Water Dispenser Energy Conservation

[M0rris82]

Let's consider a water dispenser which use as input water at room temperature and the electrical energy to make the water cold and on the other side, hot. If you combine the cold and the heated water to obtain the water at the room temperature again, in which the electrical energy transform into?

Marius

 

[russ_watters]

All you are doing here is passing energy back and forth between the two streams of water. What you'll find, though, is that the compressor energy goes into the warmer water, so the delta-T of the warmer water is larger than that of the colder water. So if you bring them back to gether, the new temp is above the original temp and the energy change from the start is equal to the input electrical eneregy of the compressor motor.

 

[Naty1]

Let's consider a water dispenser which use as input water at room temperature and the electrical energy to make the water cold and on the other side, hot. If you combine the cold and the heated water to obtain the water at the room temperature again, in which the electrical energy transform into?

This statement is self contradictory: you can't start with room temperature water, heat some, and then recombine with more room temp water and expect to obtain room temp water again...clearly the final water will be warmer...

And Russwatters reply covers the basic idea, forgetting about any kinetic energy change of water flow, frictional losses in pipes and so forth...

 

[M0rris82]

Ok, understood. You are right, the resulted recombined water will be warmer for a classic Whater dispenser. But, let's say that a warmer dispenser is in equilibrium(It will output 1L of cold water (0 degree) and 1L of warm water (50 degree). It will spend energy for making this. Right? And when you recombine the two recipients of cold and warm water, you will have 2L of 25 degree( the room temperature). What is the explanation in this case? Let's presume that the Water dispenser does not make any exchange of heat energy with the room.

 

[Dale]

Let's presume that the Water dispenser does not make any exchange of heat energy with the room.

That is a non-physical assumption. Obviously, if you assume impossible things like this then you can get impossible results like non-conservation of energy.

The fact is that it takes energy to transfer heat up a temperature gradient. The process heats up the hot reservoir more than it cools the cold reservoir. When the reservoirs are allowed to come to thermal equilibrium the total temperature is increased proportional to the amount of electrical energy used. Energy is conserved, and it eventually goes to heat.

 

[russ_watters]

The more efficient your air conditioner/heat pump setup, the less energy it uses and the closer the final temperature is to the original, but as Dale said, there must always be an input of energy in order to move heat up a temperature gradient.

If you don't understand what that last phrase really means, consider what an air conditioner or heat pump compressor is really doing: it is pushing a fluid around a closed circuit, just like a pump moves water. The difference is that in a pump, your goal is to move water, whereas with an air conditioner, the goal is to extract the heat that the fluid is carrying.

 

[M0rris82]

Ok. Let's change a little bit the problem. We have 2 reservoirs with delta T. If we will use a carnot engine or stirling engine which use as source of energy the delta T, we can produce work. But if we combine the two fluids, we don't obtain nothing. Or? Some radiations are emitted when we combine two fluids with delta T?

 

[Dale]

If you close both systems from the environment then they will reach the same temperature at equilibrium. If you open both systems to the environment and ensure that the same amount of energy leaves (via work or heat) then they will again reach the same temperature at equilibrium. If you open one system such that energy leaves and close the other system such that energy does not leave then the closed system will wind up at a higher temperature than the open system.

 

[russ_watters]

Ok. Let's change a little bit the problem. We have 2 reservoirs with delta T. If we will use a carnot engine or stirling engine which use as source of energy the delta T, we can produce work. But if we combine the two fluids, we don't obtain nothing. Or? Some radiations are emitted when we combine two fluids with delta T?

You don't have a delta-T until you create one with your chiller, so if you run a carnot engine off the delta-T you create with your chiller, you're just running the two against each other. It's basically the same as using a pump to raise some water to drop through a turbine to power a pump. By conservation of energy, the energy of the generator is equal to the energy of the pump, minus losses: you always end up with extra waste heat.

What is the purpose of this inquiry? Are you trying to develop a practical application of some kind?

 

[AceParkle]

There's no need to combine the hot and cold water to obtain the temperature of the room. You don't have to think for a long term process in gaining the evaluation. The water will become warmer if you combine the two element.