# experiment/ word problem help

• Oct 6th 2010, 04:05 PM
Tweety
experiment/ word problem help
Quote:

2.6 A seed crystal of diameter D (mm) is placed in a solution of dissolved salt, and
new crystals are observed to nucleate (form) at a constant rate $\displaystyle r (crystals \\\\ min^{-1})$ .
Experiments with seed crystals of differentt sizes show that the rate of nucleation
varies with the seed crystal diameter as

$\displaystyle r(crystals\\ min^{-1}) = AD + BD^{2}$

where D is in mm, and where A and B are constants (which must be determined).

(a) What are the units of the constants A and B? Suppose that a D = 1 mm
seed crystal produces a nucleation rate of r = 210 crystals min��1, whereas a
D = 2 mm seed crystal produces a nucleation rate of r = 440 crystals min��1.
Use this information to evaluate A and B.

(b) Calculate the crystal nucleation rate in crystals min��1 for a seed crystal di-
ameter of 0.05 inch.

(c) Derive a formula for r (crystals min��1) in terms of D (in). What are the
values and units of the constants in the expression? Check the formula with
the result of 2.6(b). Comment on your results.

(d) Derive a formula for $\displaystyle r (crystals s^{-1})$ . what are the values snd units of the constants in the expression.
for part 'a', I got the units of A and B, as $\displaystyle crystals\\\\\\ min^{-1} mm^{-1}$ for A

$\displaystyle crystals\\\\\\\\ min^{-1} mm^{-2}$ for B

so the mm would cancels out, with the D units, and the equation balances. is this correct?

Also worked out A = 200 and B =10

so equation would look something like this,

$\displaystyle r( crystals \\\\\ min^{-1} ) = \frac{200 Dmm\\\\\ crystals}{min\\\\ mm} + \frac{10Dmm^{2} \\\\\\crystals}{min\\\ mm^{2}}$

b) = $\displaystyle r = 200(1.27) + 10(1.27)^{2}$

c) ? Really Need help with part c. Do I just replace the units for diameter in my equation?
• Oct 7th 2010, 03:38 AM
emakarov
Quote:

$\displaystyle r( crystals \\\\\ min^{-1} ) = \frac{200 Dmm\\\\\ crystals}{min\\\\ mm} + \frac{10Dmm^{2} \\\\\\crystals}{min\\\ mm^{2}}$
It should be $\displaystyle 10D^2mm^{2}$. Otherwise, you are correct so far.

Concerning c), suppose that the diameter of the seed crystal in inches is $\displaystyle D'$, while the diameter in millimeters is still $\displaystyle D$. Then $\displaystyle D=25.5D'$. So it suffices to substitute $\displaystyle 25.5D'$ for $\displaystyle D$ in your formula for $\displaystyle r$ to express $\displaystyle r$ through $\displaystyle D'$.
• Oct 7th 2010, 08:44 AM
Tweety
Quote:

Originally Posted by emakarov
It should be $\displaystyle 10D^2mm^{2}$. Otherwise, you are correct so far.

Concerning c), suppose that the diameter of the seed crystal in inches is $\displaystyle D'$, while the diameter in millimeters is still $\displaystyle D$. Then $\displaystyle D=25.5D'$. So it suffices to substitute $\displaystyle 25.5D'$ for $\displaystyle D$ in your formula for $\displaystyle r$ to express $\displaystyle r$ through $\displaystyle D'$.

Thank you, could you please explain how you got D = 25.5D' ? I don't see how the two equal each other?
• Oct 7th 2010, 08:58 AM
emakarov
Sorry, I meant D mm = 25.4 mm/in * D' in.
• Oct 7th 2010, 11:45 AM
Tweety
Quote:

Originally Posted by emakarov
Sorry, I meant D mm = 25.4 mm/in * D' in.

How did you get this?
• Oct 7th 2010, 12:00 PM
emakarov
An inch has 25.4 millimeters.
• Oct 7th 2010, 12:25 PM
Tweety
Quote:

Originally Posted by emakarov
An inch has 25.4 millimeters.

Thank you, but how does this relate to the question? I understand that I have to find an expression for Dmm in terms of inches, and than substitute this into my equation, but I dont understand how to do it from the expression you have come up with.

why does $\displaystyle Dmm = \frac{25.4mm Din}{in}$
• Oct 7th 2010, 11:03 PM
emakarov
Question: How many millimeters are there in 5 inches?
Answer: 25.4 mm per inch * 5 inches = 127 mm.

Question: How many millimeters are there in D' inches?
Answer: 25.4 mm/in * D' in = 25.4 * D' mm. Call the result D.

r(D) = 200 * D + 10 * D^2. Also, D = 25.4 * D'. Therefore, r(D') = 200 * 25.4 * D' + 10 * (25.4 * D')^2.

Edit.
Quote:

why does $\displaystyle Dmm = \frac{25.4mm Din}{in}$
Note that D and D' are two different numbers.
• Oct 9th 2010, 07:01 AM
Tweety
Thank you, I kind of get it now.

I also tried doing part 'd', and need some help.

$\displaystyle r(crystals min^{-1}) = 200Dmm + 10D^{2}mm^{2}$

So we have to change from minutes to seconds, 1 minute = 60 seconds. Using your same method,

$\displaystyle r = \frac{crystals}{min}$

$\displaystyle r'= \frac{crystals}{sec}$

not sure what to do from here.
• Oct 9th 2010, 07:52 AM
Tweety
Actually I have worked out part d, so ignore my previous post.

can someone please clarify part 'c', for me as I still not able to grasp how to do it.

if $\displaystyle 1 inch = 25.4 mm$

than $\displaystyle Dmm = \frac{1}{25.4} \times D inch$

so substitute in $\displaystyle \frac{D(in)}{25.4}$ for Dmm

Why is this method not correct?
• Oct 9th 2010, 08:19 AM
emakarov
Quote:

Originally Posted by Tweety
if $\displaystyle 1 inch = 25.4 mm$

than $\displaystyle Dmm = \frac{1}{25.4} \times D inch$

Conversion between two systems is always tricky for me because I can never remember whether one has to add or subtract, or whether to multiply or divide. For example, if $\displaystyle f:\mathbb{R}\to\mathbb{R}$ is a function, the graph of f(x-5) is the graph of f(x) moved 5 units to the right. Similarly, I remember that we are supposed to get up earlier when the shift to summer time happens, but I can never remember whether the watch has to be adjusted forward or backward.

Nevertheless, this issue is trivial when considered formally. Again, I prefer to use different symbols D and D' to avoid confusion. Let D be the size in millimeters and let D' be the same size in inches.

Quote:

Question: How many millimeters are there in D' inches?
Answer: 25.4 mm/in * D' in = 25.4 * D' mm. Call the result D.
So D = 25.4 * D', not D = 1/25.4 * D', as you write.