Calculating Volume of tank through temperature?

AustinVines · 11/03/2006, 10:40 AM

First, I am no math magician so please be gentle

I have an estimate of my total system volume but it may not be very accrurate. I was thinking that if I knew the temp of the tank after removing a specific volume of water during a water change and knew the temp of the new saltwater and added the same amount back, the new temperature could be used to calculate the volume.

My problem is coming up with the formula. It may not be any more accurate than my guess since my thermometer only read into the tenths but was wondering if anyone could come up with a formula and if it was even valid. What do you think (and what would your equation look like)?

Thanks

Bono · 11/03/2006, 11:50 AM

To calculate tank volume, there is a calculator on RC's home page that will work.

You completely lost me on the temperature thing.

Amador · 11/03/2006, 12:06 PM

Not sure how you'd come up with volume from temperature - they're not really related in this case. I think you'd be better off estimating the water displacement (i.e. volume) of your rock and sand and subtracting that from your empty tank volume.

My method would be something like: "Ok, I have a rock structure that is 16" long, by 12" deep and 18" high. I can approximate its shape by assuming it's square (it isn't, but close enough). That works out to 16 x 12 x 18 = 3,456 cu. in. = 14.96 gallons." Repeat as necessary and subtract from empty tank volume.

kiknchikn · 11/03/2006, 12:10 PM

I think AustinVines was referring to using the specific heat of water to calculate the volume, which is an interesting idea, but I'm not sure it will work unless you can find out the specific heat of your salt water. I'm not chemist, but I'd think that it would be different than pure distilled water, for which all the specific heats for water I've seen were calculated on.

You might want to try this thread in the chemistry forum.

conefree · 11/03/2006, 12:23 PM

Without using the specific heat capacity of water, you could do a very simple math calculation without getting overly complicated

(V1*T1)+(V2*T2)=VtT3

V1=V2 since you are removing and adding the same amount of water so we will change it to Vc (water change volume)
T1=Tank temp
T2=Water change temp
Vt=Total tank volume
T3=final tank temp

Therefore, you get: Vc(T1+T2)=VtT3

Solving for Vt, you get:

&nbsp &nbsp &nbsp &nbsp &nbsp &nbsp Vc(T1+T2)
Vt= ---------------------
&nbsp &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp T3

Whew!

Hope that helps and you are not scratching your head!

kiknchikn · 11/03/2006, 12:38 PM

That is very cool. I'll have to try that out sometime if I remember. The problem becomes having a thermometer that is accurate enough. My coralife digital thermometers probably aren't good enough.

daveverdo · 11/03/2006, 12:46 PM

I do this all the time in a different way for a different purpose. I don't have a heater for my water change bucket. Measure the temperature of the water in the change water and the tank and calculate how much the temp will change when I make the water change. Same basic equation different application. I also use it determine how much water and at what temperature I need to gradually adjust the water temp in the main tank.

Dave

AustinVines · 11/03/2006, 01:20 PM

That is great Conefree! I did a test with some figs but came up with way off numbers. Let me show you how I estimated the tank temp change and maybe you can tell me where I am going wrong.

Assume 85g total water volume at 80 degrees

Remove 10g of 80d water
Add 10g of 70d water (from my garage storage bin)

What is the new temp going to be?

10 gallons of 70 degree water is about 12% of the total estimated volume of 85g
75 g of 80 degree water = about 88% of the total estimated volume

12% of 70 degree water is 8.4 "degrees"
88% of 80 degree water is 70.4 "degrees"

8.4 + 70.4 = 78.8 F is what I estimate my water temp to be after the waterchange. That seems right to me.

On your formula, if my final temp is 78d and the other variables remain the same(10g change with 80 and 70 degree water) I get total volume =10(80 + 70)/ 78 which gives me 20. What am I doing wrong?

RichConley · 11/03/2006, 01:38 PM

Conefree's math is wrong

V1 does not equal V2.

V2=Vt-V1

do the same manipulations and solve.

RichConley · 11/03/2006, 01:49 PM

Vt=-V2(T1-T2)/(Tt-T1)

RichConley · 11/03/2006, 01:53 PM

Vt = -10(80-70)/(X-80)

If x=78.8, Vt =83.3 gl

conefree · 11/03/2006, 02:01 PM

yeah, you're right Rich. Forgot to compensate for inflow and outflow. Thanks for the correction

AustinVines · 11/03/2006, 02:01 PM

Thanks, Rich. As an aside, I assume you have a barebottom tank with all that flow?

RichConley · 11/03/2006, 02:22 PM

Quote:

Originally posted by AustinVines
Thanks, Rich. As an aside, I assume you have a barebottom tank with all that flow?

Nope, kolorscape sand. I just have to be really careful.

Travis L. Stevens · 11/03/2006, 02:27 PM

Okay, so, for the not-so-mathematically-inclined, will you post a new forumla and example? Conefree's initial post was easier to understand, but we started to make little adjustments here and there.

TekCat · 11/03/2006, 02:38 PM

Not to raid on your paraide, but I'd like to say that this aproach might sound awesome (which it is), however it is not going to be accurate.

First of all, after you pour new water in, you must allow some time to mix with old water before you check for temperature. During this time heater will be running, or in opposite case tank will be cooling because it gives out heat to the outside air. In any case these processes need to be accounted for. I don't think it could be done easily.

AustinVines · 11/03/2006, 03:00 PM

Oh I know it isn't going to be very accurate but I have estimated new temps and salinity with it and finally tumbled to the fact that I should be able to get an estimate of volume if I could just figure out the equation.

I loaded my tank with LR without weighing it and now that it is soaked, I cannot get a good weight on it anyway. This seemed as likely a method to produce a close approximation as my estimating the LR based on size.

Travis L. Stevens · 11/03/2006, 03:07 PM

Tek, I hate to say it, but neither is the (LxWxH)/231 formula either

RichConley · 11/03/2006, 03:08 PM

Vt=-V2(T1-T2)/(Tt-T1) <--- formula.

If the tank was at 80 degrees, and you put in 50 gallons of 30 degree water, and the final temp was 70 degrees, it would look like this

Vt=-50(80-30)/(70-80)=250g.

50g = 20% of 250g, so 20% of 30' = 6'
200g = 80% of 80' = 64'
---------------------------------------------------
70'

RichConley · 11/03/2006, 03:10 PM

If you're going to do this, magnesium levels would probably actually be the most accurate way, because you can swing it around a little and not kill anything.

IE your tank is 1200 ppm, mix up 10g at 1500ppm, and then figure it out. Just substitue M for T in the equations.

TekCat · 11/03/2006, 03:17 PM

Tek, I hate to say it, but neither is the (LxWxH)/231 formula either
Yes, it's been a long day already, my brains fried

If you're going to do this, magnesium levels would probably actually be the most accurate way, because you can swing it around a little and not kill anything.

IE your tank is 1200 ppm, mix up 10g at 1500ppm, and then figure it out. Just substitue M for T in the equations.
I like it way better than the temperature. Praise AustinVines and all you math geeks

11/03/2006, 10:40 AM	#1
AustinVines Premium Member Join Date: Jan 2006 Location: One guess Posts: 219	Calculating Volume of tank through temperature? First, I am no math magician so please be gentle I have an estimate of my total system volume but it may not be very accrurate. I was thinking that if I knew the temp of the tank after removing a specific volume of water during a water change and knew the temp of the new saltwater and added the same amount back, the new temperature could be used to calculate the volume. My problem is coming up with the formula. It may not be any more accurate than my guess since my thermometer only read into the tenths but was wondering if anyone could come up with a formula and if it was even valid. What do you think (and what would your equation look like)? Thanks

11/03/2006, 11:50 AM	#2
Bono Registered Member Join Date: Aug 2002 Location: Lakewood Ranch Fl. Posts: 872	To calculate tank volume, there is a calculator on RC's home page that will work. You completely lost me on the temperature thing. __________________ Living it up in Florida Current Tank Info: 210 gal, 300lbs LR, Aquamaxx Skimmer, 3x Reefbreeders LED's & T5's.

11/03/2006, 12:06 PM	#3
Amador Registered Member Join Date: Oct 2005 Location: Jenks, OK Posts: 320	Not sure how you'd come up with volume from temperature - they're not really related in this case. I think you'd be better off estimating the water displacement (i.e. volume) of your rock and sand and subtracting that from your empty tank volume. My method would be something like: "Ok, I have a rock structure that is 16" long, by 12" deep and 18" high. I can approximate its shape by assuming it's square (it isn't, but close enough). That works out to 16 x 12 x 18 = 3,456 cu. in. = 14.96 gallons." Repeat as necessary and subtract from empty tank volume. __________________ Chris Current Tank Info: 125g AGA, 55g sump, 2x 250W MH, Mag 12, MSX 160, BB since 5/14/07

11/03/2006, 12:10 PM	#4
kiknchikn Registered Member Join Date: Aug 2006 Location: Bel Air, MD Posts: 1,556	I think AustinVines was referring to using the specific heat of water to calculate the volume, which is an interesting idea, but I'm not sure it will work unless you can find out the specific heat of your salt water. I'm not chemist, but I'd think that it would be different than pure distilled water, for which all the specific heats for water I've seen were calculated on. You might want to try this thread in the chemistry forum. __________________ I am a proud member of the Chesapeake Marine Aquaria Society (CMAS). If you're in the Maryland area check us out! Current Tank Info: 30g nano reef

11/03/2006, 12:23 PM	#5
conefree Gene Pool Lifeguard Join Date: Jan 2004 Location: Downingtown, PA Posts: 758	Without using the specific heat capacity of water, you could do a very simple math calculation without getting overly complicated (V1T1)+(V2T2)=VtT3 V1=V2 since you are removing and adding the same amount of water so we will change it to Vc (water change volume) T1=Tank temp T2=Water change temp Vt=Total tank volume T3=final tank temp Therefore, you get: Vc(T1+T2)=VtT3 Solving for Vt, you get: &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp Vc(T1+T2) Vt= --------------------- &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp &nbsp T3 Whew! Hope that helps and you are not scratching your head! __________________ Chris "A Closed Mouth Gathers No Foot" - Unknown Some people are like Slinkies . . . not really good for anything, but you still can't help but smile when you see one tumble down the stairs. Current Tank Info: 120 gallon, Ecosystem 3612, 6X54 TEK light, Tunze Osmolator, Lifereef dual overflow all down

11/03/2006, 12:38 PM	#6
kiknchikn Registered Member Join Date: Aug 2006 Location: Bel Air, MD Posts: 1,556	That is very cool. I'll have to try that out sometime if I remember. The problem becomes having a thermometer that is accurate enough. My coralife digital thermometers probably aren't good enough. __________________ I am a proud member of the Chesapeake Marine Aquaria Society (CMAS). If you're in the Maryland area check us out! Current Tank Info: 30g nano reef

11/03/2006, 12:46 PM	#7
daveverdo Registered Member Join Date: Aug 2006 Location: Horseheads, NY Posts: 1,881	I do this all the time in a different way for a different purpose. I don't have a heater for my water change bucket. Measure the temperature of the water in the change water and the tank and calculate how much the temp will change when I make the water change. Same basic equation different application. I also use it determine how much water and at what temperature I need to gradually adjust the water temp in the main tank. Dave __________________ 90g Display with fish, LR and a few soft corals, 65g basement sump, eBay LEDs

11/03/2006, 01:20 PM	#8
AustinVines Premium Member Join Date: Jan 2006 Location: One guess Posts: 219	That is great Conefree! I did a test with some figs but came up with way off numbers. Let me show you how I estimated the tank temp change and maybe you can tell me where I am going wrong. Assume 85g total water volume at 80 degrees Remove 10g of 80d water Add 10g of 70d water (from my garage storage bin) What is the new temp going to be? 10 gallons of 70 degree water is about 12% of the total estimated volume of 85g 75 g of 80 degree water = about 88% of the total estimated volume 12% of 70 degree water is 8.4 "degrees" 88% of 80 degree water is 70.4 "degrees" 8.4 + 70.4 = 78.8 F is what I estimate my water temp to be after the waterchange. That seems right to me. On your formula, if my final temp is 78d and the other variables remain the same(10g change with 80 and 70 degree water) I get total volume =10(80 + 70)/ 78 which gives me 20. What am I doing wrong?

11/03/2006, 01:38 PM	#9
RichConley Registered Member Join Date: Sep 2004 Location: Bostonian in Chicago going to DC Posts: 9,908	Conefree's math is wrong V1 does not equal V2. V2=Vt-V1 do the same manipulations and solve. __________________ NO TANKS!!!

11/03/2006, 01:49 PM	#10
RichConley Registered Member Join Date: Sep 2004 Location: Bostonian in Chicago going to DC Posts: 9,908	Vt=-V2(T1-T2)/(Tt-T1) __________________ NO TANKS!!!

11/03/2006, 01:53 PM	#11
RichConley Registered Member Join Date: Sep 2004 Location: Bostonian in Chicago going to DC Posts: 9,908	Vt = -10(80-70)/(X-80) If x=78.8, Vt =83.3 gl __________________ NO TANKS!!!

11/03/2006, 02:01 PM	#12
conefree Gene Pool Lifeguard Join Date: Jan 2004 Location: Downingtown, PA Posts: 758	yeah, you're right Rich. Forgot to compensate for inflow and outflow. Thanks for the correction __________________ Chris "A Closed Mouth Gathers No Foot" - Unknown Some people are like Slinkies . . . not really good for anything, but you still can't help but smile when you see one tumble down the stairs. Current Tank Info: 120 gallon, Ecosystem 3612, 6X54 TEK light, Tunze Osmolator, Lifereef dual overflow all down

11/03/2006, 02:27 PM	#15
Travis L. Stevens Registered Member Join Date: Nov 2004 Location: Perry, OK Posts: 13,946	Okay, so, for the not-so-mathematically-inclined, will you post a new forumla and example? Conefree's initial post was easier to understand, but we started to make little adjustments here and there. __________________ Travis Stevens Current Tank Info: Restarting 28g Bowfront

11/03/2006, 02:38 PM	#16
TekCat Registered Member Join Date: Sep 2005 Location: MN Posts: 3,130	Not to raid on your paraide, but I'd like to say that this aproach might sound awesome (which it is), however it is not going to be accurate. First of all, after you pour new water in, you must allow some time to mix with old water before you check for temperature. During this time heater will be running, or in opposite case tank will be cooling because it gives out heat to the outside air. In any case these processes need to be accounted for. I don't think it could be done easily.

11/03/2006, 03:00 PM	#17
AustinVines Premium Member Join Date: Jan 2006 Location: One guess Posts: 219	Oh I know it isn't going to be very accurate but I have estimated new temps and salinity with it and finally tumbled to the fact that I should be able to get an estimate of volume if I could just figure out the equation. I loaded my tank with LR without weighing it and now that it is soaked, I cannot get a good weight on it anyway. This seemed as likely a method to produce a close approximation as my estimating the LR based on size.

11/03/2006, 03:07 PM	#18
Travis L. Stevens Registered Member Join Date: Nov 2004 Location: Perry, OK Posts: 13,946	Tek, I hate to say it, but neither is the (LxWxH)/231 formula either __________________ Travis Stevens Current Tank Info: Restarting 28g Bowfront

11/03/2006, 03:08 PM	#19
RichConley Registered Member Join Date: Sep 2004 Location: Bostonian in Chicago going to DC Posts: 9,908	Vt=-V2(T1-T2)/(Tt-T1) <--- formula. If the tank was at 80 degrees, and you put in 50 gallons of 30 degree water, and the final temp was 70 degrees, it would look like this Vt=-50(80-30)/(70-80)=250g. 50g = 20% of 250g, so 20% of 30' = 6' 200g = 80% of 80' = 64' --------------------------------------------------- 70' __________________ NO TANKS!!!

11/03/2006, 03:10 PM	#20
RichConley Registered Member Join Date: Sep 2004 Location: Bostonian in Chicago going to DC Posts: 9,908	If you're going to do this, magnesium levels would probably actually be the most accurate way, because you can swing it around a little and not kill anything. IE your tank is 1200 ppm, mix up 10g at 1500ppm, and then figure it out. Just substitue M for T in the equations. __________________ NO TANKS!!!

11/03/2006, 03:17 PM	#21
TekCat Registered Member Join Date: Sep 2005 Location: MN Posts: 3,130	Tek, I hate to say it, but neither is the (LxWxH)/231 formula either Yes, it's been a long day already, my brains fried If you're going to do this, magnesium levels would probably actually be the most accurate way, because you can swing it around a little and not kill anything. IE your tank is 1200 ppm, mix up 10g at 1500ppm, and then figure it out. Just substitue M for T in the equations. I like it way better than the temperature. Praise AustinVines and all you math geeks