4 ms·

English is not my native language. Did I use the word "ambient" properly?

by drran 6y ago

OJFord 6y ago
Assuming I understood your argument correctly, yes. :) (ambient foo = the foo all around; usually used w.r.t. temperature, e.g. food cooling to ambient temperature in the room) The air outside that is the input to your heat pump is already there, already at whatever temperature, that's your argument right? So, yes, it is, but my point is that you can't ignore it as an input just because you don't have to order (and pay for) some air to supply. In calculating efficiency, you have to put an imaginary box around the heat pump, and look at what's going in, and what (that is the desired useful thing) is coming out. If inputs don't cost you anything, that means you might not care about lower efficiency, but it doesn't mean that the efficiency isn't lower.
- drran 6y ago
  1) If we put a very small box around the heat pump, then we will see that temperature in the box increases at constant rate, because of energy loses at the pump. I.e., it will work as a resistive heater, by converting 100% of input electricity into heat, which is the goal. Eventually, the heat pump will stop, because the goal is reached, thus we will have 0 input energy and 0 useful work. 2) To keep the pump running and keep temperature at steady level, we will need to leak heat at constant rate to the second box. Thus, the pump will work at "impossible" 100% of efficiency again: it will convert all input electricity into heat. If we leak heat at high rate, then the pump will not be able to keep up, but it will still work at 100% efficiency. If we redirect some leaked heat from the second box back to the pump as input, so it will pump it back into the box, then it will work at 100% of efficiency + additional heat pumped from the second box. Moreover, temperature in the second box will continue to increase at constant rate, so it will reach equilibrium at some point of time again. 3) For the pump to work indefinitely at steady rate, we need something to dump energy into, with infinite capacity. Also, we can use a fraction of this infinite capacity as infinite input into the system, thus the pump will work at 100% + additional heat. IMHO, first two systems are worthless, because they doomed to reach equilibrium and stop at 0/0.

English is not my native language. Did I use the word "ambient" properly?

by drran 6y ago

English is not my native language. Did I use the word "ambient" properly?

Assuming I understood your argument correctly, yes. :) (ambient foo = the foo all around; usually used w.r.t. temperature, e.g. food cooling to ambient temperature in the room) The air outside that is the input to your heat pump is already there, already at whatever temperature, that's your argument right? So, yes, it is, but my point is that you can't ignore it as an input just because you don't have to order (and pay for) some air to supply. In calculating efficiency, you have to put an imaginary box around the heat pump, and look at what's going in, and what (that is the desired useful thing) is coming out. If inputs don't cost you anything, that means you might not care about lower efficiency, but it doesn't mean that the efficiency isn't lower.

drran 6y ago
1) If we put a very small box around the heat pump, then we will see that temperature in the box increases at constant rate, because of energy loses at the pump. I.e., it will work as a resistive heater, by converting 100% of input electricity into heat, which is the goal. Eventually, the heat pump will stop, because the goal is reached, thus we will have 0 input energy and 0 useful work. 2) To keep the pump running and keep temperature at steady level, we will need to leak heat at constant rate to the second box. Thus, the pump will work at "impossible" 100% of efficiency again: it will convert all input electricity into heat. If we leak heat at high rate, then the pump will not be able to keep up, but it will still work at 100% efficiency. If we redirect some leaked heat from the second box back to the pump as input, so it will pump it back into the box, then it will work at 100% of efficiency + additional heat pumped from the second box. Moreover, temperature in the second box will continue to increase at constant rate, so it will reach equilibrium at some point of time again. 3) For the pump to work indefinitely at steady rate, we need something to dump energy into, with infinite capacity. Also, we can use a fraction of this infinite capacity as infinite input into the system, thus the pump will work at 100% + additional heat. IMHO, first two systems are worthless, because they doomed to reach equilibrium and stop at 0/0.