Liquid receiver

[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

[yhruhteb]

Hi,

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

Regards,
Guapo

DEPENDS 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.

[Peter_1]

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

[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 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 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??