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

Answer:

The velocity the air will enter the room through an opening is 20.9 $\frac{ft}{s}$

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

$\frac{p1}{d_a_i_r g} + \frac{V1^{2} }{2g} + z1 = \frac{p2}{d_a_i_r g} + \frac{V2^{2} }{2g} + z2$

Assuming

p1 = patm

p2 = patm + plab

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

$0 = \frac{plab}{d_a_i_r g} + \frac{V2^{2} }{2g}\\ \frac{V2^{2} }{2g} = \sqrt[]{\frac{-2plab}{d_a_i_r} }$

$d_a_i_r$ = - у H2O * hH2O

        $= \frac{lb}{ft^{3}} * \frac{0.1}{2} ft\\= 0.52 \frac{lb}{ft^{2}}$

Substituting the results

$V2 = (\frac{- 2 * (- 0.520) \frac{lb}{ft^{2} } }{2.38 * 10^{-3} \frac{sl}{ft^{3} } })^{1/2}$

V2 = 20.9 $\frac{ft}{s}$