Train speeds at 80kph i.e 22m/s
The train is moving towards west (22m/s) and wind to south which is (5m/s)
So resultant wind through window
√((22)^2 + (5)^2)
Which is a 13m/sA
At angle of {tan^-1(5/22+5)} which is 10.49 degrees.
Assuming Window is fully open
So 13m/s multiplied by dimensions of window
Then 13m/s × 1.5 cu. m = 19.5 cu.m
But the dimensions of compartment is a 12 cu.m ....so 19.5/12 = 1.625....so the space will be refreshed in 1.625 seconds
The train is moving towards west (22m/s) and wind to south which is (5m/s)
So resultant wind through window
√((22)^2 + (5)^2)
Which is a 13m/sA
At angle of {tan^-1(5/22+5)} which is 10.49 degrees.
Assuming Window is fully open
So 13m/s multiplied by dimensions of window
Then 13m/s × 1.5 cu. m = 19.5 cu.m
But the dimensions of compartment is a 12 cu.m ....so 19.5/12 = 1.625....so the space will be refreshed in 1.625 seconds