Can we calculate the load current for Sizing the Busbar and incomer breaker with Motor Shaft power? Process and Mechanical Team have given us the motor Shaft power. Can I use this data instead of the Rated Power of the motor to calculate the Current and size of the Breaker? Please advise.
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MCC Bus and Incomer Breaker Sizing 1
- Thread starter Histor
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1
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Dear Mr. Histor (Electrical)(OP)16 Mar 22 10:51
" Can we calculate the load current for Sizing the Busbar and incomer breaker with Motor Shaft power? Process and Mechanical Team have given us the motor Shaft power. Can I use this data instead of the Rated Power of the motor to calculate the Current and size of the Breaker? Please advise."
1. No. " Motor Shaft power " is the motor [output power] needed in (kW) to drive the mechanical load.
2. The [electrical input] to the motor in (kVA) > Motor Shaft power in (kW). This is due to the (power-factor) and (efficiency) of any motor are always < 1. e.g. pf=0.8 and eff=0.9...
3. If the breaker and busbar is supplying only one motor, is shall be at least sized to say >1.25 of the calculated kVA. Refer to local regulation which may differ depending on IEC or NEC applicable at the location.
4. However, if the breaker and busbar is supplying [a group of motors], they can be < than the total [sum of the Motor Shaft power] in (kW). This is taking into consideration that NOT all the motors are loaded continuously to the maximum (kW) at the same time. Refer to local regulation which may differ depending on IEC or NEC applicable at the location.
Che Kuan Yau (Singapore)
" Can we calculate the load current for Sizing the Busbar and incomer breaker with Motor Shaft power? Process and Mechanical Team have given us the motor Shaft power. Can I use this data instead of the Rated Power of the motor to calculate the Current and size of the Breaker? Please advise."
1. No. " Motor Shaft power " is the motor [output power] needed in (kW) to drive the mechanical load.
2. The [electrical input] to the motor in (kVA) > Motor Shaft power in (kW). This is due to the (power-factor) and (efficiency) of any motor are always < 1. e.g. pf=0.8 and eff=0.9...
3. If the breaker and busbar is supplying only one motor, is shall be at least sized to say >1.25 of the calculated kVA. Refer to local regulation which may differ depending on IEC or NEC applicable at the location.
4. However, if the breaker and busbar is supplying [a group of motors], they can be < than the total [sum of the Motor Shaft power] in (kW). This is taking into consideration that NOT all the motors are loaded continuously to the maximum (kW) at the same time. Refer to local regulation which may differ depending on IEC or NEC applicable at the location.
Che Kuan Yau (Singapore)
- Thread starter
- #3
Dear Mr. Kuan Yu, For Example there are 3 Motors connected to an MCC.
Motor A : Rated power 3kW, Shaft Power 2.6kW
Motor B : Rated power 15kW, Shaft Power 12.8kW
Motor C : Rated power 37kW, Shaft Power 32.8kW
All motors are on continuous duty. 400V, 0.85 PF
in This case for Sizing the MCC Bus and Shaft Power, Can I use Shaft power (Actual required power) for Calculation Circuit overall current and Size the Circuit Breaker?
Procedure 1: Sizing the Circuit Breaker using Shaft Power
Overall Shaft power = 48.2kW
Load Current = (48.2x1000)/ (1.732x400x0.85) = 81.8A
Size of Circuit Breaker = 1.25 x 81.8A = 102.25AT / 160AF
Procedure 2: Sizing the Circuit Breaker using Rated Power
Overall Rated power = 55kW
Load Current = (55x1000)/ (1.732x400x0.85) = 93.4A
Size of Circuit Breaker = 1.25 x 93.4A = 116.75AT / 160AF
I have been doing Procedure 2. But some of my colleagues told me to follow Procedure 1. Please advise if I can use Procedure 1?
in the given example the number of loads is less. But in reality, we have more loads so I get a significant amount of difference in Load Current when I use these two Procedures.
Motor A : Rated power 3kW, Shaft Power 2.6kW
Motor B : Rated power 15kW, Shaft Power 12.8kW
Motor C : Rated power 37kW, Shaft Power 32.8kW
All motors are on continuous duty. 400V, 0.85 PF
in This case for Sizing the MCC Bus and Shaft Power, Can I use Shaft power (Actual required power) for Calculation Circuit overall current and Size the Circuit Breaker?
Procedure 1: Sizing the Circuit Breaker using Shaft Power
Overall Shaft power = 48.2kW
Load Current = (48.2x1000)/ (1.732x400x0.85) = 81.8A
Size of Circuit Breaker = 1.25 x 81.8A = 102.25AT / 160AF
Procedure 2: Sizing the Circuit Breaker using Rated Power
Overall Rated power = 55kW
Load Current = (55x1000)/ (1.732x400x0.85) = 93.4A
Size of Circuit Breaker = 1.25 x 93.4A = 116.75AT / 160AF
I have been doing Procedure 2. But some of my colleagues told me to follow Procedure 1. Please advise if I can use Procedure 1?
in the given example the number of loads is less. But in reality, we have more loads so I get a significant amount of difference in Load Current when I use these two Procedures.
Dear Mr. Histor (Electrical)(OP)
"....#1. I have been doing Procedure 2. But some of my colleagues told me to follow Procedure 1. Please advise if I can use Procedure 1? .... #2....in the given example the number of loads is less. But in reality, we have more loads so I get a significant amount of difference in Load Current when I use these two Procedures...."
#1. Your procedure 1 is based on "overall Shaft power" and arrived at [load current] 81.8A with pf=0.85 is incomplete. You have to consider the efficiency factor say (0.9 ?). See my post dated 17th instant Your procedure 2 is based on "Overall Rated power" and arrived at [load current] 93.4A with pf=0.85 is also incomplete. You have to consider the efficiency factor say (0.9 ?). See my post dated 17th instant.
# 2. In reality a MCC would have more loads. You have to refer to the [local regulation/law] which differ; depending on IEC or NEC that is applicable at the location.
Che Kuan Yau (Singapore)
"....#1. I have been doing Procedure 2. But some of my colleagues told me to follow Procedure 1. Please advise if I can use Procedure 1? .... #2....in the given example the number of loads is less. But in reality, we have more loads so I get a significant amount of difference in Load Current when I use these two Procedures...."
#1. Your procedure 1 is based on "overall Shaft power" and arrived at [load current] 81.8A with pf=0.85 is incomplete. You have to consider the efficiency factor say (0.9 ?). See my post dated 17th instant Your procedure 2 is based on "Overall Rated power" and arrived at [load current] 93.4A with pf=0.85 is also incomplete. You have to consider the efficiency factor say (0.9 ?). See my post dated 17th instant.
# 2. In reality a MCC would have more loads. You have to refer to the [local regulation/law] which differ; depending on IEC or NEC that is applicable at the location.
Che Kuan Yau (Singapore)
- Thread starter
- #5
Dear Mr. Kuan Yu,
Thanks for your reply. What I want to understand is, Can use Shaft power for sizing incoming breakers? I mean the consumed power = Shaft power/efficiency Can I use this consumed power of a number of loads for calculating the incomer breaker current? or do I need to use the rated power of the motors to calculate the current? Thanks.
Thanks for your reply. What I want to understand is, Can use Shaft power for sizing incoming breakers? I mean the consumed power = Shaft power/efficiency Can I use this consumed power of a number of loads for calculating the incomer breaker current? or do I need to use the rated power of the motors to calculate the current? Thanks.
Dear Mr. Histor (Electrical)(OP)21 Mar 22 08:59
".... What I want to understand is, Can use Shaft power for sizing incoming breakers? I mean the consumed power = Shaft power/efficiency Can I use this consumed power of a number of loads for calculating the incomer breaker current? or do I need to use the rated power of the motors to calculate the current? ...."
1. Your [ consumed power]* is the [ mechanical load power in (kW) at 100% loading].
2. Your [ rated power of the motor in (kW) ]** is > [ consumed power]. This is the usual case taking possible "short time over-load" into consideration or that is the next higher standard kW size.
3. Based on above [ ]* and [ ]** ; the breaker and busbar shall be based on [ ]** i.e. taking care of possible over-loading.
4. Attention: The breaker and busbar may be < sum of all [ ]** . You have to refer to the [local regulation/law] which differ; depending on IEC or NEC that is applicable at the location.
Che Kuan Yau (Singapore)
".... What I want to understand is, Can use Shaft power for sizing incoming breakers? I mean the consumed power = Shaft power/efficiency Can I use this consumed power of a number of loads for calculating the incomer breaker current? or do I need to use the rated power of the motors to calculate the current? ...."
1. Your [ consumed power]* is the [ mechanical load power in (kW) at 100% loading].
2. Your [ rated power of the motor in (kW) ]** is > [ consumed power]. This is the usual case taking possible "short time over-load" into consideration or that is the next higher standard kW size.
3. Based on above [ ]* and [ ]** ; the breaker and busbar shall be based on [ ]** i.e. taking care of possible over-loading.
4. Attention: The breaker and busbar may be < sum of all [ ]** . You have to refer to the [local regulation/law] which differ; depending on IEC or NEC that is applicable at the location.
Che Kuan Yau (Singapore)
LionelHutz
Electrical
No, you can not size power wire or bus that way. You must use the rated power. Generally, using 125% of largest motor plus all the other motors is the smallest you should use. Also generally, you use the ANSI or NEC or UL rated motor current tables as the minimum motor current.
The breaker could be whatever you want, there is only a maximum on that. Go too small though and you won't be able to start the motors or will get trips randomly.
The breaker could be whatever you want, there is only a maximum on that. Go too small though and you won't be able to start the motors or will get trips randomly.
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