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# Help me for selecting a AC motor!!!!!!

Discuss Help me for selecting a AC motor!!!!!! in the Commercial Electrical Advice area at ElectriciansForums.net

T

#### thanhtongminh

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Dear all,
I'm doing my project in university. My project is to replace a DC
drive by a AC drive using inverter.
The old DC drive parameters are:
- Power 450kW
- Speed 50 to 1000r/min
(so Tn = 4298Nm) (Tn: nominal torque)
and the requirements are 250 % during 60 seconds acceleration and a
periodical overload of 160 percent during 30 seconds. That means I
need a starting torque is 250% nominal torque during 60 seconds.
(Ts = 2.5*Ts = 10745Nm)(Ts: starting torque)
Now I need to select an asynchronous motor to satisfy these
requirements but the problems are:
- The starting torque of AC motor is lesser than DC motor so I must
select a AC motor with larger power
AC motor : Ts = (1.5-2.5)Tn
So I must select AC motor power larger than DC motor to satisfy the
starting requirement.
I selected : P(AC motor) = 1.5 * P(DC motor) =675kW
So in AC drive I have:
motor power : P(AC) = 675kW
speed 1000r/min
Tn(AC) = 1.5 Tn(DC) = 6447Nm
Ts(AC) approximates 2*Tn(AC) = 12894Nm
They are my first draft calculations. Am I right?
Can you help me? and give me some advise.
I have searched on the internet an AC motor like that but I haven't
found any motor. If you have any information or help me find a motor
like that.
Look forward to hearing you soon.
Thanks
Thanh
my email is [email protected]

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R

#### rumrunner

Dear all,
I'm doing my project in university. My project is to replace a DC
drive by a AC drive using inverter.
The old DC drive parameters are:
- Power 450kW
- Speed 50 to 1000r/min
(so Tn = 4298Nm) (Tn: nominal torque)
and the requirements are 250 % during 60 seconds acceleration and a
periodical overload of 160 percent during 30 seconds. That means I
need a starting torque is 250% nominal torque during 60 seconds.
(Ts = 2.5*Ts = 10745Nm)(Ts: starting torque)
Now I need to select an asynchronous motor to satisfy these
requirements but the problems are:
- The starting torque of AC motor is lesser than DC motor so I must
select a AC motor with larger power
AC motor : Ts = (1.5-2.5)Tn
So I must select AC motor power larger than DC motor to satisfy the
starting requirement.
I selected : P(AC motor) = 1.5 * P(DC motor) =675kW
So in AC drive I have:
motor power : P(AC) = 675kW
speed 1000r/min
Tn(AC) = 1.5 Tn(DC) = 6447Nm
Ts(AC) approximates 2*Tn(AC) = 12894Nm
They are my first draft calculations. Am I right?
Can you help me? and give me some advise.
I have searched on the internet an AC motor like that but I haven't
found any motor. If you have any information or help me find a motor
like that.
Look forward to hearing you soon.
Thanks
Thanh
my email is [email protected]
Whats it to power ?,is it a ship ?,because 900 hp motors are not simply available off the shelf ,what to do is look at wind generators ,they use asyncronous motors,as generators ,heres a link ,it should tell you all you need to know Wind Energy Reference Manual ,theres some info on the maths of the thing as well,hope its usefull .if not im stumped you will have to ask the shake.
best of luck with your project.

Last edited by a moderator:
S

#### Shakey

Does the speed need need to vary when it is running?

If so, you will need an invertor drive

If not, I i would use an auto-synchronous induction motor

This is basically a wound rotor induction motor with slip rings. It starts as a wound rotor, and the Back EMF is limited by banks of resistors connected to the rotor windings connected through the slip rings

As the speed builds up, the resistors are taken out by banks of contactors in sequence

when it is up to slip speed, the final resistors are taken out and DC is applied to the rotor to make it synchronous. They are very high power (and torque) and sound ideal for your theoretical application

They have the added advantage that when syncronous, you can over-excite the rotor and turn it into a synchronous capacitor, which, if you are using it in a large industrial application, can be used for power factor correction

they are identified by the fact that one slip ring is larger than the other two, because when DC is fed to the rotor in synchronous mode, power goes in on one slip ring, and back out on two, so when is larger to handle the current flow

perhaps this has helped, perhaps it hasnt, but hey, its Bank Holiday Sunday and i'm on the 'Bow!!!!

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