| Topic: |
Science > Physics |
| User: |
"Old Man" |
| Date: |
07 Oct 2004 10:07:49 PM |
| Object: |
Engineering Linear Radiative Heat Loss |
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
"C * T_a" is the potential (t -> infinity) heat content per unit
volume of the solid sphere.
The specific heat conductivity can be taken as,
[ 12 * S * ( T_a )^4 ] / R = [ 12 / R ] [ dQ / dt @ T = 0 K]
which is proportional to the gross rate of ambient heat flow per
unit sphere volume flowing into the sphere at a sphere temperature
of T = 0 K.
The radius of the sphere, R, appears as "resistance to heat loss"
because the ratio of a sphere's volume, V, to its surface area, A,
is V / A = R / 3. That is, the ratio of the sphere's heat content to
its heat loss rate is proportional to the sphere's radius.
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
.
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| User: "Professor Gauss" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 03:46:30 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:WfidnQsj54AVmfvcRVn-sA@prairiewave.com...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
"C * T_a" is the potential (t -> infinity) heat content per unit
volume of the solid sphere.
The specific heat conductivity can be taken as,
[ 12 * S * ( T_a )^4 ] / R = [ 12 / R ] [ dQ / dt @ T = 0 K]
which is proportional to the gross rate of ambient heat flow per
unit sphere volume flowing into the sphere at a sphere temperature
of T = 0 K.
The radius of the sphere, R, appears as "resistance to heat loss"
because the ratio of a sphere's volume, V, to its surface area, A,
is V / A = R / 3. That is, the ratio of the sphere's heat content to
its heat loss rate is proportional to the sphere's radius.
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
The time constant, which increases as the temperature difference decreases,
can be computed from t_c=-T/(dT/dt). Integrating the Stefan-Boltzmann law
for T_a=0 yields: T=[T_i^(-3)+3SAt/MC_p]^(-1/3), where M is the mass of the
sphere (constant) and C_p is its heat capacity at constant pressure (assumed
constant). Differentiating readily gives us t_c=MC_p/SAT_i^3+3t, i.e., the
time constant starts off at a value based on the instantaneous rate of heat
loss and then increases by 3 seconds for every second that elapses.
--
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand.
-- Ancient Chinese proverb
Remove caps when replying.
-- Modern American expression
.
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| User: "Old Man" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 08:53:34 PM |
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"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message
news:g_ednYxdAoq00_XcRVn-hA@comcast.com...
"Old Man" <nomail@nomail.net> wrote in message
news:WfidnQsj54AVmfvcRVn-sA@prairiewave.com...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
"C * T_a" is the potential (t -> infinity) heat content per unit
volume of the solid sphere.
The specific heat conductivity can be taken as,
[ 12 * S * ( T_a )^4 ] / R = [ 12 / R ] [ dQ / dt @ T = 0 K]
which is proportional to the gross rate of ambient heat flow per
unit sphere volume flowing into the sphere at a sphere temperature
of T = 0 K.
The radius of the sphere, R, appears as "resistance to heat loss"
because the ratio of a sphere's volume, V, to its surface area, A,
is V / A = R / 3. That is, the ratio of the sphere's heat content to
its heat loss rate is proportional to the sphere's radius.
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
The time constant, which increases as the temperature difference
decreases,
can be computed from t_c=-T/(dT/dt). Integrating the Stefan-Boltzmann law
for T_a=0 yields: T=[T_i^(-3)+3SAt/MC_p]^(-1/3), where M is the mass of
the
sphere (constant) and C_p is its heat capacity at constant pressure
(assumed
constant). Differentiating readily gives us t_c=MC_p/SAT_i^3+3t, i.e.,
the
time constant starts off at a value based on the instantaneous rate of
heat
loss and then increases by 3 seconds for every second that elapses.
--
~~~~~~~~~~~
Professor Gauss
Excellent ! Elegantly intuitive. Of special note, is that
the rate of chage of the time constant, t_c, with time , t,
is independent of all spfere parameters and of the initial
temperature, ( d / dt ) t_c = 3. It's absolute.
[Old Man]
.
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| User: "Professor Gauss" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
10 Oct 2004 03:14:37 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:kLydncNiebO_C_XcRVn-ig@prairiewave.com...
"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message
news:g_ednYxdAoq00_XcRVn-hA@comcast.com...
"Old Man" <nomail@nomail.net> wrote in message
news:WfidnQsj54AVmfvcRVn-sA@prairiewave.com...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
"C * T_a" is the potential (t -> infinity) heat content per unit
volume of the solid sphere.
The specific heat conductivity can be taken as,
[ 12 * S * ( T_a )^4 ] / R = [ 12 / R ] [ dQ / dt @ T = 0 K]
which is proportional to the gross rate of ambient heat flow per
unit sphere volume flowing into the sphere at a sphere temperature
of T = 0 K.
The radius of the sphere, R, appears as "resistance to heat loss"
because the ratio of a sphere's volume, V, to its surface area, A,
is V / A = R / 3. That is, the ratio of the sphere's heat content to
its heat loss rate is proportional to the sphere's radius.
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
The time constant, which increases as the temperature difference
decreases,
can be computed from t_c=-T/(dT/dt). Integrating the Stefan-Boltzmann
law
for T_a=0 yields: T=[T_i^(-3)+3SAt/MC_p]^(-1/3), where M is the mass of
the
sphere (constant) and C_p is its heat capacity at constant pressure
(assumed
constant). Differentiating readily gives us t_c=MC_p/SAT_i^3+3t, i.e.,
the
time constant starts off at a value based on the instantaneous rate of
heat
loss and then increases by 3 seconds for every second that elapses.
--
~~~~~~~~~~~
Professor Gauss
Excellent ! Elegantly intuitive. Of special note, is that
the rate of chage of the time constant, t_c, with time , t,
is independent of all spfere parameters and of the initial
temperature, ( d / dt ) t_c = 3. It's absolute.
[Old Man]
Yes, it is an interesting result. And it is because the rate of heat loss
is proportional to the temperature difference to the 4th power. Had it been
proportional to the 3rd power, then we would have had (d/dt)t_c=2; or to the
2nd power, (d/dt)t_c=1. For the 1st power (Newton's Law of Cooling,
radioactive decay, etc.), (d/dt)t_c=0, i.e., it really is a decay
*constant*.
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand.
-- Ancient Chinese proverb
Remove caps when replying.
-- Modern American saying
.
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| User: "Steve Harris" |
|
| Title: Re: Engineering Linear Radiative Heat Loss |
08 Oct 2004 08:52:24 PM |
|
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"Old Man" <nomail@nomail.net> wrote in message news:<WfidnQsj54AVmfvcRVn-sA@prairiewave.com>...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
COMMENT:
You need to be able to derive this. Basically, it uses the
differential form of the Steffan-Boltzmann equation, and when you
differentiate P/A = S (T_1)^4 = (T_2)^4 with regard to temperature,
and assume T1 is "close" to T2 meaning T_1-T_2 is much smaller than
either T-1 or T_2, then you get P/A = 4 T^3 deltaT where T is the
generic temp which can be T_1 or T_2 (which ever one is T_ambient, can
be replaced by T_a here).. In your equation they've subsumed the 4
into the "12" , where the other factor of 3 comes from the ratio of a
sphere's volume to surface area. Here's how:
P = dQ/dt = CdT/dt = VCv * dT/dt = A S T^3 deltaT where Cv is the
volumetric specific heat capacity and V is the sphere volume
So dT/dt /(deltaT) = (A/V)*(1/Cv) * 4S T^3 = (3/R)*(1/Cv)*4S*T^3 = 12
ST^3 /(RCv)
Inverting:
deltaT/(dT/dt) = RCv/ (12ST^3) [Eq 1]
Given the differential form of Newton's law of cooling
d(deltaT)/dt = dT/dt = k * deltaT
Where k has units of 1/t and is 1/t_c where t_c is the cooling time
constant,
we recognise that the left side of [Eq 1] is just deltaT/ d(deltaT)/dt
= deltaT/[dT/dt] = t_c
So t_c for the sphere is the right side of Eq 1, QED. That's the
derivation (I've used Cv which is a better symbol for specific
volumetric heat capacity, where Old Man has used C).
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
The short answer is because T_a, the ambient temp, in the equation
above, is in the denominator. If it goes to zero, then t_c on the
left blows up.
SBH
.
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| User: "Old Man" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
08 Oct 2004 11:38:41 PM |
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"Steve Harris sbharris@ROMAN9.netcom.com" <sbharris@ix.netcom.com> wrote in
message news:79cf0a8.0410081752.7b8dfffa@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<WfidnQsj54AVmfvcRVn-sA@prairiewave.com>...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
COMMENT:
You need to be able to derive this. Basically, it uses the
differential form of the Steffan-Boltzmann equation, and when you
differentiate P/A = S (T_1)^4 = (T_2)^4 with regard to temperature,
and assume T1 is "close" to T2 meaning T_1-T_2 is much smaller than
either T-1 or T_2, then you get P/A = 4 T^3 deltaT where T is the
generic temp which can be T_1 or T_2 (which ever one is T_ambient, can
be replaced by T_a here).. In your equation they've subsumed the 4
into the "12" , where the other factor of 3 comes from the ratio of a
sphere's volume to surface area. Here's how:
P = dQ/dt = CdT/dt = VCv * dT/dt = A S T^3 deltaT where Cv is the
volumetric specific heat capacity and V is the sphere volume
So dT/dt /(deltaT) = (A/V)*(1/Cv) * 4S T^3 = (3/R)*(1/Cv)*4S*T^3 = 12
ST^3 /(RCv)
Inverting:
deltaT/(dT/dt) = RCv/ (12ST^3) [Eq 1]
Given the differential form of Newton's law of cooling
d(deltaT)/dt = dT/dt = k * deltaT
Where k has units of 1/t and is 1/t_c where t_c is the cooling time
constant,
we recognise that the left side of [Eq 1] is just deltaT/ d(deltaT)/dt
= deltaT/[dT/dt] = t_c
So t_c for the sphere is the right side of Eq 1, QED. That's the
derivation (I've used Cv which is a better symbol for specific
volumetric heat capacity, where Old Man has used C).
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
The short answer is because T_a, the ambient temp, in the equation
above, is in the denominator. If it goes to zero, then t_c on the
left blows up.
SBH
Nice derivation. Neat physics. Thanks.
Old Man was attempting to demonstrate that, in linearizing a
basically nonlinear process, as engineers are accustomed to
doing, a residue of non-linearity remains.
This is illustrated by the dependence of the time constant, t_c,
on the inverse cube of the ambient temperature, T_a:
t_c = ( C * V * T_a ) / [ 4 * A * S * ( T_a )^4 ]
However, the non-linearity is hidden by substitution of the exact
Stefan-Boltzmann law:
t_c = [ Q(T_a) ] / [ 4 ( dQ / dt ) @ T = 0 K ]
Well, ... that is, ... its hidden, ... except for that factor of 4.
[Old Man]
.
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| User: "Philip Holman" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 12:15:58 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:r5qdnZOJOMHO9vrcRVn-hg@prairiewave.com...
"Steve Harris sbharris@ROMAN9.netcom.com" <sbharris@ix.netcom.com>
wrote in
message news:79cf0a8.0410081752.7b8dfffa@posting.google.com...
"Old Man" <nomail@nomail.net> wrote in message
news:<WfidnQsj54AVmfvcRVn-sA@prairiewave.com>...
"Engineering Linear Radiative Heat Loss"
The radiative heat loss rate, dQ / dt, from a solid sphere
of surface of area, A, and temperature, T, surrounded by
an ambient environment of temperature, T_a, is given by
the Stefan-Boltzmann law,
dQ / dt = A * S [ ( T_a )^4 - ( T )^4 ]
where "S" is the Stefan-Boltzmann constant.
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
where "delta{T_i}" is the initial temperature difference at time,
t = 0, and "t_c" is the decay time constant.
If temperature were replaced by voltage, the above equation
looks similar to the discharge of a capacitor wherein the
time constant is given by
t_c = R * C
where "R" is the circuit resistance and "C" is its capacitance.
However, the decay time constant for a capacitor is independent
of voltage, but the time constant for radiative heat loss depends
upon the inverse cube of ambient temperature:
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
Where here, "R" is the sphere's radius and C is its specific heat
constant.
COMMENT:
You need to be able to derive this. Basically, it uses the
differential form of the Steffan-Boltzmann equation, and when you
differentiate P/A = S (T_1)^4 = (T_2)^4 with regard to temperature,
and assume T1 is "close" to T2 meaning T_1-T_2 is much smaller than
either T-1 or T_2, then you get P/A = 4 T^3 deltaT where T is the
generic temp which can be T_1 or T_2 (which ever one is T_ambient,
can
be replaced by T_a here).. In your equation they've subsumed the 4
into the "12" , where the other factor of 3 comes from the ratio of a
sphere's volume to surface area. Here's how:
P = dQ/dt = CdT/dt = VCv * dT/dt = A S T^3 deltaT where Cv is the
volumetric specific heat capacity and V is the sphere volume
So dT/dt /(deltaT) = (A/V)*(1/Cv) * 4S T^3 = (3/R)*(1/Cv)*4S*T^3 = 12
ST^3 /(RCv)
Inverting:
deltaT/(dT/dt) = RCv/ (12ST^3) [Eq 1]
Given the differential form of Newton's law of cooling
d(deltaT)/dt = dT/dt = k * deltaT
Where k has units of 1/t and is 1/t_c where t_c is the cooling time
constant,
we recognise that the left side of [Eq 1] is just deltaT/
d(deltaT)/dt
= deltaT/[dT/dt] = t_c
So t_c for the sphere is the right side of Eq 1, QED. That's the
derivation (I've used Cv which is a better symbol for specific
volumetric heat capacity, where Old Man has used C).
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
The short answer is because T_a, the ambient temp, in the equation
above, is in the denominator. If it goes to zero, then t_c on the
left blows up.
SBH
Nice derivation. Neat physics. Thanks.
Old Man was attempting to demonstrate that, in linearizing a
basically nonlinear process, as engineers are accustomed to
doing, a residue of non-linearity remains.
Yes they are aren't they. Structural FEM is based on a system of linear
equations to solve load distributions. However, structural deflections
affect the distributions but this is not normally taken into account
because it is more complicated (costly) and less conservative.
PH
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| User: "Steve Harris" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 07:52:25 PM |
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"Old Man" <nomail@nomail.net> wrote in message news:<r5qdnZOJOMHO9vrcRVn-hg@prairiewave.com>...
Nice derivation. Neat physics. Thanks.
Old Man was attempting to demonstrate that, in linearizing a
basically nonlinear process, as engineers are accustomed to
doing, a residue of non-linearity remains.
This is illustrated by the dependence of the time constant, t_c,
on the inverse cube of the ambient temperature, T_a:
t_c = ( C * V * T_a ) / [ 4 * A * S * ( T_a )^4 ]
However, the non-linearity is hidden by substitution of the exact
Stefan-Boltzmann law:
t_c = [ Q(T_a) ] / [ 4 ( dQ / dt ) @ T = 0 K ]
Well, ... that is, ... its hidden, ... except for that factor of 4.
[Old Man]
COMMENT:
No, the problem isn't due to linearization of a nonlinear term in
radiative cooling, for it's generic to all cooling problems, radiative
or not. As was pointed out, t_c also goes to infinite as ambient T
goes to zero, even with exact simple exponential Newtonian conductive
cooling.
BTW, here's another way to linearize the Stefan-Bolzmann equation:
P = S (Ts^4 –Ta^4). where Ts and Ta are the surface and ambient
environmental temperatures, respectively, in degrees kelvin. Now we
just factor out the polynomial:
P = S (Ts^2 + Ta^2 ) (Ts^2-Ts^2)
P = S (Ts^2 + Ta^2 ) (Ts + Ta) (Ts-Ta)
P = (Ts^2 + Ta^2 ) (Ts + Ta) deltaT
For cases where Ts and Ta are close in value (i.e., deltaT is small
with respect to either Ts or Ta, so either can be used as a generic
T), then just replace all Ts with Ta to get
P = (Ta^2 + Ta^2 ) (Ta + Ta) deltaT
P = (2Ta^2) (2Ta) deltaT
P = 4 Ta^3 deltaT
Ta could equally well be replaced by Ts, of course.
.
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| User: "" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
08 Oct 2004 09:49:02 AM |
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I didn't follow things completely in your post. (Bit of a headache this AM.)
But, I think I see a problem.
"Old Man" <nomail@nomail.net> wrote in message news:<WfidnQsj54AVmfvcRVn-sA@prairiewave.com>...
[snip]
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
Keep in mind your definition of when this applies.
[snips]
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
When T_a -> 0, can you still apply (T - T_a) << T ?
Socks
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| User: "Old Man" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
08 Oct 2004 05:41:31 PM |
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<puppet_sock@hotmail.com> wrote in message
news:c7976c46.0410080649.30ab0c19@posting.google.com...
I didn't follow things completely in your post. (Bit of a headache this
AM.)
But, I think I see a problem.
"Old Man" <nomail@nomail.net> wrote in message
news:<WfidnQsj54AVmfvcRVn-sA@prairiewave.com>...
[snip]
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
Keep in mind your definition of when this applies.
[snips]
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
When T_a -> 0, can you still apply (T - T_a) << T ?
Socks
Yes, that;s the problem.
If T_a => 0, then T - T_a = delta_T => T
Which doesn't satisfy delta_T << T. The approximation is
stated more properly by the stipulation |T_i - T_a| << T_a.
Incidentally, a more intuitive expression for the time constant is
from t_c = ( C * V * T_a ) / [ 4 * A * S * ( T_a )^4 ]
we get t_c = [ Q(T_a) ] / [ 4 ( dQ / dt ) @ T = 0 K ]
Its exactly correct, but violates the domain of the approximation.
[Old Man]
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| User: "Andy Resnick" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
08 Oct 2004 09:20:19 AM |
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Old Man wrote:
"Engineering Linear Radiative Heat Loss"
<snip>
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
<snip>
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
<snip>
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
Off the top of my head, it's because of your assumption- exponential
change in the temperature. It takes an infinite amount of time for the
temperature to cool to 0 K. This seems to be ok, as passively cooled
objects should not be able to radiate their way to 0 K in a finite
time. (Entropy considerations)
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
CWRU School of Medicine
tanspose 'op' for mail
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| User: "The Ghost In The Machine" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 12:00:48 PM |
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In sci.physics, Andy Resnick
<axr67@op.cwru.edu>
wrote
on Fri, 08 Oct 2004 10:20:19 -0400
<ck67rr$drd$1@eeyore.INS.cwru.edu>:
Old Man wrote:
"Engineering Linear Radiative Heat Loss"
<snip>
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
<snip>
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
<snip>
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
Off the top of my head, it's because of your assumption- exponential
change in the temperature. It takes an infinite amount of time for the
temperature to cool to 0 K. This seems to be ok, as passively cooled
objects should not be able to radiate their way to 0 K in a finite
time. (Entropy considerations)
The object can't even cool below the about 2.5 - 3K background radiation. :-)
--
#191,
It's still legal to go .sigless.
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| User: "Professor Gauss" |
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| Title: Re: Engineering Linear Radiative Heat Loss |
09 Oct 2004 03:55:17 PM |
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Modeling passive cooling as an exponential decay also requires an infinite
amount of time to reach the environment's temperature, so this does not
explain why Old Man's time constant goes to infinity as time goes to
infinity.
--
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand.
-- Ancient Chinese proverb
Remove caps when replying.
-- Modern American saying
"Andy Resnick" <axr67@op.cwru.edu> wrote in message
news:ck67rr$drd$1@eeyore.INS.cwru.edu...
Old Man wrote:
"Engineering Linear Radiative Heat Loss"
<snip>
if the temperature difference between sphere and environment
is small, delta{T} << T, then the sphere's temperature declines
exponentially with time, t (we assume an infinite heat conductivity
for the solid sphere):
delta{T} / delta{T_i} = exp[ - t / t_c ]
<snip>
t_c = ( R * C * T_a ) / [ 12 * S * ( T_a )^4 ]
<snip>
Old Man wonders why t_c -> infinity as T_a -> 0 K ?
[Old Man]
Off the top of my head, it's because of your assumption- exponential
change in the temperature. It takes an infinite amount of time for the
temperature to cool to 0 K. This seems to be ok, as passively cooled
objects should not be able to radiate their way to 0 K in a finite time.
(Entropy considerations)
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
CWRU School of Medicine
tanspose 'op' for mail
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