Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations TugboatEng on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Motor voltage 380-480v
- Thread starter Indy
- Start date
- Moderator
- #2
The simple answer is yes.
4.6 A x 415 V x 1.73 = 3303 VA
3 kW x 746 W/HP = 2238 VA
2238 VA / 3303 VA = 68%
The long answer is:
If the combined efficiency and power factor is above 68%,
AND if the motor is running at rated frequency and speed the motor will be capable of 3kW.
Bill
--------------------
"Why not the best?"
Jimmy Carter
4.6 A x 415 V x 1.73 = 3303 VA
3 kW x 746 W/HP = 2238 VA
2238 VA / 3303 VA = 68%
The long answer is:
If the combined efficiency and power factor is above 68%,
AND if the motor is running at rated frequency and speed the motor will be capable of 3kW.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Can you please explain this equation @waross
4.6 A x 415 V x 1.73 = 3303 VA
3 kW x 746 W/HP = 2238 VA
2238 VA / 3303 VA = 68%
Thanks!
4.6 A x 415 V x 1.73 = 3303 VA
3 kW x 746 W/HP = 2238 VA
2238 VA / 3303 VA = 68%
Thanks!
- Moderator
- #6
If the motor stator is provided with 6 separated windings [9 terminals] 380 to 480[460 V] it is possible.
a single one winding rated current is 4.6 A [connected series] and 480/2/sqrt(3)=138.6 V. In parallel for 380 V the motor could draw double current but the voltage will be 380/sqrt(3)=220 V per bobbin.
A bobbin is then rated for 220 V and 4.6 A.
At 415 V you have to connect series the bobbins since in parallel the voltage per bobbin will be 10% more than rated.
The current will decrease only by 13.6% but the torque 25.3% .Then the slip has to rise proportionally If at rated voltage was 0.5% now will be 0.5/.253=~2%.
As the maximum torque slip it is about 6*srat=3% it will be still o.k.
a single one winding rated current is 4.6 A [connected series] and 480/2/sqrt(3)=138.6 V. In parallel for 380 V the motor could draw double current but the voltage will be 380/sqrt(3)=220 V per bobbin.
A bobbin is then rated for 220 V and 4.6 A.
At 415 V you have to connect series the bobbins since in parallel the voltage per bobbin will be 10% more than rated.
The current will decrease only by 13.6% but the torque 25.3% .Then the slip has to rise proportionally If at rated voltage was 0.5% now will be 0.5/.253=~2%.
As the maximum torque slip it is about 6*srat=3% it will be still o.k.
Sorry. Wrong numbers. On an actual motor of 3 kW 460 V Tm/tr=2.9 , rated slip about 3% and the maximum torque slip was 16.8%
At 415 V the slip will be 3%/.253=11.86% and the sm=48.6%
The old rpm was n460V=nsyn*(1-3%) and the new n415V=nsyn*(1-11.86%).
The power required by the motor-for the same rated torque- will be 3*(1-11.86%)/(1-3%)=2.73 kW![[blush]](/data/assets/smilies/blush.gif "[blush] [blush]")
At 415 V the slip will be 3%/.253=11.86% and the sm=48.6%
The old rpm was n460V=nsyn*(1-3%) and the new n415V=nsyn*(1-11.86%).
The power required by the motor-for the same rated torque- will be 3*(1-11.86%)/(1-3%)=2.73 kW
However, something is still wrong here. This calculation does not match the experience, yet.
Let's start from the induction motor diagram.
The power transferred from stator to rotor it is:
Pi=m2*R2*I2^2/s where 2 it means "rotor" and m2=rotor number of phases.
s=(nsyn-n)/nsyn the slip where:
nsyn=synchronous rpm ; n=the actual rotor rpm.
I2=Vstator/SQRT(R1+R2/s)^2+X2^2) neglecting Io and X1[stator leakage reactance]. For simplification's sake I'll take the R1=0 as well- where 1 it means “stator".
The torque it is the power divided by velocity in rad/sec.
T=m2*R2*V^2/(R1+R2/s)^2+X2^2)/s/2/pi()/nsyn/(1-s)
If K=m2*R2/(2*pi()*nsyn) and R2^2=A X2^2=B Neglecting R1
T=K*V^2*s/(A+B*s^2-A*s-B*s^3)
Since s is very low we can say s^3=s^2=0
T=K*V^2*s/A(1-s)
For different V[Vi initial and Vf after change]:
Ti/Tf=(Vi/Vf)^2*si/sf/(1-si)*(1-sf)
We may consider (1-s1)/(1-s2)=1 and for a same required torque in different supply voltages Ti/Tf=1 then:
sf=(Vi/Vf)^2*si
s(415V)=(480/415)^2*3%=4%!!!
That means it is not 11.86% but only 4%. This result it is more possible.
Let's start from the induction motor diagram.
The power transferred from stator to rotor it is:
Pi=m2*R2*I2^2/s where 2 it means "rotor" and m2=rotor number of phases.
s=(nsyn-n)/nsyn the slip where:
nsyn=synchronous rpm ; n=the actual rotor rpm.
I2=Vstator/SQRT(R1+R2/s)^2+X2^2) neglecting Io and X1[stator leakage reactance]. For simplification's sake I'll take the R1=0 as well- where 1 it means “stator".
The torque it is the power divided by velocity in rad/sec.
T=m2*R2*V^2/(R1+R2/s)^2+X2^2)/s/2/pi()/nsyn/(1-s)
If K=m2*R2/(2*pi()*nsyn) and R2^2=A X2^2=B Neglecting R1
T=K*V^2*s/(A+B*s^2-A*s-B*s^3)
Since s is very low we can say s^3=s^2=0
T=K*V^2*s/A(1-s)
For different V[Vi initial and Vf after change]:
Ti/Tf=(Vi/Vf)^2*si/sf/(1-si)*(1-sf)
We may consider (1-s1)/(1-s2)=1 and for a same required torque in different supply voltages Ti/Tf=1 then:
sf=(Vi/Vf)^2*si
s(415V)=(480/415)^2*3%=4%!!!
That means it is not 11.86% but only 4%. This result it is more possible.
- Moderator
- #10
One possibility:
3 kW x 380V/480V = 3.8 kW The motor is designed as a 3.8 kW motor at 480 Volts. It may be safely run at 480 Volts saturation.
On 380 Volts there is enough iron to develop 3 kW without overheating.
Another possibility:
Could this be a harmonization issue?
It is common in North America to run 460 Volt rated motors on 480 Volt supplies.
480 Volts three phase is a common voltage. 240 Volt three phase is not a common voltage. 120/208 Volts is a common supply voltage.
At one time 230-460 Volt rated motors were run on 208 volts with slightly reduced capacity. This is still the case but in addition triple voltage motors are becomming common.
A common rating is 208-230/460 Volts. These motors are suitable for use in shopping centers, malls and large apartment buildings where 120/208 Volts is a common supply voltage and for use in industrial applications where 480 Volts is a common supply.
Refrigeration and air conditioning equipment is generally rated for 208-230/460 volts.
A picture of the nameplate will be helpful.
Bill
--------------------
"Why not the best?"
Jimmy Carter
3 kW x 380V/480V = 3.8 kW The motor is designed as a 3.8 kW motor at 480 Volts. It may be safely run at 480 Volts saturation.
On 380 Volts there is enough iron to develop 3 kW without overheating.
Another possibility:
ScottyUK said:
Could this be a harmonization issue?
It is common in North America to run 460 Volt rated motors on 480 Volt supplies.
480 Volts three phase is a common voltage. 240 Volt three phase is not a common voltage. 120/208 Volts is a common supply voltage.
At one time 230-460 Volt rated motors were run on 208 volts with slightly reduced capacity. This is still the case but in addition triple voltage motors are becomming common.
A common rating is 208-230/460 Volts. These motors are suitable for use in shopping centers, malls and large apartment buildings where 120/208 Volts is a common supply voltage and for use in industrial applications where 480 Volts is a common supply.
Refrigeration and air conditioning equipment is generally rated for 208-230/460 volts.
A picture of the nameplate will be helpful.
Bill
--------------------
"Why not the best?"
Jimmy Carter
LionelHutz
Electrical
Is this 50Hz vs 60Hz?
In North America. a motor running on a 480V power system is typically nameplated as 460V and 380V/460V =~ 50Hz/60Hz.
Even if it is 480V nameplated, I say that 380V/480V =~ 50Hz/60Hz is still close enough to be possible.
So, did the frequency of each voltage get left out of the question?
In North America. a motor running on a 480V power system is typically nameplated as 460V and 380V/460V =~ 50Hz/60Hz.
Even if it is 480V nameplated, I say that 380V/480V =~ 50Hz/60Hz is still close enough to be possible.
So, did the frequency of each voltage get left out of the question?
- Status
Similar threads
- Locked
- Question
- Question