voltage drop for motor load

[luckylerado]

I am calculating voltage drop for an overhead crane. 3 phase 480 total length from source including conductive rails and traveling cable to the furthest point is 600 feet. When I run the numbers for voltage drop based on the overcurrent device <50 amp breaker> I am getting a #3 to be within 3% however when I use the total FLA of the 3 motors <19 amps> I am figuring a #6.

Is it acceptable to use the combined FLA when calculating VD for a motor load circuit or do I need to consider the starting current?

The formula I am using is 1.732*K*D*I/cm

I must back up my numbers with the government before they will accept the crane installation.

 

[gar]

101107-0852 EST

luckylerado:

What does 3% have to do with your question?

The first question should be what is the minimum voltage to the motor such that the motor can effectively do its job? This probably needs to be broken into two parts --- start-up, the time of initial inrush --- peak power required after start-up. Initial inrush may be your limiting factor.

What kind of motors and controls are on the overhead crane? Do they run simultaneously or never together?

Suppose you had an electronically controlled drive, then inrush in the usual sense might not be important. On the other hand consider my DeWalt saw that I have previously mentioned. Operated on 120 V and at the end of about 100+ feet of #12 copper and with a nearly maximum diameter blade it is very marginal at start-up. Voltage at the saw drops 30 to 40 V while starting up and lasts for about 3 to 5 seconds. This is very marginal and if done several times in sequence without too much time difference the QO20 will trip.

Once you know what you want to achieve during start-up and under peak normal loading, then you have a minimum voltage figure with which to work. Use this to size the wire under minimum supply voltage conditions. Next does that wire size meet code requirements. If not, then increase the size to meet code.

.

 

[Smart $]

I am calculating voltage drop for an overhead crane. 3 phase 480 total length from source including conductive rails and traveling cable to the furthest point is 600 feet. When I run the numbers for voltage drop based on the overcurrent device <50 amp breaker> I am getting a #3 to be within 3% however when I use the total FLA of the 3 motors <19 amps> I am figuring a #6.

Is it acceptable to use the combined FLA when calculating VD for a motor load circuit or do I need to consider the starting current?

The formula I am using is 1.732*K*D*I/cm

I must back up my numbers with the government before they will accept the crane installation.

Your ocpd rating plays no direct role in voltage drop. You can(should?) use FLA of motors plus any non-motor combination load current. If you want to go larger, that is strictly [another] design decision. Generally, in-rush current is not a limiting factor because calculated voltage drop is not instantaneous. Voltage drop is caused by the resistance of the conductor and the current generating heat in the conductor. The result is increased resistance and slighty less current. This is action continues until equilibrium is achieved. I don't know of any formula to calculate how long it takes to change the temperature, thus the resistance encountered... but I do know it is not instantaneous, at least not within normal operating temperature ratings.

 

[iamintn]

I've had to supply similar calcs for governement approval before. Smart $ is correct. Use the FLA. #6 should be fine.

 

[luckylerado]

Thanks Guys

 

[Electric-Light]

101107-0852 EST

luckylerado:

What does 3% have to do with your question?

The first question should be what is the minimum voltage to the motor such that the motor can effectively do its job? This probably needs to be broken into two parts --- start-up, the time of initial inrush --- peak power required after start-up. Initial inrush may be your limiting factor.

What kind of motors and controls are on the overhead crane? Do they run simultaneously or never together?

Suppose you had an electronically controlled drive, then inrush in the usual sense might not be important. On the other hand consider my DeWalt saw that I have previously mentioned. Operated on 120 V and at the end of about 100+ feet of #12 copper and with a nearly maximum diameter blade it is very marginal at start-up. Voltage at the saw drops 30 to 40 V while starting up and lasts for about 3 to 5 seconds. This is very marginal and if done several times in sequence without too much time difference the QO20 will trip.

Saws and fans are easy since you're not working against a dead weight and reduced available current simply extends the time it takes to reach full speed.

Positive displacement compressors and mechanical hoists are the hardest to start since you have to overcome the full rated weight from dead stop unless you're using a mechanical clutch & flywheel or a torque converter.

Refrigerator is a good example. They use something like 75-100W once running, but the starting kVA is very high and they're hard to start on an inverter. It's not unheard of to need a 1,000W rated inverter to start a fridge.

Once you know what you want to achieve during start-up and under peak normal loading, then you have a minimum voltage figure with which to work. Use this to size the wire under minimum supply voltage conditions. Next does that wire size meet code requirements. If not, then increase the size to meet code.
.

Code is minimum, but for a crane, worst possible conditions should be considered (likely possible minimum supply voltage, rated maximum load, or the maximum load OP will ever place on this crane).

You could end up with a situation where you could lift by exploiting the slack in the cable at the beginning and using the rotor inertia at the beginning but not able to start back up if you were to stop the hoisting with the load mid-air.

Ideally, it would be able to start up under worst possible utility voltage against rated max weight mid air.