Rotors de-icing power question
 

While reading Mil Mi-8 tech docs I surprisingly saw that its TR de-icing system consumes way more electric power than MR one. To be exact it is 38 kW for a TR and only 26 kW for a MR. What could be the reason(s) that to de-ice times smaller surfaces needs much more power?

Perhaps it is like the AW139 ice protection system. The TR is anti-ice. All the blades are on all the time in icing conditions. The MR however is de-iced. Ice is allowed to accrue and then power is directed to individual heater mats on individual blades at different times to shed the ice. So only one heater mat on a blade is on at any one time. Each 139 MR blade has 6 heater mats. So only 1/6th of the heated surface is actually being heated at any one time.

Just a thought.

noooby, your thought is correct.

Perhaps the m/r is better able to handle a bit of assymetric weight than the much more delicate t/r, requiring that to be deiced all at once rather than in sections?

I'd suposed that probable reason is that TR is more prone to efficiency loss just because it has smaller chord and therefore same ice thickness ruins its aerodynamics faster than it does for bigger MR chord. Just inspired by a comparison between any MR chord and fixed wing chord.

212man, could you please post the correct link to that other forum thread?

Sorry, not sure why that happened - should be correct now

Yes, it works now, thank you!

The enigma revealed! Found an old manual on Mi-8 equipment where it is clearly written: Ampermeter will show 160 A for TR current but in fact real current is 8 times less, i.e. 20-24 A. They just wanted to use one ampermeter for two very different levels of current in the simpliest way. Thanks to all replying!