Cost to run a 30HP motor, for an hour?

[dirk martin]

For the heck of it, I asked ChatGPT the following:

How much does a 30 horse power, 73 amp motor cost to run per hour at 220 volts?

It replied with this:To determine the cost of running a 30 horsepower, 73 amp motor per hour at 220 volts, we need to calculate the power consumption and then multiply it by the electricity rate.

First, let's convert the horsepower to watts.

1 horsepower (HP) is approximately equal to 746 watts.

So, the power consumption of a 30 horsepower motor would be:

Power (in watts) = 30 HP × 746 = 22,380 watts = 22.38 kilowatts (kW)

Now, we can calculate the cost per hour using the power consumption and the electricity rate. Assuming a hypothetical electricity rate of 12 cents per kilowatt-hour:

Cost per hour = Power (in kW) × Electricity Rate (in c/kWh)
Cost per hour = 22.38 kW × 12 c/kWh = 268.56 cents or $2.6856

Therefore, at 220 volts and assuming an electricity rate of 12 cents per kilowatt-hour, the cost of running a 30 horsepower, 73 amp motor per hour would be approximately $2.6856. Please note that actual costs may vary depending on the specific electricity rate in your location.

Now, I'm not an electrician, but $2.68 per hour just seems way too low.
Is that reply accurate?

[David Buchhauser]

This seems correct to me. Simple calculations to make. But just think, if you run that motor 8 hours/day, 5 days/week, then you're looking at around $100 per month added to your electric bill. I will say that I use a 30 hp cnc rotary phase converter in my shop which does see quite a bit of use. So there is actually a 30 hp motor running as part of the phase converter. My electric bills typically run in the neighborhood of $300 to $400 per month, so what you say does not surprise me.

David

[Tom Bender]

David

Just a small adjustment to your math:

$2.68per hour x 40 hours per week x 4 weeks per month = $469

And a closer look at the loading:

A motor or a phase converter will only consume electricity proportional to the loading, which is rarely anywhere near full load on woodworking machinery. Fans are the most consistently loaded (dust collector).

[Jason Roehl]

There are two things that stand out to me.

1. 73A seems low for a 30hp motor. I poked around online, and saw a few rated in that ballpark for a 230V, 3-phase motor, but 73A x 230V = 17kW. Saw some others that were rated around 100A, which seems more realistic. For reference, I have some 208V, 3-phase motors at work, on VFDs, running at 45-50hz, rated 15hp and 20hp, and they may draw around 40A.

2. There’s no accounting for efficiency. A very good motor might get 95% efficiency, so that should be factored in (divide by .95).

All that said, you’d still probably be in the neighborhood of $3/hr to run a 30hp motor at $0.12/kWh—which is low for many residential areas in the U.S., but it could be even lower than that if it’s on a commercial 3-phase service.

[roger wiegand]

Well, to start with, our electric rates are about 34 cents/kWh; not sure who gets power for 12 cents these days after the half dozen "fees" get added on to the base rate. My solar panels keep looking like a better and better investment.

[Alan Lightstone]

Well, to start with, our electric rates are about 34 cents/kWh; not sure who gets power for 12 cents these days after the half dozen "fees" get added on to the base rate. My solar panels keep looking like a better and better investment.
Yup. Those panels just got much more valuable at my house as our local utility raised their rates 20% to $.156/kWh. Much more important to me, they also just raised what they pay me back to $0.06631/kWh from $0.0325/kWh, so since I produce a surplus, I'm doing great this year (the drought and endless sunny days isn't hurting). So, despite their despicable minimum monthly charge of $30.00 plus taxes, my panels just got substantially more valuable to me.

And those calculations seem correct, but you do have to account for efficiency and I agree that 73A seemed low for a 30HP motor, but looking up one Baldor 30A, 230V 3-Phase motor it spec'd at 72 amps. The motor had a power factor of 82, so you really do need to take that into account.

[David Buchhauser]

David

Just a small adjustment to your math:

$2.68per hour x 40 hours per week x 4 weeks per month = $469

And a closer look at the loading:

A motor or a phase converter will only consume electricity proportional to the loading, which is rarely anywhere near full load on woodworking machinery. Fans are the most consistently loaded (dust collector).

Let's see. 2.68 x 40 x 4 = 428.80. Not sure how you calculated your $469. But in any case -

Right you are!! I was thinking per week. My mistake. I think it also goes without saying that it depends on the load. Much like a car - better mileage going down hill than up hill. With the phase converter, it is acting as the "3rd leg" in converting single phase to 3 phase power. So it acts like an "idler". I was not suggesting that it is the equivalent of running a 30 hp motor at full load. In my case, the phase converter is powering several cnc machines with spindle HP ratings of 15hp, 10hp, and 5hp (at full load). Of course the loading on these depends on the work being performed.

But in essence, the OP is correct with his math. The price per kwh is for example only. This also I would think goes without saying.

David

[Edwin Santos]

David



A motor or a phase converter will only consume electricity proportional to the loading, which is rarely anywhere near full load on woodworking machinery. Fans are the most consistently loaded (dust collector).

If you wouldn't mind educating a simpleton - let's take a machine like a table saw. Does this comment mean there is a difference in power consumption between the saw simply running versus when it is actually cutting wood i.e. under load?
Is it a significant proportional difference or a relatively tight range of deviation?
Thanks

[Bert McMahan]

If you wouldn't mind educating a simpleton - let's take a machine like a table saw. Does this comment mean there is a difference in power consumption between the saw simply running versus when it is actually cutting wood i.e. under load?
Is it a significant proportional difference or a relatively tight range of deviation?
Thanks

I don't have actual numbers for a table saw, but yes, there will be a large difference. The motor will barely pull any current when running with no load, and will pull more current as you add more load. Cutting a thin piece of plywood will draw much less power than cutting a 2" thick piece of oak.

A more common example would be to compare your gas mileage if you're towing a heavy trailer- you'll use a lot more gas pulling something heavy than you do without. Another common example is a cordless power tool battery. You will drain your batteries a LOT faster drilling 2" holes in oak than drilling 1/8" holes in balsa.

Matthias Wandel did some good power testing comparing jointer heads here: https://www.youtube.com/watch?v=rehZ8SqszAk

His main focus is on the comparison of the heads, but he has "no load" values in there too. He idles at around 500 W, and it jumps up to about 1000 W when planing 1 mm off some wood. He basically found the load doubled in his case, but that's just for that particular piece of wood that was only 5" or 6" wide. A piece of wood half that wide would need half as much additional power (~750 W total, or 1.5x more), and a piece twice as wide would need twice the additional power (1500 W total, or 3x more).

I would suspect that his numbers are skewed since unloaded planer blades are (in my experience) much louder than unloaded table saw blades, meaning the wind load on the teeth alone makes up a lot of the power draw when idling.

(Side note: Matthias seems to have some issues with his Shelix head's bearings which he talked about in a followup video, so don't use this video by itself to compare helical and straight knives.)

[Malcolm McLeod]

...our electric rates are about 34 cents/kWh... not sure who gets power for 12 cents these days ...
Sorry. Average for TX is ~13.4 cents, depending on which source you believe (or don't). ETA - We're still paying a NG fuel surcharge (~2-3 cents, last I looked), but over last few months we're basically selling NG for cost ($2.30-ish/MCF), so surcharge should come off - and get us back to ~11 cents.


.. their despicable minimum monthly charge of $30.00 plus taxes, ...
Sounds pretty reasonable for a back-up 'generator' that someone else provides and maintains. And they even buy your intermittently generated surplus power. Some might call your surplus 'unreliable'. YMMV.

I'd check into a whole-house Generac with NG or LP fuel system as less despicable.

[Cameron Wood]

Then there's start up draw vs run draw. Literally running for an hour would average the start up current draw- a large multiple- into the current use for the hour, which at a rough guess would 1.5 or double the cost.

[Steve Demuth]

Well, to start with, our electric rates are about 34 cents/kWh; not sure who gets power for 12 cents these days after the half dozen "fees" get added on to the base rate. My solar panels keep looking like a better and better investment.

We pay 11.3 cents/kWh for power used. That's after $1.67/day basic service fee, which we pay regardless of how much power we use. We've got the largest solar array the REC will allow us to have on the line, so last year we used only roughly 1000kWh net draw from the grid. All in, we paid just about $750 for the electric service, which figured against our actual net draw, is $.75/kWh.

[Dan Friedrichs]

Then there's start up draw vs run draw. Literally running for an hour would average the start up current draw- a large multiple- into the current use for the hour, which at a rough guess would 1.5 or double the cost.

Say the motor draws 10x normal current for 1 sec while coming up to speed. The average of the first minute is then 1.15x, and the first hour 1.0025x.

Inrush current makes no difference.

[Joe Calhoon]

Your numbers seem about right. My rate is about 13 cents per and about $50 per month service charge. The kicker could be if due to more usage they put you on a demand meter. If that is the case you will be paying a lot more. My DC is 25 hp and a couple other machines are over 50 hp with multiple motors. Working by myself now part time and the bill is usually $180 to $220 per month. When I had a crew and working more it would be 3 to 400$ per month. A few years ago we hit the threshold for demand and the bull jumped to about $1500 per month. Just working one day a month would set the demand. Myself and a couple local ranchers negotiated with the utility to raise the threshold for demand and we would pay more per KW. I don’t mind paying for what I use.
I feel like what I pay now is very fair but fully expect electricity is going to get very expensive in the future.

[Cameron Wood]

Say the motor draws 10x normal current for 1 sec while coming up to speed. The average of the first minute is then 1.15x, and the first hour 1.0025x.

Inrush current makes no difference.



Hm, my rough guess was not good.

No idea what "inrush current" means.

[Rod Sheridan]

That’s not correct, assume 10 X running current for 10 seconds during startup.

That’s 100 seconds of running current for startup versus 3,600 seconds of running current for an hours use….Regards, Rod

[Will Blick]

This reminds me of my energy engineering days, when working with a team of energy engineers who all have different native languages. Of course, none fully fluent in English, although they always thought they were. ;) The biggest challenge was never the engineering / math, but rather, understanding what the true question, or issue was trying to be solved. The engineering / math was always the easy part. Same in your example here, you tricked AI with your question. But in all fairness, AI still did a good job of answering considering the confusing nature of your question. So kudos to AI, its response was reasonable, and the details provided demonstrated the basis for its answer.

If you asked AIs how much per hour does it cost to run a 30HP motor, and nothing else, it's response is accurate based on its assumption of $.013 kwh charge, a variable you can alter to fit your cost basis. However, your question provided the actual load in Volts and Amperage, which is more specific than the nominal rated HP of the motor. Since V*A = watts (lets leave out Power Factor here for this simple example), the stated motor load you asked in your question would cost to operate per hour:

220 Volts
73 amps
16,060 watts
16 KW (*1, for 1 hr operation)
0.12 $/KWH
1.92 $/hour to operate at your stated load.

Of course, your 73amp value might be the FLA value, not the actual load during operation, so costs should be even less.

A fully loaded TRUE 30HP load on a motor at 220V would not draw 73amps, it would draw 102 amps. This is what AI based its answer on, i.e. theoretical load, not a practical load. Hence the much higher cost per hour it calculated.

30 HP
0.75 KW / HP
22.5 KW Total load
22500 Watts
220 Volts
102 Amps

However, its very rare (almost NEVER) does the nominal HP rating of a motor actually represent the FLA, Full Load Amps. Hence why the FLA is always listed on the motor nameplate. A more accurate method to determine actual cost, if possible, is to get an amp meter reading from one of the legs of an operational motor under load. If designed properly, a motor should be oversized.

Also, motor efficiency factor is already taken into consideration when taking a real world amp reading, or operating from % of FLA value. Many large motors I tested in commercial and industrial settings for energy audits would often operate at 40-60% of FLA when fully loaded.

And yes, inrush, is not even a glitch on the overall consumption, as it occurs in a fraction of a second, so when averaged over time, it a complete non issue in nearly all applications.

[Edwin Santos]

This reminds me of my energy engineering days......

Your reply is perhaps a more interesting commentary on AI than on the underlying question of motor energy consumption.
I'm amazed (humbled?) daily at how reasoned and accurate the answers are from AI on even the most technical and niche of subjects.

[Will Blick]

Your reply is perhaps a more interesting commentary on AI than on the underlying question of motor energy consumption.
I'm amazed (humbled?) daily at how reasoned and accurate the answers are from AI on even the most technical and niche of subjects.


Considering AI is at its infancy, at least to us, the general public, I fully agree with you. I have run a lot of AI searches, and sometimes I marvel at the responses, and other times, I notice the flaws. I would run AI queries on things I knew the answer to, and sometimes the responses were spot on, other times they were mostly right, and a smaller % of the time, the response missed the boat completely. Of course, that will change with learning. What I do like, as in the OP query, is how the response is delivered, i.e. detailed enough so you can ascertain the basis of the response, that is critical, vs. just giving an answer.

However, when the subject matter gets more technical and/or complex (beyond the scope of the OP question) of course the responses become much more limiting, and sometimes, the AI simply admits, it can not answer the question, which I prefer vs. an erroneous response. I consider AI a GREAT first pass on any query, and sometimes its all you need....but when critical decisions need to be made, a knowledgeable human is often required to decipher the response on more technical issues. This not being a good example, as the AI response was not wrong, but at the same time it was not accurate.... it struggled with a loaded question ;)

[Bill Dufour]

It did not mention that 220 volts is incorrect and therefor the motor will draw more amps in the brownout then it would at 240 volts.
My swimming pool motor is variable speed with a display that shows rpm,watts, etc. Watts at full load shows the motor is well under 100% efficient. Half speed is about 1/10 the watts. I assume the flow rate is roughly 1/2 to maybe 1/4 since output pressure ratio is around that.
Bill D

[Will Blick]

220 V is not incorrect. It's much less common than 240V, but still exists for various reasons in the USA. It's just a function of the hot leg (hot to neutral), which in USA we still see, 110V (the old norm from the 50's and prior, but very uncommon today), 115V, 120V (most common) and 125V. This is why motors have tolerances of incoming operating voltages. In the past 30yrs, I have lived in places where my incoming voltage was as low as 112V, and as high as 124V. From memory, I think the current standard in USA is 114V to 126V.

But yes, amp draw would increase proportionally as voltage drops, but total energy (watts) remains unchanged, as the load determines total energy draw (Ohm's law). Significant brownouts can be very problematic, as the voltage can drop below a devices min. voltage requirement. Many higher end machines have brown protection set at a level that will prevent damage to any downstream devices vulnerable to a voltage below the set point level.

Efficiency is can of worms beyond the scope of this thread ;) (Especially VFD's) All motors, and or motor systems (such as pumps, fans, etc) are tested and illustrated in their curve profiles. The curve(s) takes into consideration all the variables of the system as a whole. The Clinton administration instituted motor efficiency mandates... IIRC, they set a min. of 90% efficiency at their rated load. (motors are not always loaded at their optimum efficiency, hence the value of the curves)

[Mike Henderson]

Just to add a complexity in case anyone puts an ampmeter on their table saw motor. When a motor is not loaded, it acts like a big inductor. It will draw a fair amount of amps but the amps will be out of phase with the voltage. That phase difference is known as "power factor". When the motor is loaded, the power factor is close to one. When it's unloaded, the power factor is low, maybe 0.4.

The only thing you get charged for is watts, and watts are not just the amps times the voltage. It's the amps times the voltage, times the power factor.

Let's take a small motor, maybe a 3HP on a table saw. It might be rated at 10 amps at 240 volts, and that's for full loading (3HP). When unloaded, it may be pulling 6 amps but those amps will not be in phase with the voltage.

Anyway, if you measure the amps on an idling motor, you'll get a higher reading than you might expect.

Mike

[Ken Combs]

Say the motor draws 10x normal current for 1 sec while coming up to speed. The average of the first minute is then 1.15x, and the first hour 1.0025x.

Inrush current makes no difference.

A few years back the local utility (city owned/operated) replaced our old dumb electro-mechanical meters with the latest and greatest smart meter. One of the reasons for that was that they were purchasing more bulk electricity from the provider than was being metered. That and labor-free reading were the reasons given for the swap.

Coincident with the change-out, everyone complained about an increase in bills. There was no rate increase, just an increase in measured usage. I've always suspected that the smart meters see the inrush of all the motors, (A/c, refrigerators, compressors, saws, etc) as billable usage while the inertia of the electro-mechanical version never registered it.

The techs working on the system agree, off the record, but will not say so publicly.

What do the experts think?

[Will Blick]

I am certainly no expert, as I retired 25 yrs ago, technology changes fast on everything. OTOH, some basic issues remain fundamentally the same.

You mentioned the rates stayed the same, only the consumption has increased. Be sure the billing tariff did not change, to be certain they did not change the billing model in addition to the consumption.

As for your suspicion of added consumption, this is very commonplace. It's rare an old meter and new meter will read the same. Considering the utility was buying more power than it was selling, might be a result of old, under-reporting meters. I assume its a very small utility from your description. Has the utility found with new meters installed, the net buying / selling KWH's are more in parity now, so the problem resolved itself? If its a small utility, these things can often be found out with a phone call, which is not the case with a $20B a year utility ;). Of course, this assumes all users received the new smart meters. Also, if all the new meter recipients are noticing the same, then one would think, the new meters are reading proper, and the old meters were under reporting. But, this does not rule out the new meters are over reporting.

I was an expert witness in several of these cases, where the meter was over recording consumption, which equated to million$ in excess charges over many years. To determine this, we hire 3rd party certified metering companies whose data and testimony stands up in court. They install transducers on the incoming lines as well as the grounds, searching for ground leaks as well. They have sophisticated computers on site which report back to their offices. This is done for at least 3 months, if over recording is noticed, it is typically left on for 6 months. When the variance is noticed, not in your favor, then you call the utility, or if enough money is at stake such as the industrial facilities I was involved with... legal gets involved before contacting the utility.

If the extra billing is significant enough, you can all split the cost of 3rd party monitoring of a few of the meters to see if you are now being billed fairly, or the new meters are in fact, over recording. Just beware, there will never be perfect parity between the two, but if you are within a couple %, that's the best that can be expected. If the dollar amount is marginal, prob. not worth all this effort, and just try an old fashion phone call...maybe the utility will do their own checks, as I am pretty certain there goal is not to intentionally over charge you. Hope this helps...

[Rod Sheridan]

Start up energy use isn’t significant, it’s more likely that the new meters are more accurate…….Regards, Rod