Determining Eddy Current Coupling Losses

[Steventyj]

I am trying to determine how much power a pump is using. The pump is attached to a 15HP motor through an eddy current coupling. I can find out how much power the motor is using at any given moment as well as the shaft speed on the pump side of the ECC. Is there a way to determine how much power is being lost through the ECC. There are plans in place to replace the ECC with a VFD, but at the moment we need to know how much power the pump is using.

Thanks,

Steven

 

[automatic2]

slippage, or shaft differential speed, would be your best quess. Manufacturer should also put out data that charts clutch voltage to hp. Eaton has some great info on its Dynamatic line.

 

[Shortstub]

Use a strobe light to measure motor slip. It will be a reasonably proportional to the motor's output power if the motor is not severly overloaded.

However, using a strobe to determine actual speed will introduce appreciable error. Therefore, use a strobe synch-ed to line frequency and "count" slip rotation for a reasonable time period, i.e., 1 minute, with the approprate correction for the number of pole-pairs of the motor. Also note, motor output will include ECC loss!

Before hi-tech was discovered it was done with a fluorescent lamp, and the slip rotation was divided by the number of poles!

 

[Steventyj]

I am pretty sure I can get an accurate idea of how much power the motor is using because I have the amperage and I can determine the speed using a strobe or other methods. My problem is that what I want to know is how much power is getting through to the pump. I need to know how much power is lost through the ECC. I will try and dig up data on the ECC, it is an Eaton Dynamatic so maybe there is some good information available.

 

[Shortstub]

Aren't you able to calculate bhp of the pump, based on operating parameters, and characteristic curves from the pump manufacturer?

I recall Eaton supplying loss data based on its operating temperature-rise above ambient.

 

[gcaudill]

IFF the motor is running less than 75% loaded (compared to FLT or HP) the losses in the clutch will be negligble compared to the costs incurred by 1.) reduced induction motor efficiency 2.) reduced induction motor power factor.

You may be looking at the problem from the wrong angle.

 

[gcaudill]

IFF the motor is running less than 75% loaded (compared to FLT or HP) the losses in the clutch will be negligble compared to the costs incurred by 1.) reduced induction motor efficiency 2.) reduced induction motor power factor.

In any case, I'm certain the losses in the clutch are very small.

You may be looking at the problem from the wrong angle if you are tyring to determine a payback for switching to a VSD.

 

[Steventyj]

For example, if the motor shaft is spinning at 1800 RPM and the motor us producing 12HP, and the output shaft of the ECC is spinning at only 900 RPM then what kind of a relationship is there between input and output power on the ECC.

 

[gcaudill]

You could measure the clutch supply DC voltage and DC current to calculate the power being consumed by the clutch, this is the input power. The output power = 12HP.

If the DC drive/controller has considerable ripple the clutch efficiency could be reduced. But still, the clutch efficiency should be relatively high compared to that of the motor.

For a 15HP AC induction motor the with an attached 12HP load the motor is only 80% loaded. The Efficiency is probably around 90% and the PF is probably around 0.80. In this condition the motor is pulling more reactive power from the line than it would at rated load (15HP).

Is this a Dynamatic drive by chance?

 

[howardbarr]

If you determine the power output of the motor, and can determine the motor speed and the pump speed, then the power to the pump is equal to the power output of the motor multiplied by the ratio of PUMP SPEED/MOTOR SPEED. The loss in the ECC is equal to the difference between the motor output power and the pump power.

ECC's are notorious power wasters. If you run this pump more than a couple hours a day, then almost guarantee you will save money by buying an inverter and new motor. MAKE SURE THAT THE MOTOR YOU USE WITH THE INVERTER IS INVERTER-RATED.

 

[Steventyj]

Thank you very much. So if I had a 100 HP motor spinning at ~1800 RPM and producing 80 HP attached to an ECC and the output shaft of the ECC was spinning at 900 RPM then the pump would be using 40 HP? and 40HP would be lost through the ECC?

We are planning on replacing the ECC but this pump fails often and it may not be correct for the application so before we size a new VFD and motor we plan to find the correct pump for the application.

Thanks

 

[CB2]

ECC's have a linear efficiency curve and are more efficient than a VFD only at full speed. Much less efficient than a VFD at lower speeds.

Eaton Corp. sold the Dynamatic division off about 5 years ago, it still exists in Wisconsin I think.

 

[gcaudill]

Dynamatic is still in business, but is no longer owned by Eaton as CB2 indicated. I forget who the new owner is, I think Dynamatic is now owned by the same people that own Torspec.

 

[jraef]

If your 1800RPM motor is running at 80% load with the pump at 900 shaft RPM, then your pump is still requiring 80% load. The only losses are in the ECC itself, not based on the speed. The speed is just determining your flow rate through the pump. I would be suprised if your losses throught he ECC are greater than 5% total. But your savings from using a VFD will still be greater. The power curve on the ECC in linear, while the power curve on a centrifugal pump powered by a VFD follows the afinity law. The power required is the cube of the speed, so at 50% speed, your pump will draw .5 x .5 x .5 power, or 1/8th that of full speed. If you want a true measure of savings expected, plot the running time with the operating speed, and keep track of the KWH over a specific time, say a month. You can then use a payback analysis software program that will calculate your operating costs on the VFD for that same time period and compare it to the costs of running the ECC.

However...

If your pump is not working right, it seems to me that this excersise is pointless. What you really need to do is re-engineer your pump requirement from the bottom up. Start with head, flow and pressure, then pick a pump curve that works for you, then determine maximum BHP, then select a motor based on that, then buy a VFD for that motor. Who cares what the old system did if it was failing? Quando Omni Flunkus Moritati

 

[Steventyj]

Yes, we are essentially starting from the bottom up. Rather than just throw in a VFD in this application we are looking at different pump options so that we can choose a proper motor/VFD/pump combination. However it is not that simple because we should first know why the current system is not working well. Is the fluid of a greater specific gravity than planned for?, Are the pipes getting clogged?, Are there valves that are causing a much greater increase in head than planned for, etc.. As part of this process we want to know where on the curve the pump currently is and so we wanted to know what amount of horsepower the pump is using.

Is it correct to say that an ECC can change the RPM, while keeping the power transfered constant by also changing the torque?

- Thanks

Steven

 

[gcaudill]

The ECC controls the torque transferred from the motor to the clutch output shaft (load). You should refer to the manufacture's torque-speed curve for the clutch output torque at varying voltages at different slip RPMs (% excitation vs. slip RPM ==> torque output). Unless you are using a speed controller on the clutch the load actually determines the speed at the given output HP.

RPM = (HP x 5252)/(torque)

If you know motor kW (V and I, hence HP) and RPM you can calculate motor torque. If you know the clutch output shaft RPM you can calculate the load HP attached to the clutch shaft.

I stick by my original claim that the ECC is more efficient than the motor by many times over due to the Eff. and PF ratings. If you know the motor HP you essentially know the load HP.

In my opinion the only negatives of the motor/ECC combination are:
1.) Poor PF and Eff. at light motor loads.
2.) Speed (pump volume/velocity/pressure) control without the correct clutch controller. I do not know what kind of pump or process you have.

 

[CB2]

qcaudill

In the last two paragraphs of your post I feel you are ignoring the efficiency of the ECC. The ECC has a linear Eff. curve, 50% speed = 50% Eff. The motor HP would be the sum of the pump BHP plus the ECC losses at a given speed.

 

[Steventyj]

CB2,

Are you suggesting that if I have a motor that turns at 1800 RPM and I want to turn my load at 900RPM then the motor is going to have to provide twice the needed horsepower of the load because half of that will be lost in the ECC?

 

[gcaudill]

The torque output of the ECC is proportional to the voltage and current applied to the coil terminals. The coil does not operate at full voltage continuously. The voltage is adjusted to vary the torque that is transmitted from the motor shaft to the output shaft of the clutch. The current will vary up and down with the voltage. Zero voltage = zero current = zero clutch output torque.

If the motor is running at full RPM and the clutch (ECC) voltage is zero, the current will be zero and the clutch speed will be zero. Where is the loss? It is in the AC motor running at less than rated load. Do you follow me CB2?

 

[Steventyj]

gcaudill,

If you look at that example then the loss is 100%. The efficiency of the system is power in/power out right? We are looking at the entire system from motor to ECC to pump, so if the 100 HP motor is using 10 HP of electrical power to stay at 1800 RPM without a load, and the pump is not using any power to move liquid than your power in is 10 and out is 0. I am not concerned with how much power the ECC is using to control the speed of the output shaft, just in the power transfered to the pump.