Heat gains

Hi all, I'm trying to work out how to cool 250 litres of water from 30°C to 25°C for an aquarium fish tank but im struggling to work out the cooling load.

I'm using the formula Q = m c dt

m = 250litres = 250kg
c = SHC for water= 4.19kj/kg
dt= 30-25 =5°C


Thinking of using a plate heat exchanger and a pump that will deliver 0.25kg/s through the exchanger ( this figure will change once i fully know i have the formula correct, its theoretical for now)


So we have :
0.25kg/s
4.19kj/kg
5k

So Q = 0.25 x 4.19 x 5 = 5.2kw.

This will complete the water cycle in 250/0.25 = 1000 seconds or 16.6mins

I understand that by reducing the mass flow rate this will decrease the cooling capacity but even still i thought these aquarium coolers are normally in the Watts range and not the KW, like i have calculated.

Why do you want to cool in 16 min? why not in 160 min, for instead?

Hi Mark
Q = m [250kg ] x 4.19 x [TD ]
Q = 250 x 4.19 x 5 / 3600 seconds
= 5237.50 Kj / 3600 = 1.45 Kw [ Kj/s ]

determine flow rate 0.25Kg/sec or litres /sec = 3.3 gallons per second TOO high, more like 0.02 L/sec.

Q= 0.02 x 4.19 x 5'c td = 0.42 Kw duty

Hi Sandro, the 16minutes pull down time was theoretical and could indeed be increased. I wanted to make sure my workings out were correct and then I could tweak the outcome.

Quote:

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Originally Posted by Magoo

Hi Mark
Q = m [250kg ] x 4.19 x [TD ]
Q = 250 x 4.19 x 5 / 3600 seconds
= 5237.50 Kj / 3600 = 1.45 Kw [ Kj/s ]
determine flow rate 0.25Kg/sec or litres /sec = 3.3 gallons per second TOO high, more like 0.02 L/sec.
Q= 0.02 x 4.19 x 5'c td = 0.42 Kw duty

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Hi Magoo thanks for that.

When I was working out the cooling load originally I didn't divide by the 3600 seconds (1 hour) as I wasn't sure wether I needed to do this or even wether I should have divided by 24hours (86,400).

The flow rate of 0.02 and the KW seems much better now although the pulldown time has increased to 208mins.

So is that 1.45kw the cooling load? I'm still unsure as to why we need to divide by 3600.

If that is indeed the cooling load how would I now begin to build the system ?

Would I now select an evaporator that is capable of absorbing 1.45kw? *And then whats the next step?

This isn't something I have ever done before but I'm interested to learn about it.

There's so many people that have no idea how to work these questions out so hopefully this thread will also give them some idea.

I know I could work it out using coolpack but that's too easy.

Is this a real job or just a bit of theory.

If it is real, then pull down load is not so important.
The first question is what temp to do they want. Lets say 25C.
So why would the tank warm up. Is it the fish swimmimng about, not really. It is likely being heated the circulation pump (find the moter KW rating), plus there is likely to be a light, find the wattage, and finally what is the ambient, need to find out gains/losses. add all the gains. then you have the maximum refrigeration load. When you have this, you then can determine how long to pull down a warm tank.
You then need to find your required flow rate, what split do you want, it needs to be small, as you do not want to shock the fish. So high flow rate is required (sparge in tank, to stop it turning into a whirl pool).

Hi Mad Fridgie,

A bit of both really. *I spoke to a friend of mine yesterday and he was wondering if it would be cheaper to make cooling system for his fish tank rather than buy one.

I said I'd look into it for him.

I've wanted to try and design and build something (on a small scale) for a number of years possibly even use it as some sort of a training rig so I thought this may be a good first project.

Once I have the figures you suggested how can I use them in my equation?

It is more than one equation.

pull down energy (no time) mass (kg)* SHC * temp diff (C) = Kj. "remember that a watt is a joule/second."
so for your original calcs 250*4.19*5 = 5238Kj,

Now lets bring time into "1" second or we could say flow kg/sec

Mass * SHC * temp diff = Kw

So lets say you have 1kw pump and 0.2Kw light and 0.4kw of heat ingress (warm building) total load is

1.6Kw.

we want a small split 2C

so we need to know the required flow rate

Duty / Temp diff / SHC = flow

1.6 / 2 / 4.19 = 0.19kgs/sec.

I'm at work today so will look through the calculations when I get home tonight. One thing I did notice was that you said a 2C split? By split do you mean SST and return water temp and why so low at 2c?

I've spoken to my friend and he gave me these figures;
He has
*1 UV bulb that is 16w
Led lights totalling 30w
5 pumps totalling 15w

The room the tank is in can get up to 28c when the weather is hot worst case scenario.

The fish are marine fish so the water temp shouldn't go below 25C, at least not too quickly.

Is it possible to have a minimum flow rate temp of 24/25C?

He is concerned with havingsupply watertemp entering the tank *too cold which could harm the fish.

*How could this be achieved? Possibly by increasing the flow rate?

Quote:

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Originally Posted by marc5180

I'm at work today so will look through the calculations when I get home tonight. One thing I did notice was that you said a 2C split? By split do you mean SST and return water temp and why so low at 2c?

I've spoken to my friend and he gave me these figures;
He has
*1 UV bulb that is 16w
Led lights totalling 30w
5 pumps totalling 15w

The room the tank is in can get up to 28c when the weather is hot worst case scenario.

The fish are marine fish so the water temp shouldn't go below 25C, at least not too quickly.

Is it possible to have a minimum flow rate temp of 24/25C?

He is concerned with havingsupply watertemp entering the tank *too cold which could harm the fish.

*How could this be achieved? Possibly by increasing the flow rate?

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Sorry split is the difference between flow and return water temp ( and yes it is about high flows)

You must also remember that you have a pump circulating through your chiller.

To ensure that the water jets from the sparge are not to cold, you insert a coil into the tank, the chilled water return passes through the coil before being sparged into the tank.
This is a very small unit, perhaps look at using a primary and secondary circuit, uses a inline beer cooler for primary water in beer cooler, and pump the water from the fish tank through the coils in the beer cooler. Your fish tank stat to control water water pump, let the beer cooler look after itself

Hi Mark.
similar enquiry from a fish tank nut with no money, suggested passing flitration water circuit through a bucket of water in a standard refrigerator. Plastic pipe coiled in bucket, add some ice for peak loads. Worked a treat. Punched a couple of holes in side of old refrig., was only necessary in summer peak loads. He spent $50.00 on an old refrig., kept his beer in it as well.

Quote:

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Originally Posted by Magoo

Hi Mark.
similar enquiry from a fish tank nut with no money, suggested passing flitration water circuit through a bucket of water in a standard refrigerator. Plastic pipe coiled in bucket, add some ice for peak loads. Worked a treat. Punched a couple of holes in side of old refrig., was only necessary in summer peak loads. He spent $50.00 on an old refrig., kept his beer in it as well.

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Yeh, the simple solution, Magoo. If the fridge has external condenser then there is no concern of damaging hidden pipes.. Mike.

Quote:

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Originally Posted by mad fridgie

It is more than one equation.
*
pull down energy (no time) mass (kg)* SHC * temp diff (C) = Kj. "remember that a watt is a joule/second."
*so for your original calcs 250*4.19*5 = 5238Kj,*
*
Now lets bring time into "1" second or we could say flow kg/sec
*

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Sorry for the late reply I didn't finish work till late last night.

I think I understand it. When I get the figure which is in Kj if I divide by 3600 then that gives me the figure in KW or Kj/s?

Quote:

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Originally Posted by mad fridgie

Mass * SHC * temp diff = Kw
*
So lets say you have 1kw pump and 0.2Kw light and 0.4kw of heat ingress (warm building) total load is
*
1.6Kw.
*
we want a small split 2C
*
so we need to know the required flow rate
*
Duty / Temp diff / SHC = flow
*
1.6 / 2 / 4.19 = 0.19kgs/sec.

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Is that the same formula Q=m.c.dt that your using to work that out?
If so can mass or mass flow rate be used as the 'm'?

So instead of trying to work out how to cool the 250litres of water i only needed to remove the heat being created by the lights, pumps and external temp?

Quote:

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Originally Posted by Magoo

Hi Mark.
similar enquiry from a fish tank nut with no money, suggested passing flitration water circuit through a bucket of water in a standard refrigerator. Plastic pipe coiled in bucket, add some ice for peak loads. Worked a treat. Punched a couple of holes in side of old refrig., was only necessary in summer peak loads. He spent $50.00 on an old refrig., kept his beer in it as well.

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The simple ideas are always the best aren't they :-).

In this case as with any other your need to understand where the load is coming from. So at present you have a tank full of warm water, so it would seem resonable that you should look at the tank mass as the load, but the only reason it is warm is due to external factors (pump ect). The tank may have warmed over a long period.
Basically it is no different to sizing an AC system or a standard coldroom, you do not take into consideration the original mass of the structure, (you know when you turn on a coolstore it will take awhile to come down in temperature) Only when you move into batch cooling do you really require the duty of the mass (for example if the tank was emptied and filled all the time would the duty of the mass be required into the calculations)