My 9v battery reads 20v ac in multimeter
Why the result is essentially meaningless:

Using a measurement mode to measure something it is not intended for can be expected to give a meaningless result - or to damage or destroy the system, or the user.
(eg trying to measure volts on an amps range.)
In this case the result is essentially meaningless, but relatively harmless. 
The AC range is intended for measuring AC voltages. IF you get a reading when DC is applied it does not mean that AC is being produced - while this is conceivably possible, the most obvious conclusion (and in this case the correct one) is that the meter is giving a spurious answer because it is being used in a manner that was not intended by its makers.    

If you apply DC to the Ohms range you will also get a reading. It relates to how the DC produces currents in the meter which can also be obtained by placing a resistor across the probes - BUT the reading is essentially unrelated to the resistance of the DC supply. 



What is happening:

It is normal for low cost multimeter circuits to read about twice the DC value when DC is measured on an AC volts range.
I have never traced the circuits used to be certain why BUT the following explanation  - seems the likely one.   


I assume that the AC circuit half wave rectifies and filters the AC to DC and applies it to a DC range with suitable scaling.   
An AC waveform that is say 10V peak-peak will give 5V DC when half wave rectified with an ideal diode.  
As the half wave circuit only passes every second half cycle the value when smoothed is 50% of what the full wave value will be.   
So a DC signal produce an effect that a ~~= 2 x AC signal would produce.   
However, the AC signal RMS value is ~= 71% of AC peak.   
And if a silicon diode is used the diode fwd voltage drop need to be allowed for.  


All this leads to AC readings of ABOUT 2 x DC value when DC is applied.