Heat Pump Refrigerator Analysis 2

[Lazyjack]

Is this the correct way to calculate the ideal and real COP of the evaporator (Beta) and max efficiency of a table top ice maker using the heat transfer method???

Description: 110V ac compression refrigeration ice machine using refrigerant 134a.
An electronic controller with display controls the machine. An 110V ac electric motor runs a fan cooling the motor/compressor and condenser. The machine is plugged into a Watt meter which is plugged into an 110V ac receptacle.
1. 3 L of room temp tap water fills a reservoir
2. 500 mL of water is pumped from the reservoir into a tray containing the evaporator
3. 140 g of ice forms on the evaporator
4. The ice is scraped off into a second tray
5. Time it takes to form the ice is 600 s
6. The remainder of the water is returned to the reservoir
7. This cycle continues until the reservoir is empty.
8. Heat of fusion of ice = 333 kJ/kg
9. Cp of water = 4 kJ/kg K
10. Watt meter 190 W
11. Temp of condenser is 50 C
12. Temp of evaporator is -5 C

Solution for COP real = mass of ice X (heat of fusion of ice + Cp of water X (Temp of water in – temp water freeze)) / Work from a watt meter in Watts X time it takes to form the ice

Here are my concerns about this method.
1. The controller is responsible for 1 – 2 Watts and the fan motor 20ish Watts. This is not work done on the R-134a and should be subtracted from the Watt meter
2. The fact that there is a fan and you can’t comfortably keep your hand on the motor/compressor tells me that it’s not 100% efficient. Losses, ohmic, inductive, frictional. So an assumption has to be made about W to balance Qin+W-Qout=0
3. 500g of 21 C water is brought to the brink of freezing. Why count only the solid?
4. The above statement implies that including the heat of fusion of ice in the calculation is as relevant as the number of cars in the parking lot
5. The above statement, if true, implies that the time to make the ice is also irrelevant
6. Qout must also be calculated to balance the equation in statement 2

Typical COP for a compression refrigeration cycle is 2 to 6 Qin to Qout respectively.
Efficiency is on the order of 30 – 40 %

 

[natepiercy]

I would say COP is your outputs divided by your inputs. Your output is the heat transfer that works in the direction you want. The input is electricity (which your watt-meter measures the rate of use of).
[1] COP = relevant_heat_transfer / work_in

There's two parts to the relevant heat transfer.
1. The heat removed that cools the liquid from the input temperature to the freezing temperature
2. The heat removed that actually freezes the goods from liquid to solid
Thus we can say
[2] relevant_heat_transfer = energy_cooling + energy_freezing

Recognizing that all of the fluid has to come to the freezing temperature, you should account for the mass of all of it for the energy_cooling portion, as it's still heat transfer in the direction you want on the system. **SEE NOTE**
[3] energy_cooling = mass_of_water_in X Cp_water X (Temp_water_in - Temp_water_freeze)
Recall that the density of pure water is 1g/mL, so 500 mL of water is ~ 500g = 0.6kg.

The energy_freezing portion is as simple as mass of frozen goods times energy required per unit mass frozen (the latter is what your heat of fusion is).
[4] energy_freezing = mass_of_ice X heat_of_fusion_ice

Therefore, from [2], [3], and [4]:
[5] relevant_heat_transfer = mass_of_water_in X Cp_water X (Temp_water_in - Temp_water_freeze) + mass_of_ice X heat_of_fusion_ice

Your input energy is measured by Power X time. With power in Watts and time in seconds, this is going to give you Joules.
[6] work_in = watt_meter_reading X time_on

Therefore, from [1], [5], and [6]:
[7] COP = [mass_of_water_in X Cp_water X (Temp_water_in - Temp_water_freeze) + mass_of_ice X heat_of_fusion_ice] / watt_meter_reading X time_on

In response to your concerns:
1. This is still work required to put into the system to achieve your end result. It counts.
2. Measuring your watt meter is measuring your power in to the system. Efficiency of the machine is not necessary to determine COP.
3. See explanation regarding equation [2]. If you truly want to think of it in terms of output / input, and you don't care what happens to that water that has been brought to the brink of freezing, then you would only use the mass of the ice, because it's your output, that is, the part you care about. The remaining water that has been cooled would then be considered a byproduct and not taken into account.
4. Not so. See explanation regarding equation [4].
5. Correct. To get the work in, you need the power draw X the time the power is being drawn. If the compressor runs for 5 minutes at 50W but it still takes the ice 10 minutes to cool, you still use 5 min X 60 s/min X 50 W = 15kJ to calculate your power input.
6. See response 2.

Nate

 

[MatthewL]

COP is defined as the ratio of heat transferred from the system (in for heating, out for refrigeration) to the energy consumed by the device doing the heating or cooling. To answer your concerns:

1. All energy consumption by the unit must be included. If you don't have the controller and fan, the system won't run, therefore you have to include it.
2. COP is NOT a balanced Qin+W-Qout=0 calculation. Most heat pumps operate at a COP of greater than one, because they are using external air/water/ground as a heat source/sink. To balance the equation, you have to include the heat gained (lost) by the external environment, and calculate W by knowing the efficiency of the equipment and multiplying the efficiency by the power consumed.
3. Tis is an unsteady state process, since you recirculate the (already chilled) excess water back to the storage tank. To accurately find the power consumption, you should set your wattmeter in totalizer mode and run the system for a full cycle, because the recirculated water is at ~0C, and will cool the water in the tank, reducing the amount of cooling required for the next cycle-by the time the tank is empty, the feed will be close to 0C if there are no significant losses to the environment (probably not true, but...). Assuming that you freeze all the water in the storage tank and have minimal losses to the environment, your total heat transferred would be 3kg*(333kJ/kg + 21C*4kJ/kgC) or 1251kJ. I would expect that the cycle time for the later cycles should be less than the 600s or the ice produced will be greater than 140g. If your cycle time, watts, and mass of ice per cycle remain constant, your total energy consumption would be (3kg/0.140g)*600s*190W=2443kJ. Your COP would be 1251kJ/2443kJ or 0.512
4. Incorrect, forming the ice requires about 4 times as much cooling as the cooling required to reduce the water temperature from 21C to 0C.
5. True, but your system can only "move" so much heat in a given time. If the system had more capacity, the time would be shorter but the watts would be higher.
6. See #2 above.

Regards,

Matt

Quality, quantity, cost. Pick two.

 

[MatthewL]

Hah, one minute late! Nice work Nate!

Matt

Quality, quantity, cost. Pick two.

 

[natepiercy]

Hi-5, Matt! I particularly enjoyed the similarities of our answer #6. That was a good mid-afternoon chuckle for me.

 

[Lazyjack]

Here are my thoughts:
Qout Condenser (Hot)
Qout = Mass flow x Cp x Tin - Tout

Qin Evaporator (cold)
Qin = Mass flow x Cp x Tin - Tout

Refrigeration Cycle
Beta = Qin/Wcycle = Qin/Qout - Qin

Heat Pump Cycle
Gama = Qout/Wcycle = Qout/Qout-Qin

Initial assumption is that W is 40% of the Watt meter after subtracting parasitic power consumption (cooling fan, computer, any other bells and whistles) Again, not all of the W on the meter is acting on the working fluid.

If Qin + W - Qout does not = 0 an adjustment to the assumption of W (lower) must be made. At the point where the equation is balanced (Qin + W - Qout does = 0 the efficiency of the heat transfer system is obtained.

I have done this calculation many times achieving classic results. Beta of 3ish Gama 4ish efficiency 30ish%. Ice does form around the evaporator but so what. The heat transfer system is in thermal equilibrium. Recorded temperatures and pressures line up close to perfect on the Pressure Enthalpy, Temperature Entropy diagrams.

 

[Lazyjack]

I'm having problems uploading my docx settings, calculations, and charts

 

[Lazyjack]

Settings and Calculations

 

[Lazyjack]

Pressure Enthalpy

 

[Lazyjack]

Temperature Entropy

 

[MatthewL]

Ok, let's try again:

Your ideal COP (by definition is Qin/W from the enthalpy plot, where Qin is heat absorbed by the fluid and W is the work done to compress the resultant gas, check your thermo textbook) is (from your P/H diagram):

(Point1 - Point4)/(Point2 - Point1)

Reading the chart (I hate these things, makes my eyes buggy!), I get (200-55)/(255-200) = 2.636

Your REAL COP is Ql/W, where Ql is the heat lost by the water to make ice (I didn't see anything about a flow meter on the refrigerant line, which you would need to use the enthalpy plot to determine the real Qin) and W is the TOTAL WORK USED BY THE SYSTEM! Once again, YOU CANNOT IGNORE THE PARASITIC LOSSES, they are what makes the REAL COP a REAL number. So, from the work above, your REAL COP is ~0.5

If you were able to use the system for the heating cycle, you would use (Point2 - Point3)/(Point2 - Point1), or (255-55)/(255-200) = 3.636

Also, the proper form for the work equation is Qin+W+Q2=0, where the sign of Qin is positive and Q2 is negative.

In your refrigeration/heat pump calculations, you can only substitute Qout+Qin for W for the ideal cycle (Qout is P2-P3, Qin is P4-P1, and P3=P4, therefore Qout + Qin = P2-P1, which is W), because as stated before, it is NOT a realistic assumption, there is more work being done on the system due to efficiency issues.



Matt

Quality, quantity, cost. Pick two.

 

[Lazyjack]

Ok, as far as parasitic losses, if I connect a toaster oven to the system because I enjoy toast with my ice, I should include that power in my evaluation of the heat pump/refrigeration cycle, and try to plot it??

 

[Lazyjack]

How about this... If I power the compressor making ice with a gas turbine engine supplying 400 MW, and my compressor uses 190W to make 140g of ice and and out of that 140g I'm supplying a vanishingly small amount of ice to 400000 homes, I should count 400MW in my evaluation of the refrigeration cycle???? Is that what your saying about parasitic losses??

 

[Lazyjack]

Let's try again...If I had an electric motor that is 100% efficient connected to a compressor that is 100% frictionless and the tubing, evaporator, and condenser were in 100% adiabatic enclosures, what % of the Watt meter could I use to determine the the cooling heating efficiency??

 

[Lazyjack]

How about this...If I want to determine the efficiency of a gasoline piston engine in a car with square wheels should I test the car on a chassis dyno having the square wheels drive the absorber?? Cuze after all the square wheels are part of the engine, Oh, and maybe I'll turn on the air conditioner while I'm at it. I wonder what Mr.Otto would say about that??

 

[Lazyjack]

Maybe we're arguing over what is a reasonable assumption. Thought provoking...I am assuming the flow rate and shape of the throttle valve enables laminar flow of the refrigerant. What if I increase the flow by changing the motor speed or change the shape of the throttle both to transition to turbulent flow. What would happen to Efficiency??? Would compression still be adiabatic. I would make the assumption the efficiency would decrease. BUT, I would also include the valve in the evaluation of the system. It's intimate with the the refrigerant!!!! The person mowing my lawn at the same time I'm making ice is not intimate with the system. So I'm not going to count the efficiency of his lawn mower in my evaluation of the heat pump refrigeration system.