I need to know how much vapour will be formed by 1 litre of 53 L 13
HIGH BUILD ZINC PHOSPHATE PRIMER when sprayed the solvent content is
60%

Ummm -- 600 ml will be vaporized as the paint dries.   Depending on
what the solvent is in the paint will determine the grams of material
vaporized and the volume of air that is required to vaporize it.

A reasonable starting point is that the solvent may be mineral spirits
or related to it.  For 600 ml of mineral spirits you will have 0.6 *
786 gm/l, VOC, volatile organic compounds = 471 gm.

Check out the MSDS here - http://www.nelsonpaint.com/msds/T_19.html

I don't have a good value for the amount of vapor that is formed from
a gram of mineral spirits so I'll put using the Ideal gas law and
answer that each mole produces around 22.4 liters of vapor (gas) and
92.1 gm / mole so you are looking at 471gm 1 * 1 mol/92.1gm * 22.4
l/mol = 114.5 l  of vapor.

Something similar to this is covered in a paper by the EPA.
 
http://www.epa.gov/dfe/pubs/tools/ctsa/appends/app-b.pdf
 or 
http://64.233.161.104/search?q=cache:VTp8JKTnE1IJ:www.epa.gov/dfe/pubs/tools/ctsa/appends/app-b.pdf+mineral+spirits+molar+weight&hl=en

The occupational exposure maximum for mineral spirits is 100 ppm
(parts per million).  If not working
outside with this paint you should be in a room that has a volume of
114 l / 0.0001 or 1,140,000 l which is a room measuring 25 x 20 x 2
meters.  Quite a room.

Thank you, the room is actually 48 metres x 9 metres x 4 metres high
but I am using 6.66 litres per hour so now I can work out the required
extraction rate