130kW cooling capacity is one of our smallest chillers, Peter.
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130kW cooling capacity is one of our smallest chillers, Peter.
Can yo build those in a factory of only 6000 sqm?:D
My place is already 4000 sqm.
That is right.Quote:
Originally Posted by Peter_1
That is true if you have designed the evaporators for sensible heat transfer, like a chilled water coil.Quote:
There's no necessity that we reach the boiling point.
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.
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.Quote:
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.
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?;)
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.
OK. :DQuote:
It's not that I don't believe you but I want to read something more about all this.
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.
This is perhaps one USIceman
http://www.wlv.com/products/databook...ta/db3ch15.pdf
Perhaps the Mod of this section can cut and paste them in that new section
surge receiver or through recriver
surge receiver or through receiver
Commisioned last Friday a Blue Box chiller with...of course...a Carel unit in it.
Very nice machine.
Peace of cake of course :D
Quote:
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:
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:
I would be interested to discuss some of the simulations you did for the:Quote:
alpha_liquid_tube ~ 2000-3000 W/(m2.K)
alpha_boiling liquid_tube ~ 8000 W/(m2.K)
Sounds like something I would enjoy. :cool:Quote:
a) processes inside the reciprocating compressor,
b) dynamic heat transfer between tube with refrigerant flow and environment
Welcome to the RE forum. :D
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.
Peter,
I'm not sure who might be able to do this except the original poster or WebRam.
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:
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.
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.
Peter, i think that my language is terrible, so my post is self-balanced :)Quote:
Originally Posted by Peter_1
It's ok in quoted state.
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.
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...Quote:
Originally Posted by Peter_1
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 [s]333[/s] (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)
:)
if it will be any difficulties with translating russian symbols, i'll help you - just note it
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.
some code from compressor simulating program:Quote:
Originally Posted by US Iceman
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...Code:# ---------------------------------------------------------------------------
#
# 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;
McGrue, is this C, Assembler or Basic?
Are you also a refrigeration engineer?
FEW;)
And that nobody dares to delete nor modify it :p
DEPENDS ON THE CAPACITY OF THE UNITQuote:
Originally Posted by guapo
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.
Code: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");
}
it is Perl, but i'm free also with C/C++, Pascal.Quote:
Originally Posted by Peter_1
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).
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
As usual, it is not neccessary to count seconds when the receiver will be emptied due to high temporary loads.Quote:
Originally Posted by yhruhteb
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...
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.Quote:
Originally Posted by mcgru-
Or page 3 of http://www.bitzer.de/_doc/d/dp-300-5-rus.pdf
McGru, you're a magician with numbers.
I have to read eacht post of you at least 3 times.:o
McGru, still no answer where we can see your program or a demo of it.:confused:
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?Quote:
Originally Posted by bangoman
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.
I just did estimations :) Do not judge me strongly if you will find any errorsQuote:
Originally Posted by Peter_1
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" :)
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
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.
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 necro threaders run!
renato (you must first consider the size of your systemQuote:
Originally Posted by Renato RR
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:)Quote:
Originally Posted by welder315
That is what we are talking about here:)
Kind Regards Andy:)