Heat Rejection Factor

[nh3simman]

To do a quick estimate of heat rejection of a refrigeration system, you could guess COP=3.

From that, Qe/(Qc-Qe) = 3
and therefore Qc = Qe (1 + 1/3)

The heat rejection factor = 1.3

But this is incorrect because the COP depends on the cyle temperatures.

The unknown is only the compressor efficiency.
Text book figures for Isentropic efficiency are 0.72 for a recip.

Surely this is not realistic? I would expect the same compressor with a very low suction to have a lower isentropic efficiency.

Has anybody seen a relationship for isentropic efficiency and pressure ratio?

[NoNickName]

Energy is heat. Work is heat. The work expended on refrigerant becomes heat, the energy used against internal friction of the motor is heat.
Therefore the whole energy absorbed by the motor is heat, and isentropic efficiency is not relevant on heat rejection.

[Abby Normal]

I used to take the load from the evaporator and divide by 0.75 as a starting point to come up with a heat of rejection, in order to begin balancing out an evap coil, compressor and condensing coil from scratch in a custom system.

[lana]

Hi there,

In the compressor we assume that every electrical power converts to heat and therefore, :

Qc= Qe +W

Of course this is not completely true because some of the electrical energy converts to mechanical.

I think it is very difficult to calculate the actual compression heat (if it is not impossible:confused: ).

Also I think if we could calculate the actual heat of compression then it wouldn't make any huge difference.

Just remember : 99% of the systems are over sized:D

Cheers:)

[nh3simman]

isentropic efficiency is not relevant on heat rejection.

According to the laws of thermodynamics, it does.

by definition
actual Power = isentropic power / isentropic efficiency

1st law energy conservation
Heat rejection = cooling duty + actual power

At specified system temperatures, the unknown parameter that affects the heat rejection is the
isentropic efficiency.

Remember, the temperatures (other than the discharge) are known.

[nh3simman]

I found this on the Coolpac web site.

Isentropic efficiency
Isentropic efficiency is used to express the energy efficiency of a compressor by comparing its energy consumption with the energy consumption necessary for an ideal (reversible and adiabatic) compression process. Isentropic efficiency can have values between 0 and 1.

Typical values are
0.4 to 0.5 for small hermetic compressors,
0.5 to 0.7 for semihermetic compressors and
0.5 to 0.8 for large open compressors.

I'm sure that these values must vary with suction and discharge temepratures.

[lana]

Hi nh3simman,

We used this equation to calculate the actual power consumption and discharge temperature in our tesis.
BUT I don't remember what we did :D .
I will try to find it and get back to you.
Cheers

[NoNickName]

It is not relevant, as the actual power consumption at any condition is a result of the compressor design, and as such IT IS FIXED per type of compressor.

[lana]

It is not relevant, as the actual power consumption at any condition is a result of the compressor design, and as such IT IS FIXED per type of compressor.

Hi there,

It is not fixed I am afraid. As you said it is dependent on operating condition and compressor design.

Cheers

[US Iceman]

I think what you are looking for is the apparent change in efficiency at various pressure ratios for any given type of compressor.

For any compressor, a general statement can be made; the efficiency decreases as the pressure ratio increases.

If you take the rated power input and divide it by the theoretical calculated power you will have the apparent isentropic efficiency.

In effect, the suction and discharge temperatures control the pressure ratio so it is conceivable to say there is a correlation between the two.

Calculating the real isentropic efficiency is a lot like trying to determine the actual discharge temperature. it is a frustrating and unrewarding exercise in futility.

[nh3simman]

I think what you are looking for is the apparent change in efficiency at various pressure ratios for any given type of compressor.

Spot on.

I have just found this exact graph in the ASHRAE equipment handbook.

The value drops almost linearly from 80% at Pr=3 to 60% at Pr=15.

I'm using it to estimate the required capacity of a matching condensing unit, given the unit cooler.

Thanks for the help guys.

[wkd]

Because most cycles are irreversible and mostly adiabatic they are not Isentropic.The main difference is the state at which the gas enters the compressor ,in most systems superheated ,for obvious reasons,hence discharge gas is superheated also.This means the compressor actually does more work during compression than is necessary in an ideal cycle.
Heat rejected is h2-h3 kj/kg and work done is h2-h1 kj/kg so when designing the system the heat per second is required divide this by h2 - h3 kj/kg gives mass flow in kg/s therefore h2-h3 x mass flow = Heat rejected in kW.If you divide h2-h1 x kg/s by the mechanical efficiency and then again by the electrical efficiency this should give you the required power for the compressor.
Practically I have also measured water cooled chillers under laboratory conditions and found that the electrical power + the evaporator duty divided by the condenser heat rejected is within 2%.Typically manufacturers publish THR as -5%.
Hope this helps

[nh3simman]

Typically manufacturers publish THR as -5%.

-5% of what?

If THR is total heat rejection, it should be in the order of 1.3 times the cooling duty.

[nh3simman]

This seems to be a reasonable 1st guess

http://tinypic.com/view.php?pic=4dbw6jc
The curve is from ASHRAE Systems 2004, based on R22 in a single screw compressor.

Linear curve fit
isentropic efficiency n = 75 - 1.4 (Pr-1)
so at Pr=4, e=70.8 which seems reasonable.

I don't think that it's bad for a 1st guess.

[NoNickName]

Hi there,

It is not fixed I am afraid. As you said it is dependent on operating condition and compressor design.

Cheers

It is fixed per compressor design, depends on operating condition and does not depend on isentropic efficiency.

[lana]

Hi there,

NoNickName , we are saying the same thing here. ;)
I meant that Isentropic efficiency is not fixed.
Isentropic efficiency changes with pressure ratio which in turn changes with operating conditions.

I have checked what we did. Isentropic efficiency is defined as :

Isen. Effi. = Isentropic power / Actual power

When a compressor is selected, then actual power at each operating condition can be obtained from the manufacturer catalogue. Operating conditions are also known so by assuming isentropic compression we can calculate the isentropic power.

Isentropic and volumetric efficiencies are usually expressed against pressure ratio. In graphs, the horizontal axis is pressure ration and the vertical axis is efficiency.

For a given compressor, when pressure ratio is 1 then isentropic efficiency is 0. Efficiency increases as the pressure ratio increases. At one pressure ratio the efficiency is maximum. When pressure ratio is increases beyond this point, the isentropic efficiency decreases after that. This is why in many cases the maximum operating pressure ratio for reciprocating compressors is limited by temperature restriction.

Cheers :)

[wolfie]

To do a quick estimate of heat rejection of a refrigeration system, you could guess COP=3.

From that, Qe/(Qc-Qe) = 3
and therefore Qc = Qe (1 + 1/3)

The heat rejection factor = 1.3

But this is incorrect because the COP depends on the cyle temperatures.

The unknown is only the compressor efficiency.
Text book figures for Isentropic efficiency are 0.72 for a recip.

Surely this is not realistic? I would expect the same compressor with a very low suction to have a lower isentropic efficiency.

Has anybody seen a relationship for isentropic efficiency and pressure ratio?
Please put on a set of gauges that you know are calaberated right and test your hp/lp pressures then and ony then can you ajust your super heats to inprove your Isentropic settings
Your Expansion Valve Is your main control of room temps

[nh3simman]

Please put on a set of gauges that you know are calaberated right and test your hp/lp pressures then and ony then can you ajust your super heats to inprove your Isentropic settings
Your Expansion Valve Is your main control of room temps

You cannot change the isentropic efficiency by adjusting superheat.

I would not advise people to tamper with tx valve pre-settings. Especially those who do not understand its function. If anyone changes the superheat on my equipment, I would be inclined void the warrantee.

[US Iceman]

I would not advise people to tamper with tx valve pre-settings.


Darn good advice. My message is the same.

Leave the factory settings as they work just fine. I normally see more problems created by adjusting the evaporator superheat than corrected by adjusting the evaporator superheat.