Math question - How much ice.... Chris

Chris

Is there a formula for the following equation: it takes 'x' btus to lower 1 gallon of water by 1 degree?

Is it true that 12000 btus can melt 1 ton of ice?

If so, can you calculate how much ice would need to be added to a 400 gallon hot tub to cool it from 103*f down to 95*f?

Would it take the same number of BTUs to cool 10 gallons of water from 100*f to 90*f as it would to cool 10 gallons of water from 150*f to 140*f?

Thanks!

Dan Peel

Inverted and converted

Up or down you have 1BTU per 1LB per 1Deg F until you start changing state with your water. SO... Your 1F temp change in your 8.33lb (1 usg) of water requires 8.33BTU.

The 12,000BTU you are refering to is the LATENT heat of conversion of one ton of Ice at 32F to one ton of water at 32F. AKA 6 btu/lb.

Your hot tub question. Yes you can. Your 400 gal US @ 8.33lb is 3332lb. You are asking for an 8F drop in temperature or 26656BTU however you are also going to bring the temperature of your melt water up by (95 - 32)63 Deg F for a total of 69BTU gain per pound of ice @ 32F used - if the ice is colder than 32F you gain cooling power at the linear rate of 1BTU per pound per degF.

Same Number of BTU's are required for any 10F change in the same fluid - except where you are faced with conversion losses.

Enjoy......Dan


To Learn More About This Contractor, Click Here to Visit Their Ad in "Find A Contractor"

Larry_7

Equations

Sensible heat equation:

Q=Weight x Specific heat x Temperature Difference

Q=Quantity of heat needed to be added...or removed

Specific Heat of water = 1 btu/lb.

1 gal. water = 8.33 lb.

So...

3332*1*8=26656 btu loss required to drop 400 gal. water 8 degF.
I'm at a loss so far in determining the lb. of ice required. I will assume the ice to be at 32 deg F. The temp of ice is important because it only takes about 0.5 btu to change the temp of ice 1 deg F. (0.5 btu required to change the temp of 1 lb. of water vapor 1 deg F also.)

1 lb. of ice at 32 deg F requires 144 btu to change to 1 lb of water at 32 deg. (Latent heat of freezing)

1 lb. of water at 212* F requires 970 btu to change to 1 lb. of water vapor at 212* F. (Latent heat of evaporation)

I must brush up on my math to solve your problem..heh..heh.

12000 btu PER HOUR will change 1 ton of 32*F ice to one ton of 32*F water in 24 hours.

144 (btu/lb) x 2000 (lb.)= 288,000 btu required to melt 1 ton of 32*F ice. 288,000 btu devided by 24 hours = 12,000 btu/hr. or one ton of refrigeration.

Clear as mud, huh?

Larry

Gary Fereday

amount of ice needed

I calculate 218.5 pounds of ice would be needed to cool 400 gallons @ 103*F to 95*F I'll have to work on the fromula a little to show where I got this! bigugh

Gary Fereday

amount of ice needed

starting with Dan's book (Pumping away) pg. 91. pounds of waterx103* - pounds of waterx95* = 33200pounds of water needed. Divided by 152 ( total of 144 + 8 btus avalible of ice ) it comes to 218.5 pounds of ice needed. there is still the factor of the holding tub temp. but that is close!
bigugh

Gary Fereday

correction!

160.5 pounds of ice needed! 33200 pounds btus divided by 207 pound btus. (144 btus + 63btus =207 btus)

Caselli

I don't follow the 33200 btu

I have followed this interesting thread, and agree with the 207 btu figure per pound of ice.
I do not follow the 33200 btu needed figure. Why not 26656?
Another posting had this figure earlier.
400 gal x 8.33 lb per gal x 8 degrees temp diff = 26656 ??
Why not therefore 129 lbs of ice, 26656 / 207 = 129 ?

Assumptions have to be made throughout. For instance 8.33 lb.per gallon is not correct at this temp. More like 8.29, at the average temp. That can be ignored. Only makes a difference of about 16 lbs.
Another question is: Do we have 400 gallons of water at 103F, and how much ice will we add to end up with a greater quantity of water at 95F? OR do we have an unknown quantity of water in 400 gallon capacity tank, to which we will add ice to end up with exactly 400 gallons of 95F water. Practically, I think this difference is only about 16 gallons too.

Fun exercise.

Gary Fereday

It is fun!

Ok you win. I just redid my ideas and got the same as you! I must have hit an extra key some where! on the math calculater! It was very late when I entered my correction! Yes It is fun! and besides that, that much ice costs a little less! Now when the ice was added did it cause the tub to overflow? boy the parameters just seem to keep coming! Kinda like Congress!

Gary Fereday

Note!

Perhaps the girl friend was very cool! Many are about worth 129 lbs. of ice! lol

Larry_7

This thing has become a real mind bender

For me. After richocheting around in this feeble mind for a couple days I came up with 237.64 lb. of 32*F ice to bring down the 400 gal. water 8*F to 95*. This translated into an additional 28.53 gal. of water. I admit I didn't consider the lower lb./gal for the temp of the water. I figured 8.33 lb/gal or 3332 lb. 26656 btu to drop 8*. Of course, my hot tub was in a controlled environment so I had zero btu loss or gain to the surrounding atmosphere.

I may attempt to show my math tomorrow. my head is beginning to hurt for some reason...

Larry

Dale W.

Another answer

Hey guy's, I took a differant slant and tried to put those years of Algebra to work. Understand it's been thirty years since the last class.
400 gallon of water=3332#
Beginning temp=103degreesF Ending temp=95degreesF 144btu to change 1# of 32degree water to 1# of 32degree ice

Let's pretend water does not change to ice at 32degreeF. That same 1# of ice at 32degreeF would be 1# of water at
-104degreeF. 32-(-144)=-104

With this assumtion I think I can figure out the ice needed even as the water volume increases.

x=#s of ice added to the tub

We start with 3332# of water at 103degreeF and add x amount of ice(-104degree water) and this will become a tub of 95 degree water. The tub will increase in volume from 3332#s to 3332+x.

(3332*103)+(x*[-104])=(3332*95)+(x*95)

343196-(104x)=316540+95x

343196 = 316540 + 199x

26656 = 199x

x=133.949

Need to add about 134# of ice to bring the water down to 95 degrees. This will raise the volume to about 416 gallons

Ok, my mind had it's exercise for the day, now I get to finish raking a yard from yesterdays water service installation.

I don't post here alot, but I read just about every day.
See you guys at the wethead meeting
Dale Wegman
Bear, DE

chris smith_2

btu

okay hows this
400 gals x 8.33lbs per gallon[i know spgrav varies with temp but its only gonna be 3+or- lbs ice]=3332lbs of hot water
3332lbs x8btus=26,656 to be removed

1lb ice at 32f to 1lb water at 32f=144btu
1lb water at 32f to 95f=63btu
144+63=207 btu heat removed per lb of ice
26656/207=128.773lbs of ice
chris smith
porter maine

Larry_7

Oops

I agree with the 128.8 lb of 32* ice needed. 26,656 btu required. Each lb. of ice must take in 144 btu to change to 32* water. Each lb. must then take on 63 btu to raise temp. to 95* (95-32=63). Each lb. will take (144+63)or 207 btu. 26,656/207=128.773 Lb. of ice. I still haven't allowed for other than 8.33 lb./gal.

As long as I'm pushing the buttons, there is plenty of room for error. One day maybe someone will invent a calculator you just ask the question and it belches out the answer.

Two half-nuthins is a whole nuthin. (Foghorn Leghorn)

Larry

Gary Fereday

ambient temp today (bigugh)

Ambient temp today reached 108.5*F degrees at an official station near here. Salt lake. Now I just think I'd place about 300#'s of ice in it and hope to not par boil myself to a lobster red!