Question

Laboratories containing dangerous materials are often kept at a pressure slightly less than ambient pressure so that contaminants can be filtered through an exhaust system rather that leaked through cracks around doors, etc. If the pressure in such a room is 0.1 in. of water below that of the surrounding rooms, with what velocity will air enter the room through an opening? Assume viscous effects are negligible.

Answer 1

The velocity the air will enter the room through an opening is 20.9
`(ft)/(s)`

Step-by-step explanation:

We have to consider 2 points

Point 1 = Being away from the laboratory door

Point 2 = Gab between the floor and the laboratory door

Using the Bernoulli equation

`(p1)/(d_a_i_r g) + (V1^(2) )/(2g) + z1 = (p2)/(d_a_i_r g) + (V2^(2) )/(2g) + z2`

Assuming

p1 = patm

p2 = patm + plab

patm will be cancel and both points will be at the ground level

`0 = (plab)/(d_a_i_r g) + (V2^(2) )/(2g)\\ (V2^(2) )/(2g) = \sqrt[]{(-2plab)/(d_a_i_r) }`

`d_a_i_r` = - у H2O * hH2O

`= (lb)/(ft^(3)) * (0.1)/(2) ft\\= 0.52 (lb)/(ft^(2))`

Substituting the results

`V2 = ((- 2 * (- 0.520) (lb)/(ft^(2) ) )/(2.38 * 10^(-3) (sl)/(ft^(3) ) ))^(1/2)`

V2 = 20.9
`(ft)/(s)`