- This topic has 7 replies, 6 voices, and was last updated 18 years, 4 months ago by
.
-
Posts
-
Need to fit a 7uf capacitor to creda T/D only have a 8 uf in car will this be ok, not sure of the tolerance on these. cheers john
if anyone can also let me know what tolerance is advised.Re: Capacitor values
bobokines wrote:Be a bit careful using an 8 MFD capacitor. I had one a few weeks ago that I had previously fitted an 8mfd to and it then kept overheating the motor. I now only fit a 7 to these machines.
There is a 7mfd pattern part available now fromConnect CAP244
The motor current rises to the square of the increase in capacitance. ie if you fit an 8 instead of a 7 you are increasing the running current by over 30{e5d1b7155a01ef1f3b9c9968eaba33524ee81600d00d4be2b4d93ac2e58cec2d}
Bob
Chris.
Re: Capacitor values
Tolerence pretty wide on these, Surpirsed it would make a difference. I know Candy always used to tell us to fit the correct one but we have all got away with 22’s on 20’s and 10’s on 8’s havent we?
Re: Capacitor values
To be honest I have virtually always used an 8 mfd replacement on creda machines.
However, I had a condenser dryer a few weeks ago that overheated the motor after I had fitted a 8mfd. It might have been a one off but I am not going to risk it again.
:lesson:
Here’s the theory … (btw I used to work as a design engineer in an Electric Motor factory … over 25 years ago now and I am going by memory so forgive me if the figures are wrong)The value of the capacitance is very carefully calculated to give the correct starting torque and ideal running current for the expected maximum load conditions.
The temperature rise of the windings has to be within the criteria given for the insulation class. In those days, most motors were class B insulation and a max temperature rise of 80 degrees C was allowed.
Class B motors are normally fitted with a TOC rated at 80 degrees above ambient temperature (usually 105 C).
Any increase in the current through the windings , (either by increasing the load or by increasing the capacititance ) gives a higher temperature rise… Losses that give rise to internal heating are proportional to the resistance of the windings and the square of the current (known as I squared R losses)
Increasing the capacitance increases the current (through the start windings) by the square of the difference in capacitance. ie if an 8mfd is used instead of a 7 the increase is (8/7) X (8/7) = 1.31
The increase in temperature rise is proportional to the square of the increase in current ie 1.31 X 1.31 = 1.72 That’s 72{e5d1b7155a01ef1f3b9c9968eaba33524ee81600d00d4be2b4d93ac2e58cec2d} higher temperature rise (in the start windings) just by increasing the capacitance by 1mfd 😯
Remember also that in a tumble dryer, the ambient temperature is likely to be much higher than 25C.
Bob
Re: Capacitor values
bobokines wrote:The value of the capacitance is very carefully calculated to give the correct starting torque and ideal running current for the expected maximum load conditions.
Any increase in the current through the windings , (either by increasing the load or by increasing the capacititance ) gives a higher temperature rise…
Increasing the capacitance increases the current (through the start windings) by the square of the difference in capacitance.
I thing that this subject was already threat.
http://www.ukwhitegoods.co.uk/modules.p … ic&t=27712
I will say that Bob has right. I do not know so deeply calculations about temperature variation and other too… but is logical: capacitor is there for providing neccesary current (dephased) for starting point only… but because he is not disconected after that he will work all the time.
What happen when capacitor is lower then calculated value: he will not supply sufficient time current to starting widing (crosspoint to “zero” of start magnetic field will be much earlier than stator field…) so motor will no rich entire resolution at calculated speed (depends of number of poles and power frequence) so he will need in the next sinusoidal again help from start winding….results motor will overheated because all time will be disipated much energy from starting winding and if capacitor is too low he will not manage to turned at all…What happen when capacitor is higher that calculated value: like I explain before capacitor in this case will supply current more time (dephase/delay) so at the begining of semi-sinusoidal wave of current on stator winding will be remaining current on starting winding… so it will work like breaking sistem (will never go thru zero point) which will generate more current consumption= heating results.
P.S. If I remember good calculation: depending of motor coupling needs and numbers of poles=> starting winding impedance (current consumption)=>capacitance … all this will be verified by the dissipating power to be inside nominal value otherwise winding will become heating resistor.
- You must be logged in to reply to this topic.