- #1

Aldnoahz

- 37

- 1

I encountered a fairly straightforward problem involving fluid dynamics, but the result I got is very confusing and I seem not to understand where I got wrong.

This is the question:

A rectangular flat-bottom barge with a bottom area A=50 m2A=50~\text{m}^2A=50 m2 is loaded so that the bottom is at H=1 m below the surface. The density of water is ρ=103 kg/m3\rho=10^3~\text{kg/m}^3ρ=103 kg/m3 , and the water surface is perfectly still. A round hole with radius r=1 cm is made in the bottom of the barge, and the water starts leaking in. When the water level reaches h=5 cm, a bilge alarm will alert the barge operator. How long will it take for the water to reach the level 5 cm? (Assume that the Bernoulli’s equation is applicable.)

My attempt to the question:

First, at the surface of the water outside the boat: P_0 + rho*g* H = const

Second, inside the small hole at the barge where I call point 2: P_2 + 1/2*rho*(v_2)^2 = const

Third, at the surface of water inside the boat above the barge, where I call point 1:

P_0 + 1/2*rho*(v_1)^2 + rho*g*h = constApplying Bernoulli's equation, I now set first equation equal to the second,

P_0 + rho*g* H = P_2 + 1/2*rho*(v_2)^2

and for second and third:

P_2 + 1/2*rho*(v_2)^2 = P_0 + 1/2*rho*(v_1)^2 + rho*g*h

but what I find strange is that there is no way to get P_2 and v_2 and they get substituted as an intermediate value, and solving this equation set I can only get equivalence of first and third equations:

P_0 + rho*g* H = P_0 + 1/2*rho*(v_1)^2 + rho*g*h

rho*g*H = 1/2*rho*(v_1)^2 + rho*g*h

solving for v_1 = dh/dt = sqrt(2*g*(H-h))

and this gives me an extremely small time, smaller than 1, which is obviously physically not plausible.

I asked a lot of people and they don't seem to figure out why this is not correct, so I am posting to ask for help.

Thanks