Thermal Divider

Is it purely a function of the size of the metal piece, or is the type of
metal part of the equation?

The other thing I don't understand is that if I do a good job of insulating
the metal piece from ambient how it won't eventually come to the same
temperature as the head.

Thermo and I never did get along very well.

Jim



"John Kimmel" wrote in message
...
RST Engineering (jw) wrote:
a thermal divider that won't be subject to the day to day shuffle
of the airflow over the cylinders?

Jim
Attach the sensor to a piece of metal. Bolt the metal to something on the
cylinder head. Wrap the metal/sensor with insulating tape to keep cooling
air off it. Change the size of the metal piece to vary the temperature
ratio.

--
John Kimmel

remove x

"He's dead, Jim."

RST Engineering (jw) wrote:
Is it purely a function of the size of the metal piece, or is the type of
metal part of the equation?

The other thing I don't understand is that if I do a good job of insulating
the metal piece from ambient how it won't eventually come to the same
temperature as the head.

Thermo and I never did get along very well.

Jim

The insulation is to shield the sensor and heat sink from the effects of
transient airflow. The insulation won't be perfect in any case, but you
can vary the amount of insulation and the size of the heat sink to vary
your results. If the insulation is perfect, then you will need to bolt
the heat sink to something that doesn't get very hot.

This idea might not work at all, but it is something that would be easy
to try, and easy to change variables on.


--
J Kimmel

www.metalinnovations.com

"Cuius testiculos habes, habeas cardia et cerebellum." - When you have
their full attention in your grip, their hearts and minds will follow.

RST Engineering (jw) wrote:
Is it purely a function of the size of the metal piece, or is the type of
metal part of the equation?

The other thing I don't understand is that if I do a good job of
insulating the metal piece from ambient how it won't eventually come to
the same temperature as the head.


If you mount the sensor in a bar that is picking up heat at one end, and
losing heat at the other - you sensor will read somewhere between the two
temperatures. But, if you don't know a lot about the heat transfer rates at
each end, you will not know how the temperature your sensor sees is related
to the temperature you want to measure.

Think of a series voltage divider with two known resistors in the middle and
an unknown resistor attached to each end. What does the votage measured at
the junction between the two known resistors tell you about the source
voltage?

--
Geoff
The Sea Hawk at Wow Way d0t Com
remove spaces and make the obvious substitutions to reply by mail
When immigration is outlawed, only outlaws will immigrate.

If you mount the sensor in a bar that is picking up heat at one end, and
losing heat at the other - you sensor will read somewhere between the two
temperatures. But, if you don't know a lot about the heat transfer rates
at each end, you will not know how the temperature your sensor sees is
related to the temperature you want to measure.

That's exactly the point, Geoff. When I do a resistive divider, it is always
between two "hard" voltages, either a variable voltage and a reference
voltage or a variable voltage and ground. I know what the reference is and
can easily calculate the variable. The thermal reference is not quite so
trivial. It is a function of airflow, ambient temperature, and phase of the
moon.



Think of a series voltage divider with two known resistors in the middle
and an unknown resistor attached to each end. What does the votage
measured at the junction between the two known resistors tell you about
the source voltage?


Absolutely nothing. An equation in one unknown with two degrees of freedom
is insoluble. There are an infinite number of correct answers and an
infinite number of incorrect answers. HOWEVER, if you let me measure the
voltage ACROSS one of those known resistors and THEN the voltage at the
junction, I've got a fighting chance if you know what the bottom end
resistor is tied to.

Jim

"RST Engineering" wrote in message
news
Think of a series voltage divider with two known resistors in the middle
and an unknown resistor attached to each end. What does the votage
measured at the junction between the two known resistors tell you about
the source voltage?


Absolutely nothing. An equation in one unknown with two degrees of
freedom is insoluble. There are an infinite number of correct answers and
an infinite number of incorrect answers. HOWEVER, if you let me measure
the voltage ACROSS one of those known resistors and THEN the voltage at
the junction, I've got a fighting chance if you know what the bottom end
resistor is tied to.


Think of heat flow as current, temperature as voltage, the actual connecton
between your divider and the heat source / sink like unknown resistors
(area, contact, material all make a difference as in a high current circuit)
your bar with the sensor in the middle is like the voltage drop in a
transmission line - flow is a function of area, material, potential. Heat
loss from the bar is a little harder - I guess in a high tension
transmission line there is some leakage to ground across the insulators?

And you thought you didn't know thermodynamics...

fwiw, I think I would just buy a different sensor, eh?

--
Geoff
The Sea Hawk at Wow Way d0t Com
remove spaces and make the obvious substitutions to reply by mail
When immigration is outlawed, only outlaws will immigrate.

"Capt. Geoffrey Thorpe" The Sea Hawk at wow way d0t com wrote in message
...
"RST Engineering" wrote in message news
Think of a series voltage divider with two known resistors in the middle and an unknown resistor attached to each
end. What does the votage measured at the junction between the two known resistors tell you about the source
voltage?


Absolutely nothing. An equation in one unknown with two degrees of freedom is insoluble. There are an infinite
number of correct answers and an infinite number of incorrect answers. HOWEVER, if you let me measure the voltage
ACROSS one of those known resistors and THEN the voltage at the junction, I've got a fighting chance if you know what
the bottom end resistor is tied to.


Think of heat flow as current, temperature as voltage, the actual connecton between your divider and the heat source /
sink like unknown resistors (area, contact, material all make a difference as in a high current circuit) your bar with
the sensor in the middle is like the voltage drop in a transmission line - flow is a function of area, material,
potential. Heat loss from the bar is a little harder - I guess in a high tension transmission line there is some
leakage to ground across the insulators?

And you thought you didn't know thermodynamics...

fwiw, I think I would just buy a different sensor, eh?

--
Geoff
The Sea Hawk at Wow Way d0t Com
remove spaces and make the obvious substitutions to reply by mail
When immigration is outlawed, only outlaws will immigrate.


Maybe use two sensors, one way out at the cold end, and the other 'half way out'. Use some sort of logic to sort out the
cold end and adjust from there...

"Capt. Geoffrey Thorpe" The Sea Hawk at wow way d0t com wrote in message
news
RST Engineering (jw) wrote:
Is it purely a function of the size of the metal piece, or is the type
of
metal part of the equation?

The other thing I don't understand is that if I do a good job of
insulating the metal piece from ambient how it won't eventually come to
the same temperature as the head.


If you mount the sensor in a bar that is picking up heat at one end, and
losing heat at the other - you sensor will read somewhere between the two
temperatures. But, if you don't know a lot about the heat transfer rates
at
each end, you will not know how the temperature your sensor sees is
related
to the temperature you want to measure.

Think of a series voltage divider with two known resistors in the middle
and
an unknown resistor attached to each end. What does the votage measured at
the junction between the two known resistors tell you about the source
voltage?

how about two temp sensors at measured intervals along the insulated bar,
then assuming the temperature gradient is constant along the bar, you can
work out the temp at the cylinder head.

Colin =^.^=

Now THERE'S a hell of a thought. Consider a "bar" three inches long with
sensors at 1 and 2 inches. Betcha the temperature at the head (zero inches)
is (temp 1) plus delta (temp 1 minus temp 2).

Jim


how about two temp sensors at measured intervals along the insulated bar,
then assuming the temperature gradient is constant along the bar, you can
work out the temp at the cylinder head.

Colin =^.^=

"RST Engineering (jw)" wrote in message ...
Now THERE'S a hell of a thought. Consider a "bar" three inches long with sensors at 1 and 2 inches. Betcha the
temperature at the head (zero inches) is (temp 1) plus delta (temp 1 minus temp 2).

Jim


how about two temp sensors at measured intervals along the insulated bar,
then assuming the temperature gradient is constant along the bar, you can
work out the temp at the cylinder head.

Colin =^.^=





Hmmm, just posted that...must be a fairly good way. Think cold junction compensation....

On Mon, 29 May 2006 17:48:07 -0700, the renowned "RST Engineering
\(jw\)" wrote:

Now THERE'S a hell of a thought. Consider a "bar" three inches long with
sensors at 1 and 2 inches. Betcha the temperature at the head (zero inches)
is (temp 1) plus delta (temp 1 minus temp 2).

Jim

Are you assuming negligible heat loss from the bar itself?




Best regards,
Spehro Pefhany
--
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