HP swap (Horsepower)

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NE (9.06 miles @5.9 Degrees from Winged Horses)
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EC - retired
We have two loads with different motors. The 5HP is too small and the 7.5 is to large. Simple, swap the motors.

We are monitoring the load of the 7.5 to determine just how "loaded" it is. (0-100%) I am using a simple current transducer and another vendor is using a "power monitor".
From what I can tell the average loaded current is 5.09 amps vs. an unloaded average of 4.37. This makes it very difficult for my equipment to notice the difference. The vendor using the power monitoring equipment is complaining of the same difficulty. A max load of 5.27 amp is well within the 7.6 FLA of a 5HP
@480.

What am I missing?
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141008-0752 EDT

Don't measure current. Measure power.

An old small 1/3 HP motor reads 146 W and 640 VA unloaded. Full load would be over 400 W and by nameplate 5.9 A at 115 V = 680 VA.

.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
141008-0752 EDT

Don't measure current. Measure power.

An old small 1/3 HP motor reads 146 W and 640 VA unloaded. Full load would be over 400 W and by nameplate 5.9 A at 115 V = 680 VA.

.

So, your trying to tell me that the vendor with the power monitoring equipment should be able to get the information they need. We have looked at their data displays, but without them being there to say "Yes/No" I wasn't going to enter trip points or change the scaling. Empty looked close at 4000 units vs the 4010 units it was when running empty. 20000 units for top end, well IDK.

They have a spare 0-10v output on their power monitor I can use. It won't take but a little time to switch from my current transducer to that and see what is going on. That and some logic change.
 
Last edited:

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141008-0930 EDT

Make sure the power monitor is actually giving you a signal proportional to power.

The motor I measured is not a very efficient motor. A more efficient motor will provide a greater ratio of full load power to noload power.

To test the power meter output signal connect a voltmeter to it.

Assuming the power meter is a three phase unit you can probably test it with a single phase load.

Apply a known resistive load. Power reading should correlate.

Add in parallel with the resistive load a high quality capacitive load of equal current (polypropylene). The power reading should not change.

Remove the resistive load leaving only the capacitive load. The power reading should go to near zero.

.
 

Jraef

Moderator, OTD
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
Might be that their power meter scale is too far off to be useful. For instance if their CT ratio is 1000:5 because the meter is intended to be used on a switchboard main, the difference between loaded and unloaded on a load that small will fall within the error allowance on the sensors.

But stick with power monitoring if you can, it's much more accurate for this sort of thing. A simple current monitor cannot tell the difference between active or reactive current, so when the motor is unloaded, the power factor is at .2 or something like that. The reactive current might still appear high, but is not at all representing the shaft power of the motor. Current monitoring alone is also subject to voltage changes from the utility, whereas kW is a direct measure of the work being done by the motor.

If 5HP was too small, that means you must have at least 6HP of load on that 7-1/2HP motor, so 85% load, far higher than where the unloaded current should no longer matter. Your numbers however seem to indicate that the loaded current is still much lower than I would expect. So how was it determined that 5HP was too small?
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
....
If 5HP was too small, that means you must have at least 6HP of load on that 7-1/2HP motor, so 85% load, far higher than where the unloaded current should no longer matter. Your numbers however seem to indicate that the loaded current is still much lower than I would expect. So how was it determined that 5HP was too small?

The two motors are on different loads. The 5HP is overloaded. The 7.5 is under.
First thought was to swap the motors. Minor detail of different frame sizes, shaft sizes & probable belt lengths. The 100' crane to move them around is just a small hitch.

The Other vendors power monitor is one I have seen before. I can not remember who makes it. Small, about 10 inches long, 5 wide and maybe 1 deep. Has three holes in it to run the wires thru for amps, three terminals for voltage, dip switches to set HP range IIRC, a 4-20ma and 1-10v output.

I have the PDF manual somewhere on this computer. CRS.

The 4000 to 20000 is scaled off the 4-20 ma to get some decimal places. 4000=0 amps, 20,000=18 amps. I need that blooming manual.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141008-2400 EDT

ptonsparky:

I just took my PH-3A supplied it with 120 V and operated it single phase. It is rated full scale 10 HP at 460 V (10 V DC). !0 HP is 7460 W. A 100 W bulb read 0.047 V (0.029 - (-0.018) ). I needed to use two holes but used only one. Two holes would double the voltage. This doesn't quite calculate to correct power. But excitation voltage is much below the nominal, and current is only about 1/10 of full scale. Probably not too bad for the excitation voltage this far off nominal.

The important point is the next test. Used a 30 ufd capacitor in place of the 100 W bulb, and the output remained at -0.018 V. Thus, the reactive current ( about 1.4 A) was completely rejected.

Always use a resistor to discharge the capacitor after running this test. Good capacitors will hold a charge for a long time. A 100 W bulb can be a satisfactory discharge resistor for this size capacitor.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141009-0937 EDT

ptonsparky:

If you have a power monitor similiar to my Load Controls PH-3A, then I suggest the 460 V and 10 HP scaling networks.

Notes from the installation instructions:
Calibration networks work accurately from 0 to 130% of rating.
Accuracy is within +/-2.5% of full scale.
Not damaged by current overload.
Output signal is clamped at about 20% over full scale rating. At about 12 V on the 0-10 V output.
Not specified, but the unit can measure reverse power flow.

.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141009-1650 EDT

ptonsparky:

Can you clarify.

Which motor did you measure? I assume it is the 5 HP unit.

What were the actual no-load and full-load current measurements on one line?
Are all three lines about the same?

Assuming currents are fairly well balanced, then what are the no-load and full-load line-to-line voltages for at least one pair of lines?

Is there any information on the power meter calibration? Your earlier post mentioned 18 A was a 20 mA reading. If a 3 phase resistive load this would be about 15 kW. If we assume 1000 W into a motor for 1 HP mechanical output, then the 5 HP motor fully loaded might be about 5 kW, 1/3 of full scale, and thus 16/3 + 4 = about 9.3 mA.

In your last post, if 100 = 15 kW, then 36.2 = 5.43 kW, and 13 = 1.95 kW. We don't know the motor efficiency and therefore the input to output power ratio is not known.

If you don't exceed the full load rated current of the motor you are not likely to burn out the motor if at a reasonable ambient temperature and adequate air circulation. This comment is just for others as you already know this.

How does the motor current at full load on the 5 HP unit compare with the motor rating? And what is the voltage at the motor at fill load?

.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
141009-1650 EDT

ptonsparky:

Can you clarify.

Which motor did you measure? I assume it is the 5 HP unit. The 7.5

What were the actual no-load and full-load current measurements on one line? AVG min 4.37, Max 5.09
Are all three lines about the same? Haven't checked lately but they have been in the past.

Assuming currents are fairly well balanced, then what are the no-load and full-load line-to-line voltages for at least one pair of lines? Good question, but off hand, 480.

Is there any information on the power meter calibration? Your earlier post mentioned 18 A was a 20 mA reading. If a 3 phase resistive load this would be about 15 kW. If we assume 1000 W into a motor for 1 HP mechanical output, then the 5 HP motor fully loaded might be about 5 kW, 1/3 of full scale, and thus 16/3 + 4 = about 9.3 mA. The "other vendor" had the scale set to 18000, IIRC, for 20mA. Amps, power, popcorn, IDK. We have changed it to 22000 to make things come out right on his side. That vendor is a no show for assistance.

In your last post, if 100 = 15 kW, then 36.2 = 5.43 kW, and 13 = 1.95 kW. We don't know the motor efficiency and therefore the input to output power ratio is not known. The 7.5 hp has a KW of 5.5 with 1.9 kW and .72 kW respectively. I did not look at the efficiency.:slaphead: I am now using the 0-10v output of the power monitor to get information for my logic.

If you don't exceed the full load rated current of the motor you are not likely to burn out the motor if at a reasonable ambient temperature and adequate air circulation. This comment is just for others as you already know this.

How does the motor current at full load on the 5 HP unit compare with the motor rating? And what is the voltage at the motor at fill load?

.

What started all this is the OV cannot get his equipment to match with what I feel is the amount of corn going through a Mixing Auger. The system my customer installed is smaller than what the OV is used to and they do not know how to make it work with what the customer has. The OV uses the power monitor to decide how much corn is in the mixing auger. They add chemicals accordingly. The MA motor, however, was oversized. There is no way we can ever fill it to the FLA of the motor. Can't happen. The max is the 1.9 kW. I do not know the logic the OV uses and they have been around a long time so something must work most of the time. Just not here.

I learned that the PM is doing what it can do. IMO I do not believe measuring KW will tell you how much product is being moved. I like the idea of counting full boxes being delivered at whatever rate. KW increase just means a bearing may be going out, something is binding up, or the product is heavier this load.

Interesting day if nothing else. We did get things to work by increasing the flow rate the OV can recognize. Instead of a smooth even flow of product it is now Start/Stop, repeat.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141009-1953 EDT

ptonsparky:

So the real problem is to measure the flow rate of corn into a mixing area. The criteria could be unit mass, weight, or volume per unit time. What is most important --- mass, weight, or volume? Mass and weight probably can be considered to be the same.

Is it correct that your customer is the supplier of the mixing auger and you provide the controls for the mixing auger system? And that a different supplier provides a feed auger or some other mechanism to put corn into your mixing auger?

Assuming this, then I think you are saying the feed system to the mixing auger can not provide accurately known feed rate information to your system so that you can control the additives correctly. It looks like you are really looking for volume feed rate information as the corn flows into your mixing auger, and not mass rate.

Do you have pictures or drawings that illustrate how these systems are related to each other?

If the feeder to your mixing auger is also an auger and it could be considered to be full while rotating, then RPM of the feed auger might be a good measure of volume feed rate.

.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
It gets even more fun. Vendor 3 is responsible for the mixing auger. He is also responsible for the aforementioned 5 Hp. Plus, my question some months ago, about a drag needing a higher RPM motor with more HP.

Production starts with dry corn at a rate that does not exceed 960 +- bush per hour max via a chain drag. I can vary that rate from 6 to 60hz. Works well. I know what it delivers. Chemical is added by vendor 2 at the rate determined by the power monitor and moisture sensing of dry corn. Mixtue is then steamed for a time in 480 bushel kettles then processed into flakes where it falls onto the 5 hp drag that was designed to move 1000 bushel per hour of dry corn. It is now at least 30% and flat. No round corners...hence overloaded at well less than 1000 bph. Vendor 3 is going to put a larger sheave on the 5hp to help this problem. Ok. I think a 7.5 would be better, but have no idea if his drag is capable.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
141009-1953 EDT

ptonsparky:

So the real problem is to measure the flow rate of corn into a mixing area. The criteria could be unit mass, weight, or volume per unit time. What is most important --- mass, weight, or volume? Mass and weight probably can be considered to be the same. I think in volume. .804 cu ft = 1 bushel. Those dealing in corn think of weight. 1 bushel corn = 56 lbs @ 15% moisture. Less moisture = more bushel, more moisture = less bushel.

Is it correct that your customer is the supplier of the mixing auger and you provide the controls for the mixing auger system? And that a different supplier provides a feed auger or some other mechanism to put corn into your mixing auger? Customer via Vendor 3, supplied all bins, drags, augers, motors etc. I designed, implemented, all control other than the addition of chemical treatment. Application of steam is by others.

Assuming this, then I think you are saying the feed system to the mixing auger can not provide accurately known feed rate information to your system so that you can control the additives correctly. It looks like you are really looking for volume feed rate information as the corn flows into your mixing auger, and not mass rate.

Do you have pictures or drawings that illustrate how these systems are related to each other?

If the feeder to your mixing auger is also an auger and it could be considered to be full while rotating, then RPM of the feed auger might be a good measure of volume feed rate.

.
Riveting.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
Not about HP anymore

Not about HP anymore

Ok, if 0-100 scale was interesting what about 0-5500. Motor is rated @ 5.5 kw.

Originally I had my limits set for amps, so feeling pretty confident there was no way in hell we could overload that motor, I bypassed my logic limits. Ooops. You can overload that auger if you plug up the discharge end. The operator noticed the display reading of over 5000 so he shut things down before damage. The logic limits are now in place as of 7 last night.

All my little PLC projects are custom and I have this one set up to email me data on 16 different items I want to track. I take readings every second, average 5 of those and save it to a file which is then sent to me at 11 PM. Just because I can.

While there last night I watched the process and numbers. The feedlot is processing wet corn at the moment, one semi every 8 minutes, but they forgot to order dry. The OV system shut down because of no corn flowing past their moisture sensor.

Scanning thru the info this morning I see that at 3:18:35 yesterday afternoon the holding bin was low enough to start affecting the flow of corn. Hours before "no corn" slapped us up side of the head.

I will be installing power monitors on future projects.

Still learning.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141011-1122 EDT

ptonsparky:

To calibrate the output of you power monitor 0-10 V signal you need a scaling factor for the power monitor.

On my Load Controls PH-3A unit the label says 460 V, 10 A, 1-10 V. But this does not really define the power calibration. The manual defines this as 10 HP for full scale presumably meaning 7457 W = 10 V with 460 V and 9.359 A. Why does Load Controls list 9.54 A? Probably a typo.

Either by dip switches or plug-in scaling units your load monitor will have some scaling factor. Find the scaling factor and use that in your system for your 10 V full scale reading.

I doubt that 5500 W is what your 10 V represents. 10 V on mine is 7457 W.

More later.


.
 
Location
NE (9.06 miles @5.9 Degrees from Winged Horses)
Occupation
EC - retired
141011-1122 EDT

ptonsparky:

To calibrate the output of you power monitor 0-10 V signal you need a scaling factor for the power monitor.

On my Load Controls PH-3A unit the label says 460 V, 10 A, 1-10 V. But this does not really define the power calibration. The manual defines this as 10 HP for full scale presumably meaning 7457 W = 10 V with 460 V and 9.359 A. Why does Load Controls list 9.54 A? Probably a typo.

Either by dip switches or plug-in scaling units your load monitor will have some scaling factor. Find the scaling factor and use that in your system for your 10 V full scale reading.

I doubt that 5500 W is what your 10 V represents. 10 V on mine is 7457 W.

More later.


.

The unit we are using lets you dial in the HP with some variable resistors. 5K = 5 HP, 7.5K =7.5 HP etc, on up to 150K =150 hp. I see it as 0- 10v can equal almost any thing I want, as long as it is linear. 0 could equal no corn (bushels or lbs) I don't care as long as 10v equals the max it can deliver in bushels or lbs be it 60 or 735. I just made it equal 5500 in my logic of the PLC. No different than the 0-100 I had used before.

All opinions subject to further education and thought... of which just happened. Full Scale should be 5595.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141011=2210 EDT

Your 5595 W (10 V signal) means you are on the 7.5 HP selection. I believe you have this connected to the 5 HP motor. 5 HP = 3730 W. If the motor efficiency is 85%, then input power for 5 HP output would be about 4388 W. Well within full scale of the 7.5 HP scaling factor.

See "Efficiency" at http://en.wikipedia.org/wiki/Induction_motor

Full mechanical motor load of 5 HP is probably within the 85 to 95 % efficiency range, and thus, input power at full load is likely between 4388 and 3926 W.

You need to know motor efficiency for your particular motor to get a good estimate of the input wattage at full load. For a continuous process at full load you would want this as your upper limit.

Once you specify the calibration constant for the power monitor, then you want to run the machine when all bearings and and other components are in good condition, and determine this residual power. This power measurement then becomes your zero (tare reading) for product (corn) moving thru the machine. It also becomes a reference for checking on machine deteroriation.

At full motor power input (5 HP out) or something near this experimentally determine the output in bushels per hour for a calibration point. This may vary by moisture content or other factors. Then do a test at about 1/2 the volume rate. Use this information to see if change in power from the residual power level to full power level is a linear relationship to volume rate.

.
 
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