How much power do the comp add to the system?

[Henry.]

How much of the power (in % or anything) will a compressor generally add at full load to the system and hence to the condenser? I need to know so i will be able to calculate the right size condenser i need. Any rule of thumb for this perhaps?

One said i need to add all the power the comp use but that doesnt seem right for me since as i see quite a good portion should be cooled of with the fans. Another said half the power it use and that might seem more ok but it still feels a little high and a third said 1/10th but that seem a little to low so im lost. I would like to really know so i dont buy a unnecessary big condenser.

I have built many slightly smaller systems before where this havent been an issue i have just used a slightly bigger one based on estimations and this has worked perfectly under all the years. But now its time for a 10 times bigger system and of course also a bigger comp so this time i need to know.

Regards Henry

[Peter_1]

Semi hermetic? Then all the power the compressor uses must be added.
Added to what? Corrected for what?
All those who says something different don't know what they're talking about.

Apparently, they never understood the basics.

I can explain it also pure thermodynamic with theoretical or isentropic compression, corrected for heat loss of the compressor, corrected for additional superheat, drawing all this in a log p/h and prove that it is all the power. But it will make you not wiser.
Roy Dossat's book is a good start, a very good start.

[Henry.]

No its a fully hermetic system, and im willing to learn.

Since the inlet gas to the compressor are always colder then the engine inside, well at least it should be, so will this cool the comp and draw and accumulate heat from it because of this.

At compression this extra outdrawn power from the comp had to be added to the overall power that the system will be designed for and then will be transferred away from the condenser.

You then have to have this in account when choosing the right condenser as i see it?

So the question in my case then is how much will the heated comp add to overall system when the compressor engine, according to datasheet, draw ~500W att -15C evap. It is cooled by 3m/s.

A rule of thumb of this would be most welcome since i guess this doesnt have to be that exact in my case since the system will handle only about 1KW load and the planned to use condenser is quite big as it is. But better safe the sorry this is the reason i ask but havent really planned to buy a book only for this. But i guess the guys here would need one..

--


And no it feels like they dont really know what they are talking about these days. Older guys that are very close to retirement and it almost seem like that they doesnt really care about calculating the right numbers anymore but use things that have worked fine the last 30 years, but this is not how i work.

[Brian_UK]

Semi or full hermetic means that the electric motor is in the refrigerant gas stream and as such will add the power used to drive the motor to the operating cooling load.

The fact that the gas cools the motor etc does not change the power input to the motor as far as calculations go.

Simple example, (teaches man to suck eggs and runs for cover) - 2kw cooling load plus 1kw motor equals 3kw condensing load.

[taz24]

How much of the power (in % or anything) will a compressor generally add at full load to the system and hence to the condenser? I need to know so i will be able to calculate the right size condenser i need. Any rule of thumb for this perhaps?

One said i need to add all the power the comp use but that doesnt seem right for me since as i see quite a good portion should be cooled of with the fans. Another said half the power it use and that might seem more ok but it still feels a little high and a third said 1/10th but that seem a little to low so im lost. I would like to really know so i dont buy a unnecessary big condenser.

I have built many slightly smaller systems before where this havent been an issue i have just used a slightly bigger one based on estimations and this has worked perfectly under all the years. But now its time for a 10 times bigger system and of course also a bigger comp so this time i need to know.

Regards Henry

There is a way of working out how much energy the compressor add to the refrigerant.

The easiest way is by using cool pack (Refrigeration utilities) do a google search. Cool pack is a free software that calculates refrigeration cycles.

The compressor effiencey will vary from make to make and type to type and to work it out you
need the data from the makers plate on the comp itself.

One way of working out refrigeration systems effiencies is the Co effiency Of Performance (COP).
this is a recognised way of working out the systems effiency ratio.

A COP of 3 means the comp basicly adds 1/3 of the energy to the refrigerant (that would be 33% ??? you do the maths ).
A COP of 6 means the comp adds a 1/6 of the energy in the refrigerant ( again maths about 15% - 17% ).

Use the Ph charts from cool pack and all will be explaned.

taz

.

[D.D.KORANNE]

The heat energy from the motor to be added will vary depending upon the application .
1) example ,low temp application say -25 deg c evap temp / 40 deg c condensing / r 404a . In this case you can take a 10 deg c difference to calculate the heat input to the condenser . In other words, for the application , take -15 deg c evaporating , other parameters as per yr design and find out power input . Compressors manufacturers softwares has the data called total heat rejection i.e. Size of condenser requd . This way you can select an optimized condenser size .

In case of open drive compressor , this method can also help you select optimum motor size too .
In this case the suction pressure regulating valve or mop will be needed which will not allow overloading of condenser as well as motor.

Hope that helps !
2) example , 0 deg c evapo temp / 40 deg c condensing , r-404a.
The difference in evaporating temp can be 10 deg c & max sst permitted by comp manufacturer whichever is lower .

[taz24]

The heat energy from the motor to be added will vary depending upon the application .
1) example ,low temp application say -25 deg c evap temp / 40 deg c condensing / r 404a . In this case you can take a 10 deg c difference to calculate the heat input to the condenser . In other words, for the application , take -15 deg c evaporating , other parameters as per yr design and find out power input . Compressors manufacturers softwares has the data called total heat rejection i.e. Size of condenser requd . This way you can select an optimized condenser size .

In case of open drive compressor , this method can also help you select optimum motor size too .
In this case the suction pressure regulating valve or mop will be needed which will not allow overloading of condenser as well as motor.

Hope that helps !
2) example , 0 deg c evapo temp / 40 deg c condensing , r-404a.
The difference in evaporating temp can be 10 deg c & max sst permitted by comp manufacturer whichever is lower .

Hello.

In principle I do not disagree with what you say, but you are over complicating things.

His original question was how much in % does the comp add. There is no point in talking
about temp differances over the evap or condenser and 10 deg C differance is too simplistic
to be used as a rule of thumb.

Where I come from we have a saying - KISS (keep it simple stupid) don't make things complicated just for the sake of it.

All the best

taz

.

[D.D.KORANNE]

Motor power addition is different for different conditions of system design .

I have shown above how best to optimize the power that adds up .

[Bachuss]

If by power you mean heat energy then a rule of thumb is arround 80% of the heat energy present in the refrigerant in your compressors discharge line is due to the compression process, the other 20% is due to the the mechanical process of the compressor. Or so I'v been led to believe.

[Bachuss]

There is a way of working out how much energy the compressor add to the refrigerant.

The easiest way is by using cool pack (Refrigeration utilities) do a google search. Cool pack is a free software that calculates refrigeration cycles.

The compressor effiencey will vary from make to make and type to type and to work it out you
need the data from the makers plate on the comp itself.

One way of working out refrigeration systems effiencies is the Co effiency Of Performance (COP).
this is a recognised way of working out the systems effiency ratio.

A COP of 3 means the comp basicly adds 1/3 of the energy to the refrigerant (that would be 33% ??? you do the maths ).
A COP of 6 means the comp adds a 1/6 of the energy in the refrigerant ( again maths about 15% - 17% ).

Use the Ph charts from cool pack and all will be explaned.

taz

.

COP is the ratio of the heat removed in KW to energy used in Kw, I am unsure how this relates to energy added to refrigerant could you clarify?

[taz24]

Motor power addition is different for different conditions of system design .

I have shown above how best to optimize the power that adds up .

D.D I agree with you and you make very valid points.

All I ried to say was the guy wants basic information and
at the moment he has not come back asking for extra help
to move him forward, so don't over complicate his first question.

All the best

taz

.

[taz24]

COP is the ratio of the heat removed in KW to energy used in Kw, I am unsure how this relates to energy added to refrigerant could you clarify?

Hi Bachuss..

Don't confuse COP with EER.

COP is a pure equation of energy imparted into the refrigerant through
the compresion process.

COP is the energy measured in KJ/Kg (Enthalpy) and can
only be truly worked out matamaticaly on Ph charts and such like.

You are relating to Energy Efficiency Ratio (EER) which is
heating or cooling out put divided by total system input power measured in KW.

Similar result but both completely differant to each other.

Cheers

taz

.

[taz24]

COP is the ratio of the heat removed in KW to energy used in Kw, I am unsure how this relates to energy added to refrigerant could you clarify?

Another way of looking at it is this..

COP is a calculation that needs working out and
quite a few factors can effect COP, but it only deals with heat energy imparted
into the refrigerant (Enthalpy)

EER takes into account all the running costs of the system in KW so if you use 3 KW
in power and you get 12 KW of heating you divide 12 by 3 and the EER would be 4...

Or 3 KW in power and 9 KW of cooling = 9 divided by 3 = 3, EER would be 3.

EER is used to find out how efficient the system is in real life not just on paper.

Hope that helps

taz

.

[Henry.]

If by power you mean heat energy then a rule of thumb is arround 80% of the heat energy present in the refrigerant in your compressors discharge line is due to the compression process, the other 20% is due to the mechanical process of the compressor. Or so I'v been led to believe.

That sound correct to me and of course i had totally forgot about these parts, so then i have to look up the comps power and calculate the condenser according to this but i think the condenser i planned to use will work anyway but better to calculate this to be sure because its a quite big compressor compared to what i usually use.


"The fact that the gas cools the motor etc does not change the power input to the motor as far as calculations go."

True, i meant that the motor draw different amount of power at different evap temps at least according to the datasheet on the comp and therefore i had to se what it draws at the exact evap temp i planning to use, nothing else. When it runs later one then the power the motor draws is the same all time.


"at the moment he has not come back asking for extra help to move him forward"

Tagged this thread to give me a mail of any new replys as soon as they were put here but havent got a single mail since the two latest so i have missed all the later replys.

Although i have got my answer; the whole total comps power need to be added to be able to calculate the right condenser size. And there were other interesting information in this thread which was highly interesting to read, so thanks a lot all you guys for the help it was highly appreciated!