Forum Replies Created
-
Posts
-
Sasikumar said:
we are having the centrifugal backward van blower. flow is 22000 nm3/hr. pr. 3300 mmWc. to run this blower what is the capacity motor required. as per supplier recomendation 315 KW with soft starter. but we started the blower frequently triping at 1700 rpm. speed not picking up with in 35 seconds. can you help me to short out the problem.
please tell us what is starter u r using now ? is it a soft starrter , Star delta, soft starter or what ?
see there is a basic calculaiton for a fan speen motor find the power requirement and take the eqivalent size of appropriate motor.
Now starter plays a vital role , set the paramer as per the manufacutere requirement if it is not still in solutin , please tell us in detail ….of your problem..
saraswatapalit said:
Please help by providing a detail note on generator transformer sizing. What are the factors that governs this rating?
thanks in advance
best regard
saraswata
please tell us in details, where u want to use them . untill and unless get the idea how u want to use them. location of the transformer and generator in the SLS how we are going to tell u.
definaltely there are some basic rules:
u can get the max efficiency of the dg set @75 percantage loading. Transformer loading charactersticsh should be in considerrations.
connected load is also a consideration , coz as per the load it will provide the dips of the voltage during starting.
if this is the main t/f of the incomer substation then u have consider lots more other parameter.
Jim said:
Hello, Owen, Garage and Kerstein,
I don't know where this discussion come from but I will offer some comments.
Direct on line (DOL) AC motors are the worldwide standard to produce mechanical work from electricity. No dispute. Its huge.
DOL AC motors provde fixed speed operation. They are dependable and last a long time if environmental conditions are maintained to standard.
DC motors are only used when variable speed is needed and a DC controller (DC drive) is required.
So any discussion about comparing DC and AC motors requires a discussion about the motor controllers also.
DC motors are more efficient than AC motors. Typically DC motors are 97-99% efficient. AC motors are not more than 94% efficient (check the manufacturers' tables).
DC drives are 99% efficient. AC motor speed controls (PWM VFD) are at best 97-98% efficient. Typically about 97%.
But efficiency is only one part of the discussion. Maintainence is another issue. DC motors require maintenence for the brushes. AC motors need grease and they run forever.
DC drives are simple and less costly and easy to maintain. PWM VFD are more complex and do require more maintenece than DC drives.
DC motors are considered special and have relatively long deliveries. AC motors for DOL use are stocked and readily available. For a given size rating, the DC motor is larger physically.
AC motors to be used with PWM VFD require a different design than do standard DOL AC motors. NEMA and other bodies have defined how these differ. Needless to say the AC motor that operates on VFD should be special.
Trade offs abound.
Generally, PWM VFD with AC motors are used as a preference over DC dirve and motors – today.
Regards,
nice discussion , u can use the general manafactued motor with the VFD drive. But now a days motor is specially desigened for the VFD and it will give you better result. specially ABB motor.
But brother DOL is a unversal applicaiton and still it is used in the industry due to natuie of its cost and maintenance naturre but modern factories who has good budget they are using the soft starter.
Well where u need a precession of control , like paper industry , still u need those dc drive anyway good discussion
khushal said:
Dear fellows,
as all ovr the world three phase power
distribution is used , then why a capicitor is used in electric fans to
produce
a phase sequence to meet the requirement.
khushal.
ha ha ha ha khushal , well said , anyway sorry to laugh ! it is a very basic question a first yr student ask. The answer is , in single phase fan the capacity does not act as a power factor, Basically it splits the phase , so that it can rotate. if you take out the capacitor from the fand and start it , it will pulsate and will not rotate but once u provide the rotation by a hand and it will start rotating.
ha ha ha , see the quadrature theory in the basic electrical engineering.
Mohan , i think the 375 u r , talking abou the H.T side and amps 1100 u r talking about in the L/T side. Anyway it can be solved by the basic power P=VI cos of Phi formula. calculate it from one side , either HT or LT side find the amps and do the basic math. voltage u , know it , p.f u know it then find the p value and u can conver it by th basic formula and convert it in KVA
SHAKIL said:
Some engineers ask for the combining of transformer neutral and body earth together … while some experts say that distribution transformer should have separate earthing for neutral and body.
What are merits and demerits of both.
There is a standard to make them seperate. please find it and u will get it. okay now lets answer your question easily , see if you have a mixed node of earthing , you will not be able to find out the circulation current path of the earthing. In this case few complexity may arise , specially if you have complex grid or in a machine yard and in swithchyard. you may face various complexity , specially yout earthing relay triping parameter.
therealabdo said:
Dear Engineer;
Actually i was working on installing a power factor correcting on a factory and i heard an info i just want to confirm it:
We all know that power factor correcting panels reduces the Kvar. and you get some discounts on the bill
What I heard is that with each 1 Kvar u SAVE, 10 Watta increases in ur consumption. Which means reducing Kvar will increase ur Watta which results a higher bill..but with the discounts u gain …the bill will be approximately the same…
What do u think?
I am not agreeing with you. If you increase the p.f It will definitely
reduce the bill. Basically P.f means capacitor. It just needs initial current
to charge the capacitor. Once it is charged, it acts as an open circuit. Well,
surely it consumes some energy which is negligible. Now if u r determined to
find the consumption of the capacitor, then please calculate with the basic
charging current calculation, which is available in any basic electrical
circuit book.P.f basically increase the power angle of voltage and current in such a way that, it consume less active power for the
inductance load specifically and decrease the KVAR / dc component of the
machine. if u r not still convince come back to me.
