-1
$\begingroup$

I know entropy rules force my statement to be FALSE. But anyway I haven't been able to find the problem in the statement.


Suppose:

  • We have a perfectly isolated cold reservoir at $T_c = 100 K$
  • We have a room temperature of $T_r = 300 K$
  • The optimal carnot heat engine efficiency will be: $\eta = 1- T_c/T_r = 0.66 $

Suppose we manage to achieve such an engine, with some losses, so only 60% efficiency. That means, each $100 W$ of heat we put into from room temperature. We achieve 60 W of useful energy, and 40 W of heat that go to the cold reservoir (therefore heating it up).

However, if we manage to have a very efficient heat pump (70%), with 60 W we will manage to pump out ($40 W - (0.7 * 60 W) = -2 W$). Therefore making the cold reservoir colder for free. Or generating 1.3W of energy out of room temperature.

Where is the fault in my logic?

Thanks!

$\endgroup$

closed as off-topic by CuriousOne, user36790, Daniel Griscom, Ali, yuggib Jan 14 '16 at 9:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – CuriousOne, Community, Daniel Griscom, Ali, yuggib
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Where can I buy that heat pump. :-) $\endgroup$ – CuriousOne Jan 14 '16 at 0:18
1
$\begingroup$

You made a mistake in calculating how much heat a heat pump can pump from a 100K reservoir to a 300 K reservoir with 60 Watts of power. The "coefficient of performance" of a heat pump is defined to be CP=$Q_H/W$, where $Q_H$ is the amount of heat that is transferred to the higher temperature reservoir and W is the amount of work input to the heat pump. For an ideal, perfect heat pump, CP(ideal)=$T_H/(T_H-T_C)$. For the temperatures you gave, CP(ideal)=1.5 . That's not an error: Heat pumps can pump more heat energy than the input work into the pump. That's why they are an attractive option in climates where the outside temperature ($T_C$) is not too low. In contrast, if you have an electrical room heater, then if you put in 100 Joules of electrical energy then the most heat energy you can get out is 100 Joules. For a near-ideal heat pump, that same 100 Joules of electrical energy input could provide about 150 Joules of heat energy to your room.

Anyway, if you provide an ideal heat pump with 60 Watts of input power (which is the power provided by your 60% efficient engine), then for an ideal heat pump with a CP(ideal)=1.5 about 1.5*(60 W)=90 Watts of heat energy will be pumped into the upper reservoir. That means 90-60=30 Watts of heat power is removed from the lower temperature reservoir. See the diagram below. The upper reservoir is drained of a net 10 Watts of heat and the lower reservoir gains a net 10 Watts of heat even for the case of an ideal heat pump. So the upper reservoir will tend to get a little cooler with time, and the lower reservoir will tend to get a little warmer with time - just as expected.

enter image description here

$\endgroup$
  • $\begingroup$ Thats it! A 60W heat pump does not pump 60W of heat out of the reservour, because you need to add the heat of the energy you put in it!. That makes the device < 100%. And the math correct. Oh and you are right heat pumps are amazing for heating, they generate more heat than what they consume....as long as you have something else to cool down. $\endgroup$ – DarkZeros Jan 14 '16 at 1:57
0
$\begingroup$

The previous answer is right. First of all, you need to understand that a carnot engine is the most efficient heat engine possible. The fault is not in the logic, but in the initial statements.

In this case I can give you a contradictory argument for your $A$ engine. If you take your engine in the reverse, ie taking heat from a source and throwing some into the sink and connect the work extracted to make an other Carnot engine (which is taking heat from the sink and throwing some to the source) work, the combined system will falsify the second law of thermodynamics by collecting heat from the sink and throwing it to the source without doing any extra work.

$\endgroup$
0
$\begingroup$

You are apparently trying to invent another perpetual motion machine. The work that you put into the heat pump has to come from the hot reservoir, and will require that you exhaust heat to the cold reservoir for this device as well. Assuming that such a device theoretically existed, and didn't exhaust heat to the cold reservoir, your claim of getting 1.3 W of energy from room temperature would seem to violate the 1st law, as this 1.3 W appears from nowhere.

$\endgroup$
-1
$\begingroup$

Your answer is in the question text:

the optimal carnot engine has an efficiency of 0.66

if we have a very efficient heat pump with 0.7...

Obviously, such a heat pump is not possible. If you assume you have something impossible, you can produce impossible results.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.