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Thermal Physics

Cool goIITian

 Joined: 19 Nov 2007 Post: 53
3 Jan 2009 17:46:56 IST
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A glass window is to be fitted in an aluminium frame.The temperature on the working day is 40 degree C & the glass window measures exactly 20cm*30cm. What should be the size of the aluminium frame so that there is no stress on the glass in winter even if the temperature drops to 0 degrees C?        Coefficients of linear expansion of glass & aluminium are 0.000009 &  0.000024 respectively.

Blazing goIITian

Joined: 20 Dec 2008
Posts: 599
4 Jan 2009 15:20:49 IST
1 people liked this

Let the dimensions be x and y at 40 degrees.
Then at 0 degrees, the dimensions of the frame and the window are equal.
dimensions of window at 0 = 20 [ 1 - (0.000009)*40 ] and 30 * [ 1 - (0.000009)*40 ]
dimensions at 0 degrees of frame are x [ 1 - (0.000024)*40 ] and y [ 1 - (0.000024)*40 ]
So 20 [ 1 - (0.000009)*40 ] = x [ 1 - (0.000024)*40 ]
or x = 20 [ 1 - (0.000009)*40 ] / [ 1 - (0.000024)*40 ]
which can be approximated as 20 [ 1 - (0.000009)*40 ] [ 1 + (0.000024)*40 ]
which can further be approximated as 20 [ 1 + (0.000024 - 0.000009)*40 ] = 20 [ 1 + 0.000015 * 40 ] = 20 [ 1 + 0.0006] = 20.012 cm.
Similarily, y = 30 [ 1 + 0.0006] = 30.018 cm.
So the dimensions are 20.012 and 30.018

Blazing goIITian

Joined: 8 Oct 2008
Posts: 8064
9 Jan 2009 21:35:52 IST
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hello dear

The final length of aluminium should be equal to final length of glass.

Let the initial length o faluminium = l

l (1 - alphaal DT)  =  20(1- alpha0 D (theta))

l(1 – 24 × 10–6 × 40) = 20 (1 – 9 × 10–6 × 40)

l(1 – 0.00096) = 20 (1 – 0.00036)

l = 20 * 0.99964 / 0.99904   = 20.012 cm

Let  breadth of aluminium = b

b(1- alphaal DT) = 30 (1- alpha0Dtheta)

solvinf for b and putting the value:

b = 30 (1-9 *10-6 *40) / (1- 24 *10-6 *40)

b        =       30.018 cm

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