chiller tank

[lynark]

hi,

Im lynark, i just want to ask how to compute the number of coils you need to used in a chiller tank. Because i have seen a water tank with so many copper tube loop together thats why i wonder how the installer or designer computed for that. assuming you already have the Cooling capacity needed.

thank you

[Brian_UK]

Heat Loss per Foot/Metre should help.

[Peter_1]

Trial and error, depends on too many variables.

[lynark]

hi peter_1

so it means their is no proper computation? just by years of experience.


hi brian_uk

can you give some a brief sample?


actually i'm more on designer of coldroom coldstorage. Now i want to try deeper application of refrigeration system. i want to start on this basic one. So any help is really appreciated

[Aik]

Q=h*A*(tp-tw)
h - surface heat transfer coefficient
A - surface area of pipes
tp - surface temperature of pipes
tw - surrounding water temperature
You must know surface heat transfer coefficient (pipe/water), it's the most difficult.
Try to find ASHRAE's articles "HEAT TRANSFER" and "LIQUID COOLERS".

[Aik]

Surface heat transfer coefficient (h) depends on many factors: density, viscosity, thermal conductivity of fluid, geometrical dimensions of clad body and etc. Also coefficient depends on velocity of fluid relatively the clad body. For natural covection of fluid surface heat transfer coefficient is:
115 - 700 kW/(m^2*C)
For forced convection water surface heat transfer coefficient is:
115 - 11500 kW/(m^2*C).
There aren't general theory of finding this coefficient but there are some theories for special cases.

[lynark]

Surface heat transfer coefficient (h) depends on many factors: density, viscosity, thermal conductivity of fluid, geometrical dimensions of clad body and etc. Also coefficient depends on velocity of fluid relatively the clad body. For natural covection of fluid surface heat transfer coefficient is:
115 - 700 kW/(m^2*C)
For forced convection water surface heat transfer coefficient is:
115 - 11500 kW/(m^2*C).
There aren't general theory of finding this coefficient but there are some theories for special cases.



Hi aik,

thanks for the info. So the Q is the computed heat load needed for the pipe? So by that i can find the Surface area of pipes? And how many loops needed from the header?


thank you

lynark

[Peter_1]

We did a test some weeks ago for a heatpump, a coil 3/8 " of 50 ft submerged in water at 12°C, Te + 1°C, R134a for standstill water gives extracts 475 W.

[Aik]

Hi aik,

thanks for the info. So the Q is the computed heat load needed for the pipe? So by that i can find the Surface area of pipes? And how many loops needed from the header?


thank you

lynark
Q - is the heat [W];
A - is the surface area of pipes [m^2];
A=Pi*D*l
Pi=3.14
D - external diametr of pipe [m]
l - lenth of pipe [m]
The most hard poroblem is how to calculate h...

[Aik]

For a single pipe situated in fluid:
Nu=h*D/k - Nusselt number
h - surface heat transfer coefficient [W/(m^2*K]
D- diameter of pipe [m]
k – thermal conductivity of fluid [W/(m*K)]

Re= V*D/ν - Reynolds number
V - velocity of fluid [m/s]
ν- kinematic viscosity factor [m^2/s]

Pr=μ*c/k - Prandtl number
μ – dynamic viscosity factor of fluid [Pa*s]
c – specific heat of fluid [J/(kg*K)]


Index "f" cover to temperature of fluid, "p" cover to temperature of pipe.

[Aik]

For corridor disposition of pipes:

[Aik]

For as on a chess-board disposition of pipes:

[Aik]

You must determine Nu, than determine h, after determine A and l.
If angle between fluid flux and pipe differ than 90, than h must be multiplited by corresponding factor.

[Peter_1]

Aik, can you translate/calculate these formulas to see if they match my practical test?

[Aik]

Aik, can you translate/calculate these formulas to see if they match my practical test?
I must know velocity of water and angle between water flux and pipe in your case. Also I must know disposition of pipes: single pipe, corridor or chess-board disposition...

[lynark]

hi aik, peter_1

here's the details of the chiller tank i have seen.


tank dimension: 4 ft x 8 ft x 4 ft depth
water temp in the tank should be 10 C
water ( stand still ) no circulation


thank you

[Peter_1]

What do you want to cool with the water?
We need desired temperature in and out and required flow.
What compressor you want to use?

[Aik]

We did a test some weeks ago for a heatpump, a coil 3/8 " of 50 ft submerged in water at 12°C, Te + 1°C, R134a for standstill water gives extracts 475 W.

Aik, can you translate/calculate these formulas to see if they match my practical test?
Peter_1, can you give data of velocity and angle, or may be can you give draft (domensions and connections) of your tank and volume flow of water. Then I'll try to calculate...

[Peter_1]

We submerged a standard 3/8 spiraled coil of 50 ft in water, diameter of spirals 35 cm (1.15 ft) and 2 cm (0.8 inch) from each other.
No water flow, only agitation of the water with a small air blower at the bottom of the tank.
Thanks

[Aik]

We submerged a standard 3/8 spiraled coil of 50 ft in water, diameter of spirals 35 cm (1.15 ft) and 2 cm (0.8 inch) from each other.
No water flow, only agitation of the water with a small air blower at the bottom of the tank.
Thanks
It's natural convection:

[Aik]

Formula is:

[Aik]

I don't fully understand the calculation, but I do it and result is true...

[sanwar]

Dear Freind,
Could you please provide me "1 year hvac preventive maintenance schedule in ms-word formate

Thanks,
Sanwar

[Peter_1]

Sanwar, post his in a new thread, not hijacking this one.

[Peter_1]

Aik, I'm impressed with your result as long as you haven't changed numbers to fit the result.
What about a 1/2" and a 5/8 " of 50 ft, same conditions?

[Aik]

Aik, I'm impressed with your result as long as you haven't changed numbers to fit the result.
What about a 1/2" and a 5/8 " of 50 ft, same conditions?
Peter_1, if you change diameter of pipe "D" you'll get result (capacity) for new pipe with the same conditions. Also you may change length "L" of the pipe...

[Peter_1]

Aik, where can I find all the numbers you inserted in your formula?
Your formula was a question which was asked many times here on RE. If we're sure about it, we then don't have to test it in real life the next time.

What about a 3/8 coil buried in the ground? Problem will be that you will not find the right numbers for 'ground' which can be anything.

[Peter_1]

Aik, If I give in the correct diameter for a 3/8 pipe which is 0.009525 and not 0.09525, then I get 80 W instead of 454 W :confused: So something is wrong with the calculation

[Aik]

Aik, If I give in the correct diameter for a 3/8 pipe which is 0.009525 and not 0.09525, then I get 80 W instead of 454 W :confused: So something is wrong with the calculation
You are right, I found mistake in my calculation...Today I'll try to explane all calculations

[Aik]

What about a 3/8 coil buried in the ground? Problem will be that you will not find the right numbers for 'ground' which can be anything.
In ground there isn't convection because the ground is solid. Therefore thise calculation don't used for ground.


Aik, If I give in the correct diameter for a 3/8 pipe which is 0.009525 and not 0.09525, then I get 80 W instead of 454 W :confused: So something is wrong with the calculation
I have reform the calculation...
Peter_1, really I gave coefficient for limited space from other method of calculating, but I suppose it's correct... Also I gave operation factors at temperatures 0 and 10 C, because I don't know its at temperatures of 1 and 12 C. If you give thise operation factors at needed temperatures you'll get more precisely result.

[lynark]

hi aik and peter_1,

I think i'm a little lost with your conversation. and thinking this is only a basic one :confused: