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# Quad TEC Water Chiller Project by Ultrasonic2's - Page 26

This is good. It seems flow rate has a huge impact on temperatures here. Could you take the pressure readings like you were showing before and then use these to reduce the flow on the more restrictive versions?
Quote:
 Originally Posted by Scarlet Infidel This is good. It seems flow rate has a huge impact on temperatures here. Could you take the pressure readings like you were showing before and then use these to reduce the flow on the more restrictive versions?
With my calculations earlier I said irrespective of everything water takes 1.16w to heat 1Âº in one hour and since it has been pretty much ascertained that a closed loop is generally in temp equilibrium taking the volume of the whole loop would seem fair.

Now i don't think thats the case as flow has made a fair difference here...then it dawned on me the calculation is NOT volume of the loop per se but is actually the flow rate per second.

So you use exactly the same calculation using the flow rate in litres per second instead of the actual volume of the loop.

The calculations should then be right...I did wonder why originally it didn't all seem to fit as correctly as I thought it should.

EDIT

For those of you how have seen the calculation -

flow rate litres per second x 4190 joules (thermal capacity of water) = watts/Kelvin.

My calculation is the same but with the physical time component added.

For those who want a better explaination -

It is the same because 4190 joules = 1.16w and 1 kelvin degree is the same size as one celcius degree (so Kelvin is often used to indicate a difference between two celcius temps.) BUT 1ÂºK is not the same temerature as 1ÂºC - 1ÂºK = -272ÂºC they have completely different scales.

Flow rate in litres per second gives the weight per litre because 1kg = 1 Litre.
Although the calcs say flow rate litres per second you are actually using the weight of the water (in grams.) to match the thermal capacity calculation = 4.193 Kj/kgÂºK which in grams is 4190j/gÂºK (it is rounded down a bit.)
Edited by zipdogso - 3/28/09 at 10:23am
Quote:
 Originally Posted by Ultrasonic2 (muffy) whats 120 leters and hour to meter per second ? anyone ?
Your asking for m/sec which is velocity it is not the same as litres/hr which is volume flow

If your looking for 1mÂ³/sec that is volume flow and can be equated to litres/hour but your in real funny numbers here ....look at what I mean.

1 mÂ³/sec = 1000 dmÂ³/s and since 1 dmÂ³ = 1 litre 1 mÂ³/sec = 1000 litres/sec be amazing the day you ever a pump to do that...

I expect the prog does some calculation of the volume of the unit and 1m/sec is that volume travelling at 1m/sec.(velocity.)
Without knowing the background calculation it is difficult to see what it does.
Quote:
 Originally Posted by zipdogso Your asking for m/sec which is velocity it is not the same as litres/hr which is volume flow If your looking for 1mÂ³/sec that is volume flow and can be equated to litres/hour but your in real funny numbers here ....look at what I mean. 1 mÂ³/sec = 1000 dmÂ³/s and since 1 dmÂ³ = 1 litre 1 mÂ³/sec = 1000 litres/sec be amazing the day you ever a pump to do that... I expect the prog does some calculation of the volume of the unit and 1m/sec is that volume travelling at 1m/sec.(velocity.) Without knowing the background calculation it is difficult to see what it does.
obviously if you divide flowrate in m^3/sec by the area the water flows trough (Pi*r^2 or (pi*Dia^2)/4 for tubing) in m^2 you get the flow velocity in m/s
 My System (13 items)
 My System (13 items)
zipdogso I have not had time to read through your posts properly but immediately several things seem strange. For example:

'4190 joules = 1.16w'
doesn't mean anything, you are missing a unit of time.

'flow rate litres per second x 4190 joules (thermal capacity of water) = watts/Kelvin.'
This equations seems right but thermal capacity is not measured in joules (or kg/KÂº) .

Specific heat capacity is measured in J/(g.k) and for water it is ~4.2 depending on temperature of the liquid.

I'm not saying any of it is wrong, its just small things like this make it hard to follow what you are saying.
Off the top of my head (assuming chiller c/w of 0):

Flow rate (cm^3/s) = Flow velocity (cm/s) x Channel area (cm^2)
Mass of water/second (g/s) = Flow rate (cm^3/s)

Time for water to pass through chiller (s) = length (cm) / Flow velocity (cm/s)
Energy into water (j) = power(j/s) x time (s)

Mass of water in channel (g) = Mass of water/second (g/s) x time to pass (s)
- or Channel area (cm^2) x length (cm)

Rearrange SHC (j/g.K)
Temp of water = energy/(mass x shc)
Edited by Scarlet Infidel - 3/28/09 at 10:16am
e.g. (big approximations)

Flow velocity = 1m/s = 1000cm/s
Channel area = w x h = 6 x 0.3 = 1.8cm^2
Flow rate = 1800 cm^3/s (or 1800g/s)

Time in chiller = length / flow velocity = 27.5 / 1000 = 0.0275s
Energy into water = 800 x 0.0275 = 22 joules

Mass of water in channel = 1.8 x 27.5 = 49.5g

Temp increase of water = energy/(mass x shc) = 22/(49.5 x 4.2) = 0.105K
Edited by Scarlet Infidel - 3/28/09 at 10:17am
There are little things that make it easier if you just follow the calculations most people really don't want/need to now the ins and outs. So long as the formulas are right (you spotted a mistake I missed.)

Quote:
 Originally Posted by Scarlet Infidel '4190 joules = 1.16w' doesn't mean anything, you are missing a unit of time.
Don't know what your saying here they are both missing a unit of time therefore they are equal. we don't go round saying the TEC Qcmax is 250w per second do we ????

Quote:
 Originally Posted by Scarlet Infidel This equations seems right but thermal capacity is not measured in joules (or kg/KÂº) .
yep...missing key presses there ...post edited to correct.

it is 4.1885 Kj/KgÂºK (at 25ÂºC) which is the same as 4188.5j/gÂºk (at 25ÂºC) it is very difficult to be specific as it's temp dependant so people use all sorts of values 4.18 - 4.2 - 4.19 etc depends which side you get out of bed !!!
Changing to a more reasonable (but still high I think) flow velocity of 100cm/s
Temp increase = 1.05k

Now I need to work out if any of this is correct or useful and how you take into account the thermal resistance of the copper, surface area, how dT affects heat transfer etc.
Edited by Scarlet Infidel - 3/28/09 at 10:35am
Quote:
 Originally Posted by zipdogso Don't know what your saying here they are both missing a unit of time therefore they are equal. we don't go round saying the TEC Qcmax is 250w per second do we ????
Watts is Joules per Second

Quote:
 Originally Posted by zipdogso yep...missing key presses there ...post edited to correct. it is 4.1885 Kj/KgÂºK (at 25ÂºC) which is the same as 4188.5j/gÂºk (at 25ÂºC)
Agreed.
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