Using the following equation: d=1.5/nL*M where nLis the number of grains interse
ID: 2155162 • Letter: U
Question
Using the following equation: d=1.5/nL*M where nLis the number of grains intersected per unit length and M is the magnification of the micrograph, estimate the average grain diameter (in micrometers) of the porous microstructure shown in the feature box on page 8 of Chapter 1 of the text. Use a random line that cut across the diagonal of the figure from its lower-left corner to its upper-right corner. (pg. 8 of Chapter 1 below)
Here is what i know.... book says answers are 170microns, and 41.3 microns.
I used a digital caliper and measure the size of the gauges.
a) 6.06 mm long 50micron scale bar
b) 11.54mm long 50 micron scale bar
Setting up equations to get magnification:
a) (6.06*10^-3 m) / (50*10^6 m) = 121.2 magnification
b) (11.54*10^-3 m) / (50*10^6 m) =230.8 magnification
Then using a 'random line' "that cut across the diagonal of the figure from its lower-left corner to its upper-right corner" and counting the grains I got:
a) 6 grains intersected
b) 11 grains intersected
Plugging this infomation in to the equation given (d=1.5/nL*M)I get:
a) d= 1.5 / (6*121.2)
= 2.06*10^-3
b)d= 1.5 / (11*230.8)
= 5.908 x 10^-4
Where did I go wrong? Those are nowhere near the answer and I have no units because the meters canceled out in the magnification!
PLEASE HELP!! And use some explanations, please do not just give me an answer, I wish to understand how to do this.
Thank you!!
Explanation / Answer
The average grain diameter is d = 1 um.The grains are very small size particles and have very tiny diameter.They fit in solids like Alumina and act like doping materials.By doping the grains we increase the electrical conductivity of the metals.Grains are very useful elements in doping and have a diameter of one micrometer. 170 micometers and 41.3 micrometers
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