[Closed] Any maths experts in today? Could do with some help.

As tco appears twice in that formula, and once within a ln function, I'd use goal seek.

Incidently you have allowed for the correction factor to allow for the fact that the exchanger won't be true [s]parrallel[/s] counter current flow?

Cheers gonefishin, but the goal seek is the bit I am trying to get rid of. I presume that because it is solvable, there must be some way of rearranging to calculate directly. Long time since my A-level maths though.

Hadn't considered a correction factor (yet) - cheers, I'll look into it.

Once designed and built, this condenser will be condensing single malt whisky, which gives me a vested interest in getting this right 🙂 !

OK, here's what I would do...

Transform your variables so that z = x/(tho - tci) and y = (thi - tco)/(tho - tci)

Then your equation is z = (y-1)/ln(y)

which requires you to find the root of y - z ln(y) - 1 = 0

Mathematica has an approximate solution for this root, which is

y = -z W(-exp(-1/z)/z)

where W is the [url= http://reference.wolfram.com/language/ref/ProductLog.html ]Lambert W function[/url]. You could use the series expansion of W for this as follows;

Let a = -exp(-1/z)/z

then y = -z W(a)
W(a) = a - a^2 + 3 (a^3)/2 - 8 (a^4)/3 + 125 (a^5)/24 +...

you will then back transform y to get tco. Define a new column in XL for each transformed variable.

Or use Goal seek 😉

I doubt you'll be able to get it into a discrete solution, it's not always possible and sometimes iteration is the only way to go. I think by rearranging it you will end up with something like

tco + e^(tco + x) = Function

which isn't that much more helpful.

Whenever I've come up against this issue with heat exchagers I use goal seek (or an interative solution if excel isn't available). Well I did until I got access to exchanger sizing software at work.

OK, here's what I would do...

Transform your variables so that z = x/(tho - tci) and y = (thi - tco)/(tho - tci)

Then your equation is z = (y-1)/ln(y)

which requires you to find the root of y - z ln(y) - 1 = 0

Mathematica has an approximate solution for this root, which is

y = -z W(-exp(-1/z)/z)

where W is the Lambert W function. You could use the series expansion of W for this as follows;

Let a = -exp(-1/z)/z

then y = -z W(a)
W(a) = a - a^2 + 3 (a^3)/2 - 8 (a^4)/3 + 125 (a^5)/24 +...

you will then back transform y to get tco. Define a new column in XL for each transformed variable.

Or use Goal seek

😯 Holy batshit 😯 . Goal Seek it is then.

Cheers for the help chaps.
First two drams will be yours. Should be ready in 2027.

Post up your numbers for the four known values and I'll knock up a quick worksheet to test the solution validity. I know nothing about heat exchangers, but I do know a little maths!

Thanks RedTi,

LMTD is 14.6°C
thi = 95°C (boiling whisky)
tho = 85°C (condensed whisky)
tci = 18°C (cooling water from the river)
tco = unknown (hot cooling water)

using goal seek, I get tco = 94.285°C

Much appreciated.

Well that might work mathematically but your temperature approach is way too close to work in practice. A thi of 95 and and tco of 94 (realistically you won't be able to measure anything more accurately than 1 C) will give you a major problem.

Are you sure your LMTD is 14.6?

The LMTD is correct for the temperature values I have taken, based on an estimate of the U value and heat transfer area.
The condensers are made in one size only, so the area will be fixed and non negotiable (I just don't know what that area is yet).
I will be calculating a more accurate U value later, but for now, I am just building up a spreadsheet, and want to get the mechanics right before tweaking the numbers so that they make sense.
I used 3 decimal places, purely because that was the output from the goal seek - not because I want to achieve that level of accuracy!
I am actually expecting tco to be around 75°C in reality.

Ah right. If the exchanger is oversized for its duty, which would appear to be the case here, then high cooling water outlet temps will always be the result. On the plus side it will likely foul up pretty quickly which will bring the cold outlet temperature down a bit. 😉

Ah, they do have fouling issues (especially with the water coming straight from the Spey).

They run this way to deliberately produce hot water, so that they can use it to pre-heat the feeds to the Stills. It is not the most efficient method for them to be recovering heat, but it is their process, and getting them to consider new approaches is nigh on impossible. "It may change the character of the whisky" is the response to all process improvements I've recommended so far. Fair comment, but frustrating!

ormancheep

thread hijack

I'm working with some distilleries regarding integration of biomass steam boilers into their processes - is this anything your distillery has considered?

email in profile

/thread hijack

Hi Smudger,
Yes, they are intending to use a remote steam supply from a stand-alone biomass boiler. Possibly one one of yours I imagine!

which distillery?

emailed you.

nothing received?

@Oman, just programme up @TiRed's suggestion. Spreadhseets are good for doing this sort of thing even if it looks complicated posted here. For a "belt and braces" check you should do the calculation both with the formula and goal seek and check the results are close.

Pretty accurate 😉 . I also have a plot of tco vs. X for all values of X (1-25)

x 14.6 Log Mean temperature difference in heat exchanger
thi 95 Temp hot fluid in
tho 85 Temp hot fluid out
tci 18 Temp cold fluid in

z = 0.217910448, z = x/(tho - tci)
a = -0.046636583, a = - exp(-1/z)/z
W(a) = -0.048977467
y = 0.010672702, y = (thi - tco)/(tho - tci)
y(thi-tco) = 0.715071022
tco = 94.284929, Closed form solution
tco(goal) = 94.284927

Need your email address for the solution, mine's in my profile.

General solution for your parameters:

[img] [/img]

You'll need an X of about 40 to get to a tco of 75 degrees (note the semi-log plot).

I like them apples 🙂

I'll email you now. Thanks very much.