PDA

View Full Version : Kvar & Csd


LEM
20th Oct 2003, 22:33
1) Some people say the Rise-In CSD oil temperature selector should be left routinely to the "In" position, and it's even written so on some checklists.

Any good reason for choosing so?



2) I don't remember the difference between KW, KWA, KWAR ...

Probably some flight engineer can enlighten me...
Thanks :D

HotDog
21st Oct 2003, 10:31
On the 747 Classic, the CSD Oil Temp indicator is hard wired to CSD OIL OUT temp and there is a push to test button to monitor CSD OIL TEMP RISE. I don't know why you would want to monitor CSD Oil temp rise instead as the limit of 148 deg.C is based on Oil Out temp.

You meant to ask KW, KVA and KVAR, didn't you?

KW = The measure of real power delivered

KVA = The apparent power output of a generator which depends on the power factor %. The engine driven generators on the classic are rated at 60 KVA which is equivalent to 57 KW.

KVAR = An indication of the reactive load on the generator. Balanced readings are insurance that voltage regulators are functioning properly.

LEM
21st Oct 2003, 16:42
Thanks for replying, Hot Dog.

However, I still have a problem understanding KVA and above all "reactive load":confused:

gas path
21st Oct 2003, 17:39
If memory serves the CSD oil temp rise is an indication that the CSD (s) are sharing the load. If one is 'working harder' the oil temp rise will be greater. (Also the oil level might be low!)
The KVAR reading as HOT DOG points out is the Reactive load and gives an indication of the performance of the GCU (generator control unit). On pushing the (KVAR) button they should all read the same (+/- 5 KVAR?), if not, one or another of the generators is being back driven.
The real current is controlled by the CSD and the reactive current by the voltage regulator (GCU) as they are 3 phase generators it's to do with line to line voltages across the phases, and (very;) )basically a generator with a higher reactive load will have it's excitation lowered and a generator with a lower excitation will have it's excitation raised until it's all back in balance.



maybe an avioncs person could explain it better???

HotDog
21st Oct 2003, 17:42
KVA = Kilo Volt Amperes

Power factor = Efficiency

KW = Kilo Watts = KVA x PF

It's a bit complicated to put it into simple terms but I will try.

Power factor is the relationship (phase) of current and voltage in AC electrical distribution systems. Under ideal conditions current and voltage are "in phase" and the power factor is "100%." If inductive loads (motors) are present, power factors less than 100 % (typically 80 to 90 % can occur)

Low power factor, electrically speaking, causes heavier current to flow in power distribution lines in order to deliver a given number of kilowatts to an electrical load.

As the loads on an aircraft generator are varied and cause phase shift, the efficiency suffers. Reactive/inductive loads cause heat losses. Therefore the REAL power of the generator in kilo-watts will be less than the design rating in kilovolt amperes.

I'm sure this is clear as mud but it's the best I can do.

IORRA
21st Oct 2003, 18:29
Hi LEM

In an AC circuit, the amount of power able to be converted into heat (or used to do work) is known as TRUE POWER, measured in Watts (W). In a purely resistive circuit, 100% of the power drawn is TRUE POWER (aka EFFECTIVE POWER). This is the power we want!

All AC circuits comprise current flowing through impedance (the vector sum of Resistance, Inductive Reactance and Capacitive Reactance). In a circuit where all of the impedance is resistance (eg a 'perfect' resistor), the circuit's current and voltage will be in phase, and all the circuit's power will be TRUE POWER. If we add inductive loads to the circuit however (such as motors, coils etc as HotDog mentioned), we cause the current to lag the voltage to the point where, in a purely inductive circuit (not practically possible), the current would lag the voltage by 90 degrees (by virtue of the back emf produced). In this situation, none of the circuit's power would be useable to do work - its voltage and current are 90 degrees out of phase. This type of power is known as REACTIVE POWER, and is measured in Volt-Amps-Reactive (VARs, kVARs if you have thousands of them). Whilst REACTIVE POWER is unusable, it still imposes loads on the generating and supply network as though it were useful (TRUE) power. This is the power we don't want!

The vector sum of TRUE POWER and REACTIVE POWER is known as APPARENT POWER, and measured in Volt-Amps (VAs). Graphically speaking, one can determine the magnitude of APPARENT POWER in a circuit by plotting TRUE POWER on the Y-axis, REACTIVE POWER on the X-axis, and APPARENT POWER as the resultant of the two.

As HotDog said, the POWER FACTOR is equal to TRUE POWER/APPARENT POWER (POWER FACTOR also equals the Cos of the phase angle between voltage and current in the circuit - the phase angle is also equal to the angle between the Y-axis and the APPARENT POWER resultant in the abovementioned graph). Ideally we want a power factor of 1 - in this case, 100% of the power consumed is being used to do work. In reality this is impossible to achieve, and represents a circuit that is purely resistive, with no inductance whatsoever. More realistic is a PF of about 0.8 or 0.9, as HotDog mentioned.

I hope this helps!

IORRA

HotDog
21st Oct 2003, 21:44
IORRA you are speaking my language.

LEM I bet you wish you never asked. :{

LEM
22nd Oct 2003, 06:18
Ok, after reading at least 1000 times all your posts ;) (thanks a lot for your time!), I start to grab something...

Actually, the whole concept is quite clear: probably the best way to explain this matter is to use a metaphor.

I would explain it to my little daughter this way: look at the generator as if it were a car engine: the power we actually get is 80hp (KW).
In an ideal world, the fuel it burns would produce 100hp (KVA), but due to the heat losses (friction, cooling etc...) we loose 20hp (KVAR).

100 minus 20 = 80.

It's as simple as that.

The lost power depends on the fact that current and voltage are not perfectly on phase.
The efficiency ("power factor") seems to be 95% on the 747 classic.
This efficiency depends also on the job of the voltage regulator-GCU: the maximum allowable difference between the 3 generators is 3KVAR on the 727.


If you agree with the above, it's enough for me :D

**********


HD: on the 727 and 737 the switch can be left indifferently on the IN or RISE position.
If you put the switch according to the "all on" general direction of all switches you'll have RISE on the 727 and IN on the 737.

I was just wondering if the routine position deserves an entry in the checklist... but now I agree with you it's better to keep it in IN
to have a better general picture.


Thankyou all, standing by for your comments!

7p3i7lot
22nd Oct 2003, 06:37
IORRA
Excellent post!
I have been through 7 type rating schools, and two electrical engineering college courses. None were as clear and concise as your explanation.
I could have saved dozens of hours wrestling with the confusion on this issue.
Thanks

IORRA
22nd Oct 2003, 07:56
HD lol!

LEM yeah I spose that's a good analogy - the main thing being that that the generator has to work to produce all of the APPARENT power, even though it's only the TRUE portion that can actually be used to do work.

Thanks 737! I certainly struggled with this in particular when I started out, but after getting into it a bit found it all quite interesting.

Cheers,
IORRA

Alex Whittingham
22nd Oct 2003, 17:40
The only thing slightly wrong with your analogy, LEM, is that KVA is the vector sum of KVAR and KW, not the algebraic sum. KVAR is 90 degrees in advance or behind KW depending on whether the circuit is overall inductive or capacitive.

http://www.bristol.gs/download_files/KVAR.jpg

A final piece of trivia for you. Because the effect of capacitors is opposite to the 'back emf' effect of motors and coils an ideal circuit can, in theory, be adjusted by adding and taking away capacitors so that the KVAR reading is zero. This is called a resonant circuit.

LEM
22nd Oct 2003, 18:22
The only thing slightly wrong with your analogy, LEM, is that KVA is the vector sum of KVAR and KW, not the algebraic sum.
Thank you, Alex, I suspected so, my explanation was purposely very simplistic.
The second part of your post is arabian to me:D as I don't recall what "inductive or capacitive" means... but I won't dare asking you to explain, unless nature provided you with a lot of patience...

Alex Whittingham
22nd Oct 2003, 20:19
IORRA referred to this in his post. In a 'normal' electric circuit the current increases as voltage increases, they are in phase. If you have capacitors in a circuit they have the effect of moving current and voltage out of phase. The illustration I sometimes use in class is to compare a circuit with a capacitor in to sloshing the water backwards and forwards in a bath. The point when all the water is hanging over the tap end represents maximum potential (voltage) but no movement (current), when it's moving past your legs it has maximum movement (current) but no potential (voltage). The result of moving current and voltage out of phase is that power (voltage times current) is reduced, power is 'lost' as reactive load.

Capacitors act to shift current 90 degrees in advance of voltage. This effect is called capacitive reactance.

An opposite effect occurs when you have an alternating current passing through coils of wire, perhaps in a motor. The current creates a changing magnetic field which creates, in turn, a current which flows against the original current, this is called a 'back emf'. The effect of this is to cause the current in a circuit with coils in it, an inductive circuit, to lag 90 degrees behind the voltage. Just as in a capacitive circuit the power is reduced as the voltage and current slip out of phase. This is inductive reactance.

Most circuits have more coils in than capacitors so the two effects are out of balance and power is 'lost'.

http://www.bristol.gs/download_files/KVAR2.jpg

The obvious solution is to add capacitors to make the capacitive reactance cancel out the inductive reactance and achieve resonance. In practice this is difficult because the amounts of capacitive and inductive reactance respond differently to small changes in AC frequency, they are constantly changing.

Wraftongate
22nd Oct 2003, 20:36
Ok -it's my turn - sorry - long


W = DC power or AC equivalent (true power)
VA = AC power (no allowance for power factor)
VAr = 'Bad' or wasted power

Try it this way - This should help you visualise what happens

Imagine 2 SINEWAVES exactly CO-INCIDENT (one for VOLTAGE, one for CURRENT)
Now ignore the bottom halves (just to make it easier to see)
This is the situation when the PF is UNITY (voltage and current in phase)
Consider The POWER (VA) consumed by the LOAD as the AREA under the half sinewave
In this case W is the SAME as VA, VAr is ZERO

Now SHIFT one of the sinewaves sideways a bit - this is what happens with a NON-LINEAR or 'reactive' LOAD (motors,transformers,capacitors etc)
The voltage and current are now 'out of phase'
The TOTAL area of the sinewaves is now LARGER - this is the POWER that the GENERATOR has to PRODUCE -kVA
The OVERLAPPING portion of the sinewaves is SMALLER - this is the POWER that the LOAD usefully CONSUMES -kVA
The DIFFERENCE between the two is kVAr - WASTED POWER (needs a bit of trig. to make the numbers add up)

I.E. large phase difference between voltage and current = low PF = greater power wastage



Why does this happen and where does the wasted power go ?

The generator still has to produce the current even if it is not used efficiently.
The current just warms up the generator and the aircraft wiring (and burns more fuel).

Most aircraft systems have the means to 'correct' the power factor i.e. bring the voltage and current back in phase with each other.


Who's next ?


Wraftongate

LEM
22nd Oct 2003, 20:54
Alex,
Ok, got it! Thank you very much for your time, excellent explanation :ok:

Wraftongate,
Thankyou also, one more way to look at it, that's clear now.


HD: wrong bet :{ ;)

mono
23rd Oct 2003, 00:15
This (http://www.sea.siemens.com/step/default.html) , (I think fantastic) site helps explain even further. There's even several tests as you progress the course.

OK it's not a/c orientated but the theory is the same.

LEM
23rd Oct 2003, 03:57
Excellent!

Thank you, mono!