I am doing a study for a customer r 3

[PUMPDESIGNER]

I am doing a study for a customer regarding their pump stations. One problem became obvious from a data logger we installed to monitor the equipment.
The pumps start and stop too frequently, about 800 times per week.
All the motors involved are 3500 rpm, AC Induction motors, 5-30 horsepower, running on three phase 240 and 480 volts.

Does anyone have general information or references regarding the cost of starting a motor?

What I am looking for is something like I found for flourescent bulbs some years ago. That information stated that the cost of starting a flourescent bulb was equal to about 15 minutes of running that same bulb.

Greatly appreciated if someone has info to help. Richard Neff
Irrigation Craft

 

[acmotorengineer]

I thought that study was effective life rather than effective power???

In any case, the starting power of the three phase AC motor is not as bad as you might think - due to the low power factor during starting. Making a lot of assumptions on your 30HP motor:

Running: 24.1 kW per unit time
Starting: 37.4 kW per unit time

Note: this is power you have to pay for; not the total (apparant) power the power company must supply.

lets say those 800 starts are evenly spaced: 1 start per 756 seconds. Figure a start is 3 seconds. For one cycle:

running = 24.1 x 763 = 18127 kW*sec real power used
starting = 37.4 x 3 = 112 kW*sec real power used

In terms of effective life: lets say that the average motor will last for 80,000 starts, full voltage. This gives an effective life of 100 weeks at 800 starts per week. Lets say a motor would give 10 years in normal service ( one start / stop per 8 hour day)

The effective life would be reduced 53 minutes per start when started at the rate of 800 starts per week: for full voltage starts of a low inertia pump, equally spaced - if I did the math right.

I'm curious what kind of life this customer is actually getting. Is this information availiable? And what type of failures they were getting?

 

[electricpete]

The last post was a good answer.

I agree the cost of power is not likely significant, but possible reduction in motor life is.

NEMA MG-10 (yes 10, not 1) Table 7 gives allowable number of starts per hour for NEMA Design B general purpose induction motor.

For 30hp 2-pole motor, the starts per hour should be less than either:
4.1
OR
31 divided by WK^2 (in lb-ft^2)

There is also a statement that for duty at the upper end of these ranges some reduction in life expectancy must be accepted. As usual there is no definition of standard life expectancy or anything like that.

One other thing to look at in your recording is whether the motor meets minimum time between starts. Per the table that is 120 seconds for the 30hp 2-pole. Starts which are too close together don't allow the motor to cool between starts and this is severe short-term abuse... regardless if the long-term average meets the guidelines.

The table numbers for the smaller motors are a little less stringent than the 30hp.

 

[PUMPDESIGNER]

I would love to get you more information, I owe you.
We cannot obtain long term history that would be very useful I am sure. Here are the facts we know for sure.

Irrigation pump station with twin 20 hp, close coupled, end suction, centrifugal pumps.
Motors are 20 hp,3500 rpm, 240 volt 3 phase (Open Delta supply, yuck!).

Station was installed about 4 years ago.

We know that the two motors were replaced within the last year, one just recently. We cannot find out if they were replaced before that, but we suspect so.

Faulty pump station controls are the problem, pumps cannot be maintained on when they should remain on. Controls cannot properly sense flow rates, so the pumps must cycle on and off under certain flow rates. Under other flow rates the pumps behave properly. Richard Neff
Irrigation Craft

 

[d23]

All:

Your statement "the cost of power is not likely significant" seems true. If the power company has a demand factor involved do the power cost of the starts go up? Is a demand factor based on KVA or kW? It seems that the pumps are starting every hour, does that mean the customer is paying a higher price per kW? I'm not trying to challenge your answers I'm just asking questions that I don't know the answers for.

Richard:

I have never been around irrigation systems. I have to believe that operation cost are a lot more important in irrigation than the petroleum industry. Would a VFD or circulation line with a back pressure regulator be considered too high of an expense with an irrigation system?

In the petroleum indusrty using some form of SCADA is important to establish trend lines. Based on trends customers can schedule repairs rather than react to catastrophic failures. Do your customers in irrigation use some form of SCADA?

David

 

[jbartos]

Suggestion: In addition to the motor starting cost, the shaft pump performance effect is also important since it may take longer than the motor start time to have the pump effectively delivering fluid.
Also, it is better to analyze the motor starting in terms of energy transient. The portion of the energy transient at the motor start, which is above the average motor energy consumption when the motor is running, is the actual motor energy waste due to the motor start. The amount of the energy wasted would also be dependent on the motor starting arrangement that is not addressed in the original posting.

 

[PUMPDESIGNER]

d23
I have concern about demand factor also.
Any quality controls would be better than what they have now.
And you are correct, money is very important, but then that is what got them into this mess anyway, low bid without engineering supervision (design build project).
SCADA is not common, but it is used (we provide it).
However this project has none. This project is obviously in a crisis mode, which is why they are investigating.

harryg,
That would help except that the pumps need to be pumping water. The controls cannot operate the pumps smoothly in response to highly variable flow rates. The pumps are turned off and then started again in differing sequences attempting to match the flow rates. The controls cannot possibly be adjusted to achieve this.

jbartos,
You are correct about the starting time. This does indeed add to the problem. However this issue clears up when the controls are replaced (they cannot be corrected, insufficient and incorrect controls).
That energy wasted is what I was originally looking for with my question. I did not receive any information from anyone telling me that the starting costs were such and such in regards to the electrical bill. I would like to, but at this time cannot make a statement such as: "Each time the motor starts the electrical bill is increased by about 10 minutes of running time." Richard Neff
Irrigation Craft

 

[jbartos]

Suggestion: To be accurate, since money is involved, one must obtain the motor starting current signature, and then integrate the portion of the starting current above the steady state motor running current. Then,
kWhr=sqrt3 x motor terminal voltage x integrated current with respect to time in hours
then, multiply kWhr x $/kWhr=loss/motor start
then, number of motor starts (i.e. 800) x loss/motor start = total loss in $ per 800 starts.

 

[electricpete]

suggestion to the previous posting... I think that power factor must be considered along with voltage and current profile to determine kw and kw-hr.

PUMPDESIGNER - I think that acmotorengineer did give you a good estimate of the energy consumption.

Another approximate formulation is that power factor averages perhaps 0.2 (from IEEE red book) during the starting period and 0.8 during run. Combine this with the fact that current goes up by factor of approx 6 (compared to FLC). So starting power/full-load power is approx (0.2/0.8)*6/1 ~ 1.5 times full load power. If it takes 10 seconds to start the motor then you have used up 15 seconds worth of pump-running power. (some of that energy goes to the fluid and some goes to loss). This ratio 1.5 is very close to ac motor engineer's 38kw/24kw.

As was mentioned, the utility billing structure may end up charging you a little more than just the cost of the power itslef if peak demand or reactive power are factored into your billing.

 

[jbartos]

Suggestion: It is obvious that the kW = sqrt3 x voltage x current x power factor. One would have to measure and integrate the starting current over a shunt resistor voltage in order to account for the varying power factor and starting current. This will result in the resistive component of the starting current, Ires, rather than the total starting current, Itotal, e.g. Ires=sqrt(Itotal**2 - Iind**2).

 

[PUMPDESIGNER]

electricpete,
You are correct. acmotorengineer did in fact give me a good approximation to use. Richard Neff
Irrigation Craft

 

[jraef]

PUMPDESIGNER,
Not being big on math, I'll not dispute the validity of the posts outlining details of how to calculate any potential cost differences, but I hope you get the gist of the messages here: that being the unlikely relative significance. I also believe that the information you refered to about fluorescent lights is just an urban legend as well. Power is power no matter how you attain or consume it. Light fixtures use power when they are on, and use none when off. The brief surge that occurred when magnetizing the transformer on old syle ballasts was insignificant in the overall power consumption equation. New fixtures have electronic ballasts so even that issue has gone away. The more likly issue at hand is OPERATING COSTS, which include maintenance and downtime. Your system undoubtedly has high operating costs becuase of these issues. Unfortunately they are more difficult to prove to the "bean counters" than the simple if-then math equation you are looking for. Without historical data you will have a difficult task.

A low cost suggestion would be to replace the motor starters with soft starters. Not only would they reduce the damaging mechanical effects of the rapid cycling, but there are several manufacturers who provide starts-per-hour and minimum-time-between-starts lockout features built in. With these features, even if the control system was erroneously calling for the pumps, they would not turn on until the time values were satisfied. In addition, the soft starters will reduce starting torque stress, provide the ability to soft-stop your pumps and mitigate water hammer, plus reduce starting current surges which may translate to utility charges IF they hit you for instantaneous peak demand chanrges (although most do not). VF Drives would be the ultimate solution, but will cost more and would require a significant change in the overall control system. Soft Starters can be used without changing the existing controls.

Good luck. Quando Omni Flunkus Moritati

 

[d23]

All:

I’m not sure that I understand demand factors. If you have the following conditions:

460V
39.5A
30 HP
0.81 PF
0.89 Eff
3500 RPM
Power cost per kW $0.10

I believe that demand factor is based on starts in a given time per 15 minutes, per hour etc. As a charge for demand factor they make you pay to improve PF to a higher number like 0.90.

With the motor above if you have enough starts to have to pay the adjusted fee all the time your expected power cost would be:

Actual kW for motor with 0.81 PF = 72.6
Adjusted kW assuming 0.90 PF = 80.6

This is 8 kW per hour. If you are charged $0.10 per kW hour then in a 30 day month the adjusted power cost is $576.00 more than it could be to operate one 30 HP motor. Over $550 per month is a lot of money where only a 30 HP motor is involved.

I’m posing this as a question, because I'm not sure I understand how demand charges work. Could someone validate or correct me on this? This may also offer PUMPDESIGNER a better way to justify his recommended changes.

Thanks
David

 

[busbar]

A number of utilities publish their tariffs/rate structures on line. There are typically several components that affect a monthly bill. Energy in kilowatthours [the listed $0.10 per kWhr] is one cost. Kilowatthour costs may not be ‘flat’ but ratchet up or down based on monthly energy totals. Also, kilowatt {or kilovoltampere} demand based on an interval of ~15-30 minutes—as kW or kVA peak—may be ~$1-5 per kW or kVA.] Additionally, there may be some form of power-factor penalty calculated from peak demand, or energy [effectively an average] for the month. Each of these three components can significantly affect electricity cost. There may also be different rates for seasons or time of day.

It is possible that the utility’s commercial department may offer spreadsheet templates to assist you in evaluating electricity costs. If not, “rolling your own” spreadsheet based on their tariff descriptions Is usually not insurmountable.

 

[PUMPDESIGNER]

You are exactly correct electricpete. acmotorengineer did in fact give me a very good simple way to approximate the starting costs for electricity.
Richard Neff
Irrigation Craft

 

[PUMPDESIGNER]

Sorry guys about the extra post, forgot to refresh my browser.

I am obviously not an electrical motor guy. But concerning power costs I do remember this graphic example.

My father operated a large truck fleet repair facility for U-Haul. Their power costs were way out of line. They called the power company who visited the site and told them what to do.

They normally opened the shop at 8:00 am. The door opened, and everyone went into the shop and turned on everything, sort of all at once.

The power company told them to start all the motors seperately about 5-15 minutes apart (I forget how far apart).

Just that one simple thing reduced the power bill so much that my father talked about it for years. Richard Neff
Irrigation Craft

 

[d23]

busbar

Thanks for the info. I tried a few of the area power company web sites and for some reason they are not bragging about their demand factor. I can get it from them on a as need basis.

David

 

[highleg]

We have Truss Plate presses here in our plant that run constantly. I am trying to determine the total cost effectiveness of starting and stopping these presses versus leaving them on for 16 hours a day, 5 days a week. There are 2 480 volt 3 hp motors per press. I want to add soft starts to negate the mechanical wear and tear induced from a start stop mode. And add timers so the operators can not leave them on. They press a truss at a rate of one every 15 minutes on average. If someone can give me some formulas, I will do the math. Im looking for energy consumption savings.

 

[jraef]

highleg,
You are probably better served starting a new post rather than tagging on to an old one, even though I recognize the relevance. You can reference this old post by copying and pasting the "thread number" at the top, but most of the time when you tag onto an old post like this you will be missed because it takes too long to read through all of the previous stuff.

"Venditori de oleum-vipera non vigere excordis populi"