Maths id like to see on the wiki (but dont have the brain power to do myslelf!)

[julianf]

Id like a caclulator where i could work out what temp 200 ltrs of oil would have to be to end at 60c if i added, say, 50 liters of oil at 10c.

I found this for water -

Calculate the final temperature when 50mL of water at 60 degrees Celsius are added to 25mL of water at 20 degrees Celsius?
Answer:
Okay, so we have some amount of water that is hot, and some amount of water that is cold. Mix em together and you will get something warm. At the end, both the hot water and cold water you started with will be the same temperature. If both will equal the same temperature at the end of the reaction, we can make them equal to each other and solve for "X".


s = specific heat (specific heat of water is 4.18 J/ degree Celsius x gram)
m = mass
(Tf - Ti) aka. "delta T" = change in temperature, Final - Initial.


However since one of them is losing heat and the other is gaining heat energy, the equation is:
qhot = - qcold


s1 x m1 x (Tf - Ti) = - s2 x m2 x (Tf - Ti)
(4.18 J / C x g) x (50.g) x (X - 60 C) = - (4.18 J / C x g) x (25g) x (X - 20 C)
DROP UNITS and solve for "X"
depending on whether you round or not, I'll put down exact numbers and round at end

209X - 12540 = - 104.5X + 2090
313.5X = 14630
X = 46.66666

In which case you can round it to 47 C.

So to sum it up, if you mix these two together, the final temperature will be 47 degrees Celsius.

but i cant follow the workings the whole way through.  I dont see where the "- 12540" and "+ 2090" come from, else i could work it out with the values for a random feedstock.

Anyhow, i thought a calculator may be a nice addition to the tools section, but, as i say, i cant even do the sum, let alone the rest!

[Glycer-rides]

That equation is much simplified by dividing each side by 25 and arranging the negatives in a more logical fashion, thus simple ratios:

2(60-x) = (x-20)

=>

140 = 3x

x = nearly 47

Edit
So, your example is even easier on the maths...

150x + (50 x 10) = 200 x 60

common denominator of 50 there, so

3x + 10 = 240

x = 76.7 C

[Rotary-Motion]

how do you do that? 

awesome

[julianf]

Andy,

Your sums leave out the specific heat figure.  Is this as its (the same) multiplier on both sides?

[Tony]

Andy,

Your sums leave out the specific heat figure.  Is this as its (the same) multiplier on both sides?

The equations in your first post massively overcomplicate it in the case where specific heat is the same for both liquids being mixed.  Since you're mixing oil with oil it all divides out.

Of course this doesn't factor in density changes with temperature but they're relatively minor.

I assume you wish to go straight from dewater temp to reaction temp using cold, previously dewatered oil?

[julianf]

Nope -

I want to react more oil than i can easily heat.  So im wondering if i heat 200 ltrs to, say, 80c, then how much i can 'water it down with' to get to, say, 65c.

This may not be the best plan anyhow, as then ill have to dewater the cold stuff, but i might be able to get it to work for me!

[Glycer-rides]

Andy,

Your sums leave out the specific heat figure.  Is this as its (the same) multiplier on both sides?

Yup, as Tony says, they cancel out. That quoted example was correct but he failed at the end by not simplifying it enough that it could be followed easily.

It's a case of KISS for our type of brewing operations. We have so many process variables and    'uncertainties of measurement' stacking up that accounting for varying densities with heat or specific heat capacites for types of feedstock pales into insignificance.