Compressor Displacement

Could someone help me making the leap from kg/s circulated to compressor displacement.

From a PH chart I can establish the kg/s required to be circulated but I need some pointers to calculate the m3/s to enable a compressor selection.

Thanks

David

On the PH chart follow the radiused line at the compressor inlet to the right hand side and read off the mass flow on the chart.

Frank

Hi David,
to change from mass to volume flow you need to know the density of the refrigerant vapour at the point which it enters the compressor. You then convert. If you give me an idea about the system, and the compressor I will help you convert.
Remember that the displacement in volume is fixed, to reduce or increase the capcity of the compressor(without changing it's speed) you need to change the system design conditions, i.e.
the condensing or evaporation, subcooling of the liquid or superheating of the suction gasses.
Hope this helps a little at least.
Regards. Andy:)

Thanks for the replies.

Andy, there is no system as such I am merely trying to fill in some of the blanks in my theory.

For arguments sake if we pressume an R22 sytem with a cooling capacity of 10kW, evaporating @ -10'C and condensing @ 40'C.

After going over some old books I came up with a Qe of 159 kj/kg and a specific volume of 0.078 m3/kg. I divided the 10kw by 159 and mulitplied the result by the specific volume.

At the moment if I assume a volumetric efficiency of 1 I am pressuming that I can select a compressor for this imaginary system based on the resulting 17.6m3/hr displacement.

So put me out of my misery, are these figures viable or do I start again.

Thanks

David

David,

Theoretically , you're absolutely correct!:)

So imagine I wanted to subcool the liquid using a condensing unit & heat exchanger, how would I convert the enthalpy difference to a condensing unit kW rating.

I would guess that I would use the enthalpy difference and multiply by the mass flow?

Thanks

David

David,

I'm assuming that you are asking how to select a condensing unit after you have selected a compressor.

Use the compressor selection manual and look for the operating or application envelope (Tevap vs Tcond). Note the highest evaporating temp. Read the compressor capacity (in kW) and add to the Power consumption (also on kW) - all at that T evap and your selected T cond ( to be extremely oversized, you could go for max. T cond). The sum of both kW's (or Btu/hr or kCal/hr) should give you a sizable condensor which will give a certain degree of sub-cooling, will not develop high condensing pressures during initial pulldown and pulldown after a defrost termination. If your ambient day/night split is high, you may need some form of fan control at night. But yes, it will cost more initially.

Did this in anyway help ?

What I was trying to do in theory was subcool the liquid of an existing installation ( a freezer room for example) using an additional condensing unit & heat exchanger.

I want to be able to transfer the enthalpy difference between the start liquid temperature (35'C for example) and the final liquid temperature (0'C) when using an additional cu & hx and come up with a kW rating for the subcool work.

Essentially what I want to be able to do is calculate a kW rating for subcooling liquid or de-superheating in the discharge side, for condensing unit or expansion valve selection.

Thanks

David

Consider the following:

Refrigerant R-22
evap temp: -10°C
cond temp: 40°C

Ahem, it appears the Prof must resort to SI units... :)

Using RefProp v7, and assuming no subcooling:

NRE = hg_-10°C - hf_40°C = 401.20 - 249.65 = 151.55 kJ/kg

If we assume the cu is capable of 10kW at these conditions, we then know our refrigerant flow rate must be: 10 / 151.55 = 0.066 kg/sec

Assuming 10°C superheated vapor entering the compressor, our refrigerant density becomes: 14.58 kg/m3

And volumetric flow thru the compressor becomes: 0.066 / 14.58 * 3600 = 16.3 m3/hr

Which differs a bit from David's 17.6 m3/hr.

If we were to increase subcooling 40°C, giving us 0°C liquid entering the evaporator:

NRE = hg_-10°C - hf_0°C = 401.20 - 200.00 = 201.20 kJ/kg

Assuming everything else remains the same, the compresor will still only pump 0.066 kg/sec refrigerant flow. But the cooling capacity increases to: 201.20 * 0.066 = 13.28 kW, and increase of 32.8 percent. :)

i'm sorry for bringing old topic to the top, but there are still people with blanks in their knowledge base ;)

Quote:

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Originally posted by Prof Sporlan
And volumetric flow thru the compressor becomes: 0.066 / 14.58 * 3600 = 16.3 m3/hr

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So we received a theoretical compressor's volume capacity of 16.3m3/hr.

At first approach, we can use it in getting the compressor's model. But if we'll choose the comperssor with displacement of 16.3m3/hr, and recalculate it's refrigeration capacity at given conditions, we will get lower capacity than we used to calculate volumetric flow.

As far as i know, every compressor has its volumetric efficiency, which has the nature of displacement, but depends also on discharge and suction pressure ratio, mainly, and some other conditions (temperatures, gas velocities, etc) in less manner.

I would like to get help on:

If i know the refrigerant volumetric flow, it's temperatures, it's velocities (at least in suction line), How can i calculate the REAL compressor's volume efficiency? (with accuracy of less than 3-5%)

Thank you very much for quick reply

Quote:

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Originally posted by Marc O'Brien
Vol Eff = 1 - c * (r^1/n - 1)

Where:
c = Clearance. (6% clearance pocket is 0.06).
r = Compression ratio.
n = C_p/C_v

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So, clearance differs from one model/manufacturer to another, isnt it?

The value "n" depends on pressure and temperature of the gas, should i take the C_p and C_v values for a gas in suction line?

The weight of refrigerant circulated per minute per ton is = to 12,000 btu hr/ 60 or 200 btu/min. so the weight of refrigerant circulated is 200/ Hv – HL lbs (ton)(min)
Hv = enthalpy of vapor
HL = enthalpy of liquid
The theoretical piston displacement per ton per minute: One needs to know the specific volume of the refrigerant vapor as it enters the compressor to determine the above
So if we let this factor be represented by Vc we multiply Vc by the weight of the refrigerant to be circulated per ton of refrigeration per minute. So the theoretical piston displacement = 200 Vc/ Hv – HL = cu ft/(ton)(minute) As Marc said, on reciprocating compressors not all the vapor can be exhausted from the cylinder by any one stroke against the high side pressure and the vapor inertia prevents complete filling of the cylinder on each suction stroke. Volumetric efficiencies ordinarily do not exceed 80%.
Now then the ratio of the actual weight delivered to that theoretically pumped is the volumetric efficiency Cv the difference between the too is occasioned by
A. The necessary clearance between the piston head and the cylinder head
B. Piston ring and valve leakage
C. Vapor inertia and friction of vapor at the valve entry ports and finally
D. Superheating of the refrigerant during the suction stroke from contact with the hot cylinder walls and piston. By the way this latter effect is the most serious of the factors affecting volumetric efficiency.
:confused: :D
Semper Fi

Quote:

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Originally posted by shogun7
A. The necessary clearance between the piston head and the cylinder head
B. Piston ring and valve leakage
C. Vapor inertia and friction of vapor at the valve entry ports and finally
D. Superheating of the refrigerant during the suction stroke from contact with the hot cylinder walls and piston. By the way this latter effect is the most serious of the factors affecting volumetric efficiency.

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Thus, the point "A" is evident (and easy to understand) and taken into account in the formulae given by Marc as "c" coefficient and variable "r".

The properties of gas are accounted (in the formulae given by Marc) by the ratio Cp/Cv, that very slightly (or partially) described in points "C" (concerning sound speed) and "D" (might be differences between Cp/Cv in suction and discharge processes).



I made graphs for Bitzer 2CC-3.2 compressor (all data from the original Bitzer's software) for various T_ev and T_cond at constant T_oh=10K: http://mcgru.remholod.com/rx/2cc-3-2.jpg
(values are slightly 1-2% decreased due to the attempt to take into account some dependances)

I tried to use Marc's formulae on these data (for T_ev=-5.C), but it does not fit at any "n" (1.15-1.25)
My investigations will be continued in first decade of march.

John McGru
I am not well versed in the metric system so bear with me please I will do my best to answer you but I hope I don't confuse anyone with the american system Maybe one of the other members can switch my calculations to your system

Good day

Quote:

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Originally posted by shogun7
I am not well versed in the metric system so bear with me please I will do my best to answer you but I hope I don't confuse anyone with the american system Maybe one of the other members can switch my calculations to your system

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<s>Sorry for an additional question - did you really mean "ton" instead of "t"?
1t=1000 kg
1ton(GB)=1016.05 kg
which one to use?</s>
(this difference does exist, but makes a little error, 1.6%)

So, afaik:
1 BTU = 1055 J
1 BTU/min = 17.58 W
200 BTU/min = 3517 W

Quote:

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I tried to use Marc's formulae on these data (for Tev=-5.C), but it does not fit at any "n" (1.15-1.25)

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Marc's equation is based on ideal gas laws and adiabatic compression, if the Prof isn't mistaken. To obtain a more accurate equation, more real world elements will need to be added to the equation. It may be necessary to resort to an equation of state derived for the refrigerant in question.

Prof:
That is absolutely corect! sorry corect needs another "r"
Roger

i'm going to invite a bicycle :)
in the nearest time i'll try to simulate the processes B,C,D mentioned by shogun7

Quote:

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Originally posted by Marc O'Brien
Polynomials wrapped to impirical data would be the best way to go

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I also think so, but the problem (imo) is in that compressor's manufacturers provide different kinds of polynomials, there are not any agrements in which (standard) kind to use - am i wrong?
It could be very comfortable in usage, if one could set only coefficients provided by compressor manufacturer for given standard polynomials.

As Marc has noted in past discussions on the subject of modeling a process, wrapping measured data in a polynomial is quite useful as a predictor, but it doesn't necessarily give you an understanding of the process itself. A better approach may be to start with an equation descibing the idealized version of the process, and modify it accordingly to predict real world measurements.

Quote:

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Originally posted by Prof Sporlan
to start with an equation descibing the idealized version of the process, and modify it accordingly to predict real world measurements.

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that's the way i'm going to go :)

I made a first and simple approach in compressor's simulation.
Using PH diagram (cannot find any useful and understandable ;) formulae, thus i've taken data from Atofina's FORANE program) I have get enthalpy, enthropy, density (and etc.) dependances from pressure and temperature of refrigerant.
Simulation goes in analog mode, with some simplifications.
Input data are: refrigerant, pressure and temperature infront the suction valve (density is recalculated), piston's diameter, stroke and frequency, clearance, mean temperature of cylinder walls, pressure in discharge line, some data on valve's diameters (to calculate pressure drops in valves), a gap between piston and cylinder wall to simulate the gas leakage.
Pressure drops are calculated from dp=dzeta*density*(velocity^2)/2 formula.
Heat transfer between walls and gas is simplificated by constant cylinder wall's temperature (70-90 celsium degrees), and it is concerned all volume of gas in cylinder undergoes the heating/cooling, and coefficient of heat transfer between wall and gas is almost constant. Then Q=k*S*dT at each moment.
I divided the phase of crankshaft rotation into 50-100 portions, thus increase the angle at each step on 1-2 degrees.
At each step i calculate:
1. heat transfer between wall and gas
2. leakage of gas through the gap
3. expansion/shrinking due piston's movement
4. suction/discharge at proper conditions (with corresponded pressure drops), mass change

Cannot still get error estimation of calculations, so at this moment i can get only values without inaccuracies.

First conclusion i made (remembering about simplifications) - is that with approach of constant wall temperature, the influence of heat transfer between wall and gas is not noticeable - due to gas heating at suction and gas cooling at discharge, the enthropy remains almost unchanged.

(to be continued)

Quote:

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Originally Posted by Marc O'Brien

c = Clearance. (6% clearance pocket is 0.06).

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I watched the DWM Copeland compressor (with broken gas distributor), and it seems that value of clearance is much smaller than 0.06

Is there any other data on clearances?

Hi!

BTW, there is a free software you can download from the internet called CoolPack (from DTU), where you can make your plots of different cycles with different data and get required equipment information (even simulating system's behavior).

Best regards,

-Manuel.