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Motor input power versus output cooling effect

Motor input power versus output cooling effect

Motor input power versus output cooling effect

(OP)
Our organisation is installing (2) Carrier chillers on a ship,(30HXC310-PH3) with a gross cooling capacity of 1043kW each.  The motor input power for each unit is 248 kW.

What I can't get my head around is how we get more out than we put in - the fact that kW is used for both confuses me.

I've checked through "Modern Refrigeration and Air Conditioning", but can't seem to find an answer.

Thanks,
Mark

RE: Motor input power versus output cooling effect

Thread403-83874

Suggestion : Check basic Thermodynamics books

Regards,

RE: Motor input power versus output cooling effect

The compressor input is not equivalent to the cooling effect. Instead it is used to transfer heat from the space to the heat rejection source. Thus the thermodynamic formula is:

compressor input + cooling effect = heat of rejection to air if aircooled, to water if water cooled.

RE: Motor input power versus output cooling effect

markcoleman: The formula for a heat engine is Eff=T1-T2 ^ T1
For a refrigeration Eff=T2 ^ T1-T2
What you are looking at is called the coefficient of performance of a refrigeration machine. Efficiencies of heat engines are always less then 1.00,while the (cop) of refrigeration machines are always greater then 1.00
The cop for a refrigeration machine is the "net refrigeration effect" ^ "heat equivilent of mechanical work"

RE: Motor input power versus output cooling effect

(OP)
Thanks very much, gentlemen.
Forgive me, but I need to understand the concept in very basic terms before proceeding to the math:  
I understand you saying that the input energy is used to MOVE heat energy from one place to another.
In this particular case, 248 kW of electrical energy is used to transfer 1043 kW of heat energy from inside the ship to the seawater.
Am I on the right track now?

RE: Motor input power versus output cooling effect

Hi MARKCOLEMAN!

It's urgent you go back to the THERMODYNAMICS Basics/Fundamentals, as Quark says.

Being more detailed:

- 248 and 1043 kW being energies, are not of the same kind, even you should to sum them to balance the chiller refrigeration cycle. The First Thermodynamic Law! I think you know what is it? Otherwise you should do!

- That kind of cycle means refrigerating gas state changes, latent heat transfers at the level of the evaporating and condensing coils in balance with the electrical energy consumed by the equipment named by  compressor which dialogs with another equipment called by the name of the expansion device/valve.

- And there are, as well, some fundamental definitions/concepts as COP, cycle efficiencies (2 nd Law!), subcooling, superheat, system exergy anlysis (3 rd Law!), you should know for we talk the same language...

Good study and good luck.
zzzo   

RE: Motor input power versus output cooling effect

Mark!

You may have the idea that energy is not conserved here, but
235zzzo is right on and now I will explain it a bit with basic fundamentals (as this topic is raised couple of times in this forum and so no harm in going back to basics).

First Law of Thermodynamics states 'In a system which undergoes thermodynamic cycle, the algebraic sum of all the energy interactions across the system boundary is zero'. Or in other words the total quantity of heat transfer across the system boundary should equal the total work transfer across the boundary. (no accumulation of internal energy as this is a cycle)

Let us consider heat input to the refrigerant as Q1 (evaporator), heat output from refrigerant as Q2 (condenser), work input to the refrigerant as W1 (compressor) and work output from refrigerant as W2 (expansion device).

So by first law, Q2-Q1 = W1-W2, but there is no work done if you use expansion valve the equation becomes
Q2-Q1 = W1 (ofcourse, for larger machines the expansion valve can be replaced by a turbine and we can get useful work, atleast theoretically)

So you can through out more energy (heat) than you require to power(work) your compressor. The power input will be minimum when the difference of source and sink temperatures is less. (just check the COP equation)

And as for power drawn by the compressor, I described it in the thread I mentioned.

PS: In contrary to your thinking, any machine which takes x kW power to remove x kW of heat in a cycle violates the first law and leads to the invention of Perpetual Motion Machine of First Kind.

Regards,

RE: Motor input power versus output cooling effect

COP = Cooling Load/Input Energy

Kw/Ton = 12000/(COP x 3413)

Heat rejected to sink = Cooling capacity + compressor input

RE: Motor input power versus output cooling effect

markcoleman: these guys are fine engineers, however I have the feeling that you might still be confused Sooo, let me try to simplify it this way.
REFRIGERATION:
The process of the absorption of heat from one location and its transfer to another for rejection or recuperation.
THE MOTER INPUT IS THE TRANSFER MEDIUM..DRIVING THE COMPRESSOR OF COURSE.

RE: Motor input power versus output cooling effect

(OP)
You guys are great.

I understand the concept -  the math is no problem.  
I started out as a millwright, and migrated into machine design, and so never studied thermodynamics. Now, I find myself in this odd position of overseeing the HVAC and reefer installation on a ship.

Thanks again to you all, .....
I'm learning a bunch from all the other threads as well!
Mark

RE: Motor input power versus output cooling effect

I am sorry for one grave mistake in my earlier post. Excuse me for quoting violation of the first law in my note. That is wrong and it generated due to a little algebraic miscalculation. It doesn't violate first law eventhough the motor power equals heat addition to the system.

Please ignore my statement on PMM1.

Regards,

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