What happens to the extra power when specifying a motor for a pump 3

[ahika]

Hi there,

This is my first post so hopefully you're able to answer this.

A pump supplier recently specified a pump for a duty of 105 l/s and 60m Head- Abs pump power 72.5 kW, NOL Power 75.3 kW. All good.
The next step is to match a motor to this pump and the closet matching they have available is 90 kW. VSDs are not being used so motor is operating at 50Hz

I have uploaded the pump data sheet

My question is this:
The pump duty 72.5 kW and a 90 kW motor is being utilised.

May be a stupid question but what happens to the extra power being supplied?

Is it being dissapated as heat? Should I try and shop around to find a motor that would be closer matching?

From an energy efficiency perspective I am effective losing 19% (90kW-72.5kW)/90kW by over specifying the motor.

What is your experience with specifying the motor?

Your thoughts appreciated.

 

[TD2K]

The output of a motor isn't fixed. If you have a 90 kW motor and you need 72.5 kW, that is what it will produce. The efficiency of the motor is affected by load, they are less efficient at low loads so you don't want to oversize them. There is also the issue of slip and how close you operate to the rated speed and that's a function of the load on the motor. The motor will even produce more than 90 kW if needed, subject to its potentially burning out windings, tripping the overload protection, etc.

Likely one of the electrical folks is screaming right now at my explanation.

 

[Artisi]

TD2K: a reasonable explanation without getting too deep into the tech. side of things, however, I would think that efficiency would only change 1 or 2% for the operation of 72.5 / 90 but as with all pump installations who knows where it will operate in its real life application - the theory / curve is only a starting point. I wish that every pump I ever selected operated right on the "duty' given by the customer.

ahika: A reasonable pump selection, a little right of BEP and wouldn't like to see it run out on the curve much further but the old TKL hydro Titan range is pretty solid pump. Think I would go back to the supplier and discuss the application a bit further - especially if the pump is likely to operate at a higher flowrate / head.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[ahika]

Thanks for that Artisi and TD2K

I think I will take your advice and go back to them to discuss further.

 

[KllrWolf]

The motor will provide (or try to) the exact amount of power required by the load. Efficiency and Power Factor will change, but the motor output will be what the load demands, provided you are not overloading the motor and burn it up.

 

[ahika]

Thanks Kllrwolf,

This is where my electrical knowledge is lacking so trying to join the dots to get a better understanding.

Isnt a motor rated at a certain power based on the frequency? So isnt a 90kW motor delivering 90kW at 50Hz? A VFD can increase/decrease the power accordingly by altering the frequency which makes sense and I understand. This application doesnt have a VFD so power is supplied at 50Hz.

Next question:

If I was to connect a power meter to this load after commissioning would it read 72.5kW or 90kW?

 

[stanier]

The following websites provide a wealth of infornmation on pumps and their selections.
To these you can add the websites of manufacturers such as Warman, KSB, Sulzer, ITT Flygt, Toroshima, Grundfos etc etc

http://www.mcnallyinstitute.com

http://www.pump-zone.com

http://www.pumpfundamentals.com

http://www.pumpsystemsmatter.org

If you were a memeber of ASME or IMechE you would get access to their virtual libraries and there is an enormous amount of material in the books therein.

“The beautiful thing about learning is that no one can take it away from you.”
---B.B. King

http://waterhammer.hopout.com.au/

 

[bimr]

"May be a stupid question but what happens to the extra power being supplied?"

Nothing, if you forget about electric motor efficiency. The electrical motor output will simply match the pump load.

If you have over estimated the system head loss, then your pump may run out on the pump curve and exceed the power available from a 75 kw motor. So it looks like a reasonable motor selection.

Rgarding motot efficiency. Electric motors operate at their best power factor and efficiency when fully loaded, so you do not want to purchase a motor that is too big, and common sense dictates that one that is too small is even worse

http://www.mcnallyinstitute.com/06-html/6-04.html

Most industrial motors are most efficient when running from 65% to 100% of the rated power. The maximum efficiency is normally 75% of the load. Most motors tend to dramatically lose efficiency at loads below 50%. Power factor also deteriorates as loads decrease. Motors are
considered under loaded when running below 65% load. On the other hand, motors are considered overloaded when running for long periods of time above their rated power. Overloaded motors will overheat and lose efficiency.

http://www.ciras.iastate.edu/publications/EnergyBP-ChemicalIndustry/Sourcebook_Chapter7.pdf

Since your electric motor is operating above 65% of the rated load, you don't have to worry about the electical motor efficiency either.

 

[Oldhydroman]

Could this be what has happened?

You want a flow of 105 L/s and you've worked out your worst case head and rounded it up to 60.000 metres. Then you sent these numbers through to your chosen pump supplier and they've sorted out for you a pump which can deliver your 105 L/s. In practice, the pump they've chosen will deliver the flow you want against a 60.5 m head and, if these were your actual conditions, you would need a motor output power of 72.5 kW.

Your required head is a little less than 60.5 m, and, chances are it will actually be less than the 60.000 metres you asked for. So the pump flow will be a little more and, as a consequence, the power requirement will also be a little more (the characteristics of the pump type mean that the proportion of flow increase is greater than the proportion of head decrease so the overall power requirement increases when the head decreases). And, for that particular pump model, with that particular diameter impeller, with this particular fluid, and running at a nominal 3000 rpm (a 2 pole induction motor on a 50 Hz supply), the worst case combination of head and resultant flow means that you will never need more than 75.3 kW; this is the NOL Power. Your pump supplier’s simplistic selection program has picked the next size [standard] motor power – hence they have chosen for you a 90 kW motor.

Unless you are buying a gazillion motors each month from a single manufacturer, you’re pretty much stuck with picking your motor from a restricted menu of standard sizes. Generally speaking, the bigger the motors are, the bigger the steps between each size: 22 kW, 30 kW, 37 kW, 45 kW, 55 kW, 75 kW, 90 kW, 110 kW etc. It is SO annoying when the power you need is just a tiny bit more than one of the standard sizes and you’re forced to choose the next size up. As well as the increase in initial cost [and size envelope and bell housing size and shaft coupling size and the weight of all three bits] you also pay for it for evermore because the larger motor will be running at significantly less than its “rated” maximum power so will have a slightly poorer efficiency and power factor than a motor running much closer to its rated power.

What can you do? If it were a positive displacement pump (more my specialist area) then I would look again at the defined requirements. A 60 metre head is a lovely round number; is it actually something more like 58.123 metres and everybody that’s been involved in the selection process has added a bit on? With a positive displacement pump any reduction in head would cause a similar reduction in power. But that’s not your situation, and if the real head is less than 60.5 metres then the required power will be more than 72.5 kW (but never more than 75.3 kW).

You could fit the 75 kW electric motor and then apply some additional throttling to increase the apparent head (and hence reduce the actual flow) in order to bring the absorbed power to no more than 75 kW, i.e., increase the throttling until the motor current is no more than the full load current figure given on the motor label. It seems a bit wasteful but the power lost across the throttle would probably be less than that lost in the electronic circuits of a variable speed drive unit (the other common technique for reducing the flow to the required value). And the energy savings, if any at all, would be so minor that the VSD would never pay for itself.

What would I do? I would cheat! (Or at least use some basic engineering arguments to skew things so that they worked in my favour.)

Ask yourself what the motor manufacturer means when describing a motor as “75 kW”. They mean that when you put enough load on the shaft to cause the mathematical product of torque and speed to equal 75 kW then the motor will draw a certain amount of current – and the [mainly electrical] losses in the motor are such that the amount of waste heat generated when the shaft power is 75 kW can be completely dissipated by the air flow coming from the fan at the back of the motor. For a flow of air to be able to cool the motor it should be obvious that the motor must be hotter than the ambient air temperature. The greater the temperature differential, the greater will be the heat dissipation. The “rated power” of the motor is the power at which the cooling requirement can be satisfied without the windings getting so hot that it starts to impact badly on the motor’s service life. If you massively overheat the motor then the windings will burn out very quickly. If you run it just a bit hotter than you’re supposed to then the winding insulation will not last as long as it otherwise would.

The most common [cheapest] way of protecting the motor from overload (or, more correctly, over-temperature) is the simple thermal overload relay. Tiny heating elements in the overload relay are connected in series with the motor windings – this means they will have the same current as the motor. As the motor load increases, the motor current will increase and the heating effect inside the overload relay will increase with it. A bimetallic strip close to the heaters will start to bend when the heating effect becomes more severe, and, when the strip has bent enough, it will trip a set of contacts wired into the motor starter circuit and this will stop the motor. You can trim the setting of the overload relay so that it trips at the right current – but it’s all a bit approximate: the simple relay is trying to model the thermal inertia of the motor (the time it takes to trip depends on the extent of over-current) and the relay makes no allowance for the initial state of the motor when the overload started (you can overload a cold motor more and for longer than one that was already at its steady state running temperature). TD2K is spot on in saying that the overloaded motor will burn out or the overload protection will trip but these two events are essentially the same thing. An overloaded motor will draw more current than it can cope with and will eventually get so hot that the windings will burn out. So you install a device to detect this over-current and ensure it trips before you actually damage the motor.

Yes, yes, yes – but where is this diatribe going and how does it help you? The “rated power” of the motor depends on how much cooling is available and on how hot you can allow the windings to get. Most motor manufacturers ask you to de-rate the motors a little if the ambient temperature is higher than 40 deg C, you also have to de-rate if the altitude is more than 1000 metres. The manufacturers don’t usually offer a facility to run at a higher power for lower ambient temperatures or for lesser altitudes but that doesn't mean the cooling effect of the colder and denser air flow isn't better (it is). What it actually means is that the manufacturers just want you to buy a bigger motor than you really need (tongue in cheek here).

Are you, perhaps, running at sea level with a probable maximum ambient temperature of 30 deg C? See how much your motor manufacturer wants you to de-rate the motors and work the numbers backwards: you typically de-rate by 2% for operation above 1000 meters and by another 2% for each additional 330 metres. It’s harder to extract the ambient temperature correction from manufacturers’ literature because the relationship between ambient temperature and winding temperature is non-linear, but for a 30 deg C ambient instead of 40 deg C you could reasonably expect sufficient increase in cooling effectiveness for you to up-rate the motor by another 5%. After all, you’re only looking for an extra 0.3 kW from your “75 kW” motor (that’s just an extra 0.4%).

This is how you can make a 75 kW motor work for you and not void the motor manufacturer’s warranty. Ditch the simple thermal overload relay and use a system which actually measures the motor winding temperature. A 75 kW motor will almost certainly have a PTC thermistor buried in each winding. A thermistor is a semiconductor device whose resistance varies massively with temperature. The PTC [Positive Temperature Coefficient] thermistors put in the motor windings have a characteristic such that their resistance suddenly goes high just as the windings reach the critical temperature which you want to avoid. The three thermistors are wired in series and can be connected to a special “thermistor relay”, the contacts of which will interrupt the starter circuit and stop the motor if the winding temperature reaches the critical shutdown temperature.

Now do a few simple things to ensure the motor’s existing cooling efficiency is maximised: paint the motor black, don’t let the sun shine on it and don’t let the cooling air re-circulate back into the motor fan inlet (keep the local ambient temperature down). I’m sure you will find that your “75 kW” motor will be perfectly happy at 75.3 kW and, if it isn't, you will have protected it adequately from being damaged by using the thermistors instead of the simple thermal overload relay.

But what if the motor does overheat? There is an aftermarket product aimed principally at the variable speed drive market to allow the motors to run at a very low speed. You take the shaft driven fan off the standard motor and fit a new cowling which contains a small electric motor driving a new fan. These forced ventilation units give you the opportunity to increase the cooling of the motor so that you can up-rate it a little: for example, you could fit the fan for a 90 kW motor onto your 75 kW motor (they will both be a 280 frame size). In fact, just using a separate cooling fan might give you that little bit of extra shaft power anyway because the effort needed to drive the fan will now be taken from a separate motor and the equivalent amount of torque from the main shaft can be re-directed into driving your pump.

If you decide to stick with the 75 kW motor then it might be a good idea to ask for the motor to be supplied with some form of temperature detector in the windings. A thermocouple or RTD in the windings can be connected to a simple display to let you know how the motor is coping. The motor supplier can tell you what temperature constitutes an “alarm” and what should be considered the shut-down temperature (these values depend on the class of insulation in your motor). Even if you feel you probably won’t need to do this it is always much easier to get the motor with the detector built in from new rather than trying to retrofit one to an existing motor. If your display were a little more sophisticated and had built in relays then you could use this device in place of the thermistor relay - but be careful to make it fail-safe (relays have to be energised for the motor to be allowed to run, relays de-energise on detector failure etc.)

 

[Artisi]

Oldhydroman:

All sounds pretty good in theory, but I would think that the extra costs of 75 versus 90kW and factoring in the any efficiency or power factor changes wouldn't make your suggestion viable which would always be subject to disaster for any small change of operating conditions. However, I am with you on the over specified duty, after everyone involved in the project has added their own "safety" margin - "just in case" the final duty is nearly always over specified.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[Oldhydroman]

Yes, Artisi, you're right. It so happens that the cost differential between 75 kW and 90 kW is quite small because both motors are the same frame size. Had it been that the choice had to be made between 55 kW and 75 kW or between 90 kW and 110 kW then the change in motor frame size would make quite a substantial difference to the initial cost. There's likely to be little saving in whole life running costs. There may, however, be some on-going savings if this were a relatively small enterprise and the electricity standing charges were based on installed power and peak demand (the starting current of the larger motor will be greater).

But here's the question (and I'm learning here too because this type of pump isn't my forte): Is there ANYTHING that can happen in the application of this pump which can cause the absorbed power to exceed 75.3 kW (the published NOL power)?

If the specific design of the pump is such that it can never, ever demand more than 75.3 kW from its motor then I believe a motor rated at a "nominal" 75 kW would suffice as long as the altitude were significantly less than 1000 metres and the ambient temperature were significantly below 40 deg C. It may well be that this is not a continuous duty application (even more reason to choose the smaller motor).

Were it my project, the only concessions I would make to running a motor so close to it's actual maximum power output would be to paint it black and use a thermistor overload relay (with short circuit detection) rather than a simple thermal overload relay. That way I could protect the motor based on the actual temperature of the windings rather than a simplistic modelling of temperature based on a theoretical maximum current. And, on a 100-150 Amp three phase circuit, a [small] thermistor overload relay actually works out cheaper than a [big] thermal overload relay (although, admittedly, you do need some extra wiring to the electric motor terminal box).

DOL

 

[ahika]

Hi All,
Thanks for all your ideas and discussion. Very interesting indeed.

I think I will go back and question the motor size and try to utilise an energy efficient motor.

 

[Artisi]

Really no need to question the motor size, 90kW is the correct selection for that impeller trim.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[bimr]

Agree with Artisi. The motor size is appropriate. It may be more productive to double check the head loss calculations of the system.

You may consider spending the extra money to obtain a high efficiency motor.

 

[ahika]

OK, I will do that.

This is a great forum and look forward to offer my advice to you guys in the future!