Chosing proper CB for single phase induction motor

[sameee781]

How can I chose proper circuit breaker to be used in a single phase induction motor?
Example case:
Power: 5hp
Voltage: 230v AC

 

[dpc]

This will depend somewhat on what electrical codes you must comply with. In the US, you must comply with NEC Article 430. Table 430.248 shows a full load current of 28 A. The maximum circuit breaker size (for a standard thermal-mag breaker) would be 250% of this.

 

[sameee781]

P=5hp=746*5w=3730w
V=230v
I=P/V=3730/(230*0.75)=21.6A [where pf=0.75]

How the full load current of this motor be 28A? Why CB would be 250% of full load current? Usually, do the CBs not allow 3 to 5 time of rated current?

 

[ScottyUK]

Factor in the efficiency of the motor - you're calculating output power, not input power. Small motors aren't very efficient, and single phase types are less efficient than three phase types.

 

[sameee781]

Thanks...

Now, my question is Why CB would be 250% of full load current? Usually, do the CBs not allow 3 to 5 time of rated current?

 

[davidbeach]

Motor rules are different. Read the code, the NEC if in the US, whatever is applicable if elsewhere. At least in the NEC, almost every rule is different for motors than for general loads.

 

[dpc]

Usually, do the CBs not allow 3 to 5 time of rated current?

Only for a short period of time. Circuit breakers have two tripping mechanisms - thermal and magnetic. If you apply continuous current above the rating of the breaker, it will eventually trip on the thermal element. The thermal element has an inverse time characteristic.

250% is the maximum rating allowed. It can be smaller if you like.

 

[jraef]

That rule in the NEC by the way presumes other things, such as that you have some OTHER form of thermal overload protection for the motor set closer to the motor FLA, not just a breaker. Again, if you are in the USA*, read and fully understand article 430 of the National Electric Code.

*(presumed by virtue of you using HP to describe your motor and the absence of the word "Eh?" in your posting).

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[sameee781]

No I'm not in USA. I'm in Bangladesh. In our country voltage is 230v with frequency 50Hz. Motor is rated in hp.

 

[jraef]

Ok, then find and read whatever code or standard applies there. The point is, there are rules to follow that may not necessarily track with simple engineering calculations. They generally don't contradict, but they often add to enginnering requirements in the name of safety. Ignore them at your own peril.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376

 

[iceworm]

... Now, my question is Why CB would be 250% of full load current? ...

Here's my understanding - perhaps it will help:
The motor is protected from overload by design. One selects the motor so that it will adequately provide all the power required for all normal loading, with out unduly overheating. The thermal overloads protect the motor windings from thermal damage should the mechanical system fail, say like bearings dragging, and push the motor into overload. The circuit conductors are also protected from overload by the thermal overloads. (Assuming you are under the NEC - but also likely true for the rest of the world) The overloads are selected to trip inside of the motor thermal damage curve. The conductors are selected such that the thermal damage curve is also outside of the overload trip.

The primary purpose of the circuit breaker is not to protect the motor or the conductors in the event of a mechanical failure. That's the job for the overloads.

If the motor faults internally amd disentigrates into a molten pile of slag, there is no motor to protect - only a fire to put out. If the cable suffers a forklift attack, again, no cable to save, only a fire to put out.

The primary purpose of the CB is to protect the structure - limit the electrical heat input keeping the fire going.

The CB is selected such that the motor will reliablily start. From the OP, I suspect the CB will be a UL489. You should be able to find a trip curve and download. Print out an 11x17. Hand plot an expected motor starting curve. The motor could draw 4 to 6 times FLA for the starting time. The CB has to be outside of the motor starting curve.

Hope this helps

So sayeth the worm

 

[sameee781]

In our country phase voltage is 230v but in worse case it may goes below 200v. Sometimes during irrigation period, water pump motor burnt out. Of course current increases with the decrease of terminal voltage. I wanted to know the rating of safest CB to avoid this case.

 

[iceworm]

If you change the CB out to one small enough to protect the motor from over curent, the motor will most likely trip the CB on startup. A CB will not protect the motor from over current.

You will need to install thermal overloads.

You may also wish to install a UV trip - although that would be somewhat uncommon on a 5hp single phase

ice

 

[waross]

The circuit breaker will NOT protect from overloads. You need an overload relay or similar device to protect from overloads.
A breaker that protects against short circuits will not protect against overloads.
Most overload devices will not operate fast enough to protect against short circuits.
Exception, there may be some devices that have both functions in one device.

Bill
--------------------
"Why not the best?"
Jimmy Carter