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flow of water in a cooling system 3

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somacast

Mechanical
Aug 10, 2010
88
Hello everybody,

At first kindly refer to this topic for earlier discussions as I was seeking help to calculate the heat transfer from the tank of a cooling system we want to enhance
Further to this (kindly refer to attachment), looking at the full system overhere, it appears that a big amount of BTU is required to cool this knowing that the pumps are 27 m3/hr each, so the first intention to follow our old practice of cooling the water tank with an AC unit evaporator as shown in previous thread wont do, I did calculate based on q(dot)= m(dot) * Cp * delta T , and the flow of water is known and assumed that we want to bring water temp from 55 C to 25 C ....

my questions are :

1- as far as I know that pumps in parallel means we add their flow rates, which means in our case here it would be 3 x 27= 81m3/h ... however by going to site and looking through the small observation opening at the top of the tank (1.5 m3) I would say that there is no way its 81 m3/hr .. because that would have meant that this tank would have emptied in a minute which does not happen... can you please tell me if Iam write ? is it 27 x 3 ? or the 27 is divided equally between the 3 pumps over here as they take from the same branch?

2- well q (dot) = m (dot) * Cp * delta T , in addition to heat losses from tank calculated earlier cover my BTU needs to cool ? assuming that the cooling happens in the tank or just after it by means of a chiller?

if required the following are the parameters of the system ... Pumps discharge pressure = 3.0 Kg/cm2 , tank pressure = Atm . pipes sizes are shown in the attached P&ID
 
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Ok,

From what I can see here you have a closed loop water cooling system consisting of a buffer tank, three pumps and two air cooled heat exchangers, with two pumps rather oddly feeding one HX, then the combined water feeds three what appear to be instrument air packages where this is the cooling water supply before returning the flow back to the buffer tank. Therefore your comment about the flow emptying the tank is just plain wrong as all the water which exits the tanks returns to it at the same rate. This is a closed loop system with nowhere for the water to go other than back to the start point.

In essence three parallel pumps will increase flow. However the amount three pumps will run at could be anywhere from 81m3/hr to say <40. The issue is what are the pressure losses in each of the components and how does it respond to more flow and how does the pump curve look. This is known as the system curve. If it is fairly flat then 2 or three pumps in parallel can make a flow of between 2 to 3 times. If the curve is quite steep it could be 1.5 times the pump duty flow or less when operated in parallel.

E.g. One pump through your air cooled HX will give you a flow of X. However two identical will not give you 2X because the pressure losses are proportional to the flow^2. for a loop system though it gets more complicated as each element, whether it be a common section (the 4" pipe) or individual sections will give you different pressure loses at different flowrates. On something like this it's probably easier to do some trials and tests rather than try and work it out.

By the way, you need to place your coolers downstream of your air operated heat exchanger NOT in the tank. All you are doing if you do this is replacing the air driven cooling system with another one. The highest temperature water should enter the air driven heat exchanger so it looses the most heat to the air, but you wull never get it cooler than about 7deg C above the ambient air temperature (which is?). If you want the water cooler then cool it further AFTER the HX, but before the air compressor HX.

Hope this helps.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
@LittleInch ,

Thanks again , the problem with this close cooling water system is that it is a package unit, and that includes the compressors, the tank, the pumps, and the air cooled heat exchangers, and in such packages they don't provide good details about the components (the pumps in this case) so I don't have the performance curves or anything for the pumps.. (I will try to dig for the third pump which is oddly placed with one of the coolers as it was added at some later stages)

so is there any smart way to measure the flow? my section which are the concerned party does not want to invest in a cheap portable flow meter although we need it always... because 40 to 80 is a big difference, and that would certainly affect the result of q(dot)= m(dot) Cp Delta T, and hence would mean an over sized unit which we don't really need..

also I can see many elbows and reducers in addition to flow in the tube side of the shell & tube exchangers which means significant pressure drop is expected, so does this help to expect (higher/lower) flow rate? given that the curves are not available, but all what I know is that its a centrifugal multi stage pump ...

 
If you had the pump curves you might be able to approximate the flow from some power measurements (volts, amps)etc.

A clamp on UT meter would be good as well but I guess that costs money. Not that much though - this is an example. There's lots of other examples probably a bit less than this.
If you could divert the return flow somewhere and time how long the tank took to empty or go down a known level might work, then turn it off pretty fast and re-fill the tank.....

The greater the length and fittings of the common piping, then your curve will probably be sharper, but that second pump on one unit will also be an issue.

Did you get the point about not putting the cooler in the tank? It might be "what everyone does", but that doesn't make it right in this case. You won't actually lower the air temperature too much or at all of you do that, only shift the heat outflow from the air coolers to the AC unit.

Why are you doing this anyway? Is the air temperature too high? Are the air coolers too small? If its a packaged unit then surely it is either working within its design limits or it isn't

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
@LittleInch

Thanks for the link , its not expensive too but I will have to convince some funny people that we need this all the time,

back to the topic , yes I did get your point about placing it downstream the air coolers, because if I cool it before the air cooled H.E. it wont help anymore over there, in fact I might end up heating it as the ambient approaches 45 C at least mid day .. so yes thanks for this important point.

we are doing this because this old unit (1979) used to work alright , or lets say operators used to live with it , but with recent increment in ambient temperatures they started facing elevated temperature problem, if you don't cool the water well, the air leaving first stage compression to second stage does not get cooled in the inter-stage hence causing high temperature air to enter second stage causing it to trip, (it trips on 376 F), also they complain that water leaving the air cooled H.E. is about 120F which does not help at all ...

in fact a few degrees C down would do, but just need the right unit, I had the same idea which you just suggested about closing tank inlet and run the pump for seconds and c .. but I don't think the operations would agree to do that in this weather mainly that they might end up having to re-prime all of the pump..

from the readings I have here recorded by operations I can see that the air cooled H.E. brings the temperature down about 8 to 9 degs F in general ...
 
There's probably cheaper ones around - I just plucked that one out of thin air.

A 37 year old unit still going strong eh.

They might just need a new or cleaned air cooler??, but 45 C isn't going to yield much cooling.

Have you thought about cooling the air going through the air cooled HX? A little inefficient maybe but very easy to place the evaporator in front of the HX or after the fan.

Doesn't help with trying to work out what your water flow is...

If you can measure the pump electrical inlet power you will get a clue on flow rate if you use a pump efficiency of say 70% and can measure the pumps outlet pressure (assume inlet is atmospheric?)

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
@LittleInch ,

Yes its quite old, but they don't belive in old here, as long as it works, let it work :D

about the evaporater idea it was already studied to be implemented to other coolers, and for this one they already placing a water hose to spray it all the time :) , I wonder if its helping them ,but overall it wont get the temperature to what we want..

so if I may ask please : after getting the current (A) , how do I find the flow from current and pressure?? (inlet pressure is atm yes)

thanks alot
 
Power in is effectivel Volts x amps to get watts.

Assume say a 0.97% efficiency for your motor to get shaft power. (multiply electrical power by 0.97)

shaft power for a pump in kW is [ Q(m3/hr) x H(m) x SG(1) ] / [367 x eff (0.7)]

Solve for Q. H is the differential head.

This is far from exact as the efficiency might be 55%, but it'll tell you if you're pumping 10m3/hr or 20.

Spraying the air cooled HX might be Ok - you'll get some evaporative cooling.

Just because they've done it before doesn't mean it's correct or good or will work for your set-up.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Thank you very much , I will try to use that , and noted your point about the evaporative cooling also...

one thing I came to know recently is that the air coolers do cut in when temperature at the outlet of the after coolers of the compressor reaches some set point (cant recall it) .. so in case the chiller is to be added at the tank downstream, the air coolers wont start I think as the water will be below its auto start set point.
 
No problem.

Difficult to know what the control system looks for, but your description sounds about right. As said before it sounds to me like adding a cooler in the tank is simply replacing the air cooler for a different HX that can get you a lower temperature of the water as it's not limited so much by the ambient air temperature. However it costs more to run.

Good luck and let us know how it goes.

LI

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
You can also put the refrigerant evaporator in the tank but relocate the air cooled coolers downstream of the intercoolers so they cool the heated water downstream of the intercooler but before the water is returned to the tank.
 
Not a bad idea. Anything but the original plan will work better.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
@Littleinch : thanks a lot , I will see what happens as modifications go slow here due to procedures.

@lilliput1 : thank you , I shall consider that idea as well ...
 
@littleInch : I will see about that too , although the aim by my supervisors to install a chiller/cooler to cool the water, they wont go for other options easily ... but thanks again any info is helpful
 
Make sure you get from the manufacturer the cooling requirements of the compressor aftercooler. Get the recommended water flows and temperature in and out of each. They may be designed for 85F condenser water produced by cooling tower instead. Confirm lowest entering water temperature acceptable and corresponding flow. Check requirement for moisture separator at the compressor discharge downstream of the aftercooler. What is the make and model of the compressor. Is this going to be implemented for one or for all the compressors? If all, list all the compressors & do above for each.
 
@lilliput1 : thanks again, getting anything from the OEM wont be easy for a unit which was commissioned in 1979 :) , and for moisture separator it already exists and a dryer as well , but I agree about lowest allowable temperature for the inter & after coolers I have to get those, thank you.
 
Hi LittleInch,

Your optimism about our motors may be a little misplaced. :) Not many motors achieve 97% efficiency, and not all of the current flowing does useful work.

With only voltage and current available and without knowing power factor or efficiency, a rough rule of thumb is to use 1kVA input = 1 HP shaft output, i.e. power in HP = voltage * current * 10-3.
 
@ScottyUK : thank you very much, that was helpful too
 
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