Another voltage drop??
- Thread starter eetwo
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- Wisconsin
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- PE (Retired) - Power Systems
Both should be considered.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101203-1247 EST
eetwo:
Do you have an actual installation, or a hypothetical problem?
Certainly you want to meet some reasonable criteria for voltage drop at full load.
If you have an application where there is a large voltage drop from inrush current, then you need to question what effect this has on the performance of the motor and its connected system.
My DeWalt radial arm saw has a problem on 120 V at the end of something over a 100 run of #12 copper. The start-up time is unsatisfactory, but the saw gets very little use, and therefore no reason to change. If this was in a lumber yard, then it would get immediate attention. You could not live with this as a problem.
You need to figure out what are the correct questions to ask to solve whatever problems may exist.
If I had a fixed location for my DeWalt, then I would change it over to 240 V and stay with #12 wire.
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eetwo:
Do you have an actual installation, or a hypothetical problem?
Certainly you want to meet some reasonable criteria for voltage drop at full load.
If you have an application where there is a large voltage drop from inrush current, then you need to question what effect this has on the performance of the motor and its connected system.
My DeWalt radial arm saw has a problem on 120 V at the end of something over a 100 run of #12 copper. The start-up time is unsatisfactory, but the saw gets very little use, and therefore no reason to change. If this was in a lumber yard, then it would get immediate attention. You could not live with this as a problem.
You need to figure out what are the correct questions to ask to solve whatever problems may exist.
If I had a fixed location for my DeWalt, then I would change it over to 240 V and stay with #12 wire.
.
brian john
Senior Member
- Location
- Leesburg, VA
kingpb
Senior Member
- Location
- SE USA as far as you can go
- Occupation
- Engineer, Registered
For motor voltage drop during starting, you need to take into consideration performance requirements, VD is just one aspect.
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
A lot will depend upon the type of motor and its use as to if VD will be a problem.
Since a saw was brought up, I will say this most hand tool motors and some table type saws have brushes in them this means they are a DC motor and VD just makes them run slower and have less torque, synchronous motors on the other hand don't like voltage drop too much and can over heat and burn up, this has to do with how they operate, for the most part 5% VD will allow most motors to startup and run with no problems, but High torque application type motors for loads like an air compressor can stall at very little voltage drop.
I have seen many portable air compressors burned up on the end of a 100' 16/3 extension cord. but hand tools with DC motors will run all day.
Look up "all about electric motors" and you should find allot of info on how different motors and uses will determine how voltage drop will effect them.
Since a saw was brought up, I will say this most hand tool motors and some table type saws have brushes in them this means they are a DC motor and VD just makes them run slower and have less torque, synchronous motors on the other hand don't like voltage drop too much and can over heat and burn up, this has to do with how they operate, for the most part 5% VD will allow most motors to startup and run with no problems, but High torque application type motors for loads like an air compressor can stall at very little voltage drop.
I have seen many portable air compressors burned up on the end of a 100' 16/3 extension cord. but hand tools with DC motors will run all day.
Look up "all about electric motors" and you should find allot of info on how different motors and uses will determine how voltage drop will effect them.
Ok here is the problem. I am working on the installation of a maple sap pump the water pump 1 hp, and will be running +/-30 cycles per hour, each cycle in the area of 30 seconds. The pump is located 1100' from the power supply which is it self 1300' from the service. The existing location is fed with #6 copper, and has a 3/4hp pump as its only load. The customer buried 1 1/2" PVC to the new location. So I need to size the lines and pull them in. When doing voltage drop calcs I am unsure if should be using the FLA of the motor (10 amps) or the inrush (22 amps) for the calculations. The difference in the two numbers is small, but it translates to huge $'s as copper is high again.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101203-2211 EST
eetwo:
Your were ask what is the motor type. That is important. I doubt that the inrush current is only about 2 times the full load current. But the type of motor could influence it. My guess is 4 to 1 or more.
I do not understand what "and will be running +/-30 cycles per hour," means. How can you have minus cycles per hour?
Your cycle time makes more sense. So if the on-off period is 30 seconds, for example 15 on and 15 off, then there may be a lot of motor heating from the inrush. This would be 120 on-off cycles per hour.
The heating in the wire makes no difference other than you want that heating to be below code requirements.
What does matter is ability of the motor to reliably start (get up to a reasonable full speed), and motor heating resulting from many startup inrush cycles in a short time relative to the thermal time constant of the motor.
It appears you have total run of 2400 ft. Part of this run is 1300 ft of an unspecified wire size, and 1100 ft is possibly #6 copper. Consider the 1100 ft of #6 at 20 deg C, too low, but that is where my wire table is pegged. The resistance is 0.395 ohms per 1000 ft. Thus, your resistance is 2*1.1*0.395 = 0.869 ohms. There is also inductance but I will ignore that. At 50 A, 5 times full load, the loop voltage drop is at least 43 V. If there was 120 V at what you define as the source, then at the motor the voltage would be 120 -43 = 77 V. Since your have another 1300 ft of something you can expect your source voltage to drop and therefore the motor voltage would be lower yet. Also note inductance was neglected. This is just an illustration of a potential problem. Better information is needed. Also does the power company voltage drop below 120, like maybe 105. What I am referring to is the equivalent source voltage within the supply transformer.
Start with more details on the motor. What is its locked rotor current? Maybe you need to run some experiments on the motor under operating conditions. Then apply a generous safety factor to any judgements you make.
You really need to consider higher voltage for long runs like this.
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eetwo:
Your were ask what is the motor type. That is important. I doubt that the inrush current is only about 2 times the full load current. But the type of motor could influence it. My guess is 4 to 1 or more.
I do not understand what "and will be running +/-30 cycles per hour," means. How can you have minus cycles per hour?
Your cycle time makes more sense. So if the on-off period is 30 seconds, for example 15 on and 15 off, then there may be a lot of motor heating from the inrush. This would be 120 on-off cycles per hour.
The heating in the wire makes no difference other than you want that heating to be below code requirements.
What does matter is ability of the motor to reliably start (get up to a reasonable full speed), and motor heating resulting from many startup inrush cycles in a short time relative to the thermal time constant of the motor.
It appears you have total run of 2400 ft. Part of this run is 1300 ft of an unspecified wire size, and 1100 ft is possibly #6 copper. Consider the 1100 ft of #6 at 20 deg C, too low, but that is where my wire table is pegged. The resistance is 0.395 ohms per 1000 ft. Thus, your resistance is 2*1.1*0.395 = 0.869 ohms. There is also inductance but I will ignore that. At 50 A, 5 times full load, the loop voltage drop is at least 43 V. If there was 120 V at what you define as the source, then at the motor the voltage would be 120 -43 = 77 V. Since your have another 1300 ft of something you can expect your source voltage to drop and therefore the motor voltage would be lower yet. Also note inductance was neglected. This is just an illustration of a potential problem. Better information is needed. Also does the power company voltage drop below 120, like maybe 105. What I am referring to is the equivalent source voltage within the supply transformer.
Start with more details on the motor. What is its locked rotor current? Maybe you need to run some experiments on the motor under operating conditions. Then apply a generous safety factor to any judgements you make.
You really need to consider higher voltage for long runs like this.
.
kwired
Electron manager
- Location
- NE Nebraska
Is there any way possible to use a smaller pump - like maybe pumping at a slower rate for a longer time - may reduce current even more and you will have less voltage drop issues.
I do agree you need to run higher voltage for the long feed, with them kind of lengths the size of transformers needed probably cost less than the larger conductors and raceways that would otherwise be needed. Use of three phase instead of single phase if that is available is also worth consideration.
Soft start or VFD will reduce starting current so you will have less voltage drop at starting.
I do agree you need to run higher voltage for the long feed, with them kind of lengths the size of transformers needed probably cost less than the larger conductors and raceways that would otherwise be needed. Use of three phase instead of single phase if that is available is also worth consideration.
Soft start or VFD will reduce starting current so you will have less voltage drop at starting.
Last edited:
kwired
Electron manager
- Location
- NE Nebraska
Another possible idea is a battery powered DC motor and run an AC line to run the battery charger - hopefully the battery charger will not need to draw much current to replenish the charge before the motor will need to run again. Or go green and solar or wind power to charge a battery.
SAC
Senior Member
- Location
- Massachusetts
"I think what he probably means is that it will start 30 times per hour, give or take, and when it does run, it will run for about 30 seconds. So 90 seconds off, 30 seconds on. " This is a sap transfer pump, basically sap from the trees collects in a small tank , 5 gallons or so, when it is full the pump starts to pump the sap, the 1100' to the next collection tank, which then pumps the sap 1300 feet to the roadside for pick-up. So some days it may run near constantly, and others it may run once an hour.
exactly what I meant, I am not sure what kind of motor the pump has, as it hasn't arrived yet. The motor will run 240 volts not 120. Yes it is correct that the total circuit length will be 2400' from the utility transformer. 3-phase is not an option. Neither is larger conduit for the new pump location, as it is long buried. Ambient temps will be in the 32-40F range when this is in operation. I am considering step up/stepdown for the new run, but am also working with americas farmer, who doesn't have a lot of money to spend.
exactly what I meant, I am not sure what kind of motor the pump has, as it hasn't arrived yet. The motor will run 240 volts not 120. Yes it is correct that the total circuit length will be 2400' from the utility transformer. 3-phase is not an option. Neither is larger conduit for the new pump location, as it is long buried. Ambient temps will be in the 32-40F range when this is in operation. I am considering step up/stepdown for the new run, but am also working with americas farmer, who doesn't have a lot of money to spend.
kwired
Electron manager
- Location
- NE Nebraska
Sounds to me like he possibly could save more by having a better plan before burying a raceway without any clue as to what will need to be pulled through it.
Also as I mentioned before, different design could cost less to install because of less load over more time. I don't know all the details of the process involved, some of my ideas may not work.
You may possibly be commited to whatever the plan is whether it is the most practical or not.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101204-1055 EST
eetwo:
7 to 8 A is a more likely value for a 240 V 1 HP motor. Your 10 A seemed to imply a 120 V motor. If we assume 8 A for full load of a 240 motor, then the inrush is probably around 40 A for an induction motor.
Here are two definitions for the word cycle from dictionary.com
"any complete round or series of occurrences that repeats or is repeated. ", or
"a sequence of changing states that, upon completion, produces a final state identical to the original one."
So a cycle is from --- the point in time of motor start, thru turn off, and on to the point of the next motor start ---
A more useful statement for your number of cycles per hour might have been 30 +10/-30. The 30-30 implies no syrup, and that is a possibility at some point. Maybe words like --- the typical number of cycles per hour is 30 but might range from 40 to 0 --- would more accurately describe the operation.
From some of the above figures assume the pump can pump 10 gallons/minute, and the tank is 5 gallons, then saturation occurs when the inflow is 10 gallons per minute and the pump runs continuously.
Drop the inflow to 5 gallons per minute. Start from an empty tank. It takes 1 minute to fill the tank. The pump turns on. It removes syrup faster than it is coming in, but it takes longer than 1/2 minute to empty the tank because additional syrup is flowing in. The equation for this is
g(t) = 5 + 5t - 10t = 5 - 5t
where t = 0 is the time when the tank reaches full and t is in minutes
g(t) is the number of gallons in the tank at time t
g(t) becomes 0 when t = 1 minute. At this point in time the pump turns off.
It takes 1 minute to refill, then the cycle starts over.
For these assumptions the duty cycle is 50%, and the cycle time is 120 seconds ( 2 minutes or 30 cycles per hour ). This is sort of intuitive that if the inflow is half of the pump capacity that the duty cycle would be 50 %.
The motor cools during the 50% off time. 2 to 3 seconds of high inrush current to the motor is probably easily dissipated in the 120 second cycle time. Any slower inflow rate provides more cooling time for the motor.
Consider 40 A ( 5 * 8 = 40 ) inrush and 0.87 ohms resistance and the voltage drop is 34.8 or about 40. 40/240 = about 17 % drop. Probably tolerable but you really need to know how the motor performs under low voltage conditions and what is the lowest supply voltage you can normally expect at this location.
.
eetwo:
7 to 8 A is a more likely value for a 240 V 1 HP motor. Your 10 A seemed to imply a 120 V motor. If we assume 8 A for full load of a 240 motor, then the inrush is probably around 40 A for an induction motor.
Here are two definitions for the word cycle from dictionary.com
"any complete round or series of occurrences that repeats or is repeated. ", or
"a sequence of changing states that, upon completion, produces a final state identical to the original one."
So a cycle is from --- the point in time of motor start, thru turn off, and on to the point of the next motor start ---
A more useful statement for your number of cycles per hour might have been 30 +10/-30. The 30-30 implies no syrup, and that is a possibility at some point. Maybe words like --- the typical number of cycles per hour is 30 but might range from 40 to 0 --- would more accurately describe the operation.
From some of the above figures assume the pump can pump 10 gallons/minute, and the tank is 5 gallons, then saturation occurs when the inflow is 10 gallons per minute and the pump runs continuously.
Drop the inflow to 5 gallons per minute. Start from an empty tank. It takes 1 minute to fill the tank. The pump turns on. It removes syrup faster than it is coming in, but it takes longer than 1/2 minute to empty the tank because additional syrup is flowing in. The equation for this is
g(t) = 5 + 5t - 10t = 5 - 5t
where t = 0 is the time when the tank reaches full and t is in minutes
g(t) is the number of gallons in the tank at time t
g(t) becomes 0 when t = 1 minute. At this point in time the pump turns off.
It takes 1 minute to refill, then the cycle starts over.
For these assumptions the duty cycle is 50%, and the cycle time is 120 seconds ( 2 minutes or 30 cycles per hour ). This is sort of intuitive that if the inflow is half of the pump capacity that the duty cycle would be 50 %.
The motor cools during the 50% off time. 2 to 3 seconds of high inrush current to the motor is probably easily dissipated in the 120 second cycle time. Any slower inflow rate provides more cooling time for the motor.
Consider 40 A ( 5 * 8 = 40 ) inrush and 0.87 ohms resistance and the voltage drop is 34.8 or about 40. 40/240 = about 17 % drop. Probably tolerable but you really need to know how the motor performs under low voltage conditions and what is the lowest supply voltage you can normally expect at this location.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
101204-1132 EST
eetwo:
Percent voltage drop is not your real concern.
The important considerations are:
1. The minimum voltage that will result in satisfactory startup of the motor.
2. The minimum voltage that will result in satisfactory operation of the motor under continuous steady-state load.
What does satisfactory mean for startup?
The motor will start and come up to normal speed.
If there is a high cycling rate, then that the motor does not overheat.
That the circuit breaker and/or fuse protecting the motor or circuit does not trip.
What does satisfactory mean for steady-state operation?
The desired function can be performed.
The motor does not overheat.
These criteria are a function of absolute voltage to the motor, not a percentage drop from some voltage. Suppose you have a nominal 240 V supply, and the percentage criteria is a 10% drop from 240, or an absolute of 240-24 = 216 V at the motor for a minimum. Now what happens if the supply is actually 220 V and you still have the 24 V drop? The current did not drop and may actually have increased. In this case you would be below a satisfactory operating point.
In general some of the rough design guide lines and the way motors are designed does provide some variation in supply voltage and still provide satisfactory performance.
Your case where you want to minimize the copper cost means you need a fair amount of information on the motor characteristics and your main supply voltage variation.
My point is you can not just use some arbitrary current and percentage drop as a criteria for your design.
Review the history of the development of the electric cash register, and the automotive starter to see how some of these factors are important. Search for Charles F. Kettering and see what you can learn.
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eetwo:
Percent voltage drop is not your real concern.
The important considerations are:
1. The minimum voltage that will result in satisfactory startup of the motor.
2. The minimum voltage that will result in satisfactory operation of the motor under continuous steady-state load.
What does satisfactory mean for startup?
The motor will start and come up to normal speed.
If there is a high cycling rate, then that the motor does not overheat.
That the circuit breaker and/or fuse protecting the motor or circuit does not trip.
What does satisfactory mean for steady-state operation?
The desired function can be performed.
The motor does not overheat.
These criteria are a function of absolute voltage to the motor, not a percentage drop from some voltage. Suppose you have a nominal 240 V supply, and the percentage criteria is a 10% drop from 240, or an absolute of 240-24 = 216 V at the motor for a minimum. Now what happens if the supply is actually 220 V and you still have the 24 V drop? The current did not drop and may actually have increased. In this case you would be below a satisfactory operating point.
In general some of the rough design guide lines and the way motors are designed does provide some variation in supply voltage and still provide satisfactory performance.
Your case where you want to minimize the copper cost means you need a fair amount of information on the motor characteristics and your main supply voltage variation.
My point is you can not just use some arbitrary current and percentage drop as a criteria for your design.
Review the history of the development of the electric cash register, and the automotive starter to see how some of these factors are important. Search for Charles F. Kettering and see what you can learn.
.
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