Single phase motor loading query

In summary, if you increase the drive pulley size on a belt driven air compressor to 3.5", the motor will turn 16.6% faster and use the same amount of power. However, the torque and speed of the two pulleys will be different.
  • #1
65fastback
3
0
Hello Guys,

I have a small belt driven air compressor with a 3HP (2.2kW) single phase 240VAC motor.

It is currently only running the motor at 72% of FLC (8.9A, FLC = 12.3A )

The drive pulley is 3". If I increase it to 3.5", the compressor will turn 16.6% faster.

Is it possible to calculate or estimate the increase in motor current after this pulley swap change?

Thanks,
Ben.
 
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  • #2
65fastback said:
Hello Guys,

I have a small belt driven air compressor with a 3HP (2.2kW) single phase 240VAC motor.

It is currently only running the motor at 72% of FLC (8.9A, FLC = 12.3A )

The drive pulley is 3". If I increase it to 3.5", the compressor will turn 16.6% faster.

Is it possible to calculate or estimate the increase in motor current after this pulley swap change?

Thanks,
Ben.

If I have interpreted you question correctly, the motor that is driving the belt has a constant rating - 3HP at some rated speed in this case - regardless of what size drive pulley you mount on the end of it. Thus, the power the motor uses will be the same. Hence, the current is the same.

The trade off will come with the torque and speed of the two pulleys. Since you increased the drive pulley speed, there will be less torque (consistent with the conservation of energy law).

Power = torque x speed

This assumes you are not overloading the motor and are just running it at the designed load. In other words, your load (air compressor) is still the same so the motor will still use the same amount of power.

CS
 
  • #3
In other words, your load (air compressor) is still the same so the motor will still use the same amount of power.

I disagree...you will be compressing about 16.6% faster...doing more work...

the dynamics of motors are too complicated for me...but if the compressor is rated at sufficient RPM change the pulley and give it a try...likely you'll draw about 16.6% more power to run the compressor...and but it will compress faster...
 
  • #4
Naty1 said:
I disagree...you will be compressing about 16.6% faster...doing more work...

the dynamics of motors are too complicated for me...but if the compressor is rated at sufficient RPM change the pulley and give it a try...likely you'll draw about 16.6% more power to run the compressor...and but it will compress faster...

That's why I said assuming the compressor isn't overloading the motor. If the motor is running at its rated load, then the compressor cannot possibly be working more.

If you are saying that the compressor will require that the motor be overloaded (which I have no opinion as to whether or not that is true) then certainly the motor will draw more current.

CS
 
  • #5
Naty1 said:
I disagree...you will be compressing about 16.6% faster...doing more work...

the dynamics of motors are too complicated for me...but if the compressor is rated at sufficient RPM change the pulley and give it a try...likely you'll draw about 16.6% more power to run the compressor...and but it will compress faster...

You are correct.

It is a 2 pole, 50Hz motor, so it will turn 2850rpm at all times.

With no load, ie. nothing on the shaft, it pulls 2.2A.
With the current belt setup, it is pulling ~8.8A. Torque (and current) increase to meet the load. MOTOR speed remains the same.

I want to squeeze a few more CFM from the compressor by turning it faster with a larger drive pulley. It will pull more current, but I wanted to calculate the pulley size and resultant current before going to the trouble of swapping it.

I am an electrician, but I have long forgotten the calcs like this.
 
  • #6
I take it this motor is an induction motor. The greater the load, the higher the slip frequency, thus the lower the RPMs, but it's not a large change under you're load change.

The 2.2A is mostly 90 degrees out of phase with the load current.
The 8.8A represents a load current of about 7.7 Amps.

All things being otherwise equal, the RPMs will be increased by 17% to 9.0 Amps real current.

On the meter this should read 9.25A, as the magnetization current of 2.2A is included at 90 degrees out of phase.
 
Last edited:
  • #7
65fastback said:
You are correct.

It is a 2 pole, 50Hz motor, so it will turn 2850rpm at all times.

With no load, ie. nothing on the shaft, it pulls 2.2A.
With the current belt setup, it is pulling ~8.8A. Torque (and current) increase to meet the load. MOTOR speed remains the same.

I want to squeeze a few more CFM from the compressor by turning it faster with a larger drive pulley. It will pull more current, but I wanted to calculate the pulley size and resultant current before going to the trouble of swapping it.

I am an electrician, but I have long forgotten the calcs like this.

Be mindful that increasing the compressor RPM as well as motor running current creates increased heat in both devices, which reduces their duty cycle, so you'll have to reduce usage per the normal manufacturer's recommended cycle times. Additionally, due to the greater physical loading imposed by the drive pulley's increased circumference, initial start-up current will increase somewhat, so you may pop the circuit breaker upon compressor start-up especially if any other electrical devices are on the same circuit breaker.
 
  • #8
65fastback said:
You are correct.

It is a 2 pole, 50Hz motor, so it will turn 2850rpm at all times.

With no load, ie. nothing on the shaft, it pulls 2.2A.
With the current belt setup, it is pulling ~8.8A. Torque (and current) increase to meet the load. MOTOR speed remains the same.

I want to squeeze a few more CFM from the compressor by turning it faster with a larger drive pulley. It will pull more current, but I wanted to calculate the pulley size and resultant current before going to the trouble of swapping it.

I am an electrician, but I have long forgotten the calcs like this.

The simplest thing to do is just look at the motor performance curve for the torque value you want and read off the line current. So you would just need to determine how much torque is on the shaft for the increased load and cross reference it on the curve.

Otherwise you'll need to know or determine the motor characteristics.

Essentially, the line current is a function of the shaft power, line voltage, motor efficiency and power factor. So if you know the efficiency and PF of the motor, the current the motor draws from the source is equal to:

[tex]I_{in} = \frac{\tau n_r}{7.08 \cdot V_{in} \cdot PF \cdot \eta} [/tex]

I_in is the line current
tau is the torque on the motor's shaft
n_r is the rotor speed
V_in is the line voltage
PF is the motor's power factor
eta is the motor's efficiency

This might give you some idea of what to expect depending on what parameters you know.

CS
 
  • #9
Unfortunately, the power factor varies with load.
 
  • #10
Phrak said:
Unfortunately, the power factor varies with load.

For a small change in load it would probably be close to the full load PF. Of course, that's why I said he might be able to get some idea. Still, I the best way is to look at the performance curves from the manufacturer.

CS
 
  • #11
He's running a power factor of about 8.5/2.2, based on the no load current given. At full rating it's about 11.8/2.2, I believe.

It is generally true, that for a given motor current/torque ~= const, and RPM/voltage ~= constant.

In retrospect, I don't think the OP has gotten such grand advice. If the power required by the compressor goes up as the square of the shaft frequency, the load current could be a calculated 11.5 Amps.
 
  • #12
stewart, these motor questions come up often enough. I have some references for DC motors, but none for AC. A quick search tells me I'm not using the right key words. Don't know what they should be...
 
  • #13
Phrak said:
stewart, these motor questions come up often enough. I have some references for DC motors, but none for AC. A quick search tells me I'm not using the right key words. Don't know what they should be...

Most of my reference material is in the form of books (I'm hesitant to trust some online sources so I prefer published work). I personally like Electric Machines by Charles Hubert.

CS
 
  • #14
If I had a text for everything I wanted to know about... :smile:

I was hoping to find something such as a model in RLC, some motor constants, and things like slip frequency as a function of load.
 
  • #15
Last edited by a moderator:
  • #16
Thanks for the replies.
Its not centrifugal, its reciprocating positive displacemant (piston) compressor.
I have found a suitable pulley, so I'll make the swap and measure the current.
 
  • #17
65fastback said:
Thanks for the replies.
Its not centrifugal, its reciprocating positive displacemant (piston) compressor.
I have found a suitable pulley, so I'll make the swap and measure the current.

Somehow, it seems that power should be proportional to speed squared for fluids in motion.
Frictional loss, however, may be proportional to speed, so it should be a bit of a mix. Heat loss from compression, I don't know.

How about a raffle; best guess gets bragging rights.
 

Related to Single phase motor loading query

1. What is a single phase motor?

A single phase motor is a type of electric motor that runs on a single phase power supply. It is commonly used in household appliances such as washing machines, refrigerators, and fans.

2. How is a single phase motor different from a three phase motor?

The main difference between a single phase motor and a three phase motor is the number of phases of power supply they require to run. A single phase motor uses one phase, while a three phase motor uses three phases. Additionally, three phase motors are typically more efficient and have a higher power output compared to single phase motors.

3. How is the loading of a single phase motor determined?

The loading of a single phase motor is determined by the amount of power it draws from the power supply. This is usually measured in watts or horsepower. The higher the power drawn, the higher the loading of the motor.

4. What is meant by "overloading" a single phase motor?

"Overloading" a single phase motor refers to running the motor at a power level higher than its rated capacity. This can lead to overheating and damage to the motor, and should be avoided to ensure the motor's longevity and efficiency.

5. How can I calculate the loading of a single phase motor?

The loading of a single phase motor can be calculated by measuring the voltage and current drawn by the motor and using the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes. This will give the power consumption of the motor, which can then be compared to its rated capacity to determine the loading.

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