I never took a thermodynamics class but I've been wondering about this: if you continuously cooled the warm refrigerant in the condenser with the...

I never took a thermodynamics class but I've been wondering about this: if you continuously cooled the warm refrigerant in the condenser with the same ice produced in the freezer, would the efficiency of the system improve? by how much?

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  1. 1 month ago
    Anonymous

    I don't think so.

  2. 1 month ago
    Anonymous

    No

  3. 1 month ago
    Anonymous

    let me get this straight
    you want to route your warm refrigerant through your freezer and you think that it will improve efficiency?
    maybe you should have taken that thermodynamics class

    • 1 month ago
      Anonymous

      >you want to route your warm refrigerant through your freezer
      no. I want to move the ice from the freezer to the condenser coils.

      >you think that it will improve efficiency?
      it wouldn't even if the purpose of the system was to obtain liquid water?
      wouldn't this thing transfer the heat produced in the condenser to the ice, thus lowering the difference between "ambient temp" and freezing temp? (and also avoiding having to dump it into the environment and cooling the refrigerant much faster, thus improving the speed of ice production)

      >maybe you should have taken that thermodynamics class
      any suggestions for a college dropout?

      • 1 month ago
        Anonymous

        >no. I want to move the ice from the freezer to the condenser coils.
        How's this different?

        • 1 month ago
          Anonymous

          The point of a refrigerator is to take heat out not put it back in.

          you wouldn't be heating the evaporator and the ice being produced if you took the ice with your hands and put it in the condenser coils...

          am I a moron for not explaining well or for not understanding thermodynamics?

          • 1 month ago
            Anonymous

            You're explaining it fine, you just don't understand the underlying thermodynamics.

            The cooling circuit does not remove heat, it just moves it around and introduces net heat to run the pump. Your scheme is equivalent to introducing heat to the cold side output (heat input as it were) and you would immediately fail your chemical engineering heat transfer class.

          • 1 month ago
            Anonymous

            I'll add that *multi-stage* refrigeration is possible, but it's not what you describe. While multi-staging, you stack the cold end of *a second cooling loop* to the heat sink of another and the temperature change you can accomplish is higher, but you generate twice as much heat and require external heat sinks all the same. It is not possible to improve the coefficient of performance in this way (and if the refrigerants used are identical, it is deterministically lower probability), it is merely a gimic used where parallel loops are incapable of achieving the temperature delta.

            Hope this helps

          • 1 month ago
            Anonymous

            >Your scheme is equivalent to introducing heat to the cold side output
            how is it equivalent? at most you would be dumping ("moving") the heat obtained from the water being frozen into the ice produced previously (to convert it to liquid), which would require exactly 0 extra energy, no? if not, then clearly I don't understand heat pumps conceptually... like, at all.

            I'll add that *multi-stage* refrigeration is possible, but it's not what you describe. While multi-staging, you stack the cold end of *a second cooling loop* to the heat sink of another and the temperature change you can accomplish is higher, but you generate twice as much heat and require external heat sinks all the same. It is not possible to improve the coefficient of performance in this way (and if the refrigerants used are identical, it is deterministically lower probability), it is merely a gimic used where parallel loops are incapable of achieving the temperature delta.

            Hope this helps

            but in that case you require 2 pumps...

          • 1 month ago
            Anonymous

            >if not, then clearly I don't understand heat pumps conceptually... like, at all.
            It certainly seems that way.

            A heat pump (all chemical [and thermoelectric/peltier] devices) move heat from one side to another by modifying two of the other thermodynamic properties of a substance (usually entropy and pressure by inducing phase changes). This action has a net zero change in the heat and energy of the system. HOWEVER, in order to induce this change, energy (which ends up as heat) must be introduced to power one or more pumps of electrical gradients.

            The ratio of heat moved to heat introduced is known as the coefficient of performance and is the main motivation for using heat pumps to begin with.

            So your scheme has you removing X units of heat from water, introducing those X units of heat to the heat sink, and picking up Y units of heat (dumped into your heat pump itself) from running the pump. But your heat is dumped *inside* the system. So for every circuit you remove zero heat from the system and introduce Y heat.

            What you describe isn't even multi-staging, it's closing the system and seeing what happens: what happens is that your freezer will continuously increase in temperature until the cooling circuit ruptures or it reaches an equilibrium of heat radiation.

            There is a single interesting thing that this *could* do, namely ghetto multi-staging, but that is NOT what you describe, is NOT more efficient, does NOT work over most temperature ranges, and is NOT used in any system because dedicated multi-staging series are much more effective for this purpose (that purpose is only to increase the temperature difference achieved by the unit, which dramatically lowers efficiency).

            If you have more questions, go read a chemical thermodynamics textbook, there are chemical design discussions specifically dealing with all possible setups. I've been very thorough but you're not going to understand this if you don't have a correct conception to begin with.

          • 1 month ago
            Anonymous

            I understand that if I cycled the water over and over then the system would obviously heat, but I'm not saying that. the purpose of this system would be to desalinate water and dump the extra salt water while.

            Not him but your solution is more efficient IF for some weird reason you are required to wait until the water has frozen. If you just want to cool down water it'd be more efficient to just put in the water until it is cold enough and not until it freezes.
            If you wait until it freezes and then use the excess heat to cool down the condenser, then if the heat pump didn't require any energy to run, then at most for every degree per gram of mass that the ice cools the condenser, you would gain 1 degree of cooldown per gram in the evaporator. But as the process is not 100% efficient, you are cooling down 1 gram by 1 degree, and then giving up that gain to cool down, say, 1 gram by 0.5 degree. The other half of the degree is the mechanical energy that the heat pump itself needs to run and is then converted to heat, which you are also dumping into the ice that you made previously.
            That is basically what makes the process less efficient. Instead of dumping as much as possible of the heat generated from friction into the environment, you are unnecessarily dumping some of it into the thing you are trying to cool down, and thus undoing some of the work that you already did.
            I don't know what the exact formula would be but it shouldn't be too hard to deduce it from the previous reasoning.

            >Not him but your solution is more efficient IF for some weird reason you are required to wait until the water has frozen
            that's exactly what I want. I got this idea while studying freeze desalination.
            the water itself would not be reused after freezing, and the leftover (salty water) would be dumped.

            >That is basically what makes the process less efficient
            right, now I get why these people are telling me it's not efficient...
            yeah, I obviously should have been more clear. duh.

          • 1 month ago
            Anonymous

            >the water itself would not be reused after freezing
            I meant to say: the (salty) water would not be frozen again and would not need to be used as ice.
            in short, the point of this would be to freeze water to separate the salty water.
            then I'd heat it by reusing the heat in the condensator to make the process faster and also more efficienct.
            now I'm wondering how much efficient it would be... since FD uses like 30x the energy per mass required in a reverse osmosis desalination process.

            excuse me for my lack of explaining, I'm ignorant at this stuff and obviously have trouble giving all the info needed...

          • 1 month ago
            Anonymous

            Yeah, there's no way you would reduce the energy usage by that much, at most I think you reduce the energy usage by half which would mean it would use 15x the energy used by the other method. Assuming the figure you have doesn't already do what we're talking about.

          • 1 month ago
            Anonymous

            In that case, I'm not sure if it would be better to put the ice in the condenser, or if it would be better simply to use it to pre-cool the incoming water.
            When pre-cooling it is easy to calculate the temperature delta, it would be | ambient temp - ice temp | / 2 assuming perfect isolation and assuming you use the same mass of ice and water.
            Calculating the delta when putting the ice in the condenser is a bit harder but I think it should be lower, using the same argument as before. You are dumping both the heat from the incoming water into the ice, but also the mechanical heat from the pump itself which could've been dumped into the environment.

          • 1 month ago
            Anonymous

            Yeah, there's no way you would reduce the energy usage by that much, at most I think you reduce the energy usage by half which would mean it would use 15x the energy used by the other method. Assuming the figure you have doesn't already do what we're talking about.

            well, I've been playing with the carnot efficiency formula, and it seems that reducing the temperature difference between the freezing temp and the "ambient" temp would improve the efficiency of the system beyond the simple transfer of energy, so taking energy from the condensator and freezing at like -3°C should improve efficiency a lot, no?
            I could be wrong, of course. I'm not really sure if I'm applying it correctly in this case.

            [...]
            The enthalpy of fusion of water isn't relevant. The definition of efficiency is exactly (Heat removed from low-temperature region)/(Net work required for heat removal).

            [...]
            If the question is reformulated as
            >"could the efficiency of a system be increased by using some fraction of 'waste products' at some rate to cool the hot-side heat sink"
            This question turns into a parallel mass and energy balance with multiple heat and mass streams. Then the answer would be "Possibly over some temperature ranges for some setups" but you need to define the situation in which this occurs. In the case of desalination, you'll have to develop the terms for the composition and thermodynamics (including enthalpy of solvation if you actually use the chilled waste on the heat sink cooling stream) for the proposed setup.

            Even formulated correctly there are too many degrees of freedom for any answer to be given right now. Give us details on the heat pump and refrigerant being used, feed water temperature and pressure, cooling water (or air) stream information, etc. and we can do the homework problem.

            I'd also like to point out that this is ALREADY done in freeze desalination setups except that the waste stream is never mixed or allowed anywhere near the cooling loop but led directly through a heat exchanger to pre-chill the feed water. No chemical engineer above room temperature IQ would ever allow a low-temperature waste stream to leave such a system without using it as a heat sink to improve efficiency. Ambient temperature water is generated at essentially no cost through evaporation towers.

            >you'll have to develop the terms for the composition and thermodynamics (including enthalpy of solvation if you actually use the chilled waste on the heat sink cooling stream) for the proposed setup.
            yeah, I guess I definitely need to study thermodynamics in a serious way. I've only read parts of a thermodynamics book.

            >you need to define the situation in which this occurs
            >Give us details on the heat pump and refrigerant being used, feed water temperature and pressure, cooling water (or air) stream information, etc. and we can do the homework problem.
            I know you would need this info, but that would take me a whole another level of research lol. I don't even have the basics well covered... I'd rather pay someone (a technician) to do that.
            I've tried watching videos on how chillers, fridges and other heat pumps work, and I still don't "get" the formulas that use all these technical details.

            >No chemical engineer above room temperature IQ would ever allow a low-temperature waste stream to leave such a system without using it as a heat sink to improve efficiency.
            almost none of the stuff I've read shows it being done, though I'm wrong (probably) and didn't pay attention to the details...

            oh well, guess I should leave the obviously heavy stuff to smart, knowledgeable people and not assume they might have missed something.

          • 1 month ago
            Anonymous

            Not him but your solution is more efficient IF for some weird reason you are required to wait until the water has frozen. If you just want to cool down water it'd be more efficient to just put in the water until it is cold enough and not until it freezes.
            If you wait until it freezes and then use the excess heat to cool down the condenser, then if the heat pump didn't require any energy to run, then at most for every degree per gram of mass that the ice cools the condenser, you would gain 1 degree of cooldown per gram in the evaporator. But as the process is not 100% efficient, you are cooling down 1 gram by 1 degree, and then giving up that gain to cool down, say, 1 gram by 0.5 degree. The other half of the degree is the mechanical energy that the heat pump itself needs to run and is then converted to heat, which you are also dumping into the ice that you made previously.
            That is basically what makes the process less efficient. Instead of dumping as much as possible of the heat generated from friction into the environment, you are unnecessarily dumping some of it into the thing you are trying to cool down, and thus undoing some of the work that you already did.
            I don't know what the exact formula would be but it shouldn't be too hard to deduce it from the previous reasoning.

          • 1 month ago
            Anonymous

            Not him but your solution is more efficient IF for some weird reason you are required to wait until the water has frozen. If you just want to cool down water it'd be more efficient to just put in the water until it is cold enough and not until it freezes.
            If you wait until it freezes and then use the excess heat to cool down the condenser, then if the heat pump didn't require any energy to run, then at most for every degree per gram of mass that the ice cools the condenser, you would gain 1 degree of cooldown per gram in the evaporator. But as the process is not 100% efficient, you are cooling down 1 gram by 1 degree, and then giving up that gain to cool down, say, 1 gram by 0.5 degree. The other half of the degree is the mechanical energy that the heat pump itself needs to run and is then converted to heat, which you are also dumping into the ice that you made previously.
            That is basically what makes the process less efficient. Instead of dumping as much as possible of the heat generated from friction into the environment, you are unnecessarily dumping some of it into the thing you are trying to cool down, and thus undoing some of the work that you already did.
            I don't know what the exact formula would be but it shouldn't be too hard to deduce it from the previous reasoning.

            The enthalpy of fusion of water isn't relevant. The definition of efficiency is exactly (Heat removed from low-temperature region)/(Net work required for heat removal).

            >the water itself would not be reused after freezing
            I meant to say: the (salty) water would not be frozen again and would not need to be used as ice.
            in short, the point of this would be to freeze water to separate the salty water.
            then I'd heat it by reusing the heat in the condensator to make the process faster and also more efficienct.
            now I'm wondering how much efficient it would be... since FD uses like 30x the energy per mass required in a reverse osmosis desalination process.

            excuse me for my lack of explaining, I'm ignorant at this stuff and obviously have trouble giving all the info needed...

            If the question is reformulated as
            >"could the efficiency of a system be increased by using some fraction of 'waste products' at some rate to cool the hot-side heat sink"
            This question turns into a parallel mass and energy balance with multiple heat and mass streams. Then the answer would be "Possibly over some temperature ranges for some setups" but you need to define the situation in which this occurs. In the case of desalination, you'll have to develop the terms for the composition and thermodynamics (including enthalpy of solvation if you actually use the chilled waste on the heat sink cooling stream) for the proposed setup.

            Even formulated correctly there are too many degrees of freedom for any answer to be given right now. Give us details on the heat pump and refrigerant being used, feed water temperature and pressure, cooling water (or air) stream information, etc. and we can do the homework problem.

            I'd also like to point out that this is ALREADY done in freeze desalination setups except that the waste stream is never mixed or allowed anywhere near the cooling loop but led directly through a heat exchanger to pre-chill the feed water. No chemical engineer above room temperature IQ would ever allow a low-temperature waste stream to leave such a system without using it as a heat sink to improve efficiency. Ambient temperature water is generated at essentially no cost through evaporation towers.

  4. 1 month ago
    Anonymous

    The point of a refrigerator is to take heat out not put it back in.

  5. 1 month ago
    Anonymous

    No it would be negative efficiency, heat in introduced in the loop.

    The same way leaving your refrigerator door open actually heats your house.

    Look up basic calculations on the Coefficient of Performance for heat transfer circuits and you'll quickly realize why this is the case.

  6. 1 month ago
    Anonymous

    No, because you would make the ice less cool in the process.

  7. 1 month ago
    Anonymous

    Put a solar concentrator on a liquid nitrogen to electricity powered spacecraft
    When the fuel passes through the turbine it will then be condensed again by the coldness of space, and the cycle will repeat

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