How do i size a thermal Storgae Vessel(Buffer Tank)

[Tamarco]

Hi everyone,

Does anyone know how to correctly size a Buffer vessesl for a chilled water application?

Tamarco

[Peter_1]

What's your machine, are they able to run in partial load, pipe length, capacity, volume of the system, constant load or many times partial load,......
Many new machines don't need a buffer any longer.

[frank]

A buffer vessel is used to smooth out the return water temperature so eliminating unnecessary start/stops of the compressor. How big is your chiller/chilled water circuit?

[Andy]

Hi everyone,

Does anyone know how to correctly size a Buffer vessesl for a chilled water application?

Tamarco

I size the buffer tank to hold 10 minutes worth of the volume flow. This is to accomodate the maximum starts/hr of a large motor.
3600 litres flow/hour would be a 600 litre tank.

Kind Regards Andy

[Peter_1]

http://www.mcquay.com/mcquaybiz/lite...r%20hunting%22

http://www.refrigeration-engineer.co...hp/t-3030.html

I once heard 10 to 15 l/kW for odler chillers.

[mickandlee]

Try this for size.

Basics formula is kW = kg/sec x Δt x sp ht, where sp ht = 4.18 (sorry for you imps)

Therefore the total chilled water system volume can be calculated from

kg (or litres) = (a x b) ÷ (4.18 x c)

where
a = smallest capacity control step of chiller times 0.5 ( max cycling frequency occurs at this load – above this the compr. runs longer with shorter of’s – below this compressor is off longer than on)

b = time in seconds between compressor starts i.e. 30 starts / hour = 2mins. Or 120 secs.

c = max allowable chilled water temperature rise. Usually the full load temp range of the chiller multiplied by the minimum percentage capacity control. i.e. . if chiller max duty is to cool from 12 to 6 deg C, then if min capacity control step is 25% then allowable chilled water rise is 6 x 0.25 = 1.5 deg C

Example
130 kw chiller cooling from 11 to 5 deg C with 3 equal capacity control steps. Compressor motor limited of 30 starts per hour.

Total system volume = [(130 x 0.33 x 0.5) x 120 secs) ÷ [4.18 x (6 x 0.33)]
= 311 litres.

If chilled water volume contained in evaporator, water pipes and cooling coils is less than 311 litres then a buffer tank may be required.

However
Can this volume be found by increasing chilled water pipe work size?
Can you accept a slightly higher chilled water temperature rise.

This is were a good system designer earns his chicken feed by applying these compromises to the best advantage.

As a matter of interest the chiller in the above example would have a flow rate of 130 / (6 x 4.18) which = 5.18 ltrs / sec. Using 65 mm heavy steel pipe at 35 ltrs / mtr, means that 8.9 mts of pipe would meet the requirement in this case.

You need to know
Chiller duty
Unloading steps
Temp rise at full load
Compressor motor max starting frequency
Max acceptable temp rise.
Armed with this info its simple.

Peters15 l/kw gives 1950 ltrs
Andies 10 min flow gives 3108 ltrs.

Not saying these are incorrect, as I said it deponds upon many factors.

Mick

[Renato RR]

You must size bufer acording to instold compressor.Bitzer reciproc. compressors demands min. 5 minuts of operating before it stops.So the size depends on colling power with respect to water flow and compressors min runing time as well as frequency of start and stop.Acording to 130kW that would be 1800lit.

Renato

[JoshuaE]

kg (or litres) = (a x b) ÷ (4.18 x c)

where
a = smallest capacity control step of chiller times 0.5 ( max cycling frequency occurs at this load – above this the compr. runs longer with shorter of’s – below this compressor is off longer than on)

b = time in seconds between compressor starts i.e. 30 starts / hour = 2mins. Or 120 secs.

c = max allowable chilled water temperature rise. Usually the full load temp range of the chiller multiplied by the minimum percentage capacity control. i.e. . if chiller max duty is to cool from 12 to 6 deg C, then if min capacity control step is 25% then allowable chilled water rise is 6 x 0.25 = 1.5 deg C

Example
130 kw chiller cooling from 11 to 5 deg C with 3 equal capacity control steps. Compressor motor limited of 30 starts per hour.

Total system volume = [(130 x 0.33 x 0.5) x 120 secs) ÷ [4.18 x (6 x 0.33)]
= 311 litres.

mickandlee, I am a little confused with your equation, Where can I find it, for example is it referenced in CIBSE or Imeche?

The reason I am confused is that it would appear that you use the “capacity control step” on both the numerator and denominator, this means that it becomes constant and has no effect?

I might have just read the equation wrong? Any help you can provide would be greatly appreciated!

:S Josh

[Brian_UK]

I think that you are four years too late for an answer maybe.

The factor he has used is part of a bracketed calculation within each side.