has a point. A lumped mass analysis involves an unsteady heat transfer condition. A lumped-heat-capacity system analysis assumes that the internal thermal resistance of a body is negligible in comparison with the external resistance, and its temperature is uniform all through.

These systems are obviously idealized since a temperature gradient must exist in a material if heat is to be conducted into or out of the material. Thus, the smaller the physical size of the body, the more realistic is the assumption of uniform temperature throughout.

Heat lost by convection from a body equals the decrease of its internal energy. Thus

q = hA(T-T[sub][&infin;][/sub]) = -C[&rho;] VdT/d[&tau;]​

Where V is the volume, A is the convection surface area, C is the heat capacity, [&rho;] is the density, T is temperature and [&tau;] is time. The initial conditions being T = T[sub]o[/sub] at [&tau;] = 0
and the solution to the equation is:

(T-T[sub][&infin;][/sub])/(T[sub]o[/sub]-T[sub][&infin;][/sub]) = e[sup]-[hA/([&rho;] CV)][&tau;][/sup]​

C[&rho;] V/hA is called the time constant since it has the dimensions of time. Holman shows how systems like this can be analysed by using electrical circuits with a similar behaviour.