can anybody help me?

[becky]

i have a few questions and thought some of you may be able to help me/re-assure me.

i have a few bits of test data and need to work out different COPs, firstly the chart COP using the work input found from the p-h chart, which i think means (h1-h4)/(h2-h1)?
then the system COP using the mechanical power which i think is (h1-h4)/mech power, but i dont have mech power directly i have rpm & torque.
then also the overall COP from using the electrical power which i think is (h1-h4)/elec work but i dont have electrical power directly either but i think that is volts x amp x power factor (which i have)

also why do all the values obtained for the COPs differ?


what is considered to be the optimum condition of a refrigerant at entry to the compressor, what are the actual conditions and the reasons for this?


what happens when the heat transfer to the evapourator is reduced, and what happens if the heat transfer from the condenser is serverly reduced or even cut off?

any help would be great

[LRAC]

Hi Becky

and welcome to the forum, they do say women are quick these days not for me at my age though.

Youv'e gone straight for the kill in your questions and boy have you some good questions, not the sort of stuff i keep on the top of my head. I'm sure other members will have the refrigeration bible at hand too enable the questions to be answered correctly.

Hope your stay is a long one and i hope to get back to you soon with some answers, my books at work.

Kind regards
Lrac

[becky]

thanks
i have a big report to do as part of my thermodynamics module, i have done a lot of it but are not sure about the above asked questions. i've been ill and so didnt start as early as i should have (but as im not on my death bed i can't have an extension) now im getting worried i'm not going to have it done in time, especially as i have 3 other reports to do aswell
thats why i've come straight out with the questions im just rooting around trying to find answers

[Andy]

i have a few questions and thought some of you may be able to help me/re-assure me.

i have a few bits of test data and need to work out different COPs, firstly the chart COP using the work input found from the p-h chart, which i think means (h1-h4)/(h2-h1)?
then the system COP using the mechanical power which i think is (h1-h4)/mech power, but i dont have mech power directly i have rpm & torque.
then also the overall COP from using the electrical power which i think is (h1-h4)/elec work but i dont have electrical power directly either but i think that is volts x amp x power factor (which i have)

also why do all the values obtained for the COPs differ?


what is considered to be the optimum condition of a refrigerant at entry to the compressor, what are the actual conditions and the reasons for this?


what happens when the heat transfer to the evapourator is reduced, and what happens if the heat transfer from the condenser is serverly reduced or even cut off?

any help would be great

Hi Becky

OK here's PH diagrams in the quick

Everthing is per kG flowing through the system., so you need to multiply the energy input to the evaporator (the horizontal line along the bottom) by the right number og kG's to give you system refrigeration capacity.

The longer horizontal line along the top is the heat rejected to the ambient air (usually air can be other media).

The difference between these two lines is the energy input by your drive motor (usually a motor, but could be other drivers)

If you lower the bottom line you lower the evaporation temperture/pressure and increase the energy required to rotate one kG of refrigerant around the system and REDUCE the COP.

If you raise the top line you increase the condensing temperature/pressure. This also requires more energy input, again lowers the COP.

The above explains the differing COP's.

Deep freeze systems have a lower evaporation pressure (lower height bottom line) so a poorer COP than chill systems which have a higher bottom line.

The torque/RPM thing you can look up It's in most engineering science books.

The 3 phase power formula can be found the same

Hope this helps

Kind Regards Andy

[becky]

thanks for that andy

i have this information:

location 1 (leaving the evaporator) pressure 90 kpa gauge, temp 7.5 c

location 2 (entering condenser) 645 kpa gauge, 58.4 c

location 3 (leaving the condenser) 645 kpa gauge, 19.4 c

location 4 (enerting evapourator) 90kpa gauge, -11.6 c

how do i plot that on the ph chart? (its r12 if i havnt already said) i know i have to convert the pressures to absolute

[Andy]

thanks for that andy

i have this information:

location 1 (leaving the evaporator) pressure 90 kpa gauge, temp 7.5 c

location 2 (entering condenser) 645 kpa gauge, 58.4 c

location 3 (leaving the condenser) 645 kpa gauge, 19.4 c

location 4 (enerting evapourator) 90kpa gauge, -11.6 c

how do i plot that on the ph chart? (its r12 if i havnt already said) i know i have to convert the pressures to absolute

Becky have you seen a plot on a PH diagram before. The shape is a rectangle with the right hand side sloping from left to right

Location 4 is bottom left corner.

Location 1 is bottom right corner.

Location 2 is top right corner.

Location 3 is top left corner.


If you go to the software section on RE look for the link to cool pack, down load it and use Ref utilities this is a software version of a PH diagram. I would also recommend digging up a copy of Dossats Principles of Refrigeration, this book is very much into pressure enthalpy diagrams

Hope this helps

Kind Regards Andy

[becky]

yeah i know what they look like but we always plot them based on the 2 temps, then considering superheating and sub cooling. when i try to plot the data i have it just doesnt seem to fit

[becky]

If you go to the software section on RE look for the link to cool pack, down load it and use Ref utilities this is a software version of a PH diagram.

i cant find this

[Andy]

i cant find this

Sorry Becky I cant find it either

please use the following linkhttp://www.et.web.mek.dtu.dk/Coolpack/UK/


Kind Regards Andy

[becky]

i google searched it and downloaded it. thanks anyway

it agrees with me that the values dont work lol

however i may be entering the data into the wrong boxes

[Josip]

Hi, Becky

welcome to RE

i google searched it and downloaded it. thanks anyway

it agrees with me that the values dont work lol

however i may be entering the data into the wrong boxes

Now you have CoolPack

go to Refrigeration utilities - click on Log(p)-h diagram

choose your refrigerant R12

go to Draw - polyline

and use your state points to draw your cycle.

To see COP

go to Cool Tools-Cycle Analysis-One stage cycle DX evaporator and start to play with Cycle specs.

Hope this will help a little

Best regards, Josip

[Andy]

i google searched it and downloaded it. thanks anyway

it agrees with me that the values dont work lol

however i may be entering the data into the wrong boxes

Hi Becky

your superheat and subcooling is shown in the numbers you have

Suction Superheat -11.6 deg c to 7.5 deg c 19.1k

Subcooling can be found by the difference between your saturation pressure/temp for 649kpa (don't forget to correct for absolute not gauge)and 19.4 deg c

Discharge superheat is found again from the difference between the 649kpa saturation and the 58.4 deg c

All you need is your input kW from your torque/RPM and you have all the data I can think of that you would need.

Kind Regards Andy

[becky]

thanks guys, big help

anybody got any ideas about these:

what is considered to be the optimum condition of a refrigerant at entry to the compressor, what are the actual conditions and the reasons for this?


what happens when the heat transfer to the evapourator is reduced, and what happens if the heat transfer from the condenser is serverly reduced or even cut off?

any help would be great

[Andy]

thanks guys, big help

anybody got any ideas about these:

Optimum conditions are saturated vapour (no suction superheat) but in reality the compressor is only able to compressor dry gas, so at least 1.5 k of superheat is required to obtain dry vapor (the liquid in the suction gas would damage the compressor).

Reduced heat transfer would reduce the evaporating pressure, such as when the fan goes off on an air cooling unit or the heat exchanger heat transfer surface is fouled on an water chiller.


Reduced heat exchange on the condenser would cause the condensing pressure to rise

Kind Regards Andy

[becky]

thanks very much

[becky]

Reduced heat transfer would reduce the evaporating pressure, such as when the fan goes off on an air cooling unit or the heat exchanger heat transfer surface is fouled on an water chiller.


Reduced heat exchange on the condenser would cause the condensing pressure to rise

Kind Regards Andy

can you expand a little on these at all (sorry to be a pain) i think the 2nd one is due to the heat not being rejected as well?

[Andy]

can you expand a little on these at all (sorry to be a pain) i think the 2nd one is due to the heat not being rejected as well?

If heat transfer is reduced in either the evaporator or the condenser the temperature difference or TD increases.

Good practice would dictate 6k td in the evaporator and 10k td in the condenser.

TD is the difference between the media cooling the condenser or heating the evaporator.

Example air at 35 deg c ambient would create a condensing temperature of 45 deg c assuming a 10 k td.

Air onto an evaporator at 2 deg c with a 6 k td would be a -4 deg c evaporation temerature.

Hope this helps. Kind Regards Andy

[wkd]

Becky
I checked this out and don't understand the figures your using either.
90 kpa coreected for absolute give a sturation temperature of -13.9 odd and with 7.5 c suction temp gives a suction superheat of 21.4 K way to high.
645 kpa corrected for absolute gives a dew temperature of 30.1 against a discharge temp of 58.4 c.This doesn't coincide with the amount of suction superheat in the system.I am not surprised you can't get coolpack to work correctly i'd suggest re-checking the figures,while I re-check yours.