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power consumption by 30 hp motor

power consumption by 30 hp motor

power consumption by 30 hp motor

(OP)
we have 8 centrifugal pumps each of 30 hp. i am interested to know that if all pumps are operated simultaniously then how to calculate power consumed by them in one hour and number of energy units generated in that hour?

RE: power consumption by 30 hp motor

I assume that measuring the power to the motor driving these pumps is what you want to do. Do you want to measure VA's, VAR's, or watts? For watts, you simply need to measure the voltage(line to neutral for Y connected) and current (true rms) going to one leg of the motor as well as the p.f. For VA's or VAR's you need to measure the power factor as well.
P(watts)=VIcos(phase angle)  Ptotal=3P
reactive power(Q)=VI sin(phase angle)   Qtotal=3Q
complex power(S)(remember vectors here)=P+jQ  Stotal=3S

These numbers all use line to neutral voltages and line currents. I am not sure what you mean by 'energy units generated'.

RE: power consumption by 30 hp motor

buzzp's formulas work OK as long is there is no wave disortion...like there may be with VFDs.

RE: power consumption by 30 hp motor

(OP)
buzzp, by energy units generated i mean that how much i have to pay to electric supply company if these pumps are operated for 1 hour simultaniously keeping in mind that  tarrif rate is 0.2 cents per unit generated by operating these pumps.

RE: power consumption by 30 hp motor

Does the utility monitor your power factor? This is usually the case in commercial or industrial facilities. I do not know much more than that about the billing. If they are concerned about power factor then you will have to use the complex power formula above(better confirm if they bill for VA or VARs on your bill as I am not sure which they would use, Im guessing complex power).

Busbar, notice the true rms in parenthesis. This is how the utility will calculate power consumed. Is there something else that should be included in this calculation for billing if drives are used? I think you are worried about IEEE 519 requirements for harmonic distortion placed back on the utility. I know companies can get fined if this is found in excess of the recommendations in 519, on some grids.

RE: power consumption by 30 hp motor

10-4.  Did you know you can post a link to your faq?  FAQ237-221

RE: power consumption by 30 hp motor

Buzzp,

I never heard of electricty bills based on the true RMS value of the current. The customer is only billed for the integral of sum of V(t)*I(t) on all three phases and there might be an additional charge if fundamental power factor is poor.

RE: power consumption by 30 hp motor

electricuwe, in Canadian utilities, apparently it's common to bill for kilowatthours BUT base the demand portion on kilovoltamperes.  I guess that 'automatically' covers a number of concerns about distortion and power-factor numbers.  It seems that kVA-demand billing provides a very equitable tariff, nowadays being fairly easy to define and measure, and based simply on rms voltage and current.  

RE: power consumption by 30 hp motor

Some clarification of my statements:
1. Use true rms values for all power measurements, this
   includes true rms voltage and true rms current. True rms
   measurements will take into account the power used by a
   drive as well, since the waveform will be distorted. I
   should say that I am not 100% sure on this but if they
   billed on an averaging rms scheme then they would be
   ripping people off because peak voltages on a distorted
   power grid can be significant. Since most of these
   devices look at the peak voltage and then use a
   little math (multiply by 0.707) to come up with the rms
   value.
2. Some utilities charge commercial or industrial
   facilities power factor penalties and some charge fines
   for harmonic distortion above and beyond IEEE 519
   recommendations for harmonic content. Some also charge
   for VA's consumes as well. Mind you, this would never
   be the metering system they use for residential use. At
   least not yet.
Thanks for the blurb about the link to my faq busbar. I will take some time to figure that out later.

RE: power consumption by 30 hp motor

busbar and buzzp,

is the billing based on the fundamental current (taking in acount phase shift) or on the true RMS current (taking in account phase shift and harmonic currents)?

I work in the field of high power converters and we export equipment to all parts of the world. Of cause there are restictions on harmonics like IEEE519 but I never heard of electricity bills taking in account harmonics or being based on true RMS currents.

psozot hasn't mentioned that his pumps operate on VSDs, so for his answering his question our discussion will not be very useful since in this case there will be only negliable harmonic currents. I'd like to help him with his problem but unfortunately I do not understand what's meant by energy units ? For kWh 0,2 cents seems to be very low.

RE: power consumption by 30 hp motor

I am not sure what his units are either. Perhaps busbar can elaborate on the true rms measuring for utility billing. All I know is there are utility meters available that measure rms and true rms by way of VA metering. I can't tell you what regions or anything that use what methods. This is an interesting topic and I plan to check it out further. I would prefer to have my house billed for true rms rather than rms for the reason mentioned in a previous post. The harmonic distortion billing is levied in the way of fines. I have not heard of anyone being billed on any unit that could be used to quantify harmonic distortion.

RE: power consumption by 30 hp motor

A far as a “true RMS” power definition—voltamperes are the product of RMS current and RMS voltage measured over at least one full AC electrical cycle.  Regardless of the shape of the voltage and current waveforms, their ‘DC thermal equivalent’ can be accurately measured and yield precise apparent power, and, time integrated, apparent energy.  The ability to slice up AC waves accurately, fast enough and in real time (much less in places where meters get installed) has only been possible about the last 15 years.  Once a periodic wave is reduced to a stream of bits, calculating various quantities becomes duck soup.  

Watts are the product of instantaneous voltage and instantaneous current, summed through many slices or fractions of of a cycle.  These definitions are inherently wide-band and not frequency dependent.  There is not any significant disagreement on how to measure real and apparent quantities.  Note the term ‘RMS watts’ is contradictory and inappropriate.

Search on NRC Canada, power measurement and authors Arseneau and Filipski.  

However reactive power is defined or measured, the Pythagorean relationship for apparent, real and reactive power does not necessarily hold.  What to call the “leftover” power is a subject of wide debate, and has tended to be conveniently defined based on the capabilities of various meter designs.  

Reactive power and distortion components get screwed up, because of the 90° phase shift needed for measurement (¼ of an electrical cycle) can’t be done for two or more simultaneous frequencies; making the fundamental definition of reactive power frequency dependent.  

In the late 1980s, Arseneau and Filipski captured nasty in-the-field [like arc furnaces] voltage and current waveforms [up to 47% THD current] and extensively tested electromechanical and solid-state meters to observe what distortion’s effect had on registration.  

Voltampere demand measures everything—as long as it is periodic—it doesn’t matter how awful the waveform looks.  So right now it is the fairest way to quantify power and energy at a given point; i.e., at the ‘meter,’ which is usually also the ‘point of common coupling.’  With all the dsp stuff, meter users, standards folks and manufacturers will hammer out new algorithms that more closely represent an accurate and fair way to buy and sell electricity.  A big step has been IEEE 1459-2000 DESCRIPTION at http://shop.ieee.org/store/product.asp?prodno=SS94823
 

RE: power consumption by 30 hp motor

Per a utility guy, utility meters pick up all power going through, regardless of frequency (i.e. harmonics), and naturally correct for power factor, to give kW.

a 30 Hp pump, if loaded to 65%, will draw 14.54kW, 8 pumps will draw 116.37kW, and in one hour that is 116 kWH.

The electric demand, kW, will also be 116.

The bill will normally be comprised of a cost related to the energy (kWH) plus  a cost equal to the demand (kW) usually measured by the peak demand in any given 15 minute interval in a particular month.  So don't run your pumps at the same time your AC system is peaking out, it will kill your bills.

PacificSteve

RE: power consumption by 30 hp motor

I understand and agree with all of your statements busbar. I will search for that doc you suggested. It will be interesting to see how they measured the peaks. Thanks for the post.

Another note: I think we are talking about peak demand metering and standard billing. Usually peak demand is not used in residential zones, I wouldn't imagine.

RE: power consumption by 30 hp motor

I agree with PacificSteve, but it should be considered that usually the power consumption transfered via the harmonics is negliable:

5% of fifth harmonic voltage with 20% of fifth harmonic current will give 1% of power transfered by the fifth harmonic. Only voltage and current of the same harmonic order will result in power if integrated over a period of the fundamental frequency. You will see this if you write down the formula given in one of my previous posts for voltages an currents containing harmonics. You will need some sheets of paper but then you will see that all products containing factors with different frequencies will integrate to zero over a fundamental period.

Regarding the term peak demand you should consider that for a utility a peak is a high demand at least for several minutes. For someone working in electronics a peak is something lasting only several microseconds. You shouldn't mix up the things.

RE: power consumption by 30 hp motor

My point is that kVA-demand metering [common in Canada] by default takes care of both real and reactive components, and doesn’t care about power factor or distortion components; provided the (analog-to-digital converter’s) sampling rate is high enough.  Effectively, this makes the user (in an indirectly stated manner) responsible for his own low (or any) power factor and nonsinusoidal power flows.  

Voltamperehour energy and voltampere demand based on Vrms•Irms [as opposed to VA determined by square-root-of-the-sum-of-squares of real and reactive readings] has been proposed since ~1992 by Raymond H Stevens, a sort of modern guru in electricity metering.  His focus has been simplification of rate structures that will inherently incorporate the consequences of poor power factor and distortion.  [Nowadays, problems usually have to cause some sort of complaint, leading to field 'spot checks' and a rather subjective negotiation process.]  It isn’t a trivial task.  For instance, apparent power is defined as a directionless quantity, so how do you resolve bidirectional metering that is getting a lot more popular with upstart IPPs, QFs and ISOs?  

Because of the limits of applicability for reactive flows, 4-quadrant metering becomes limited in tracking true system loading and the accurate limits of power delivery.  In the simplest sense, it is up to the utility client if it wants to improve the efficiency of its power use; its ‘reward’ is lower kVA demand and cheaper utility bills.  Efforts taken by the user to operate his equipment at higher PF and more sinusoidally will more readily pay off.  The buzzword of power quality will be more directly and increasingly reflected in monthly transactions.  The lowly watthour alone is no longer capable of reflecting the true cost of power delivery with all but the most passive electrical loads.  But, the concepts of differing means of accounting are slow to be accepted where power exchanges are being resolved to fractions of a mil.  

Residential rates may not have demand charges, but in larger homes where 208Y service is needed, it's typically billed at a commercial rate, so there must be some homes that have demand meters.   As for 480Y, I can’t be sure.   An associate in the HVAC/air-balance trade has told me of homes in Florida with 60-ton chillers cooling their private atriums… Air- conditioned parrots and monkeys—imagine that!
 
I’m sure that it’s possible that a few residences even have medium-voltage service.  One of the telecom newsgroups mentioned a residence with a 5ESS [central-office class] switch…

Back to the original thread, I guess that this really only addresses two of his eight 30hp motors {assuming 1 ton ≈ 1 hp}.
  

RE: power consumption by 30 hp motor

Suggestion marked by ///\\\to PacificSteve (Mechanical) Apr 12, 2002
Per a utility guy, utility meters pick up all power going through, regardless of frequency (i.e. harmonics), and naturally correct for power factor, to give kW.
a 30 Hp pump, if loaded to 65%, will draw 14.54kW,
///What about efficiency of the motor? E.g. EFF=0.9 or so?\\\
 8 pumps will draw 116.37kW, and in one hour that is 116 kWH.
///Check the efficiency and possibly jack up this figure.\\\
The electric demand, kW, will also be 116.
///It is better to go by the nameplate data and verify them by measurements, to be on the accurate side.\\\

RE: power consumption by 30 hp motor

   Sin instrumentos de medida adecuados no es posible conocer con precisión el consumo en KW de las bombas.
   Teniendo una simple tenaza amperimétrica es posible aproximarse con la siguiente formula KW=(1,732*V*A*cos)/Rend. Si la intensidad está entre el 75 y el 100% de la nominal, se puede tomar cos=0,88 y el Rend.=0,9.
   Puede hacerse otro cálculo aproximado partiendo de los datos de placa del motor de los que se obtiene el consumo a plena carga. Por la intensidad del motor en amperios determinar la potencia consumida, considerando que esta varia como un arco de circulo entre la corriente de vacío y la de plena carga.

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