Fluid Dynamics Flow Rate

Homework Statement

a 4mm hole is 1m below the surface of 2m diameter water tank.
a.what is the volume flow rate through the hole in L/min?
b.what is the rate in mm/min at which the water level in the tank will drop if the water is not replenished.
[PLAIN]http://i285.photobucket.com/albums/ll55/tebsa08/tank.jpg[/PLAIN]

Homework Equations

Q=vA
v1A1=v2A2
V1=A1$$\Delta$$(x1)=v1A1$$\Delta$$(t)
V2=A2$$\Delta$$(x2)=v2A2$$\Delta$$(t)
p+.5$$\rho$$v1^2+$$\rho$$gy=constant

The Attempt at a Solution

a. A1=$$\pi$$r^2=$$\pi$$*1^2=3.14m^2, A2=$$\pi$$r^2=$$\pi$$*.002^2=1.25x10^-5m^2
im sorry i really dont understand fluid dynamics, i have no idea how to find v, to get Q... ans part b, i dont even undestand the question.... plz someone help me,
can i find v by, $$\rho$$+$$\rho$$gy=$$\rho$$+.5$$\rho$$v^2
and all the densities cancel so v=$$\sqrt{}2gy$$???

Last edited by a moderator:

i think V1=A1*x=3.14*1=3.14m^3??

btw the answers are 3.3L/min and 1.06mm/min

can anyone help??

alphysicist
Homework Helper
Hi fredrick08,

can i find v by, $$\rho$$+$$\rho$$gy=$$\rho$$+.5$$\rho$$v^2
That's the right idea, but this isn't written quite correctly. It needs to be:

$$P_1 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2$$

after the simplifications you have already made (v_1=0 and h_2=0). What do you get?

huh??? sorry i dont understand??? if v1=0 then doesnt v2=0 aswell?????
coz v2=$$\sqrt{}v1$$??

and how does v1 and h2 both equal 0????? plz explain??

alphysicist
Homework Helper
Didn't you already set v1=0? Bernoulli's equation applied to two points is:

$$P_1 + \rho g h_1+\frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$