Liquid receiver

[Renato RR]

What i need to consider to size liquid receiver.

Renato

[Lc_shi]

How much liquid will you need at lowest capacity of system operation ? Full capacity minus lowest capacity liquid is needed to go into liquid receiver:)my understanding,only for ref

rgds
LC

[guapo]

Hi,

I think the size of liquid reciever can charge all of refrigerant in the system.

Regards,
Guapo

[NoNickName]

Or, in case you would like to recover all or most of the liquid into the receiver, calculate it for full charge.

[Renato RR]

It is Copeland logic to recover all refrigerant in receive.Well can go wrong with this.

Thanks all,
Renato

[NoNickName]

What logic? Can you please elaborate? What can go wrong?

[Renato RR]

Nothing can go Wrong.In Copeland V volume write the same.But 300l receiver it is big mother ****er.And it isnt cheep.
The trouble is that I can provide such big receiver in that short time so I need to make paralel conection of 2 receiver each of 90l.That is total 180l.

I think thet will be enough.The manufacturer of chiller advice me 50l but I am not agree with his calculation.

Renato

[NoNickName]

300 liters receiver? What a huge PV!
Is that designed under PED certification?
What compressor are you planning to use?

[Renato RR]

Bitzer CSH8571-110Y R134A 400/3/50 X 2.

Renato

[US Iceman]

If the system charge is 300 liters, the receiver needs to be about 125% larger so that you have some vapor space on top of the liquid.

One question; how did you get from 180 liters to 300 liters for the receiver volume? That is quite an increase!

[NoNickName]

That compressor would require a total charge (with a DX evaporator and condensing coil) of approx 80 to 110Kg.
I would suggest a liquid receiver around the internal volume of the coil, for a flooded condensing coil control.
Otherwise, for a fan speed controlled condensation, I would suggest a liquid reicever of the volume of the flooded circuits (if used for subcooling), let's say 15-30% of the volume of condensing coil.
If EEV is required or used, reduce the volume further by 30%.

[Renato RR]

1)I have 2 compressors in engine room.
2)Discharge pipe first go to water condenser (sanitary water)
3)Then pipe go 65m in lenght to the air condenser outside the building.(rise 4m)
4)Pipes are from water condenser to air condenser "discharge"65 mm and back "liquid" 42 mm.
5)density of R 134a is for liquid 1160kg/m3 and for wapor 300kg/m3.
6)In situation when water condenser live refrigerant in liquid the 65mm would be full of 1160kg/m3,and when vapor leaves 300kg/m3.
7)Ask for more?

Renato

[Renato RR]

It is water chiller.

Renato

[NoNickName]

Well, thanks for pointing you were designing a split chiller. This is news. If you think there's anything else you would want to write in order to help us to help you, please feel free. Otherwise we could only be guessing.

You can use the 65m pipe run as liquid receiver and reclaim it from the bottom in case of recovery need. Much better than filling the plant with unnecessary volumes.

[Renato RR]

We buy chiller from Italian suplier and we only mount it.But from the start i have bad filling with this.

Soory if i give insufisent information.But I want to heard from you basic design prenciples and then start forvard.Also if I write the novell some of you will not read.

Renato

[NoNickName]

We buy chiller from Italian suplier and we only mount it.But from the start i have bad filling with this.

Soory if i give insufisent information.But I want to heard from you basic design prenciples and then start forvard.Also if I write the novell some of you will not read.

Renato

Oh, then I'll stop helping , until you buy the chillers from us. :D

[Renato RR]

There is time for ewriting.Now is time for helping.

Renato

[US Iceman]

OK Guys, let's start over.


What i need to consider to size liquid receiver?

First you need to know the total volume of refrigerant in the system if you want to do a pump down for 100% of the system charge.

You need to know the condenser charge: this depends on the head pressure control method. A flooded condenser coil will hold a lot of liquid during the winter.

Next you need to calculate the volume of the refrigerant piping, evaporators and accessories (driers, etc).

Use the liquid and vapor density at each component's operating condition to find the mass.

Total all of the mass calculations and you have the total refrigerant charge.

The receiver is normally sized to provide a volume equal to approximately 125% of the total system charge.

If you have some components that can trap liquid (maybe the water-cooled condenser), the amount of refrigerant calculated maybe too small.

[Renato RR]

Thanks Ice.

Renato

[Peter_1]

Croatia is not a EU member yet, do they need to complie then with PED?

[Peter_1]

Oh, then I'll stop helping , until you buy the chillers from us. :D
NoNickName, he wants a chiller which gives no problems:D :p

[Andy]

Ok all:)

What internal volume is the evaporator and will it need to be pumped out to service.
I work all out in litres, say the coil has an internal volume of 300litres and it is DX, which would mean a liquid content of 200 litres, then calculate the liquid line volume, won't be much in a chiller add it together and you have the net litres required. The gross volume will be
300 divided by 6 and multiplied by 7 350 litres gross.

Hope this helps.

Generally chiller receivers are sized at 50% of the above, due to the fact that they usually don't pumpdown, if they do the first figure is the one to go for.

Kind Regards. Andy:)

[NoNickName]

NoNickName, he wants a chiller which gives no problems:D :p

Well, charging 300Kg + of refrigerant is not the way to go for "no problems".

[Andy]

Well, charging 300Kg + of refrigerant is not the way to go for "no problems".

300litres would not be 300kg with the given density it would be 360kg aproximately.

This would not necessarily be a large charge, I have just sized the receiver for what I would call a small 404a blast freezer the net volume is about 700 litres.

Receiver volume is based on 2/3 of the evaporator volume + the liquid line, NOT the condenser volume, the volume in the condenser is just that, pumping down the system will not change the volume in the condenser or the condensate line.


Better an oversized receiver with less charge in it rather than a small receiver over charged.

Kind Regards. Andy:)

[US Iceman]

Receiver volume is based on 2/3 of the evaporator volume + the liquid line, NOT the condenser volume, the volume in the condenser is just that, pumping down the system will not change the volume in the condenser or the condensate line.

If you are using head pressure controls to flood an air-cooled condenser, the volume of the condenser does matter.

You could use the entire volume of the condenser (to calculate the mass of course) if the condenser is operating in very cold weather.

The liquid has to go somewhere in the summer.;)

[Andy]

If you are using head pressure controls to flood an air-cooled condenser, the volume of the condenser does matter.

You could use the entire volume of the condenser (to calculate the mass of course) if the condenser is operating in very cold weather.

The liquid has to go somewhere in the summer.;)

Never thought of that one:D

Again a very good reason for not undersizing the receiver. Generally I would use inverters on the condenser fan rather than back large amounts of liquid into the condenser, reduces the refrigerant charge.

Kind Regards. Andy:)

[NoNickName]

Flooding the condenser in winter results in high subcooling, which in turn results in low evaporating temperatures and subsequent low compressor performance. Much better Andy's solution: a chopper or frequency converter on fans.

[US Iceman]

...high subcooling, which in turn results in low evaporating temperatures and subsequent low compressor performance

I'm not sure how you equate high subcooling values with loss of compressor performance. The lower pressure ratio due to the decreased discharge pressure and the higher net refrigerating effect provide a substantial boost in compressor performance.

I can see the evaporating temperature dropping a bit, since there is not as much flash gas. But then again, if the compressors have capacity control, this would not be an issue. The compressor would unload and the evaporating pressure would be constant.

I'm not promoting flooded condensers and discharge pressures, but if you are operating an air cooled refrigeration system in the winter with TXV's you have to do something besides VFD's.

No argument about the need to reduce the system charge.

[NoNickName]

No, it is not a matter of flash gas... suppose the liquid refrigerant is 40°C in summer, and it is 20°C in winter.
What would you think the temperature after evaporation would be in the two cases?

It would be lower in the second case, that would results in lower evaporation pressure and lower suction pressure, and lower compressor performance.

I personally do not like to flooded condensers, though at standstill liquid migrates from receiver back to condenser. Air cooled chillers here are operated with just VFD, and we have installations in Russia and Finland.

[Andy]

Doctors differ patients die:eek:

As far as I can see you are both correct

The problem is actually TXV's not the head pressure, this do not operate correctly at low pressure drop.

Flooded condensers are not much used in the circles I travel in, perhaps it's just not cold enough in Ireland in the winter, it just about freezes.

Large amounts of subcooling increase system cycle efficiency, but cause liquid distribution problems.

We use EEV's (Danfoss) and float the heat normally.

where we are employing heat recovery we use the original condenser for subcooling, controlling the heat recovery condenser at 45 deg an dthe subcooled one at 30 deg c, increases the standard comercial type system C.O.P from 2.74 in the summer to 3.38 and only slightly reducing the C.O.P from 3.48 compared to floating head in the winter.

Kind Regards. Andy:)

[US Iceman]

suppose the liquid refrigerant is 40°C in summer, and it is 20°C in winter.
What would you think the temperature after evaporation would be in the two cases?

OK, let's take a look at this.

If the mass flow at the evaporating temperature using40°C liquid refrigerant is 1 kg/min (just an example number), the 1 kg coming out of the evaporator (at the evaporating pressure) is comprised of the flash gas from 40°C to the evaporating pressure. The flash gas will be some percentage of the vapor volume leaving the evaporator. Let's say 15%.

The remainder of the mass flow (say 0.85 kg ; what's left after the flash gas) is the refrigerant that provides the refrigerating effect in the evaporator. This portion of the mass flow also provides some portion of the total vapor volume leaving the evaporator.

The total of the two above constitute the entire mass flow out of the evaporator. (0.85 kg + 0.15 kg/minute). The volume flow would then be 1 kg/minute X the vapor specific volume at the evaporating pressure.

When the liquid temperature is decreased, the amount of flash gas formed decreases. (let's say we are now down to 0.1 kg) We are assuming the evaporating pressure remains constant. (more on this later)

Since we now have 0.1 kg of flash gas per minute, the mass balance available is 0.9 kg/minute available for cooling.

The refrigeration effect is increased due to the increased enthalpy difference so we have more capacity available from the remaining 0.9 kg.

With this we now have 0.9 kg/minute (higher mass flow available that is useful) and the greater enthalpy difference. Therefore the compressor has greater capacity.

Since the volume of flash gas formed is reduced, and the compressor is a constant volume device (assuming it has no capacity control), you would have a decrease in suction pressure.

So I agree with you the suction pressure will decrease if the liquid temperature is lowered, if the compressor does not have capacity control.

If the compressor has capacity control, it should be controlling the suction pressure at a constant pressure.

The lower volume flow coming from the evaporator will balance with the decreased compressor swept volume that is active at approximately the same pressure.



The problem is actually TXV's not the head pressure, this do not operate correctly at low pressure drop.

I agree Andy. The TXV's are the reason for head pressure control in the first place.

Let me back up a minute and state the older systems I have worked on were before the balanced port TXV's were available for refrigeration. With the older TXV's you had to have good head pressure control to feed liquid on very cold days (say below 0°C). Down to this point fan cycling or fan speed control worked just fine.

However, when the air temperature got down to about -28°C, it was a different problem. In this case with the old valves you needed the flooded condenser controls to keep the system running.

With the new balanced port valves I suspect fan speed control may work just fine.

[NoNickName]

So I agree with you the suction pressure will decrease if the liquid temperature is lowered, if the compressor does not have capacity control.



Good, we came at an agreement.
Now, suppose the compressor has capacity control. In this case the COP will be increased by the lower mass flow, but it will be decreased by the disproportion of the motor size to the actual mass flow.
This will also negatively affect the power factor of the plant because


Factors affecting the power factor of an induction motor are size, speed and load. The larger the
motor and faster its speed, the higher the full-load
power factor. The power factor of a motor varies
according to its load. The higher the percentage of
the rated load, the higher the power factor.

[US Iceman]

So you agree the flash gas does have something to do with it then???

Since we are in agreement? :D


In this case the COP will be increased by the lower mass flow

Only if you consider the use of all of the heat transfer surfaces and the re-balance that occurs. Since less mass flow is circulating at part load, the suction pressure can be increased slightly and the condensing temperature reduced a little.

The refrigeration system itself can become more efficient, but when you consider the motor load and power factor with this, the amount of energy used per kg circulated can increase.

This final effect of this is also determined by the compressor type and if a VFD is used to provide speed regulation. (plus the drive losses too)

[NoNickName]

I agree on the conclusions, and on the fact that higher subcooling does not equates in better overall performance

[Peter_1]

No, it is not a matter of flash gas... suppose the liquid refrigerant is 40°C in summer, and it is 20°C in winter.
What would you think the temperature after evaporation would be in the two cases?

It would be lower in the second case, that would results in lower evaporation pressure and lower suction pressure, and lower compressor performance.


Haven't read thoroughly the whole thread:
Liquid 40°C in summer and 20°C in winter will give a lower evaporationg temperature in winter conditions at the same condensing pressure??
It will increase the net effect on the evaporator a lot because teh enthaply of the liquid has increased seriously.
The COP of the compressor will increase with +/-25 to 30% and you're talking about a lower performance?


The power factor of a motor varies
according to its load. The higher the percentage of
the rated load, the higher the power factor

This is true but it may not be exagerated to much.
What is then the advantage of the digital scroll?
And form the unloader valves on larger machines and so many small semi-hermetic machines?

The net effect you gain by increasing the suction pressure by unloading is by far much higher then the negative effects of increase of the power factor.

We measured this some 15 years ago on a open 8CC compressor of DWM/Copeland and there was as far as I can remember almost no measurable difference.

And you pay - at least in Belgium - only the power you consume for smalelr applications, not the apparent power.
Line current will be a little bit higher but that's all. So if power factor rises, you could say, who cares!?


I agree on the conclusions, and on the fact that higher subcooling does not equates in better overall performance
I'm almost sure this is a wrong statement. Will try to simulate this the next days.

[NoNickName]

Guys, I'm not talking about a theoretical refrigeration circuit. I'm talking about reality. Suppose you've got a 100kW evaporator (you choose the design temperature, I don't care, it's just for example).
Now, the liquid in the TEV will decrease its temperature, with the net result of a lower evaporating temperature. Yes, you've got a higher performance for that amount of liquid, and the evaporator will benefit, but it will give still 100kW at the design temperature... or more but at a lower Te.
Now, suppose you've got more kW to evaporate as a net result of a higher subcooling. The evaporator will react by lowering its average Te. The TEV is a slave of the evaporator and will react accordingly to keep the set SH.
You see there's no gain whatsoever from a high subcooling IN A REAL CASE.
If by chance the evaporator was selected with a safety margin, than you'll probably have a benefit, but who really installs evaporators that generous?
Who cares what the enthalpy gain is, when one cannot take advantage of it. Worst of all: too much enthalpy gain will cause a relative undersizing of the evaporator.

On another basis: why few to no manufacturers are using subcooling circuits in condensing coils?

It sounds strange to me that in Belgium you pay for active power. I think the utility company let you pay for kWh+kvarh and not just kWh, like in most parts of the world. And kvarh are more expensive than kWh.

EDIT: oh, BTW, I wanted to buy digital scrolls from Copeland. They told me they are not on sell.

EDIT2: as you see, I didn't mention flash gas, because I never ever want to see flash gas in any working conditions. And, my experience teach me that the smaller liquid receiver, the better, eventually no liquid receiver is best, when combined with a good fan speed regulator.

[US Iceman]

On another basis: why few to no manufacturers are using subcooling circuits in condensing coils?

They cost extra money and no one will buy then!

[US Iceman]

Worst of all: too much enthalpy gain will cause a relative undersizing of the evaporator.

This might be true if the liquid temperature is lower than the evaporator temperature. Then the liquid entering the evaporator must first be warmed up before it will boil.

You loose evaporator capacity because the evaporator must warm up the liquid, which takes the place of the phase change surface area and delays the superheating of the vapor.

There is such a thing as too much subcooling. One I mentioned above. The other is if the TXV capacity is not adjusted for the increase in capacity due to the lower enthalpy.

[NoNickName]

This might be true if the liquid temperature is lower than the evaporator temperature. Then the liquid entering the evaporator must first be warmed up before it will boil.

I'm not expert of flooded evaporators, on the other hand liquid cannot enter the evaporator at all in a DX operation.



There is such a thing as too much subcooling. One I mentioned above. The other is if the TXV capacity is not adjusted for the increase in capacity due to the lower enthalpy.

How can TXV capacity be adjusted on the fly between summer and winter?

[US Iceman]

...on the other hand liquid cannot enter the evaporator at all in a DX operation

Sure it does. There is about 85% of the entering liquid which is atomized into droplets after the TXV which enters the evaporator coil. This is where the evaporator capacity comes from in the first place. The balance (+/-15%) is the flash gas created during the expansion process, if the liquid temperature is warmer than the evaporating temperature.

If the liquid temperature entering the evaporator is lower than the evaporating temperature, the liquid will not boil.


How can TXV capacity be adjusted on the fly between summer and winter?

You don't. And I'm not suggesting you adjust the TXV's from summer to winter.

The valves have to be selected for the amount of subcooling that may be found. A selection that meets both operating criteria.

In a DX system designed for subcooling (which is a good thing to prevent or reduce the flash gas in the liquid lines) the valve capacity must be corrected for the amount amount of subcooling that will be seen.

The subcooling increases the valve capacity.

[NoNickName]

The subcooling increases the valve capacity.

by a negligible fraction of the TEV design capacity.
I'm afraid that we have to agree to disagree on this matter.
But I'm proud to read opinions from such competent forumists.

[Peter_1]

Thank you :D

[Lc_shi]

It's very good discussion:)
Many basic concepts are coming out from fuzzy pictures.
Enough subcooling can prevent flashgas and then affect the valve capacity.
It should be said that: subcooling prevent valve normal capacity not decrease:D

really appreciate master's opinions!
Wish there's some day to come together to have face to face talk and argument:p

rgds
LC

[Renato RR]

Yes,subcooling increase valve capacity.But how can eneywone predict the amount of subcooling for all year condition.During summer subcoling can be 5 K but in the winter 25 K.
I have to reed some Sporlan bulletin to continue, scuse me.Untill then keep this discusion live.

Renato

[Peter_1]

This might be true if the liquid temperature is lower than the evaporator temperature. Then the liquid entering the evaporator must first be warmed up before it will boil.

You loose evaporator capacity because the evaporator must warm up the liquid, which takes the place of the phase change surface area and delays the superheating of the vapor.

There is such a thing as too much subcooling. One I mentioned above. The other is if the TXV capacity is not adjusted for the increase in capacity due to the lower enthalpy.

Mike,

What about pump systems where the coil/evaporator is fed with pure liquid. Liquid arives at -10°C for a cooling room, so it is injected or fed in the coil at a temperature +/-10 K lower then the room temperature.
It must be warmed up first before it starts to boil.

What is the difference?

The liquid has also a better thermal contact with the copper tubes then a gas/liquid mixture.
The more surface you can cover with low temperature liquid, the larger the capacity will be.

And with a TEV, the portion of the liquid that is transformed to flashgas while decreasing the pressure is +/- 30%. The energy to transform this change is taken from the liquid itself.
The more we feed liquid at evaporating temperature, the less flashgas will form behind the TEV, the more nergy remains in the liquid for a given mass flow, the more surface there will be covered with liquid.

So, I still remain sceptic with the statement that subcooling not allways increases evaporator capacity.

[wambat]

Dossat/Horan say, When liquid refrigerant is subcooled before entering the metering device, the refrigeration effect per unit mass is increased. The example given is the heat removed (6.96btu/lbm). for the saturated cycle, in the example, the refrigeration effect per unit mass is 60.38btu/lbmas shown in a proof11.11 in section 11.6. the proof states for the sub cooled cycle, the refrigeration effect per unit mass, is 67.34btu/lb, a difference of 11.53% This is for actual refrigeration cycle
This in turn reduces mass flow rate by 10.27% and since the mass flow rate is less for the sub cooled cycle, it follows that the volume of vaporthat the compressor must handle per unit of capacity will also be less. Going on, the compressor displacement required for the sub cooled cycle is smaller then that required for the saturated cycle because the volume of vapor per unit of capacity is less. so the COP will increase by 11.5% in the example.

[NoNickName]

... so the COP will increase by 11.5% in the example.


This is the example of how books should not be read.
In real life we already have a compressor, a TEV and an evaporator, the three chosen for a given SC.
In case SC is increased, there will be no COP increase and no different mass flow, because the components are still the same.
The net enthalpy gain will become lower evap. pressure, which in turn becomes as a lower compressor suction pressure, and a lower COP.

If the manufacturer declares data with 3°K SC, please stay with that. Any increase in SC will result in poorer performance for a given chiller.

[wambat]

Gee when you write your book ...please let us know, I can always use a good laugh

[NoNickName]

Ok, Let's publish some producers data. I attach the image below. sorry, the catalogue is in Italian, so I will translate for you.

The capacity of the evaporator must be corrected if the subcooling is different than 4°K. the corrected capacity is obtained by DIVIDING the evaporator capacity by the correction factor.

Eg. 15°K 40kW R22, the new evaporator capacity is 40:1.11 = 36kW

So the evaporator capacity DECREASES as subcooling increases.

[Lc_shi]

My intuition told me there's some misunderstanding. I'll think it over and come back.

thanks for every one endeavour to make it clear.

rgds
LC

[Renato RR]

Yes if evaporating temp. stais at -10C.But i think (maybe wrong dont atack me emediatly) that increase in subcooling would change evaporating temperature.Then new rules are in game.

Renato

[NoNickName]

And increase in subcooling will decrease the evap temp, which will further decrease the TEV performance, all other data the same. Anyway a TEV must be selected for dp, not just for Te.

[Peter_1]

....Dossat/Horan say, ..... so the COP will increase by 11.5% in the example.
That's what I thouht and was trying to explain with no so many words.

[NoNickName]

Ok, Let's publish some producers data. I attach the image below. sorry, the catalogue is in Italian, so I will translate for you.

The capacity of the evaporator must be corrected if the subcooling is different than 4°K. the corrected capacity is obtained by DIVIDING the evaporator capacity by the correction factor.

Eg. 15°K 40kW R22, the new evaporator capacity is 40:1.11 = 36kW

So the evaporator capacity DECREASES as subcooling increases.

Everybody please have a look at the table attached in the previous page.

[Peter_1]

The reason for this correction factor is just because the TEV is sized for 'normal' liquid temperatures.

When this liquid is subcooled, it will overfeed the evaporator because on that moment, the valve will be too big and hunting will occur.

The correction factor has nothing to do with a reduced evaporator capacity but with an increased valve capacity.

[Peter_1]

Everybody please have a look at the table attached in the previous page.

Table in previous page, I don't find any table in the previous page.

[NoNickName]

Table in previous page, I don't find any table in the previous page.

http://www.refrigeration-engineer.com/forums/attachment.php?attachmentid=887&stc=1&thumb=1&d=1145004047

[NoNickName]

The correction factor has nothing to do with a reduced evaporator capacity but with an increased valve capacity.

Well, which is the same seen under a different POV. Infact overcharging (please read here: higher subcooling, or flooded condenser) will result in a higher feeding (thanks to higher enthalpy) into the evaporator and

... roll of drums....

lower evaporating pressure and lower performance.

[US Iceman]

What about pump systems where the coil/evaporator is fed with pure liquid. Liquid arrives at -10°C for a cooling room, so it is injected or fed in the coil at a temperature +/-10 K lower then the room temperature.
It must be warmed up first before it starts to boil.

What is the difference?

A lot of pumped overfeed systems use back-pressure regulators on the evaporators. In these systems there are usually several evaporators that are operating only at the pressure of the pump receiver. Some evaporators are at high pressures, some are at low pressures.

In these systems the liquid refrigerant in the pump receiver is at the saturation temperature of the lowest pressure.

An example:

We have a pumped overfeed system with three temperature requirements; -12.2C (10F), -6.6C (20F), and 0C (32F).

All evaporators are connected to the suction line going to the pump receiver operating at -12.2C. The higher temperature evaporators (-6.6C & 0C) are controlled by back-pressure regulators.

Several evaporators are piped directly into this suction line; no back-pressure regulators. Their evaporating temperature will be -12.2C with no allowance for pressure losses.

The other evaporators use back-pressure regulators to control the saturation temperature/evaporating temperature at -6.6C, and 0C.

The refrigerant pumps are supplying liquid into all of these evaporators at -12.2C.

On the -12.2C evaporators all is well, since the liquid temperature is -12.2C and the evaporating temperature is -12.2C, so any heat added to the refrigerant (by the evaporator) causes the liquid to boil.

The evaporators operating at -6.6C and at 0C also have liquid entering the coil at -12.2C. The liquid is colder than the evaporating temperature, so it does not boil. It just flows through the evaporator picking up sensible heat at the evaporating pressure.

Since the evaporator pressure is higher (due to the back-pressure regulator), the liquid remains subcooled below the saturation temperature. When the liquid has warmed up to the saturation temperature of the evaporator pressure, it begins to boil.

When the evaporator has to warm the liquid up for it to boil, the evaporator has lost some capacity due to the sensible heat transfer, rather than a phase change.

Same thing can happen in a DX system. If the liquid refrigerant is colder than the evaporating temperature, the liquid will not boil in the evaporator.

The main concern with all of this is: What is the evaporating temperature and what is the liquid refrigerant temperature?

Does that help explain my earlier comments?

[US Iceman]

And with a TEV, the portion of the liquid that is transformed to flash gas while decreasing the pressure is +/- 30%. The energy to transform this change is taken from the liquid itself.

I quite agree with you. The flash gas is the "cost" of achieving the lower temperature liquid at evaporating pressure. The percentage of flash formed is due to the condensing and evaporating temperatures.


The more we feed liquid at evaporating temperature, the less flash gas will form behind the TEV, the more energy remains in the liquid for a given mass flow, the more surface there will be covered with liquid.

Yes. The colder the liquid is before it flows into the TEV, lower amounts of flash gas will form, which increases the NRE (net refrigerating effect). This means you will have a higher percentage of mass available for useful cooling.

The total mass flow has not changed we are just using the mass flow in circulation more efficiently.

[US Iceman]

If I try to translate and understand the Danfoss data posted, the example says the evaporator capacity is 40 kW / 11.4 Tons.

Using the same example for R-22, the correction factor for 4K of subcooling is 1.0. If the subcooling is no greater than 4K, no correction to the valve capacity is required.

Further, the example uses a 15K value for subcooling that will be provided by something. It does not say where the subcooling comes from, only that 15k of subcooling will be available.

The correction factor for this is 1.11.

Using the example evaporator capacity of 40 kW, the valve capacity required with 15k of subcooling is now, 40 kW divided by 1.11 = 36 kW.

Th evaporator capacity is not reduced 11%. The valve capacity has increased 11% due to the additional subcooling.

What this example is stating is the TEV valve capacity has increased 11% with 15k of subcooling instead of the same valve capacity with 4k of subcooling.

Therefore you select a TEV with a capacity of 36 kW. When the 15k of subcooling is included, the valve will have sufficient capacity to operate a 40 kW evaporator.

Sorry my friend, you are reading the information incorrectly.

EDIT+++



The reason for this correction factor is just because the TEV is sized for 'normal' liquid temperatures.

When this liquid is subcooled, it will overfeed the evaporator because on that moment, the valve will be too big and hunting will occur.

The correction factor has nothing to do with a reduced evaporator capacity but with an increased valve capacity.

I agree with Peter. He had the correct answer the first time.

[US Iceman]

Yes, subcooling increase valve capacity. But how can anyone predict the amount of subcooling for all year condition. During summer subcooling can be 5 K but in the winter 25 K.

A good question Renato.

You have to base it on the worse point of operation, which means the lower subcooling value. This provides sufficient valve capacity during the summer when the lower subcooling is available.

If the subcooling increases, the valve capacity also increases.

The increased subcooling in the winter is also another reason the refrigeration systems do not run as long during the winter. The systems are more efficient due to the lower liquid enthalpy entering the TEV's.

[US Iceman]

This in turn reduces mass flow rate by 10.27% and since the mass flow rate is less for the sub cooled cycle, it follows that the volume of vapor that the compressor must handle per unit of capacity will also be less.

Going on, the compressor displacement required for the sub cooled cycle is smaller then that required for the saturated cycle because the volume of vapor per unit of capacity is less. so the COP will increase by 11.5% in the example.



This I can agree with. When subcooling is incorporated into the design calculations, the compressor can be smaller as the mass flow required is less with the additional subcooling.

If the subcooling is increased after the system (which was designed for little or no subcooling) has been installed, the system run time will be greatly reduced due to the increase in capacity from the additional subcooling.

[NoNickName]

Well, I won't insist any longer, everyone is convinced of its own opinions. But the arguments on this thread would be useful to me when training our engineers. I didn't think that so many engineers would be convinced that overcharging a chiller is the right thing to do in order to increase COP.
I will include the caveats in the installation manuals.

[AMARNATH JHA]

You should consider the ratio of liquid and vapour.Ratio should be 80:20. 80 percent liquid and 20 percent vapour.
Regards

[US Iceman]

I didn't think that so many engineers would be convinced that overcharging a chiller is the right thing to do in order to increase COP.

We are not suggesting any system be overcharged and that by overcharging a system you increase the COP.:(

Obviously we must be loosing something in translation.

If the system is designed to use a large amount of subcooling you can take advantage of the benefits with less compressor displacement and lower required mass flows. The evaporating temperature will not decrease, since it is fixed by the compressor displacement and evaporator surface area.

If a system is designed for a very small amount of subcooling, and later a lot more subcooling is provided (by whatever means) after the system has been designed and installed, it is possible some interesting things could occur. This I believe is your point if I am not mistaken.:)

It did provide for a very good discussion though.

Best Regards,
US Iceman

[NoNickName]

If a system is designed for a very small amount of subcooling, and later a lot more subcooling is provided (by whatever means) after the system has been designed and installed, it is possible some interesting things could occur. This I believe is your point if I am not mistaken.:)



Yes, it is exactly my point. ;)

[Peter_1]

Convinced or not, that's not the point.
I asked the opinion of Helpman and Goedhart and Danfoss. As soon as they gave me an answer, I will post their view on this issue.

Sometimes, things are difficult to explain, especially in another language, I mostly feel this on my elbow.

I will try to explain it in another way, I hope you can follow me with this one: if you inject refrigerant in a coil at -10°C or you feed glycol of -10°C in a coil, the cooling capacity of the glycol coil will be a lot higher and there is even no phase change with the glycol, only a very good contact of the glycol with the inner walls of the copper.

The reason why the refrigerant has to boil completely to vapor in an evaporator is that the compressor can't handle liquid.

I personally think that pure thermodynamically seen, it's better that the refrigerant remains as long as possible at its coldest temperature.

I'm not saying my view is the right one, just throwing some suggestions in the ring again.

[Peter_1]

You should consider the ratio of liquid and vapour.Ratio should be 80:20. 80 percent liquid and 20 percent vapour.
Regards
You mean after the TEV?
You can calculate this but it's determined by evaporating and liquid pressure and liquid temperature.
Not excatly 80/20...theoretically, it can be 100/0.

[US Iceman]

Hi Peter,

This is discussion is developing into a long one. :D


... you feed glycol of -10°C in a coil, the cooling capacity of the glycol coil will be a lot higher and there is even no phase change with the glycol, only a very good contact of the glycol with the inner walls of the copper.

If a colder glycol is used, the coil capacity will certainly increase due to an increase in LMTD. If the glycol temperature is warmer, the coil capacity will decrease since the LMTD is now smaller. I agree with you the heat transfer is better with the liquid; glycol or liquid refrigerant. Heat transfer with refrigerant vapor is not very good.

The glycol heat transfer is dependent on the glycol flow rate and temperature differentials.

On a refrigerant evaporator, the capacity depends on the mass flow and enthalpy difference between liquid and vapor. This one is a phase change, where the glycol is a sensible heat transfer process.

I think we can agree on those points.

The tricky part with refrigerant feeding into an evaporator is the condition of the refrigerant; subcooled or saturated.

Plot this on a PH diagram. (example attached)

If the liquid exists at a pressure equal to the saturation temperature of 0C for example. If you subcool the liquid more, the line at 0C is extended to the left. If the liquid is cold enough when it flows through the TEV, the pressure drops straight down on the PH diagram.

If the liquid temperature is colder than the evaporating temperature the liquid will not begin to boil until it has warmed up to the saturation line (bubble point) of the refrigerant at that pressure.

In other words, the liquid is subcooled below the saturation temperature of the evaporator pressure.

On a liquid overfeed system using back-pressure regulators it is the same. If we pump liquid refrigerant from -10C up to a pressure equal to 0C, when the liquid enters the hand expansion valve the liquid temperature is still below the evaporating temperature and will not boil. The liquid is still in the subcooled region of the PH diagram.

After the liquid begins to warm up to the saturation temperature of the evaporator, the liquid begins to immediately boil. So you do have increased heat transfer with more useful liquid, and less flash gas.

The problem is we have to use part of the evaporator surface to warm the liquid up. This may not be very much, but it does happen.

Most evaporator manufacturers will say the liquid temperature must not be lower than the evaporating temperature. This is the reason why.

Liquid temperature above the evaporating temperature are OK.


The reason why the refrigerant has to boil completely to vapor in an evaporator is that the compressor can't handle liquid.

Very true on a DX system. On liquid overfeed system the pump receiver protects the compressor. On a DX system you must have suction superheat to protect the compressor.

Does that help explain my comments? ;)

[NoNickName]

I personally think that pure thermodynamically seen, it's better that the refrigerant remains as long as possible at its coldest temperature.



That makes perfectly sense. But in a real chiller, things can't be seen under a purely thermodinamic point of view. Components are chosen and installed based on a design (essentially cost-driven).
A given evaporator will transfer a certain amount of kJ/s at a design condition. Increasing the enthalpy in kJ/Kg, given a fixed kJ heat transfer capacity, the mass flow in Kg/s must reduce to keep things balanced.
In order to achieve this result, the TEV must throttle back, with a net result of a lower evaporating pressure (same surface, less mass flow) for the same SH.

[wambat]

We are not suggesting any system be overcharged and that by overcharging a system you increase the COP.:(

Obviously we must be loosing something in translation.

If the system is designed to use a large amount of subcooling you can take advantage of the benefits with less compressor displacement and lower required mass flows. The evaporating temperature will not decrease, since it is fixed by the compressor displacement and evaporator surface area.

If a system is designed for a very small amount of subcooling, and later a lot more subcooling is provided (by whatever means) after the system has been designed and installed, it is possible some interesting things could occur. This I believe is your point if I am not mistaken.:)

It did provide for a very good discussion though.

Best Regards,
US Iceman

Either way, weather it,s designed for more subcooling or not the fact remains that all that I quoted remains true, presonnally I trust the authors credentials and for the life of me I don't understand why this is so confusing. All was written as operating under actual conditions. That is to say that given some operating condition, if the conditions were to change and more sub cooling were made avialible then the results as I quoted would result. :confused:

[US Iceman]

hi wambat,

I am not disputing any of the information you provided. In fact I agree with the logic behind the statements.

This thread originally started off as a simple question of how to size receivers and took a turn somewhere about the real and perceived effects of subcooling and the effect on the system.

Now it's taken another direction about refrigeration thermodynamics.

[Peter_1]

Well, these discussions forces you think a little bit further then you normally do, then most techs do, and you soemtimes can learn a lot of it.

It also pushes you to go back to the basic and sometimes, you need to grab a book again.

Will read it tomorrow morning, it's to late now (midnight) and someone's waiting upstairs :)

[wambat]

Peter...Lucky! :>)

[Peter_1]

It sounds strange to me that in Belgium you pay for active power. I think the utility company let you pay for kWh+kvarh and not just kWh, like in most parts of the world. And kvarh are more expensive than kWh.

Only the big consumers have a kvarh meter.
I don't have one in my shop, let's say that you don't have one till a fuse of 63 A , sometimes 80 A on a 3phase 380 V+N net.
We normally have a kWh meter (which is active power), not a kVArh meter, the reative power.
But, in the transformer cabin is a central kvarh meter which counts this lost power on the main supply. If someone on this transformer is losing too much power via reactive power, then they will try to localize it.

I never saw a reactive power meter - even in Italy - in a butcher, bakery, a small garage..

The big machines are installed in factories but they have there own transformers, at least in Belgium. In these transformer cabins, bot reactive and active power is measured and you see cos-phi capacitors.
But they pay almost 1/2 of the normal tariff/kWh, so it's worth then to measure the kVAr and they counter measurements.

[Peter_1]

In order to achieve this result, the TEV must throttle back, with a net result of a lower evaporating pressure (same surface, less mass flow) for the same SH.

I think that you should say....(same surface, less mass flow...but a mass flow with a higher enthalpy)

[Peter_1]

This is discussion is developing into a long one. :D
The problem is we have to use part of the evaporator surface to warm the liquid up. This may not be very much, but it does happen.


Long discussions learns me the most.

I agree wih you on lest's say everything, just the above extract out of your post needs some clarification for me...

You need to warm up the liquid before it begins to boil, before we reach the beginning of the phase change......OK, I'm with you......or in other words, heat has to be absorbed from the space through the tubes and added to the subcooled liquid to reach the boiling point.
I see this as a positive thing, heat is been drawn away from the space...:confused:

There 's no necessity to reach the phase change (unless for the compressor of course in a common DX system)

Perhaps, that was what you were trying to explain say and something was lost due to the language barrier.

It became here of course more and more a hypothecial thesis.

There's no necessity that we reach the boiling point. Suppose a liquid overfeed system - theoretical coil - and we subcool the liquid that much that it enters a very small coil at a high speed where there can't be added enough heat to even reach the boiling point, this coil will cool and it will work with at it's highest efficiency.
And we haven't reached yet the phase change status.

If we feed that coil with liquid at evaporating temperature, then this coil will perform much worse then the first one.

[NoNickName]

I never saw a reactive power meter - even in Italy - in a butcher, bakery, a small garage..


This is off topic, but nowadays also houses in Italy have got a kvarhmeter, though the utility does not let you pay for kvarhs until a certain power installed, and it can well be as much as 63 or 80A, which is BTW the current drawn by a small water chiller (small by our standards).

[Peter_1]

Small to your standards???
That's a chiller of +/- 130 kW and fits perfectly in Montair's higher capacity products.

[NoNickName]

130kW cooling capacity is one of our smallest chillers, Peter.

[Peter_1]

Can yo build those in a factory of only 6000 sqm?:D
My place is already 4000 sqm.

[US Iceman]

You need to warm up the liquid before it begins to boil, before we reach the beginning of the phase change......OK, I'm with you......or in other words, heat has to be absorbed from the space through the tubes and added to the subcooled liquid to reach the boiling point.



That is right.


There's no necessity that we reach the boiling point.

That is true if you have designed the evaporators for sensible heat transfer, like a chilled water coil.

The capacity of the refrigerant evaporator is based on a phase change. If the refrigerant does not boil, the evaporator capacity will be less than it was designed for. It still does some cooling since the cold liquid will absorb heat. However, if the refrigerant does not boil, it will not provide the design cooling capacity.



Suppose a liquid overfeed system - theoretical coil - and we subcool the liquid that much that it enters a very small coil at a high speed where there can't be added enough heat to even reach the boiling point, this coil will cool and it will work with at it's highest efficiency.



What I am trying to describe is not a theoretical system operation, but a practical one. I agree with you on the theory. It is possible to do what you describe with refrigerants. In this example, the evaporator is working like a chilled water coil. No phase change. However, for a liquid overfeed system this is not the most efficient.

The evaporators are designed to have cold liquid at the evaporating temperature entering the coils. As soon as the refrigerant begins to boil the highest heat transfer occurs at this time.

On a liquid overfeed evaporator the liquid refrigerant is pumped into the coil with additional liquid to provide a completely wet coil surface. Just as you described earlier.

Not all of the liquid will evaporate. Some returns back to the pump receiver. The extra liquid and the phase change result in the highest efficiency of the evaporator coil.

If you compare a DX evaporator of a specific capacity, to the same physical coil using liquid overfeed you will see the coil using liquid overfeed will have about 10-15% more capacity.

The percentage increase is due to the coil surface being wet with boiling refrigerant, since it is not being used to superheat the refrigerant vapor.

If the liquid entering the evaporator is subcooled, it does provide some useful cooling since the liquid is absorbing heat from the space.

However, since the evaporator capacity is designed for a phase change, it may not reach it's full capacity if the evaporator has to warm liquid. If the liquid does not evaporate there is no vapor formed. Then you would not need a compressor.

DX evaporators and liquid overfeed evaporators are similar in construction and purpose. Both boil liquid refrigerant.

A DX coil adds superheat by using a TEV to protect the compressor. 100% vapor leaves the coil.

A liquid overfeed coil has zero superheat for higher efficiency. 25-30% of the refrigerant leaving the evaporator is vapor, the rest of the refrigerant (70-75%) is liquid (on a mass basis).

Does that help to explain it?;)

[Peter_1]

US Iceman,

I will have to study a litlle bit on the liquid overfeed systems and the phase change designed coils.

I'm really impressed about the knowledge of you.

I think Wambat and Josip can add also something more regarding this issue.

What literature do you recommend?
It's not that I do'nt believe you but I want to read something more about all this.

Searched in Stoecker bu didn't found anything usefull.

Wambat, or do I have to say Sh..un :p pointed me to Dossat. I will try to make some time to read that specific chapter.

[US Iceman]

It's not that I don't believe you but I want to read something more about all this.

OK. :D

It is a lot of information to try to post and explain. There is another thread in the industrial section for liquid overfeed systems perhaps we should move over to that thread to keep it together??

I will look for some information also.

[Peter_1]

This is perhaps one USIceman

http://www.wlv.com/products/databook/db3/data/db3ch15.pdf

Perhaps the Mod of this section can cut and paste them in that new section

[luzhang]

surge receiver or through recriver

[luzhang]

surge receiver or through receiver

[Peter_1]

Commisioned last Friday a Blue Box chiller with...of course...a Carel unit in it.
Very nice machine.
Peace of cake of course :D

[mcgru-]

When the evaporator has to warm the liquid up for it to boil, there will be a difference between coefficients of heat transfer between the refrigerant and tube wall. In approximate numbers, the heat transfer between well-turbulented liquid and tube wall will be about 2000-3000 W/(m2.K). But in the case of boiling, this coefficient will be 2-3 times greater - 6000-8000 W/(m2.K). Buuut, when the refrigerant step into the liquid/gas ratio of about 30% and lesser (due to it's boiling), the effective coeficient of heat transfer becomes lesser - due to decreasing area of contact between (boiling) liquid and tube wall and thus increased contact area between gas and tube wall (and its greately lesser heat transfer coefficient)
alpha_liquid_tube ~ 2000-3000 W/(m2.K)
alpha_boilingliquid_tube ~ 8000 W/(m2.K)
alpha_gas_tube ~ 50..100 W/(m2.K)
Tube wall thickness is about 1mm, copper, the resistivity to transfer the heat is about 0.4 (m2.K)/W
Total heat transfer resistivity of tube wall itself and tube inner surface (contact with refrigerant) is a sum of h.t.resistivities. Let's look onto differences:
R1=0.4+1/8000=0.400125 (case of boiling refrigerant, 100% liquid)
R2=0.4+1/100=0.41 (case of 100% gas)
Difference between R1 and R2 is only about 3%.
We can see that most of resistance comes from tube wall (it's thickness and material), and that boiling process brings only a small effect to the heat transfer.
But also we should consider the hydrodynamics in the tube and the lesser specific heat capacity of gaseous refrigerant rather than liquified due to it's lesser density and ability to faster get outer temperature.



That's the explanation I was looking for and it makes sense: the difference in heat transfer coefficient between boiling and liquid state.

Is your conclusion that it doesn't make only a small negatiev difference in evaporator capacity if we feed seriously subcooled liquid?

May I ask you, where have you found this literature?
These figures were derived from tests in a laboratory environment I suppose?
Second question, where can we see/buy your work?



Wow, two posters (you and the other Chinese poster) in one day with the professional competence of programming and the needed thermodynamic knowledge to convert it in a software program.
As said with the other poster, I'm impressed.:eek: :eek:

[US Iceman]

WOW. That was a great explanation mcgru-. :eek:

I would like to add; any improvement in the tube side heat transfer makes very little difference since the air film coefficient may the controlling factor. The refrigerant heat transfer (whether it is phase change or sensible) is almost always greater than the air film coefficient.

In effect, we can improve the refrigerant film coefficient using turbulators, spray nozzles or other mechanisms in the phase change or single phase regime. However, since the air film coefficient is so low to begin with any improvement in the overall U value due to the refrigerant heat transfer is negligible.

As you suggest, the wall resistance may be the easiest factor to deal with.

You are correct. I was not allowing for the wall resistance in my discussions. My thoughts were only based on:



alpha_liquid_tube ~ 2000-3000 W/(m2.K)
alpha_boiling liquid_tube ~ 8000 W/(m2.K)


I would be interested to discuss some of the simulations you did for the:



a) processes inside the reciprocating compressor,
b) dynamic heat transfer between tube with refrigerant flow and environment


Sounds like something I would enjoy. :cool:

Welcome to the RE forum. :D

[Peter_1]

Mcgru and USIceman and other readers, I made just a terrible mistake.

I edited this marvelous post instead of quoting it.
DAMN ME.

Can someone please restore my fault asap. with teh original post.
PLEASE!!!!!!!!!!!!!!!!!

1000 times SORRY for this.

[US Iceman]

Peter,

I'm not sure who might be able to do this except the original poster or WebRam.

[Peter_1]

This was the post I should have been posted:

That's the explanation I was looking for and it makes sense: the difference in heat transfer coefficient between boiling and liquid state.

Is your conclusion that it doesn't make only a small negatiev difference in evaporator capacity if we feed seriously subcooled liquid?

May I ask you, where have you found this literature?
These figures were derived from tests in a laboratory environment I suppose?
Second question, where can we see/buy your work?



Wow, two posters (you and the other Chinese poster) in one day with the professional competence of programming and the needed thermodynamic knowledge to convert it in a software program.
As said with the other poster, I'm impressed.:eek: :eek:

[Peter_1]

Mike (or others), don't you have the original post in your private email box as send to you automatically when someone post a reply on a thread you subscribed?

What a stupid mistake.

I will contact Webram immediately now.

[US Iceman]

Peter,

I changed my email notification. :( I was getting so many emails every day I changed the setting to NO EMAILS since I visit the site every day.

[mcgru-]

I edited this marvelous post instead of quoting it.
Peter, i think that my language is terrible, so my post is self-balanced :)
It's ok in quoted state.

[Peter_1]

McGru,
If there's one thing you shouldn't care about, then it is how you write English here on RE.
Some have difficulties with my English but who cares..

If you use IExplore, then you can instal a free, goodworking English spell checker for it at http://www.iespell.com/

I bet :p you speak better Russia and other languages then I speak English, so...

The purpose of this site is providing information to techs all over the world with the same interest, let's say for some... the same passion.

What about the possibility we can see your work? I'm very interested in it and others too.

Do you mind if I try to restore your original post?

Kind regards.

[mcgru-]

That's the explanation I was looking for and it makes sense: the difference in heat transfer coefficient between boiling and liquid state.

I have to say, that my experience in thermodynamics and all other refrigeration stuff is much lesser than of Prof.Sporlan - I'm just study myself, and my studying is based probably on not so good book sources...
So, i think that there are a lot of better books, but for my _estimative_ calculations in many cases the Polmann-handbook is enough.
Complicated hydro-dynamic simulations are carried out using very complicated software. It's not in my force to create such one by myself only in a foreseeable future. So my way is an estimation now. Sorry :)

OK, the source of my estimative data on boiled liquid heat transfer coefficients is....

1. Turbulent flow of water in a tube
1.a (by Sender and Merkel)
h_ci = 2040*(1+0.015*t)*(w^0.87)/(d^0.13) [W/(m2.K)]
1.b (by Schack)
h_ci = 3370*(1+0.014*t)*(w^0.85) [W/(m2.K)]
Let we have a tube of 12mm diameter (10mm inner), the speed of liquid about 1m/s. So, for water, estimative h_ci will be:
h_ci_1a = 2040 * (1^0.87)/(0.01^0.13) = 3712
h_ci_1b = 3370 * (1^0.85) = 3370
[W/m2.K]
Imagine that instead of water we have the liquid subcooled refrigerant. Heat capacity of water is 4.18 kJ/(kg.K), and the h.c. of liquid refrigerant is about 1.2-1.6 kJ/(kg.K), about 2.5 times lesser.
There is no linear proportion, but somehow the heat capacity of liquid depends heat transfer coefficient.
So, if we divide that values (3712 or 3370) onto, for example, 1.5-2, we'll get those estimations of heat capacity coefficient for turbulent flow of refrigerant: (3370..3712)/(1.5..2)=1685..2474.
Difference from estimation of 2000-3000 goes from... tubes may have outer diameter of 10mm (8mm inner), or speeds are greater than 1m/s.

2. Boiling/Condensation
2.a (Condensation, according to Nusselt, for horizontal tubes)
h_c = 8900 * 1 / (d*dt)^0.25
estimation (for water): h_c = 8900*1/(0.01*10)^0.25 = 15800 [W/(m2.K)]
here, dt is a temperature difference, so this dependance is non-linear on temperature.
To make 1kg of water boiled, we need 333 (Sorry! my mistake :) it is about 2200) kJ. The usual enthalpy differences of refrigerant in subcooled applications are 100..150 kJ/kg - as we can see the ratio is 14..20 times.
Let's divide... 15800/(14..20)=790..1120 [W/(m2.K)] - the value for refrigerant.
(yeah, the condensation has lesser coefficients than in boiling case due to lesser turbulences in droplets being created, laminar flow)
2.b (unknown author, for water)
h_c = 1.94 * q^0.72 * p^0.24, where q is heat flow density, W/m2, p - pressure
For evaporators we have heat flow desities of about 150kW/m2 (consider surface of not fins, but inner area of tubes!), pressures - 2..3 bars
So, estimation is: h_c = 1.94 * (150000)^0.72 * (2..3)^0.24 = 12200..13460 [W/(m2.K)]
it is a bit lesser, than of Nusselt for horizontal tubes, though.
(i am already rethinking my way, sorry) :)
2.c (diagram of unknown author, experimental values for R22)
It's a picture, i'll post it here later. For heat flow densities of about 100kW/m2, R22, the h_c is about 7500-12000 W/(m2.K) for pressures 0.39..2.15 bars respectively.
(But these data is experimental, and not mine :) so you may trust it :) )

(went for preparing coffee and that picture)
:)

[mcgru-]

if it will be any difficulties with translating russian symbols, i'll help you - just note it

[mcgru-]

Dear moderator, i'm sorry for this big message. If it will not fit in forum rules, i'll (or you, please) delete it... Thank you for your patience.


I would be interested to discuss some of the simulations you did for the:

some code from compressor simulating program:


# ---------------------------------------------------------------------------
#
# START OF ROTATION
#
for ($a=0; $a<=360; $a+=$anglestep) {
$curr{a} = $a;

%prev = %curr;

# Heat transfer to wall
$curr{S} = $curr{pos} * 2*3.14159*$Radius + 2*$Scyl;

%curr = HeatTransfer(\%curr,$Twalls,$tau);

# Leakage between piston and cylinder wall

if (($curr{P}>$Pin) and (not $discharge)) {
$wleak = sqrt(2.0*($curr{P}-$Pin)*101300/$curr{D});
$wleak=180 if ($wleak>180);
$dMsuction = $wleak * $Sleak * $curr{D} * $tau;
}else{
$dMsuction = 0;
}
$curr{m} -= $dMsuction;
$curr{P}*= $curr{m}/$curr{V}/$curr{D};
$curr{D} = $curr{m}/$curr{V};
$Mleak+=$dMsuction;

#print "dMsuction:$dMsuction\n";
#PrintPoint(" leakage:");


# Expansion

%next = %curr;

$next{a} = $curr{a} + $anglestep;
$next{pos} = ($Stroke / 2.0) * (1-cos($next{a}*3.14159/180));
$dpos = $next{pos}-$curr{pos};
$dVprev = $dV;
$dV = $dpos * $Scyl;
if (($dVprev/$dV)<0) {
$suction = undef;
$discharge = undef;
}
$next{V} = $curr{V} + $dV;
$dVv = $dV/$tau;
$next{D} = $next{m} / $next{V};

$gamma = getG($curr{P},$curr{T});
$Pshouldbe = $curr{P} * ($curr{V}/$next{V})**$gamma;
if ((not $suction) and (not $discharge)) {
$next{P} = $Pshouldbe;
$next{T} = getTbyS($next{P},$next{enthr});
$next{H} = getH($next{P},$next{T});
}



%curr = %next;

#PrintPoint(" expansion:");

# Suction and discharge


if ($curr{P}<=$Pin) {
$w = $dVv / $Ssuc; ## here the crsection of input equalized to Scyl/2
# $w = $dVv / $Scyl;
$wv = $dVv / $Svalve;
# $maxw = 0.95 * getSS($curr{P},$curr{T});
# if ($wv>$maxw) { ## should not be supersonic
# $dVsuccoeff=$maxw/$wv;
# }else{
# $dVsuccoeff=1;
# }
$dPin = $DefaultPressureDropSuction + $dPsuc + $dzeta * $Din * ($w*$w)/2.0 /101300;
$Psuc = ($Pin -$dPin); # values that should be to start process
print "dPin=$dPin; Psuc=$Psuc;\n";
$suction = 1 if ($curr{P}<=$Psuc);
}

if ($curr{P}>=$Pout) {
$w = $dVv / $Ssuc/1.5;
# $w = $dVv / $Scyl;
$dPout = $DefaultPressureDropDischarge + $dPdis + $dzeta * $curr{D} * ($w*$w)/2.0 /101300;
$Pdis = ($Pout+$dPout);
print "dPout=$dPout; Pdis=$Pdis;\n";
$discharge = 1 if ($curr{P}>=$Pdis);
}


if ($suction) {
# $dVsuc = $dV * ($curr{P} - $dPin) / $Pin;
$dVsuc = $dV * ($Psuc - $dPin) / $Pin;
$dm = $Din * $dVsuc;
$Vsuc+=$dVsuc;
$next{m} += $dm;
$next{enthr} = ($curr{enthr}*$curr{m} + $Sin*$dm)/$next{m};
$next{H} = ($curr{H}*$curr{m} + $Hin*$dm) /$next{m};
$next{T} = ($curr{T}*$curr{m} + $Tin*$dm) /$next{m};
$next{D} = $next{m} / $next{V};
$next{P} = $Psuc;
print "suction @ $Psuc! dVsuc=$dVsuc; Vsuc=$Vsuc; w=$w:$wv dP=$dPin\n";
# $Pupper = $Pin;
}


if ($discharge) {
# $dVdis = $dV * ($curr{P} - $dPout) / $Pout;
$dVdis = $dV * ($Pdis - $dPout) / $Pout;
$dm = $curr{D} * $dV;
$next{m} += $dm;
$next{D} = $next{m} / $next{V};
$next{P} = $Pdis;
print "discharge @ Pdis=$Pdis! dV=$dVdis; dVv=$dVv; w=$wout; dP=$dPout\n";
# $Pupper = $Pout;
}

%curr = %next;

PrintPoint(" suc|dis:");

$arm = sin(3.14159*$a/180);#*$Stroke/2.0;
$direction = $arm/abs($arm) if (abs($arm)>0);
$arm = abs($arm);

$InnerGasWork = ($curr{P} - $Pin) * $dV * 101300 ;

$MotorWork = - $InnerGasWork + $FrictionForce * abs($dpos);
# $MotorWork *= $arm ;
$MotorWork = 0 if ($MotorWork<0);

$MotorPower = $MotorWork / $tau;
$MotorWorkTotal += $MotorWork;

# print " MotorPower: $MotorPower\n";


# print "angle:$a\n";
}


$Vreturned = $Mleak/$Din;
$Vsucreal = $Vsuc - $Vreturned;
$Veff = $Vsucreal / $Vcyl;

#printf "Vsuc=%.4e; Vsucreal=%.4e; Veff=%.4f\n", $Vsuc, $Vsucreal, $Veff;
printf "Veff=%.4f\n", $Veff;

printf "Total Veff: %.4f\n", $Veff*$Veff2;

printf "Compressor displacement (one piston): %.2f m3/h\n", $Vcyl/(60/$W)*3600;
printf "Compressor volume capacity (per piston): %.1f m3/h\n", $Vcyl/(60/$W)*3600*$Veff*$Veff2;

printf "Compressor displacement: %.2f m3/h\n", $Vcyl/(60/$W)*3600*$NumberOfCylinders;
printf "Compressor volume capacity: %.1f m3/h\n", $Vcyl/(60/$W)*3600*$Veff*$Veff2*$NumberOfCylinders;


$MotorPowerAverage = $MotorWorkTotal / ($tau*360/$anglestep);
print "Mean crankshaft power (per piston): $MotorPowerAverage\n";

$MotorHeatAverage = $MotorPowerAverage / 0.95 / $MotorPowerEfficiency;
print "Mean motor heat power (per piston): $MotorHeatAverage\n";
printf "Mean motor heat power: %.0f\n", $MotorHeatAverage*$NumberOfCylinders;


I have to say, that result of ruuning ths program is not true as i wanted :) It should be reworked, but in 2003 i went to another kind of job...

[Peter_1]

McGrue, is this C, Assembler or Basic?
Are you also a refrigeration engineer?
FEW;)
And that nobody dares to delete nor modify it :p

[yhruhteb]

Hi,

I think the size of liquid reciever can charge all of refrigerant in the system.

Regards,
GuapoDEPENDS ON THE CAPACITY OF THE UNIT

[mcgru-]

simulating of liquid cooling using a pipe.
Cooling liquid - water, in a cylinder tank, mixing, thermoinsulated walls.
Coolant liquid - brine (russian equivalent is "rassol"), flow in the dipped (immersed into the cooling liquid) tube.



void main(void) {
T=3600, dT=10, dt=0.2;
dt_trv=7;

char fname[1024];
strcpy(fname,"h.cfg");

tBAK bak;
tTUBE tube;
tRASSOL rassol(tube.M, tube.din, 1.0);

readconfig(fname, &bak, &tube, &rassol);

double Tambient = +25;

double Qmean=0, Qpeak=0, qin, qout, q;

double dL;
if (rassol.dL==0.0) {
dL=rassol.w*dt;
}else{
dt=rassol.dL/rassol.w;
dL=rassol.dL;
}
if (dL<tube.din) {
dL=tube.din;
dt=dL/rassol.w;
}
tube.dL=dL; rassol.dL=dL;


FILE* fp;
if ((fp=fopen("hh.res","w"))==NULL) {
fprintf(stderr,"Cannot open file gor writing result.\n");
exit(1);
}
int qqq=0;

double last_temp=rassol.t;
double t;
for (t=0; t<=T; t+=dt) {
if (dt_trv==0) last_temp=rassol.t;
tube.AddPacket(last_temp-dt_trv);
if (tube.last!=(tRASSOL*)NULL)
last_temp=tube.last->t;
if ( qqq != int (t/dT) ) {
if (t>0)
printf("T=%g; t_bak=%g; t_in=%g; t_out=%g; Qmean=%g\n", t, bak.t, tube.rassol->t, last_temp, Qmean/t);
fprintf(fp,"T=%g; t_bak=%g; t_in=%g; t_out=%g; Qmean=%g\n", t, bak.t, tube.rassol->t, last_temp, Qmean/t);
}
qqq=int (t/dT);
qout=tube.HeatPackets(bak.t);
qin=bak.Qin(Tambient);
q=qin-qout;
Qmean+=q*dt;
if (Qpeak>q) Qpeak=q; // otsos tepla
bak.Heating(q);
}
printf("T=%g; t_bak=%g; t_in=%g; t_out=%g; Qmean=%g\n", t, bak.t, tube.rassol->t, last_temp, Qmean/t);
fprintf(fp,"T=%g; t_bak=%g; t_in=%g; t_out=%g; Qmean=%g\n", t, bak.t, tube.rassol->t, last_temp, Qmean/t);
fclose(fp);
printf("End.\n");
}

[mcgru-]

McGrue, is this C, Assembler or Basic?
Are you also a refrigeration engineer?
it is Perl, but i'm free also with C/C++, Pascal.
Basic and fortran were tooo long ago :)

Yes, my hobby now is refrigeration, after it was impossible to gather money in science here in russia in 95-02 years (i dealt with nuclear physics, high power ion beams, its influence onto the matter).

[mcgru-]

Well, concerning the liquid receiver size (volume) I would like to say, that...
The best size of receiver for a refrigeration machine for all ambient conditions is HUGE (infinite). But the cost of RM in this case will also be infinite :)
On the other side, if we'll forget about differences in ambient conditions (say, we'll have all-time +25C near condenser), we can even delete receiver from the set of components at all - the receiver's function will play the flooded (down to 1% of flooding) condenser. The stable mass flow of refrigerant makes it unnecessary to keep the liquid refrigerant at all.
And on the third side, the refrigeration machine is composed and serviced by humans. And we should care about comfort in servicing the RM. If we would like to depressurize the contour sometime inn the future, we also would like to save money on the cost of refrigerant that will fly off in air. The density of liquid refrigerant is 50..100 times greater than of gas state. So, if will take care about only liquid refrigerant, we will save 98..99% of refrigerant.
Close outlet vent of receiver, run compressor, some time later almost all refrigerant in liquid state will gather in receiver (if it is lower than condenser, of cause). Close then inlet vent of receiver and... you may sell that receiver with a bunch of expensive refrigerant ;) (joke)
Then other contour may be depressurized, letting only 1-2% of refrigerant fly away.

What size should receiver be?
I think, as it should gather all *****:
V_rec = 1/2*V_cond + V_tubes_cond_rec + V_tubes_rec_evaporator + 1/5*V_evaporator.
In some cases when we use cheap ***** (R22) and when we know that construction of RM is very-very strong (will be no leaks), we may set smaller liquid receiver - almost all ***** will be in partially flooded condenser...

well, that's my PoV

[mcgru-]

DEPENDS ON THE CAPACITY OF THE UNIT
As usual, it is not neccessary to count seconds when the receiver will be emptied due to high temporary loads.
As usual, the differences in mass flows along the refrigerating contour are no so large to take care about it. So, larger load - larger suction - larger discharge into condenser - more condensate - larger income into receiver.
For example.
Duration of refrigerant's voyage is about 30-60 sec for usual systems - 15m discharge liquid line, 0.5m/s speed. Load 300 g/s (45kW of R22) gives the value for liquid volume in receiver of 18kg. Density is about 1.2kg/dm3, so the volume of receiver may be near 18/1.2*1.25 = 19 liters.
Having 45kW basic evaporator load at Tev=+0C, Tcond=+45C, R22, we may be interested in condensing capacity (starting conditions) of 75kW. the Guentner's GVH-067B/2-L(W) may satisfy us. It's inner tube's volume is 33 liters. Half of them - 16 liters. 10m of 22mm tube from condenser to receiver and farther 15m to evaporator has the volume of 6-7 liters. Sum of them, multiplicated on coef 1.1-1.25, is 24-29 liters. It is larger than 19 liters, so the dependance on the capacity will be not so sensitive...

[mcgru-]

That's the explanation I was looking for and it makes sense: the difference in heat transfer coefficient between boiling and liquid state.
shorter saying, there is almost no difference between turbulent liquid flow and boiling liquid flow (99% of liquid) in evaporator. Because of other thermal resistances of tube wall and, mostly, the air film (as US-Iceman said). I just discussed the heat transfer coefficient.

[Peter_1]

Or page 3 of http://www.bitzer.de/_doc/d/dp-300-5-rus.pdf

[Peter_1]

McGru, you're a magician with numbers.
I have to read eacht post of you at least 3 times.:o

[Peter_1]

McGru, still no answer where we can see your program or a demo of it.:confused:

[bangoman]

how much r134a in kgs will go into a reciever of capacity 0.02 cubic metres and how is it calculated.

[frank]

how much r134a in kgs will go into a reciever of capacity 0.02 cubic metres and how is it calculated.
Are we talking liquid or vapour?

The density of R134a vapour is 14.423kg/m3 @ 0C (32F) so you could fit 0.28846kg into a volume of 0.02m3 (14.423*0.02)

For liquid @ 0C the density is 1.2949kg/l so you could fit 25.898kg into 20litres (0.02m3).

Obviously, at different temperatures the results will be different as the density alters.

You need to look at the Saturation Tables for R134a and do the maths for your maximum system temperature.

If you are looking to put this into practice though, you must allow at least 20% volume capacity for liquid expansion, that is, reduce your final calculated liquid figure by 20% to allow for expansion.

[mcgru-]

McGru, you're a magician with numbers.

I just did estimations :) Do not judge me strongly if you will find any errors

Concerning the demo software.
my "insideCompressor.pl" module is not working now - since 2003 i changed the module on refrigerant properties, and i should fit new funcionality to the main module.
my hh-project (tubes with brine inside tank of liquid be cooled) seems to be working, i'll revise it (revised - try in attachment).

i made them only for myself yet, not for sale. so... "as is" :)

[DeB]

Normal rule of thumb is 120% of full system charge for pump down systems. If you do not wish to use a large receiver,depending on system design, you may opt for subcooling with a liquid receiver bypass which allows for a reduced operating refrigerant charge

[patrickj]

In normal practice you size the liquid receiver to have 50% excess capacity so that all refrigerant in the system can be pumped down to the receiver.

[S.K.VARDE]

It should be 3/4th full, if the entire charge is transfered to receiver. L/D ratio should be from 8 to 14 for safe design.

you did not mention the refrigerant, system application and the designed capacity.

Before asking, you should specify maximum details as you can.

[The MG Pony]

the necro threaders run!

[welder315]

What i need to consider to size liquid receiver.

Renato
renato (you must first consider the size of your system
talk to your refrigeration dealer and tell him what the system is and how many btu's it is and he will look up your system and let you know what size to use
or if it's a carrier system look up carrier and it will tell you
mike

[Andy]

renato (you must first consider the size of your system
talk to your refrigeration dealer and tell him what the system is and how many btu's it is and he will look up your system and let you know what size to use
or if it's a carrier system look up carrier and it will tell you
mike
And you would believe what a dealer has to say. Me I would work it out for myself:)

That is what we are talking about here:)

Kind Regards Andy:)

[Renato RR]

The start up was sucesfully.The machine work satisfactory.Each circle have 220 kg. of R134a.We placed liquid receiver of 90 lit.The manufacturer of chiller advice 50 lit.This is all for reference and I hope that will help to some of memebers.


Best regards,
Renato

[The MG Pony]

Good to hear mate :D COngrats :)

[TopMechanic]

Yeah, you gotta figger that 80% of the receiver can hold 100% of the refrigerant in the system.

[spr2003]

In past times we were used to calculate a receiver to recover all the refrigerant hosted in the plant. Now, the trend is to fill the plants with as less gas as possible. So the receiver must be calculated only to balance the amount of gas depending by the opening and closing of the solenoid valves on the users, and the floating of the expansion valves. On supermarkets, direct expansion, in case of condensers next to the rack, we are around 1 kw ref power with 1 kilogram gas ratio. In this case the receiver capability is 1/4 of the whole gas in the plant (around 25 liters).
Don't forget the size of the safety pressure valves is still calculated on the whole gas amount, and not on the receivers capability. In some screw chiller plants or NH3 chiller plants we took off the receiver at all. Yust a small bottle before the expansion valve to prevent flashing.

[RefrigerationVN]

Hi Friend,
Can you help me to calculate exactly about "liquid receiver - High & Low pressure" in Ammonia system?

Thanks

[patrickj]

My Dear Frend,
Normal practice in sizing the Liquid receiver is to have sufficient capacity for a complete pump down of the system plus 50% EXTRA VOLUME.For that you calculate the total volume of all the coolers (if it is flooded) plus liquid holding capacity of seperators ,. liquid lines and the volume of vapour lines converted to liquid volume

patickj

[US Iceman]

The 50% of additional volume over the calculated amount seems to be based on requirements for future growth or very loose refrigerant volume estimating.

I can't say I have ever seen this recommendation before, however, it would certainly do no harm in making the high-pressure receiver larger than it needs to be. It would just cost a little more money.

In very large systems it is almost unheard of to have 100% pump down capability due to the size of the receiver required. In these systems, it may be more cost effective to size the receiver for pump down of partial areas of the facility only too.

[Mauro]

I find that a deepened study of instalaçao must more be made, I am necessary to know the thermal load of the installation, to know the amount of gas to be circulated per minute.

[brian_chapin]

I'm not sure if anyone ever answered the original question for a large system - in a large industrial ammonia system (say 10,000# or better) we like to size our high pressure reciever to hold AT A MINIMUM the amount of liquid in the longest circuit +40%.

So, if you had a circuit with 10 evaps that held a total of 4000# of refrigerant, you would size the receiver for a MINIMUM of 5600#.

The best practice, of course, is to size the high pressure receiver to hold the entire system charge + 20%. The plant I am in right this minute has a 16" long, 5" radius high pressure receiver. That's approximately 314 cubic feet of refrigerant storage for a storage capacity of about 18,000# of liquid ammonia at 95f condensing temp with a 20% head space for vapor.

Obviously, intercooler, low pressure and other receivers would be sized by load.

[US Iceman]

Hi Brian,

Welcome to the RE site. Please feel free to contribute and ask questions.



The best practice, of course, is to size the high pressure Receiver (http://www.refrigeration-engineer.com/forums/glossary.php?do=viewglossary&term=88) to hold the entire system charge + 20%.


Why? How often would you find it necessary to pump down the entire system? A better practice is to design the system to operate with a critical charge.

In a critical charge system the liquid is held at low pressure instead of high pressure. Low pressure liquid is much safer than a high pressure system.

The standard design practices used today have not changed much in the last 150 years. I think we need to start investigating some alternatives specifically due to PSM, RMP, and total system charges.

[brian_chapin]

Why? How often would you find it necessary to pump down the entire system? A better practice is to design the system to operate with a critical charge.

I suppose the best answer (for me) is that RETA suggests it. It's actually a wonderful ability to have - the ability to pump the entire system charge into a single vessel.

As for critical charge - well, it's design is 12,000# - running 15 liquid recirc coils (including a blast tunnel) and about 20 DX devices. I imagine we could run as low as 10k but at full load in the summer we have about 9-10k out in the system and very little in the HPR.

As for frequency - since last July we've only had to pump the entire system down once - and we could have gotten away with less - we just wanted to check our full PSM SOP for it.



I think we need to start investigating some alternatives specifically due to PSM, RMP, and total system charges.


I hear you there - I'm a Process Safety Manager myself. Honestly, the system (when properly applied) really does help everyone do their jobs better and more efficiently.

Thanks so much for your warm welcome - I think I might just stick around.

[The MG Pony]

Welcome! and do most certainly stick around :) Loads of fun and tons of neat and handy people!

[US Iceman]

It's actually a wonderful ability to have - the ability to pump the entire system charge into a single vessel.


That's very true. My issue with this is it adds to the system complexity and costs. A better reason is to prevent the storage of high pressure liquid. RETA may say something like this but it's based on "that's the way we have always done this".

My reason for reducing the charge is NOT to eliminate the need for the PSM & RMP programs, but rather to make the systems more flexible and safer. Even if the system charge is under 10,000 lbs, we should STILL be doing something like the safety and environmental programs.

I think the day will soon be here when we are forced to deal with a reduced version PSM, or as I refer to it "PSM Lite".

Most of the system charge is determined by the type of equipment that is installed. For example, large surge drums on flooded coils. My question is WHY? There are other ways to achieve this.

DX of course has a much smaller charge requirement, but it also places a limit on how low you can operate the discharge pressure, so this also costs additional money.

We have a long way to go before high-performance ammonia systems will be installed.:( It's sort of like the old tag line from the Six Million Dollar Man TV series: We have the technology, we can rebuild it! Unfortunately, not very many people pursue this as it forces them to think outside of the box.

Please do stick around. We have some good ammonia people here from all over the world, but I think you are the first Safety guy, so you could add some interesting insight into issues.

[danielenrd]

What i need to consider to size liquid receiver.

Renato
In a stable circuit you can choose the liquid receiver (1/4-1/3 [l] / total charge of refrigerant [l]).

Daniele

[ramih]

Question:
Must the liquid receiver always be located lower then the condenser??