Water Temp Calculations

[BradleyW]

Hi all, I have a question regarding chilling your wort to your desired pitching temperature. Is there a calculation to know how much water at a certain temperature is needed to bring your wort down to pitching temp?
For example I make small batches and dilute my wort up to around 5l. So, I was wondering instead of messing around with ice baths to bring my 2-3l of wort down to a reasonable temp and then topping up with cold water, I put near frozen or frozen water into the hot wort could I calculate how much water would be needed to get me to around 30c for kveik or 20c for other yeasts?
In other words, to bring 3l of wort @80c down to 30c I would need x litres of water @y temperature. Is there any way of working out the values of x and y?
Thanks

 

[foxy]

I knew there was a formula, I have seen it before on brewing forums.


Measure the temperature of the hot water (H), then measure the temperature of the cold water (C). Identify the target temperature (T). You'll mix hot and cold water in the respective ratio T-C : H-T. If your hot water is at 100, your target is 80, and your cold is 40, your ratio would be 80-40 : 100-80, or 40 : 20, which equals 2:1.

Now pick your target volume and divide it into enough parts to make that ratio convenient. In this case, the ratio is 2 parts hot to 1 part cold, so you need a total of 3 parts. If you needed 3 liters, you'd need 2 of hot and 1 of cold. If you needed 4 liters, you'd need 2.667 (2 * 4/3) of hot and 1.333 (1 * 4/3) of cold.

 

[BradleyW]

Nice one! Cheers

 

[GeorgieV]

I think you can calculate it as a heat transfer. Qwort = Qwater + Q loss
Where Qwort is the heat released from the hot wort to get to the target temperature and Qwater is the heat absorbed by the water. Qloss are the heat losses which should be small if the mixing happens quickly so you can ignore it.

Qwort = Mwort*Cwort*(T initial- Ttarget),
where Mwort is the mass if your wort in kg, Cwort is the wort's specific heat capacity and (T initial- Ttarget) is the temperature difference in deg C.

Qwater = Mwater*Cwater*(Ttarget-Tinitial) ,
Mwater is the mass of water, Cwater is the specific heat capacity of water and (Ttarget-Tinitial) is the temperature difference of the water.
Cwort is difficult to know but given that the wort is mostly water you can assume it equal to Cwater. This assumption affects the precision.
Mwort = wort volume* SG , Mwater is easy as 1L = 1kg

So using x and y from the original post we get to
wort volume* SG*(80degC - 30degC) = x*(30 degC - y)

Obviously this equation has 2 unknown variables, so you need to fix one of them in order to solve it.
For example if you fix x = 2 L then you can calculate y. If y comes below freezing, then you'll need some form of external cooling.

 

[BradleyW]

I think you can calculate it as a heat transfer. Qwort = Qwater + Q loss
Where Qwort is the heat released from the hot wort to get to the target temperature and Qwater is the heat absorbed by the water. Qloss are the heat losses which should be small if the mixing happens quickly so you can ignore it.

Qwort = Mwort*Cwort*(T initial- Ttarget),
where Mwort is the mass if your wort in kg, Cwort is the wort's specific heat capacity and (T initial- Ttarget) is the temperature difference in deg C.

Qwater = Mwater*Cwater*(Ttarget-Tinitial) ,
Mwater is the mass of water, Cwater is the specific heat capacity of water and (Ttarget-Tinitial) is the temperature difference of the water.
Cwort is difficult to know but given that the wort is mostly water you can assume it equal to Cwater. This assumption affects the precision.
Mwort = wort volume* SG , Mwater is easy as 1L = 1kg

So using x and y from the original post we get to
wort volume* SG*(80degC - 30degC) = x*(30 degC - y)

Obviously this equation has 2 unknown variables, so you need to fix one of them in order to solve it.
For example if you fix x = 2 L then you can calculate y. If y comes below freezing, then you'll need some form of external cooling.

Nice one, knew I should've listened in maths more!! Just a quick question, in your post is SG specific gravity?

 

[GeorgieV]

Yes, SG is specific gravity. In physics and engineering it's commonly known as density.

 

[BradleyW]

Yes, SG is specific gravity. In physics and engineering it's commonly known as density.

Great thanks

 

[Jim Brewster]

The temperatures would sort of "average out" but weighted for their respective volumes and specific heat capacities.

3 litres of wort, 80C starting temp, 30C target temp: 80C - 30C = 50C

Say chilled water is 5C: 80C - 5C = 75C

Volume of chilled water needed
50÷75x3litres = 2 litres chilled water

Gives 5 litres total volume at an expected temp of 30C. Probably overly simplistic but if there is a difference you could measure the final temp and use the difference as a factor to adjust for the different specific heat capacities for next time. So just as an example say the wort ended up at 40C that's 0.8 of the 50C drop, next time you would use 3 litres÷0.8, so 3.75 litres.

If using ice, less volume would be needed because energy is used when melting the ice from solid to liquid (heat of fusion), so would chill it more than using the above method but an adjustment could be made for that too.

I think you'd probably want the wort slightly lower than 30C for pitching, unless using kviek yeast.

I do similar I use a small boiling pot and top up to 5 litres with cool water while chilling but never really thought about calculating it.

 

[BradleyW]

The temperatures would sort of "average out" but weighted for their respective volumes and specific heat capacities.

3 litres of wort, 80C starting temp, 30C target temp: 80C - 30C = 50C

Say chilled water is 5C: 80C - 5C = 75C

Volume of chilled water needed
50÷75x3litres = 2 litres chilled water

Gives 5 litres total volume at an expected temp of 30C. Probably overly simplistic but if there is a difference you could measure the final temp and use the difference as a factor to adjust for the different specific heat capacities for next time. So just as an example say the wort ended up at 40C that's 0.8 of the 50C drop, next time you would use 3 litres÷0.8, so 3.75 litres.

If using ice, less volume would be needed because energy is used when melting the ice from solid to liquid (heat of fusion), so would chill it more than using the above method but an adjustment could be made for that too.

I think you'd probably want the wort slightly lower than 30C for pitching, unless using kviek yeast.

I do similar I use a small boiling pot and top up to 5 litres with cool water while chilling but never really thought about calculating it.

Great thanks Jim
Yeah I'm using kveik at the moment because it's still around 21-22 in my pantry where I keep my FV, so a bit too warm still for your "normal" yeasts.
I also used to just top up (liquor back I think is the techy term?) with cold water, but the last brew I did I put some pretty large chunks of ice in there and I noticed the temp drop pretty sharpish and I was just wondering if there was a way to be a bit more accurate with the process.
Thanks

 

[Jim Brewster]

I used this video to do an estimate, its American but luckily in metric.


Substituting in your figures gives the below 1,250g of ice, so say 1.25L of water, to get 3L from 80C to 30C. You could just do it in Excel but this show the numbers and the different contributions.

(some minor rounding errors)

The specific heat capacity of wort seems to be less than that of water which is surprising, I assumed 95% that I worked out pro-rata from something I found online so may be wrong, but it's all I could find
"Specific Heat = 4.186 - 0.0293 x %sugar. At 20% sugar, the SG is 3.600. 20% sugar is equivalent to a specific gravity of 1.083"

This doesn't account for heat loss which I assumed to be zero just for simplicity, so bearing in mind heat loss is greater when the wort is hotter, it will be more accurate with a starting temp of say 70C than 90C because the losses will be less. Also if you used cubes vs a block which would take longer to melt meaning more losses. Otherwise the losses perhaps need to be worked out as well, which would be quite complicated with a lot of factors, although there might well be an estimate for heat loss for water in an open metal tank somewhere.

All probably a bit OTT but its interesting anyway!

 

[BradleyW]

I used this video to do an estimate, its American but luckily in metric.


Substituting in your figures gives the below 1,250g of ice, so say 1.25L of water, to get 3L from 80C to 30C. You could just do it in Excel but this show the numbers and the different contributions.

(some minor rounding errors)
View attachment 35529
The specific heat capacity of wort seems to be less than that of water which is surprising, I assumed 95% that I worked out pro-rata from something I found online so may be wrong, but it's all I could find
"Specific Heat = 4.186 - 0.0293 x %sugar. At 20% sugar, the SG is 3.600. 20% sugar is equivalent to a specific gravity of 1.083"

This doesn't account for heat loss which I assumed to be zero just for simplicity, so bearing in mind heat loss is greater when the wort is hotter, it will be more accurate with a starting temp of say 70C than 90C because the losses will be less. Also if you used cubes vs a block which would take longer to melt meaning more losses. Otherwise the losses perhaps need to be worked out as well, which would be quite complicated with a lot of factors, although there might well be an estimate for heat loss for water in an open metal tank somewhere.

All probably a bit OTT but its interesting anyway!

Wow! I knew I'd get a thorough answer on here!!! Thanks!!
It'll certainly help make chilling less stressful/boring. Other option was of course no chill but still not 100% on that one, especially here in summer when it's scorchio!
Thanks again