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How much glue?

oxicottin

New member
Mar 21, 2013
7
Hello everyone, This is my first post and im not very good at math and I have been trying to figure out how many bags of glue we go through or we will need.

Here is the glue we use:

UNI-REZ™ 2635 A resin with chemical and grease resistance, low temperature flexibility, excellent high/low temperature recycling, good adhesion to plastics, suitable for cabinet and furniture assembly, multipurpose assembly and adhesion to preheated metals applications.

Ok, now we run the thickness at .0015 at a width of 2.125 and the length is in linier foot lets say 216,000lf. Now our bags come in 50lb bags. How in world do I figure out how much glue we can get out of a bag for the 216,000 linier foot. I have bee reading and im hearing volocity and volume ect. and im confused on what to even calculate and how. "The glue is in beads and melted down"

Thanks!!!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We can figure this out if we have one other vital piece of information...the weight density of the glue. We need to know how much space a pound of the glue occupies (or be able to compute this if the weight density is given in other units). Is this information given?
 

oxicottin

New member
Mar 21, 2013
7
We can figure this out if we have one other vital piece of information...the weight density of the glue. We need to know how much space a pound of the glue occupies (or be able to compute this if the weight density is given in other units). Is this information given?

I will do some digging around and see if I cant find the weight density in the meantime,
How is this figured out anyway? Thanks

OK, I cant find anything on Weight Density so I will have to call the company tomorrow, But here is specks I did find.

Appearance/state: Amber solid
Odor: Mild amine
Ph: Not applicable
Vapor Pressure: <0.001 mm Hg at 20°C (68°F)
Vapor density: Not applicable
Boiling point: Not applicable
Melting point: 140°C
Solubility in water: <0.1% at 25°C (77°F)
Specific gravity: 0.96 at 25°C/25°C (water=1.00)
Evaporation rate: Approx. 0 (n-BuAc=1)
% Volatile by weight: 0


Softening Point, °C, Ring & Ball, ASTM E 28 135-145
Viscosity, LVT Brookfield, Spindle # 31, cps/mPa×s at 190°C, ASTM D 3236 3000-4700
Color, 40% in n-Butanol, Gardner 7 Maximum
TYPICAL VALUES
Acid Value, mg KOH/g 0.9
Amine Value, mg KOH/g 7.5
Color, 40% in n-Butanol, Gardner 5
Tensile Strength, psi (MPa), ASTM D 1708 280 (1.9)
Elongation, %, ASTM D-1708 200
Tensile Modulus, psi (MPa), ASTM D 638 3000 (21)
Peel Strength, pli (N/25mm), plasticized PVC to itself, ASTM D 1876 20 (88)
Mandrel Bend Flexibility, (1” diameter), pass, °C, ASTM D 3111 -30
Open Time, sec., 10 mil film, ASTM D 4497 25
Shear Adhesion Failure Temperature (SAFT), 1 kg. wt. °C, ASTM D 4498 130
Water Absorption, ASTM D-570:
after 24 hours, %
after 7 days, %
1.70
3.11
Pounds per Gallon, (g/cc), 25°C 8.0 (0.96)
Flash Point,°C, COC 271 (520°F)
 
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oxicottin

New member
Mar 21, 2013
7
Is this what your looking for?

Pounds per Gallon, (g/cc), 25°C 8.0 (0.96)


Thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I am assuming you are trying to figure this our for a real world application rather than asking for help on a homework problem, so I will work out a full solution for you.

Given the specific gravity is 0.96, we then know that 1 cubic centimeter is 0.96 grams. Now, there are 231 cubic inches in a gallon and 2.54 centimeters per inch and 453.592 grams per pound. Hence:

\(\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\cdot\frac{1\text{ lb}}{453.592\text{ g}}\cdot\left(\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{12\text{ in}}{1\text{ ft}} \right)^3\approx60\frac{\text{lb}}{\text{ft}^3}\)

Next, we need to compute the volume of glue we need for the given job. Assuming the dimensions you have given are all in ft., this is:

\(\displaystyle V=2.125\,\times\,216000\,\times\,0.0015=688.5\text{ ft}^3\)

This means we need, in pounds of glue:

\(\displaystyle 60\frac{\text{lb}}{\text{ft}^3}\cdot688.5\text{ ft}^3\approx41310\text{ lb}\).

Dividing this by 50, we find that this equates to about 826.2 50 lb. bags of glue that will be needed for the job.
 

oxicottin

New member
Mar 21, 2013
7
Yes its for real word... yes its in inch. was post #4 the weight density you were looking for?

Thanks
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
So, the thickness and width are in inches? In that case divide the results I gave above by $12^2=144$ to get $5.7375$ 50 lb. bags.

The specific gravity gave us a means of finding the pounds per cubic foot.
 

oxicottin

New member
Mar 21, 2013
7
Given the specific gravity is 0.96, we then know that 1 cubic centimeter is 0.96 grams. Now, there are 231 cubic inches in a gallon and 2.54 centimeters per inch and 453.592 grams per pound. Hence:

\(\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\cdot\frac{1\text{ lb}}{453.592\text{ g}}\cdot\left(\frac{2.54\text{ cm}}{1\text{ in}}\cdot\frac{12\text{ in}}{1\text{ ft}} \right)^3\approx60\frac{\text{lb}}{\text{ft}^3}\)
.

Mark, Instead the formula above can you write it out in "General Math" as in:

0.096 * this / this = this

I'm sorry I never took algebra or any of those classes (Never had to years ago) so I'm lost on your formula. Thanks!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Realistically, you're not going to get $100\%$ yield from each bag or box of glue. I don't know whether that's going to be an important factor or not. It might be second-order, or even third-order, in which case you might not be interested. As the surface area-to-volume ratio of your bags increases, this effect will increase.
 

oxicottin

New member
Mar 21, 2013
7
Realistically, you're not going to get $100\%$ yield from each bag or box of glue. I don't know whether that's going to be an important factor or not. It might be second-order, or even third-order, in which case you might not be interested. As the surface area-to-volume ratio of your bags increases, this effect will increase.
Yes I understand that is not aprox but that still doesnt answer my question in post 9 "how is it calculated in general terms"

Thanks!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Mark, Instead the formula above can you write it out in "General Math" as in:

0.096 * this / this = this

I'm sorry I never took algebra or any of those classes (Never had to years ago) so I'm lost on your formula. Thanks!
What I did was to take the given specific gravity of \(\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\) and then multiplied by 1 in the form of a string of equivalences such that that all units cancel except for the desired units of \(\displaystyle \frac{\text{lb}}{\text{ft}^3}\).
 

oxicottin

New member
Mar 21, 2013
7
What I did was to take the given specific gravity of \(\displaystyle 0.96\frac{\text{g}}{\text{cm}^3}\) and then multiplied by 1 in the form of a string of equivalences such that that all units cancel except for the desired units of \(\displaystyle \frac{\text{lb}}{\text{ft}^3}\).
Thank you for the help.... Is this correct then?

0.96 g/cm^3 X 62.4280115 = ___lb/ft^3

Code:
DENSITY
g/cm3. Multiply By  62.43 To Obtain lb/ft3
                    0.03613 To Obtain lb/in3
Conversion Chart Found Here..
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, those are roughly equivalent to what I posted.