r.d.m Posted August 16, 2005 Share Posted August 16, 2005 havnt got ehiem pumps but 2x cannisters , i brought home the inductive tester and the 2213 and 2217 were about spot on 8w and 20w respectivly, doesnt help much i know ,but i was curious Link to comment Share on other sites More sharing options...
CThompson Posted August 16, 2005 Author Share Posted August 16, 2005 havnt got ehiem pumps but 2x cannisters , i brought home the inductive tester and the 2213 and 2217 were about spot on 8w and 20w respectivly, doesnt help much i know ,but i was curious ← Thanks RDM, pleased to read your result was accurate to what they were rated at. Link to comment Share on other sites More sharing options...
agro77 Posted August 16, 2005 Share Posted August 16, 2005 Craig, Measure the total ciruit, then measuse all devices. I'd bet the farm they are different. them tong testers are known for the inaccurcy. It even comes down to the angle you hold them at. Just my 2c Link to comment Share on other sites More sharing options...
Gav Posted August 16, 2005 Share Posted August 16, 2005 maybe the instrument you used wasnt that accurate craig? was it calibrated properly? what was the standard deviation of error for the tong meter you used? considering your reading would have been 0.4amps and the expected reading should have been 0.2 amps, a small amount of error can mean one heck of a difference. maybe it is worth trying a different method of testing? Link to comment Share on other sites More sharing options...
Fins Posted August 16, 2005 Share Posted August 16, 2005 Is the measurement in Peak or RMS? The pump may be rated at 50W RMS The instrument may be measuring Peak. Link to comment Share on other sites More sharing options...
OziOscar Posted August 16, 2005 Share Posted August 16, 2005 Sounds pretty much on the money IMHO... that would make sense given the usual Teutonic adherence to specifications. D'Oh to me. Cheers - OziOscar. Link to comment Share on other sites More sharing options...
Pacco Posted August 16, 2005 Share Posted August 16, 2005 If the meter you are using can't even measure something that is 15W, I somehow doubt is going to be too accurate measuring 50W. I very much doubt Eheim would lie about the power usage (and certainly not for such a large amount), and I'm sure they check the power usage of the pumps with much better equipment. With the fact that it is both a pump and a filter that are having the same reading (which probably would be from different manufacturing batches) I think the reading being wrong is a likely explanation than defective units. It would probably be best to get them tested by more accurate means before jumping to the conclusion that Eheim are defrauding their customers. Cheers Pacco Link to comment Share on other sites More sharing options...
Ducksta Posted August 16, 2005 Share Posted August 16, 2005 I spoke to my dad about it this afternoon (he has worked in the electrical wholesale industry since pre-my-birth, working with anything from transformers and power poles to huge electric motors) and he said exactly what Pacco has just said. If the unit wasn't able to give an accurate (ish) reading for a 15W appliance, then it was either not properly calibrated or it is not an accurate enough one to be measuring small power draw, and he wouldn't expect a true reading for a 50W draw. For some perspective: He said he wouldn't be using anything that wouldn't give a reading with a 15W consumption on anything expected to run on normal mains power, as the reading is likely to be out by too much when dealing with low amps. Craig, as far as I can tell reading the thread a second time, you still haven't said if the meter was definately properly calibrated or if you tested anything besides the pumps and got 'normal' values for everything else? Or if the pumps were the only things you tested? Or if you tested other appliances too, and their readings were whacky also? Or if the meter was actually one normally used on massive power motors and not household appliances. Surely working as a science guy the concept of controls in experiments sits relatively easy - what were your controls in this case? Link to comment Share on other sites More sharing options...
Lee Miller Posted August 16, 2005 Share Posted August 16, 2005 Craig, You said: Lee, sorry mate, your grunt and juice "don't" help. When it comes to the ins and outs of electricity, call me Cat weasel. I know you have a point, but I don't understand it. My simply understanding is that the pump has a 50 watt power consumption. For example, if I wanted to connect this up to a UPS, I would use this rating to understand how long a battery would last (for the battery size). However, when tested it had wattage of 96 watts, and not the advertised 50 watts. I stand corrected; yes I agree there will be some friction in the pipes, so the head will not be zero. However, the runs of tubes are very short, and will be negligible, or should I say they can't affect the amount of power the pump consumes as this is fixed. The only difference I should have is in the output of water from the pump. And even if this output was dramatically reduced by head from height the pump has to move the water, or head from friction inside the pipe, I still can't affect the wattage it uses. Is this correct? Juice = power consumption grunt = the amount of head If so, then I would still say, the wattage is fixed. If I have the pump working harder, the power consumption would still be fixed, but the output from the pump would be reduced? My ‘grunt’ and ‘juice’ analogy was an attempt to differentiate between the power that an electric motor can produce and the energy it consumes in producing that amount of power (efficiencies come into play here – see the link provided by OziOscar). I originally asked what Eheim’s specific claim was. Does Eheim claim that the pump consumes 50 watts to generate a specific output flow rate working against a specific pressure or was the claim that the pump motor was capable of producing 50 watts of power? This is still unclear to me. This may seem like a moot point but I believe it to be of significance if suggesting that Eheim has made a misleading claim. You also stated: I understand your point, but no, they are operating at no head (head as in hight to pump the water Lee). If you take a siphon hose, put one end into a tank, and start the siphon from the other end, and let the water fall into a bucket. Now pick this loose end up and raise it above the level of the water in the tank (where the water is being siphoned from). The water level in the hose, where the water stops in the hose is the same height as the water in the tank. If you want to move water beyond this point, this is the head height the pump has to move the water, as gravity will make the water to equal the water height in the tank. If I had a pump plumbed into a sump, pumping water from here to the top of the tank, then the head height would be from the water level in the sump, to the height beyond this the pump has to move the water. If I had this same pump plumbed inlet/outlet via the bottom of the tank, there will be no head "height". As Lee points out though, there will be head caused by the friction from inside the pipe, but this is a different type of head, and not as impacting as head friction. This is why the shorter the tubing you have the better off you are. I trust Lee, that you will agree with this? Am I correct in my understanding that the pump is virtually attached to the bottom plate of the tank and is discharging vertically? If this is the case then the ‘head’ that the pump is working against would be the depth of the tank (approximately 1 metre?). Colfish alluded to this when he said: If the tank is FULL, I’d be of the opinion that those pumps are pushing at a very high head pressure. Therefore using all the 'juice' available to produce maximum 'grunt' to move the stated volume. To explain this more simply; if the pump outlet was attached (via a bulkhead or similar) to a hole in the bottom plate of the tank the column of water directly above the hole would be applying pressure even before the pump was turned on. Therefore the pump would have to overcome this pressure before it could move any water back into the tank. You state that the inlet and outlet of the are both connected to the bottom plate of the tank. If that is the case then you could have the state of equilibrium that you allude to in your analogy of the siphon hose, but only when the pump is not operating (pressure on the inlet side = pressure on the outlet side). However, when you start to move water by turning the pump on, the pump will have to work against the pressure on the outlet side (which could be as much as the ‘head’ of water in the tank – but this is starting to get a bit too technical for me this late at night). In summary, I believe you may have grossly underestimated the ‘head’ that the pump is working against. Ash, You asked: Question for Lee - why would they rate a pump by its output equivalent of electrical energy when its output is water? As you know, a ‘pump’ is an encased rotor driven by a motor. My original question was whether Eheim quoted the power output of the motor. This would seem to make sense because a manufacturer could be relatively confident of the power output of its product but would be unable to assess the variables (efficiencies of the system, losses due to friction, cavitation within the pump, etc) involved in determining the volume output in the range of likely applications. Cheers, Link to comment Share on other sites More sharing options...
Markus13 Posted August 16, 2005 Share Posted August 16, 2005 Out of pure curiosity, and because I can (my uncle is a sparky & electrical engineer) I gave my e-heim 1046 pump yesterday (yes, only a small pump) to my uncle to do some testing. He hooked it up between 2 shallow square buckets of water (to get an indication with some load plus so as to not run it dry) and hooked the pump up to some testing equipment he has (he specialises in medical equipment, so he can test as low as 0.5 watt). The 1046 is rated 5 watt maximum, the avergae draw current from this little experiment came back as 4.96 watts utilised, with the maximum reading being 5.04w and the minimum 4.85w. Bear in mind this is based on this single unit. Link to comment Share on other sites More sharing options...
CThompson Posted August 16, 2005 Author Share Posted August 16, 2005 Thanks for everyone’s input, I have been learning a lot. Markis13 and rdm– thanks for going to the trouble of testing your pumps. My intitial post was to find out if my readings were okay, by having others with the ability (and more knowledge on the topic), to also test their gear and find our their equipment’s accuracy. Your’s was near spot on, and I’d much rather hear this than find out my readings were correct. I have learned that while the tong meter is accurate in that it doesn’t need calibration, it is not fully reliable at lower wattages, as it is more suitable for much higher wattages which it is produced for. So I’m not happy that this style of meter was given to me to use. It works by giving you an amp reading, and this number is taken and multiplied by the volts. Thus, to get 96 watts on the Eheim 1060, the tong meter gave me a reading of .4 amps. Point 4 multiplied by 240 (volts) = 96 (watts). It can give reading down to .1 amp, so does a reading of .4 mean the reading is accurate? I took the readings several times on both pumps, and moved the meter around a bit, and the reading of .4 was stable (ie. not fluctuating). I also understand that our 240 volts can vary +/- 10, and have recently been told, depending on the time of day can vary more than this. So this can lead to further inaccuracies from a tong meter. For a tong meter to be more accurate, at the time of the amp reading, a reading needs to be taken detailing what the volts were at the time of the amp reading. To explain this more simply; if the pump outlet was attached (via a bulkhead or similar) to a hole in the bottom plate of the tank the column of water directly above the hole would be applying pressure even before the pump was turned on. Therefore the pump would have to overcome this pressure before it could move any water back into the tank. You state that the inlet and outlet of the are both connected to the bottom plate of the tank. If that is the case then you could have the state of equilibrium that you allude to in your analogy of the siphon hose, but only when the pump is not operating (pressure on the inlet side = pressure on the outlet side). However, when you start to move water by turning the pump on, the pump will have to work against the pressure on the outlet side (which could be as much as the ‘head’ of water in the tank – but this is starting to get a bit too technical for me this late at night). In summary, I believe you may have grossly underestimated the ‘head’ that the pump is working against. Lee - It is my basic understanding that with a pump plumbed into a tank in this situation, when the pump is turned on, and water is pushed out, it is initially not working at zero head height, as it has first to push the water out against the water preasure the height of the tank. However, once it pushes this water out, water is rushing in behind it to fill in the vacancy left by the water the pump is pushing out, so in effect the pump is getting a 'leg-up' in its efforts to pump the water by water pushing it from behind. As the pump is pumbed into the same depth of water (inlet and outlet), they count each other out and the pump is at zero (near) head? Note the question mark. Craig Craig Link to comment Share on other sites More sharing options...
Ducksta Posted August 16, 2005 Share Posted August 16, 2005 Craig, are you sure you work as a Science guy and not in the NSW parliament? You seem unnaturally adept at dodging questions Link to comment Share on other sites More sharing options...
Pacco Posted August 17, 2005 Share Posted August 17, 2005 Lee, the box does state that the 50W is power consumption rather than power output. Also, I don't think any head pressure will be a deciding factor in power usage (especially with very little head). However I am pretty sure if you measured it with something as accurate as what Markus used you would get a reading of 50W +/- a few percent. Cheers Pacco Link to comment Share on other sites More sharing options...
BlakeyBoyR Posted August 17, 2005 Share Posted August 17, 2005 I would be very surprised if Eheim's reputation for low running costs etc. had been based solely on what is written on the side of the box. Surely if Eheim were falsifying power consumption it would've been picked up by now? I mean, I think it's great someone's taken the time to test their pumps, but me personally, I doubt Eheim would put anything in writing that was either false or could not be backed up if they were challenged. You dont get the reputation Eheim have by making things up Link to comment Share on other sites More sharing options...
CThompson Posted August 17, 2005 Author Share Posted August 17, 2005 Craig, are you sure you work as a Science guy and not in the NSW parliament? You seem unnaturally adept at dodging questions ← I work at a scientific organisation, I am not a scientist. Sorry, what question have I dodged? I don’t have much time to do this stuff on the internet, and I should be working right now. If I’ve missed a question, sorry, but please don’t read anything into it . Craig Link to comment Share on other sites More sharing options...
Ducksta Posted August 17, 2005 Share Posted August 17, 2005 Just having a poke Craig - but a few people did kind of ask if you were sure the meter was accurate at low amps (although the use of the word calibrate may be wrong, I don't know but my dad said to make sure the meter was calibrated and he talks the talk so who knows?) The other was whether you tested other appliances to get a control for the 'experiment'. Link to comment Share on other sites More sharing options...
Ash Posted August 17, 2005 Share Posted August 17, 2005 if both the inlet & outlet have the same water pressure - doesn't this negate head height - which is more about pressure differences? Link to comment Share on other sites More sharing options...
r.d.m Posted August 17, 2005 Share Posted August 17, 2005 craig , what if you black the house out with only your 2 pumps running, then get out by the meter with a stopwatch for maybe 30 mins, take a meter reading before and after,this will give you the consumption in watts,just a thought! Link to comment Share on other sites More sharing options...
CThompson Posted August 17, 2005 Author Share Posted August 17, 2005 Just having a poke Craig - but a few people did kind of ask if you were sure the meter was accurate at low amps (although the use of the word calibrate may be wrong, I don't know but my dad said to make sure the meter was calibrated and he talks the talk so who knows?) The other was whether you tested other appliances to get a control for the 'experiment'. ← No worries mate. I didn't answer those questions straight away as I was off finding out an answer... Link to comment Share on other sites More sharing options...
CThompson Posted August 17, 2005 Author Share Posted August 17, 2005 I have had a very comprehensive response to my query. There is a diagram that Gary has used to demonstrate in his answer but I am unable to copy and past this part in. I spoke to him on the phone as well, and I guess the upshot is that the tong meter is not the meter that can be used at these low wattage I asked him to respond via email, and told him I intended to copy and past it on ACE. I have removed his surname and email address, as I didn't ask for permission to put this on the post. He was a friendly courteous, and obviously intelligent man, who as you will see by his response, knows his stuff. Enjoy the answer Craig -----Original Message----- From: Gary Sent: Wednesday, 17 August 2005 3:08 PM To: THOMPSON, Craig Robert Subject: Power Consumption Query Hello Craig, Thank you for your enquiry regarding the power consumption of your Eheim 1060 pump. Your enquiry deals with an extremely complex area of electrical theory where both resistive and inductive loads exist and cannot be easily explained due to the electrical knowledge required but I will try to keep it as simple as possible. Firstly you need to know that you are not dealing with a purely resistive device (i.e. Heating Element) for which your formula would hold true P = E x I where P (Power) = Watts, E = Volts, I = Amperes however due to the fact that we are dealing with a reactive circuit (resistive and inductive) you are measuring nothing more than apparent power and therefore other factors need to be taken into account in this case “inductance”. Any electro-magnetic device which induces a magnetic field to another component (motor winding to a rotor, transformer – primary to secondary, Bobbin in an air pump to the magnetic swing arm) needs to have the power factor taken into account before we are able to measure the True or Real Power consumed. Power Factor is best described as - The fraction of power actually used by electrical equipment compared to the total apparent power supplied, this is usually expressed as a percentage or decimal. The power factor indicates how far a customer's electrical equipment causes the electrical current delivered at the customer's site to be out of phase with the supply voltage. This can be calculated as the COSINE of the angle of lead or lag of the current vs voltage curves. P = E x I x PF where P (Power) = Watts, E = Volts, I = Amperes & PF = Power Factor We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This "phantom power" is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts, symbolized by the capital letter P. The combination of reactive power and true power is called apparent power, and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA). As a rule, true power is a function of a circuit's dissipative elements, usually resistances ®. Reactive power is a function of a circuit's reactance (X). Apparent power is a function of a circuit's total impedance (Z). Since we're dealing with scalar quantities for power calculation, any complex starting quantities such as voltage, current, and impedance must be represented by their polar magnitudes, not by real or imaginary rectangular components. For instance, if I'm calculating true power from current and resistance, I must use the polar magnitude for current, and not merely the "real" or "imaginary" portion of the current. If I'm calculating apparent power from voltage and impedance, both of these formerly complex quantities must be reduced to their polar magnitudes for the scalar arithmetic. There are several power equations relating the three types of power to resistance, reactance, and impedance (all using scalar quantities) however this is not the scope of this discussion. These three types of power -- true, reactive, and apparent -- relate to one another in trigonometric form. We call this the power triangle: Using the laws of trigonometry, we can solve for the length of any side (amount of any type of power), given the lengths of the other two sides, or the length of one side and an angle. Power dissipated by a load and what your electrical supply company charges you for is referred to as true power. True power is symbolized by the letter P and is measured in the unit of Watts (W). Power merely absorbed and returned in load due to its reactive properties of the device is referred to as reactive power. Reactive power is symbolized by the letter Q and is measured in the unit of Volt-Amps-Reactive (VAR). Total power in a reactive AC circuit, both dissipated and absorbed/returned is referred to as apparent power. Apparent power is symbolized by the letter S and is measured in the unit of Volt-Amps (VA). These three types of power are trigonometrically related to one another. In a right triangle, P = adjacent length, Q = opposite length, and S = hypotenuse length. The opposite angle is equal to the circuit's impedance (Z) phase angle. The real simple answer is that Eheim quote all power consumption as True Power because this is what your electricity supply company charges you for and it is what you as a consumer are generally interested in “how much will this equipment cost me to run over a given period”. You cannot simply use an ammeter or tong tester when dealing with reactive circuits as this will only indicate the apparent power consumed and is not accurate without knowing the Power Factor - which for your 1060 pump is about 0.5 (incidentally PF can vary considerably depending on type of circuit, technology used, quality of raw materials etc.). The only real way of measuring True Power consumption without taking into account inductive and capacitive loads within the circuit and without complex calculations is to use a watt/hour meter or a kw/h meter - the type that is used to measure true power you consume at your house. I hope this helps you better understand your dilemma and has helped to provide the answers that you seek. Best regards, Gary Link to comment Share on other sites More sharing options...
r.d.m Posted August 17, 2005 Share Posted August 17, 2005 Link to comment Share on other sites More sharing options...
Ash Posted August 17, 2005 Share Posted August 17, 2005 so with that power factor of 0.5 - does that mean that you should be multiplying your tong reading of 98W by 0.5 to get 49W? Link to comment Share on other sites More sharing options...
BlakeyBoyR Posted August 18, 2005 Share Posted August 18, 2005 Whoa, Eheim take complaints rather seriously don't they! Link to comment Share on other sites More sharing options...
CThompson Posted August 18, 2005 Author Share Posted August 18, 2005 Whoa, Eheim take complaints rather seriously don't they! ← I just see it as a fully professional response –and all credit to them for it. Craig Link to comment Share on other sites More sharing options...
Ash Posted August 18, 2005 Share Posted August 18, 2005 I guess that's a big difference - imagine asking the same question of some faceless no-name company. Link to comment Share on other sites More sharing options...
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