Pump Heat Talk
- Thread starter CoralNerd
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CoralNerd
Active member
Slightly off topic, sorry.
Q: If I install a pump internally in the sump vs externaly, which would increase the water temperature the most?
The external pump would be in the cabinet where the sump is. Currently I have my closed loop pump (waveline DC10000II) installed internally and I leave the cabinet doors open most of the time with some computer fans blowing on the sump water.
Second question, if I installed the a pump externaly, should I be concerned about water leaks?
Q: If I install a pump internally in the sump vs externaly, which would increase the water temperature the most?
The external pump would be in the cabinet where the sump is. Currently I have my closed loop pump (waveline DC10000II) installed internally and I leave the cabinet doors open most of the time with some computer fans blowing on the sump water.
Second question, if I installed the a pump externaly, should I be concerned about water leaks?
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I think I can answer this. An external pump will transfer some heat to the water but much less than a submerged pump. However, I don't think the heat in the overall scheme of things means much UNLESS you already have trouble with tank over heating. In the summer what is your room temp.? AC.?
Second question is easier, yes. Especially with Vectra. If you supplied unions with compression fittings, you are trying to pump a lot of flow through 3/4" plumbing [more heat BTW]. These fitting are rock solid. If you use the Eco Tech 1.25" adapter and adapter you will get greater flow with less engergy [less] heat AND you use good plumbing practices this is also rock solid. But, please, don't try to adapt America thread fittings NPT to the pump British Thread pipe threads and it will cause problems and long term leakage in an external setting.
RJ
Second question is easier, yes. Especially with Vectra. If you supplied unions with compression fittings, you are trying to pump a lot of flow through 3/4" plumbing [more heat BTW]. These fitting are rock solid. If you use the Eco Tech 1.25" adapter and adapter you will get greater flow with less engergy [less] heat AND you use good plumbing practices this is also rock solid. But, please, don't try to adapt America thread fittings NPT to the pump British Thread pipe threads and it will cause problems and long term leakage in an external setting.
RJ
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CoralNerd
Active member
House temp is 75. The tank runs about 79.3 with the fans. I recently changed over to two dc Vantec Stealth fans which are quite but don't move a ton of air so I have one additional fan that kicks on at 79.5.
I do plan on using the Ecotech adapter kit http://store.ecotechmarine.com/store/vectra/vectra-pump-accessories 1.25 to slip union, to 1" reducer to pvc 1" spa tubing. T.Y.
I do plan on using the Ecotech adapter kit http://store.ecotechmarine.com/store/vectra/vectra-pump-accessories 1.25 to slip union, to 1" reducer to pvc 1" spa tubing. T.Y.
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Not sure I understand plumbing. Are you talking M1? If so as I said it is 3/4". I going to use the reducer to widen back up to 1'? That is better than straight 3/4" but the restriction will still reduce flow. Keep in mind this pump's flow curve was tested with 1.25" plumbing anything less such as my straight 1" system or use of the supplied unions mean you max flow will be significantly reduced from the 2000 max gph.
Like you fans practically, even with the water given heat off. Flow is created and the water is exposed to the surface where evaporation will take away the heat too. You get same with external pump. I am comfortable with external pumps, just use good plumbing practices.
RJ
Like you fans practically, even with the water given heat off. Flow is created and the water is exposed to the surface where evaporation will take away the heat too. You get same with external pump. I am comfortable with external pumps, just use good plumbing practices.
RJ
My own anecdotal experiences (after almost 3 decades of reefing, and countless pumps across dozens of tanks) is that all things being equal, pumps run internally transfer all their heat to the water, pumps run externally do not. Fan cooled externals transfer even less heat to the water than convection cooled models. That is the main reason that I always use externals - plus they appeal more to my aesthetic sensibilities.
Maybe my observations are intuitively obvious, maybe not. Thought Id share regardless.
Maybe my observations are intuitively obvious, maybe not. Thought Id share regardless.
sleepydoc
Team RC
I started a separate thread with the post below before realizing this thread had been started. To keep the discussions together, I'll just re-post it below...
At the risk of losing some people, I'll start with the first law of thermodynamics: Conservation of Energy. Simply put, energy is neither lost or created. For most of us the only energy we are worried about is the heat in our tanks, but the motion of the water contains and height of the water are also both forms of energy.
Energy is added to the system either by transfer of heat from the surroundings (heat from the room, if it is warmer than the tank, or sunlight,) or from electrical energy from the wall via heaters, pumps and lights.
Energy is lost from the tank either by transferring to the surroundings if the room is cooler than the tank or via evaporation (this actually does the same thing since the water vapor leaves the tank and moves to the room.) Chillers transfer heat from the tank to the room via the coils.
If you tank is too cold, you aren't too concerned with all this. You add a heater which takes xxx watts of electrical energy and converts it to heat in the tank. Life's good.
What most people are concerned about is too much heat in their tank. Then we need to figure out how to remove it, or, better yet, keep it from getting there in the first place.
Ignoring heaters, the two major sources of added heat for most systems are lights and pumps. I won't go into lights in detail, but basically MHs add the most, followed by T5s and LEDs.
Focusing on pumps, while their primary purpose is to move water, they also generate heat. When we talk about how efficient a pump is, what we are really saying is how much extra energy is used or wasted by the pump beyond the energy given to the water by moving it. If we actually calculated this energy and compared that to the actual power consumption of the pump, we could calculate the efficiency, but ultimately, all of this energy becomes heat anyway.
Looking at things in more detail, there are basically 2 types of pumps and 2 configurations: AC and DC pumps, internal/submersed and external configurations. While pumps
To compare them I'll go through the 4 possibilities:
Internal AC pump - this is easy to figure out. If the pump consumes 50 watts of power, it is adding 50 watts to your tank. It is functionally the same as having a 50 watt heater on. I see people talk about pumps being more efficient or dissipating heat better and 'running cool.' For an internal pump this doesn't matter - all of the energy goes into the water.
Internal DC pump - this is a little more complicated since it has a controller. The total power consumed by the pump will be that used by the motor plus that used by the controller (TP=MP+CP). The power consumed by the motor is all transferred to the water. The power consumed by the controller is transferred to the air, which may add slightly to the tank, but for our purposes I'll ignore that.
External AC pump - this gets a little more complicated. The power used (=heat generated) by these is dissipated in two ways. As the motor heats up the heat is either transferred to the water running through the pump, or to the air surrounding it. Now we have two concerns - first the overall efficiency, and second, how much of that power is transferred to the water vs the surrounding air. If you have an efficient 100W pump that transfers 50% of the power to the water, that's functionally the same as an inefficient 200W pump that only transfers 25% to the water, assuming your air conditioning keeps the room cool.
External DC pump - This simply combines #2 & 3. The heat energy added to the water is simply that consumed by the motor minus that dissipated in the air.
For an internal pump, efficiency is key. Fewer watts consumed means fewer degrees on the thermometer. For internal DC pumps, you have to factor the controller, making things a bit more complex. It is completely possible that two 1000 gph pumps consume 100 watts but one has a 90W motor and a 10W controller, while the other has a 50W motor and a 50W controller. I haven't seen any analysis or comparison on controller power consumption; I assume that they are mostly similar, but I have no data to back that assumption.
For external pumps, the design and efficiency of the cooling system plays a big roll. When someone says a pump 'runs cool,' that could mean that it's an efficient pump that doesn't consume much power, it's got very good cooling an dissipates the heat quickly, or that it's well insulated so all the heat is going into the water rather than to the external casing. Functionally, the best way would probably be to measure temperature of the water going into the pump vs temperature of the water leaving the pump. Unfortunately, most of us don't have that capability and have to rely on empirc data (i.e. how hot our tanks get.)
At the risk of losing some people, I'll start with the first law of thermodynamics: Conservation of Energy. Simply put, energy is neither lost or created. For most of us the only energy we are worried about is the heat in our tanks, but the motion of the water contains and height of the water are also both forms of energy.
Energy is added to the system either by transfer of heat from the surroundings (heat from the room, if it is warmer than the tank, or sunlight,) or from electrical energy from the wall via heaters, pumps and lights.
Energy is lost from the tank either by transferring to the surroundings if the room is cooler than the tank or via evaporation (this actually does the same thing since the water vapor leaves the tank and moves to the room.) Chillers transfer heat from the tank to the room via the coils.
If you tank is too cold, you aren't too concerned with all this. You add a heater which takes xxx watts of electrical energy and converts it to heat in the tank. Life's good.
What most people are concerned about is too much heat in their tank. Then we need to figure out how to remove it, or, better yet, keep it from getting there in the first place.
Ignoring heaters, the two major sources of added heat for most systems are lights and pumps. I won't go into lights in detail, but basically MHs add the most, followed by T5s and LEDs.
Focusing on pumps, while their primary purpose is to move water, they also generate heat. When we talk about how efficient a pump is, what we are really saying is how much extra energy is used or wasted by the pump beyond the energy given to the water by moving it. If we actually calculated this energy and compared that to the actual power consumption of the pump, we could calculate the efficiency, but ultimately, all of this energy becomes heat anyway.
Looking at things in more detail, there are basically 2 types of pumps and 2 configurations: AC and DC pumps, internal/submersed and external configurations. While pumps
To compare them I'll go through the 4 possibilities:
Internal AC pump - this is easy to figure out. If the pump consumes 50 watts of power, it is adding 50 watts to your tank. It is functionally the same as having a 50 watt heater on. I see people talk about pumps being more efficient or dissipating heat better and 'running cool.' For an internal pump this doesn't matter - all of the energy goes into the water.
Internal DC pump - this is a little more complicated since it has a controller. The total power consumed by the pump will be that used by the motor plus that used by the controller (TP=MP+CP). The power consumed by the motor is all transferred to the water. The power consumed by the controller is transferred to the air, which may add slightly to the tank, but for our purposes I'll ignore that.
External AC pump - this gets a little more complicated. The power used (=heat generated) by these is dissipated in two ways. As the motor heats up the heat is either transferred to the water running through the pump, or to the air surrounding it. Now we have two concerns - first the overall efficiency, and second, how much of that power is transferred to the water vs the surrounding air. If you have an efficient 100W pump that transfers 50% of the power to the water, that's functionally the same as an inefficient 200W pump that only transfers 25% to the water, assuming your air conditioning keeps the room cool.
External DC pump - This simply combines #2 & 3. The heat energy added to the water is simply that consumed by the motor minus that dissipated in the air.
For an internal pump, efficiency is key. Fewer watts consumed means fewer degrees on the thermometer. For internal DC pumps, you have to factor the controller, making things a bit more complex. It is completely possible that two 1000 gph pumps consume 100 watts but one has a 90W motor and a 10W controller, while the other has a 50W motor and a 50W controller. I haven't seen any analysis or comparison on controller power consumption; I assume that they are mostly similar, but I have no data to back that assumption.
For external pumps, the design and efficiency of the cooling system plays a big roll. When someone says a pump 'runs cool,' that could mean that it's an efficient pump that doesn't consume much power, it's got very good cooling an dissipates the heat quickly, or that it's well insulated so all the heat is going into the water rather than to the external casing. Functionally, the best way would probably be to measure temperature of the water going into the pump vs temperature of the water leaving the pump. Unfortunately, most of us don't have that capability and have to rely on empirc data (i.e. how hot our tanks get.)
sleepydoc
Team RC
Your observations are exactly right!
gcarroll
Active member
Better than I could ever explain it! If you don't mind, I will refer back to this post over and over when this topic comes up. This discussion has come up over and over over the years and until now no one has given such a comprehensive explanation that is this easy to understand! Great job!
That is exactly what I did. Did so because of my existing 1" soft plumbing set up. IF I was starting from scratch I would go with 1.25" until I split off to my 3/4" returns.
You will be happy going ahead and getting the adapter. BTW, for submerged use, you DO need tape or paste. The British Standard threads are designed to be screwed down without those. I just used some FOOD GRADE silicone grease. I 3 years time the adapter will unscrew without much effort.
Greg and Sleepy Doc I am doing my best to carry on this discussion here.
I am very familiar with the principal of conservation of energy and do agree 80 watts goes in and it makes 80 watts of energy in the tank. Energy changes forms- electrical to kinetic to chemical to heat. This is the basic principal that brought us out of the stone age. But, always a but, 80 watts goes into the tank in the form of electricity. This is converted to kinetic [in this case rotary movement] and heat. All of the energy is transmitted to the tank but it is of different forms. All motors do this and the energy sometimes goes through several forms until the process at hand is finished. Unlike what our engineer stated it all does not end up as heat. The movement of water has energy that continues after it leaves the pump. It is also contains heat energy. Both of these forms immediately begin to disburse throughout the aquarium from friction of surrounding water- which does convert some of the kinetic to heat. However the currents generated cause other factors to take place. The "energized" water flowing from the pump push and even more water around. Some makes it to the surface of the water. Here two other changes in energy form take place. Some of the water evaporates giving of heat and cooling. Direct cooling also takes place along the air water interface if there is a temperature differential between the air and water. I understand the principle behind saying all 80 watts is transferred to 80 watts of heat but it does not fully explain what is happening. In fact I believe, if the pump is efficient enough in producing kinetic energy [current], the net result of the flow within the tank will actually result in a net cooling of the tank.
I wanted to try to avoid all of this about conversion of energy from one form to another but here it is. I also think it is simplistic to say all energy created by the pump creates heat as to process continues. There are other forms of energy conversion biological and chemical processes that take advantage of the heat or kinetic energy created but you know this.
RJ
Excellent write up!
I would add with the external A/C pump, (and any running equipment) where it is located has an impact on where the heat goes. An external pump in a stand or a small fish room that does not have adequate ventilation will heat up the tank as well. A stand that has an unvented area right below the tank is a heat trap, that heat is transferred time bottom of the tank and if the stand is not well ventilated, it will also add heat to the water in the sump. If in a small inadequately ventilated fish room, all of the running equipment will add heat to the room it is in. Keeping the ambient air temp air around the tank can become important. If the air temp in the stand or fish room is higher than the tank water temp, that heat will transfer to the water. Heat management is always important.
Our equipment creates heat, that is a given, how we handle that heat is important, as is using the least amount of wattage in the system as we can and still provide the best care we can. Thankfully equipment has become much more efficient in the past 10 years.
I would add with the external A/C pump, (and any running equipment) where it is located has an impact on where the heat goes. An external pump in a stand or a small fish room that does not have adequate ventilation will heat up the tank as well. A stand that has an unvented area right below the tank is a heat trap, that heat is transferred time bottom of the tank and if the stand is not well ventilated, it will also add heat to the water in the sump. If in a small inadequately ventilated fish room, all of the running equipment will add heat to the room it is in. Keeping the ambient air temp air around the tank can become important. If the air temp in the stand or fish room is higher than the tank water temp, that heat will transfer to the water. Heat management is always important.
Our equipment creates heat, that is a given, how we handle that heat is important, as is using the least amount of wattage in the system as we can and still provide the best care we can. Thankfully equipment has become much more efficient in the past 10 years.
sleepydoc
Team RC
Yes. I mentioned the effect of ambient heat briefly in the beginning, but ommitted it (or assumed that your AC worked well enough to get rid of the heat) in an attempt to keep things simple. As you clearly described, everything is interrelated, and the 'system' is not truely just the tank and sump, but the immediate surroundings as well.
sleepydoc
Team RC
RJ,
You're right, this does get quite involved, and is probably beyond what many or even most people on this forum care to consider, which is why we moved it to a separate thread, so those who care to discuss can, and those who just wanted to know about the pump could focus on that instead.
I tried to keep my explanation reasonably simple while still going into enoug detail to explain what was going on; hopefully I succeeded....While techinically correct on some points, I think you are confusing things a bit.
The kinetic energy is actually a relatively small portion of the energy used/deliverd. A quick calculation gave me approximately 15 watts for a 1000 gph pump (don't quote me on that, I could be wrong,) but regardless, all of the kinetic energy does in fact end up as heat. There is a classic freshman physics experiment where students measure the amount of heat generated in an insulated & sealed bucket of water by a paddle agitating the water. In short, just moving the water around raises the temperature, as we would expect.
You are correct that turbulence at the water surface may infact augment evaporative cooling, and the net effect may be less than that of an equivalently powered heater, but the fact remains that the evaporation is removing energy that was put into the system. If you had and equivalent amount of evaporation from a fan blowing across the surface without the pump, or even from some powerheads, your water would be much cooler because there is simply less energy being added to the tank. In addition, since the question is not whether we are going to pump water, rather how we will pump it there is the implicit assumption that any induced cooling effects from evaporation, augmented convective cooling, etc are going to be equivalent; the only change will be how much energy the pump adds to the system while generating the water movement we desire.
Even if you had a pump that was 100% efficient and generated no heat while pumping water, thermodynamics dictates that you would still have a net increase in energy to the system; the energy extracted from the system can't be greater than that added.
If you want to be a purist, you would say that the energy is still present, just transferred to the water vapor as heat of evaportation, so it is now 'above' the tank, rather than in it, but let's not go there!
sleepydoc
Team RC
Thank you - Hopefully I got everything correct. I tried to balance scientific detail with the need to keep it understandable for the average guy/gal. As far as I'm concerned anything anyone posts on reef central is quotable. If we don't want to be quoted, we sholdn't post!
shred5
Premium Member
That is it in nut shell except I don't think ti would be a 50 /50 between controller and pump... I think you said it for effect to show the difference.. A controller should use very little power ...
There are other things that could cause heat too in a pump like friction..
You an Electrical Engineer?
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sleepydoc
Team RC
No, I seriously doubt any controller/power supply would take 50% of the power. That was just to highlight the point.
Yes, friction causes heat, too, but that's part of the intrinsic inefficiency of the pump; I just lumped it all together.
I was an EE in a former life. Still comes out once in a while...
shred5
Premium Member
Turned into a short discussion.. I guess we all agree..
I work for a Electrical engineering consulting firm..
We do electrical and lighting for pretty much any type of building.
I work for a Electrical engineering consulting firm..
We do electrical and lighting for pretty much any type of building.
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