Gas warfare

[Dirty Harry]

Need some help with me maths

 

i was reading a wfa article about Ypres in 1915. They used 6000 cylinders to produce 160 tons of chlorine, which I get, but the article went on to say this produced 1part gas to 1000 part ambient air. How did they arrive at this number?

 

thanks

 

paul

[JMB1943]

WFA says concentration is 1 in 1000, this is 1000 parts per million (ppm).

My approach to determining this would be as follows.

Let’s start with what we know,

160 Tons Chlorine = 160 x 2240 x16 ounces = 160 x 2240 x 16 x 28 =  160,563,000 grams; call that 161,000,000 grams

So concentration of chlorine at the front lines = Weight chlorine / Weight air at front lines

Volume air at frontline = Length x Height x Depth [L x H x D]

We will take 1 yard = 1 meter, and you will have to look up the length over which chlorine was released = L

Now have to make some assumptions,

Assume Height = 6 feet = 2 meters to cover soldiers in trench or running on surface.

Assume gas cloud covers ground from front line trench to where?? This is the depth = D

So what is the depth? Distance between front line trench and support trench? = D1

                  OR

Distance from frontline trench to halfway to support trench? = D2

                 OR

Y yards only past the trench? = D3, 

Then V = L x 2 x D where D = D1 or D2 or D3 or any other value that you choose.

So Mass of air = V x density = L x 2 x D x 1.293 ; call that 1.3 grams per liter.

So concentration, C = 161,000,000 / (L x 2 x D x 1.3)

Suppose C = 0.0003, this is 0.03%, which is 0.03 x 10,000 ppm = 300 ppm.

Regards,

JMB

PS If we take WFA estimates as true, can calculate D if we know L.

 

 

 

 

[Dirty Harry]

brilliant piece of work, will look up the official maps and c if L can be determined once im off night shift

Edited 8 March , 2023 by Dirty Harry
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