Answer:
Q=0.0415m^3/s
Step-by-step explanation:
A large container of water is open to the air, and it develops a hole of area 12 cm2 at a point 6 m below the surface of the water. What is the flow rate (m3/s) of the water emerging from this hole?
equation of continuity
a1u1=a2u2=Q..............................1
also from bernoulli equation
P + ½ ρ v2 +ρ g h =..................................2
P1 + ½ ρ v2 +ρ g h1 = P2 + ½ ρ v2 +ρ g h2 =
since pi=p2
½ ρ v^2=ρ g h1-ρ g h2
1/2v^2=gh1-gh2
V^2=2g(h1-h2)
V^2=2*9.81*(6-0)
V^2=117.72
V=34.59m/s
from the equation 1 we know that volumetric flow rate
area*velocity
convert 12 cm^2 to m^2
12/10000
a2=0.0012m^2
a2= is area of the opening at the base
34.59m/s*0.0012m^2
Q=0.0415m^3/s
u=velocity
a2=area of the hole
Q=volumetric flow rate
g=gravity
ρ=density of water 1000kg/m^3