Equilibrium Temperature

It would increase the temperature of the reservoir!

The equation(s) you want are related to specific heats.

Q=mcΔt

Q = heat (rate)

m = mass (rate)

c = specific heat (in your case water)

Δt = temperature differential

I can not further set up the equations because I do not fully understand your problem, that is, you mention you require 70m3 avg, and use 15 m3, but have a reservoir of 670 m3.

In addition, the contaminants picked up from the cooling cycle will concentrate in the main reservoir.

milo

How much you send back? And how often? It is bad homework. Start with this: 1 m3 of water @35 degr. C and 1 m3 water @25 degr. C - delta t will be 10/2 the rest is for you: 15 m3 ......35 degr. C and 15 m3 @ 25 degr. C = the same. your result temperature will be 25 + (10. 15/670) with no additional losses counted.

If it is an actual life scenario, along with what has been explained above,

You want the heat to be dissipated somewhere ? You go in a closed loop and then where is the heat supposed to be lost ?

For this usually we don't have underground reservoirs , rather overground reservoirs and cooling towers, both of them lose heat through evaporation, but with that an amount of water also is lost.

The underground storage will lose heat, at a bit increased rate as the temperature of the water inside rises from 25^oC. However soil not beein a good conductor of heat will not be effective in pulling back the temperature, and you will have the temperature inlet to HE rising, that will cause the fluid temperature and the temperature of water at outlet to rise, till it stabilises at some temperature above what is calculated (that calculation assumes the water to be at 25^oC and 35^oC however after some time it will be 25+ and 35+)

Without getting into calculations, the heat picked up in the process has to be dumped somewhere before the water's returned to the tank, else heat will accumulate in the tank, the water therein will heat up, and you'll gradually lose cooling capability.

That being said, that heat can be dumped either to the environment (heat radiating from piping, water-to-air radiator, water-to-water heat exchanger, etc.), to some other place (say another process (including space-heating) through a heat exchanger), or out the underground tank into the surrounding ground.

First things first: your underground tank's likely insulated from the surrounding ground, else you'd always be losing heat to it; either that, or the surrounding ground is a poor thermal conductor (which means that its likely dry) or that something else is heating it (hot plant-equipment above it?). If it's not, please let us know how the water's temperature inside it is maintained at 25 C ... no ground is ever that hot naturally, unless it's near a geothermal source.

Another possibility: take the thermal insulation off of pipes going into the process if the ambient temperature around them is less than 25 C .. that way, the water will enter the process more cooly and will come out the other side more cooly too. Same thing atthe process outlet ... strip off the insulation and radiate the heat into the ambient environment if it's cooler than 35 C.

In addition, you may want to dump the heat into something useful; if you have a process that needs heat or if you need space-heating, use a heat exchanger to pass heat from your process' cooling-outlet to it.

If you've after all this you still have heat to dissipate, use a water-to-air radiator (natural draft or forced-air, depending on how much cooling you need) or a cooling tower if the air temperature is below 25 C, or use a water-to-water heat exchanger if you have a river, a lake, or a cooling pond nearby.

Hope this helps. Cheers! DZ

I never have been able, to reach water out of the ground, lower than 25 degrees celsius on the caribbean islands. This has been a costly experience, because our water cooled A/C systems did not perform. In Europe I had 16 degrees.

@ redfred.

never mind your comments, and also your harsh behavior, let me fill the space information that might be left

1. Water is continuously refilled in the reservoir and continuously withdrawn at average rate of 70 m3/hr.

2. Flow mentioned is per hour for all values.

3. there is no contamination of particles in the return water.

4 The reservoir is made of cemented concrete and top is covered with fibre glass sheet.

5. The temperature of reservoir i mentioned (25 C) is almost equal to the temperature with which water comes out of ground, i don't know the geothermal details of this.

6. Certainly the returning water will increase the temperature of the reservoir, reducing the cooling capability of water but what i wanted to know is upto what value, how can it be calculated, there must be some equilibrium temperature above which the temperature will not rise.

If you don't understand the question it doesn't mean the person asking the question is not qualified or don't have engineering knowledge, it may be other way.

Wait a minute, now I see the problem. As you mentioned you have a working cooling system that continuously draws a certain amount of water from an underground reservoir and the same water after being heated by 10C returns to this reservoir. You expect this reservoir to over time to have warmer than 25C water coming out of the ground eventually but this doesn't happen. Well obviously the amount of thermal energy you are dumping into the underground reservoir of water, container and earth that this reservoir resides in is insignificant compared to the energy you are adding. You have for your application a virtual infinite heat sink.

Remember, thermal insulation slows down the transfer of heat, it does not slow it down. Your 35C water is trying to heat all of the mass of earth you are standing on. While 15,000 kilograms of water is a lot of mass, it pale in comparison with the denser earth itself.

yes you are probably rite,

even if i consider the reservoir as thermocline as someone mentioned, the rise in temperature of 2.7 degree C will occur after approx 44 Hrs. This time is more than enough for thermal heat dumped to water reservoir to be transferred to earth and loss to atmosphere.

After all these comments and suggestions i suppose that it is better to re-circulate the cooling water inspite of wasting it without any considerable effect on system.

Is this tank 670 m3 capacity constantly refilling against use?

Or a tank emptying over time, then being refilled?

Either way the water in temperature, if different to 25 C, is also factor.

But say it's 25 C

Ignore the tank volume for a moment and imagine it as a pipe and a loop

Then the system output would rise to the ratio of 70 - 15 = 55 @ 25 C in (replenishment water) v.s. 15 @ 35 C (returned water) on initial circulation. I.e. 15 / 55 = 27% rise on 10 C = 2.7 C.

In an insulated pipe system, second circulation is 12.7 C supply to the exchanger plus 10 C gain. Or the effect instantly multiplies.

That you have 670 m3 slows the 27 % rise multiplication.

But ultimately the energy effect is equal to adding exchanger rise to the tank refilling flow.

Imagine this as forming a thermocline, and the exchanger will not see the rise until one tank volume is used. The volume riding above is now 25 + 2.7 = 27.7 C.

The next 27.7 + 27% of 12.7 rise, and so on.

Obviously you now must increase the flow through the exchanger - so the % rise on volume basis ratio changes.

Against this process is the (unknown) heat loss of the tank. Ground is a fairly good insulator. This system will rise, but how much is impossible to estimate on the data supplied.

I would think a simple process engineering move, would be to return the water via another passive exchanger. But I don't know your ambients or if this is 24/7, or if you could run the exchanger circuit at night to dump the heat rise. Or if a bit of evaporative cooling is the path to follow.

Equilibrium is impossible to estimate with out loss data or such information as above.

If I've read it wrong and it is a use and refill tank - the same process applies but obviously the last bit is going to be many cycles hotter.

Kyzine

Temperature mixed water:

670 x 25 + 15 x 35 = 17275/670 = 25.78 = 25.8C will be the reservior temperature after return of 15 m3 water at 35C.

I shall try to explain how I understand your question and give an answer according to my interpretation:

from a 670 m^3 at 25 °C you use during a given period of time (not specified) 70m^3 from which 15 m^3 are used as coolant and arrive at 35°C. If you direct those 15 m^3 back to reservoir you will compensate in order to stay at same volume only 55m^3 which are I assume at 25°C. At the end of a cycle you have in the reservoir 600m^3 at 25°C, 55m^3 at 25°C (refill) and 15 m^3 at 35°C . The remark about pollution is correct but I do not know the type of heat exchanger you use and how big the risk is.

If we make a balance the result is for the 670 m^3 the temperature will be :

(600+55)/670*25+15*35/670=25.224 °C

Is my assumption OK?

I think that you are right on- See my post #11 to get the same result.

Your issue is straight thermodynamics.

You are adding 15m3 X 10C x 1 (specific heat of water) units of energy to (670-15)m3 of water- I assume from your input that the (70-15)m3 of water returns to the tank at about the 25C starting temperature.

Basically- do the math. The (670-15)m3 of water is 97.8% of total volume. The actual NET resulting temperature will be about ((25C X 97.8%) + (2.2% X 35C)) = 25.22C.

I used "about" because the actual heating temperature difference will be the 35C minus the "final" temperature of the mix, since the mix will be warming slightly, reducing the actual DeltaT. A more accurate calculation would have been to multiply the "35C" used in the earlier calculation by 97.8% to allow for the lost differential.

70-15= 55 What do you do with this water? I tried to give an answer, based on your info. If you give a poor description, you will get a poor answer and keep us busy with guesswork. Analyze your system and provide more correct info if you want reliable answers.