Closed Loop Vs. Tunze Streams

[ahchung]

Hi goby1, I am not very strong in physics and would like to ask a question. Does the evaporation rate have any relation to water flow rate? If so, this might explain why a pump with same watt as heater does not raise the water temperature by the same degree.

[Weatherman]

If the water surface is agitated by the flow and the air’s wetbulb temperature is below the water temperature (which it almost always is in our houses), then evaporation will increase with increasing water flow.

This is one way heat is dissipated (through latent heat of vaporization).

[goby1]

ahchtung, that's addressed back a few pages. Yeah, it does. The heat transfer is the same, however. See the post in advanced too.

An equivalent statement to what I've been saying applies to external pumps. The more efficent they are, the higher the percentage of heat transfer to the water. A 100% efficient pump would conduct no heat to the air around it. All wattage would go to thermal energy in the water.

In terms of the isolated system in reference to the definition of the second law, the system is isolated with respect to mechanical interactions with the environment. The only open pathway for energy transfer is a thermal one.

I saw those ecotech pumps on a website yesterday. That is a neat idea. Nice not to have to drill holes. I wonder what the efficiency is. I couldn't find any numerical specifications.

G1

[crumbletop]

achung,

Not only does water movement transfer energy to the surrounding air, there are also vibrations from the water movement, and from movement of parts of the pump, that also goes to flex/vibrate the tank which in turn vibrates the surrounding air creating what we commonly refer to as noise. The tank itself is not a closed system. A heater does not vibrate and cause any sound (i.e. transfer of energy to the air via vibration), thus it is more efficient at heating the _tank water_ while a pump cannot _ever_ be this efficient because it is loosing more energy to the surrounding environment that is not yet _heat_. In addition, controllable tunze pumps have an _external_ transformer that presumably accounts for some of the power usage. This energy would also not be directly added to the tank as would be the case with a submersed heater. I don't have controllable tunze, so I may be totall wrong about the external transformer that converts ac to dc and then the dc drives the pump. So I don't think submersible tunze (or any powerhead) @ xW = submersible heater @ xW. Too much energy is lost to the surrounding environment before it has a chance to become heat.

Jack

[getoyute]

Wheeeeew...can you say thread hi-jackers. Well, at least I learned some things about heat and some new engineering jargon:big eyes:

[goby1]

crumble, the vibrational transfer pales in comparison to the thermal transfer.

It's only an open system with respect to thermal transfer and evaporation. Consider the boundary conditions. I don't see what ac dc ac has to do with it.

G1

[omeg]

goby1, so a fan blowing on top of the water that is warmer than the air, will in theory increase the water temperature?

How is it different than that water movement from a pump will compare with no water movement from a heater?

Sorry to hijack the thread, but this is good reading.

[goby1]

The evaporation kinetics will follow an Arrhenius (exponential) relationship, the driving force being related to the difference in activity of the water in both phases. In the solution, the activity is about 1. In the air, it's about equal to the partial pressure of water in the air. The temperatures plays role to, with faster evaporation at higher temperatures.

Increased mixing in the water increases the thermal gradient at the interface, increasing evaporation. More air movement above the water lowers the partial pressure of water in the air, which also increases evaporation.

As mentioned, the thermal energy transfer through evaporation is related to the latent heat of vaporization. Without evaporation, and a steady-temperature in the tank, the temperature is higher, but the rate of transfer is the same... otherwise the temperature would rise until the rate is matched, through conduction, to the watt draw of the pump.

G1

[goby1]

Omeg, no, that's not what I was saying. Temperature and heat are different things.

The water has an activity of roughly 1 everywhere in the tank, but externally, it is a function of the partial pressure. A fan delivers drier (lower partial pressure h20) to the interface, and the pump delivers water with roughly the same activity from the middle to the interface. The water movement inducing air convection is the primary way that water movement increases evaporation. This movement is much less than that induced by a fan of the same wattage.

I suppose we are turning to kinetics now... more interesting than the thermo anyway...

G1

[Mogrash]

goby1

The ac to dc is something specific to to the Tunze (at least the controllable ones). They have an external transformer that converts ac to dc and then dc drives the pump. In the process a considerable amount of heat is lost in the transformer. The only relevancy to our thermal/kinetic discussions is that the Tunze (assuming the rated power consumption is taken before the transformer and not just a theoretical number) would be putting even less then the rated watt draw into the tank as a portion would be lost in the external transformer.

Of course, the normal submersible pumps would still function as a heater at their given power consumption within our aquarium, although one with a secondary purpose.

On the increase in evaporation, I figured the water motion was more a factor of increasing the surface area rather than the water inducing more air convection. Obviously they could both work in conjunction but which has more impact?

[Weatherman]

Originally posted by goby1
The temperatures plays role to, with faster evaporation at higher temperatures.

You do need to be a bit careful here.

Evaporation rate is not necessarily higher at higher temps.

If the air temp is 80F, the web bulb 80F, and the water temp 60F, evaporation will be zero (in fact, you'll get condensation, in this case).

If the air temp is 70F, the wet bulb 20F, and the water temp 70F, evaporation rate will be rapid.

[goby1]

Mogrash, agreed. The air convection will have much more impact.

[crumbletop]

I think the problem here is that regardless of how much energy from a pump is turned into heat, it is still _not_ the same as adding a heater to the tank. Thermal transfer to the outside environment, via increased evaporation, via sound, via direct transfer of kinetic energy from the water to the bouncing air molecules, etc is increased with a pump vs a heater. So even if all of the energy used by the pump was converted to heat inside the tank, the loss to the external environment will be greater than it would be with a heater. In addition, this thread was about tunze vs closed loop. The external transformer on the tunze gives off some heat to the external environment and thus even less of its energy is released into the tank. So given all those factors, the implicit statement that started this debate -- "the tunze is the same as adding a heater" -- is not true.

Jack

[goby1]

Weatherman, gotcha. I was implying that that for fixed air temp and fixed dewpoint, the evaporation kinetics will be more favorable for warmer water, assuming dewpoint is not the air temperature.

G1

[goby1]

crumble, the tranformer is the source of the difference. There would be no difference if there was no tranformer. Say the system is 110 watts, with 10 lost in the transformer. The same spec pump at 120V would use 100 watts, all surrounded by water. The transformed system has 100W surrounded by water as well. I'm saying it's the same because you don't gain any benefit in flow by using a low voltage pump.

G1

[Weatherman]

The key temperature for determining evaporation rates is really the wet bulb temperature. If the wet bulb temp is higher than the water temp, you won’t get evaporation, regardless of what the air and water temp are.

Other processes, however, can allow for the dissipation of heat, such as conduction, convection or advection.

[goby1]

For a wet bulb temperature above tank temperature, the air would have to be pretty close to saturated.

With the dewpoint (or alternatively the partial pressure of water in air), air temperature, and water temperature, you can calculate the wet bulb temperature.

That knowledge will tell you the direction of the reaction, but not the rate. You still need to know the rate constants and geometry, etc..

G1

[crumbletop]

Heater = stagnant, tunze = not stagnant. Faster/more heat transfer with a turbulant surface. Heater != tunze. Same would apply for others (heater != seio). If I plopped a powerhead in a cooler and shut the lid, and plopped a heater with the same wattage into a different cooler and closed the lid, then you'd be right. It'd be kind of hard to admire the inhabitants if my tank were a closed cooler though...

Jack

[crumbletop]

Weatherman, That's one awesome picture in your avatar. Is there a larger version of that on the net somewhere?

Jack

[Weatherman]

Originally posted by goby1
For a wet bulb temperature above tank temperature, the air would have to be pretty close to saturated.

In most situations we commonly encounter, yes, that's true.

But, if the air temp is 100F and the wet bulb is 75F, the air is not close to saturation. If water temp were 73F in this case, there would be no evaporation.

[Weatherman]

Originally posted by crumbletop
Weatherman, That's one awesome picture in your avatar. Is there a larger version of that on the net somewhere?

Jack

I can't remember where I found that.

I did a google search on "lenticular Rainier". There's lots of nice pictures to choose from.

[goby1]

Faster/more heat transfer with a turbulant surface? Nope, same heat transfer. Lower temperaure though. The mode of heat transfer effects the temperature, not the rate of heat transfer.

G1

[crumbletop]

??? so same mode but more surface area (turbulant) = same rate of heat dissipation?????

[goby1]

Same total for the tank. More surface area available for evaporation, more dissipated through evaporation, and less via conduction through the walls. Same total though.

G1

[Scuba_Dave]

A pair of Tunze for me
Electric bill is what makes the decision for me
Plus I like how the single Tunze is working in my 125g