Building a Bimetallic Thermometer: Aluminium and Brass Sheets

[dasher]

I am currently working on a bimetallic thermometer using alumium and brass.

The coefficient of linear and volumetric thermal expansion of aluminium is 23 and 69 respectively. The coefficient of linear and volumetric thermal expansion of brass is 19 and 57 respectively.

I am using aluminium and brass sheets. The dimensions are 0.015 by 2 by 10 inches for brass and 0.016 by 2 by 10 inches for brass. I will be sticking them together to make the thermometer. And, does anyone know what can be used to cut the metal sheets? Because the the dumensions of the sheets I have are 0.015 by 4 by 10 inches and 0.016 by 4 by 10 inches respectively.

I am worried that the similar coefficients of thermal expansion will cause the changes to be very minor and the thermometer might fail or be inaccurate. Could someone advise me on whether the thermometer will work for a temperature range of 0.0 to 100.0 degrees C?

And, is the coefficient of linear of volumetric thermal expansion used in this case if I intend to calculate the expansion and the degree to which the thermometer will bend?

Lastly, if the bimetallic thermometer fails, what if I make a monometallic thermometer? I believe aluminium, with a greater coefficient of linear thermal expansion, will be a good choice, right? But why do most people make bimetallic instead of monometallic thermometers? Is it for aesthetic purposes?

I would appreciate any helpful response. Thanks alot! (:

 

[Redbelly98]

To cut sheet metal, you can get buy a pair of "sheet metal sheers" or "tin snips" at a hardware store.

One issue will be, how will you attach the two strips together? The two metals must not slip with respect to each other, but must be able to bend.

Use the linear expansion coefficients to calculate the bend. That's to see if your device will work in principle, not to set the usable temperature scale. You'll need to calibrate it to get the temperature scale correct.

Lastly, it must be bimetallic if you are to see any observable change with the naked eye. A single-metal strip will not expand/contract enough for you to see it.

 

[dasher]

Thanks alot! I am intending to use E6000 glue to stick the metals together. The staff at the art and crafts shop I went to recommended me that. Will that work?

I will calculate if the bimetallic strip will work in theory soon. (:

As for a monometallic strip, can't the temperature change be indicated by an increase in length? I was thinking that that might work actually.

 

[Redbelly98]

Thanks alot! I am intending to use E6000 glue to stick the metals together. The staff at the art and crafts shop I went to recommended me that. Will that work?

I'm not familiar with that glue, so I don't know how strong it is. You can try it out. I have known store staff who will simply recommend whatever it is they sell in their store, even though they have little knowledge about things like bimetallic strips and thermal expansion coefficients.

According to this Wikipedia article, it is common practice to attach the strips by riveting or welding:
http://en.wikipedia.org/wiki/Bi-metallic_strip

As for a monometallic strip, can't the temperature change be indicated by an increase in length? I was thinking that that might work actually.

In theory, yes. But for every 1 C temperature change, your 10" strips will change in length by about 0.0002". How would you propose to measure such a small difference?

FYI, in case you need the equation for a bimetallic strip's curvature, the Wikipedia article has the equation.

 

[dasher]

true. i hope they gave me correct info. i hope the glue works!

and okay. now i see the rationale behind making a bimetallic instead of monometallic thermometer! thanks a lot ((:

according to the formula in wikipedia to calculate the curvature of the bimetallic strip, (delta)T is the current temperature minus the reference temperature. what is the reference temperature in this case?

 

[Redbelly98]

Reference temperature is the temperature while you are attaching the strips together, since they will be straight at that point.

The forces involved with thermal expansion are incredibly strong, so whatever method you use to attach the strips must be strong also. If I were doing this as a home project, I'd probably try using sheet-metal screws spaced every 2" or so along the strips. But go ahead and try the glue first, if you've already bought it.

By the way, on that Wikipedia equation, I noticed it simplifies when the two thicknesses are equal (as they are in your case).

 

[dasher]

ok thanks alot. what about the normal screw and washer instead of the sheet-metal screws? will it work? wouldn't the screws affect how the bimetallic strip bend?

and yep! but after i calculated, i got ~33 /in, what does that mean? i think i might have made some calculation errors because 33 /in doesn't really make much sense to me.

 

[Redbelly98]

Screw and washer might work too, try it out. I suggested sheet metal screws because they would be a tight snug fit in the holes of the metal. On the other hand, a standard machine screw and nut might hold the pieces against each other better. If you go that route, I'd include a lock-washer since repeated movements back and forth tend to make screws loosen up.

And yes, the screw would have a small effect on the amount of bend, but I wouldn't worry about it.

The curvature calculated from the Wikipedia equation is the reciprocal of the radius of curvature. 33/in means the radius of curvature of the bend is

$$1/r = 33/\text{in}$$

or

$$r = \frac{1}{33/\text{in}} = \frac{1 \ \text{inch}}{33} = 0.030 \ \text{in}$$

That says the strip will bend into a circle whose radius equals the thickness of the combined strips! That can't be right, so let me double-check that equation ...

For a very rough ballpark figure, I'll set the Young's modulus E for the two materials equal to each other. Then the E's all cancel out in the equation. And since the thicknesses h1 and h2 are equal as well, we have

$$\kappa = \frac{6 \cdot 2 \ h^3 \ \epsilon}{(1+4+6+4+1) \ h^4} = \frac{3 \ \epsilon}{4 \ h}$$

or
$$r = \frac{1}{\kappa} = \frac{4 \ h}{3 \ \epsilon} = \frac{4 \ h}{3 \ \Delta \alpha \ \Delta T}$$

The expansion difference is 4 x 10^-6 per deg C, and h is 0.015", so for a 1 deg C change I get
r = 5000 in, or about 400 ft.
For a 10 C change it will be 1/10 of that, about 40 feet.
And for a 100 C change it will be 1/100 of that, about 4 feet.

I think the deflection is given by
(1/2)L^2 / R,
so for a 10" length we have:
0.01" deflection for 1 C change
0.1" for 10 C change
1" for 100 C change

You'll definitely notice a bend if you submerge it in boiling water, and you could make a gauge accurate to perhaps 5 or 10 C.

 

[gavin_gj]

I've been thinking about having a go at building one- how is your one going?

Brass and Alu are similar in coefficients- why not use e.g. steel, alu- with greater difference in coefficients, should get much more movement?

Gavin