An extremely large tank that is open to the atmosphere contains water (rou = 100

Question

An extremely large tank that is open to the atmosphere contains water (rou = 1000 kg/m^3) with H = 2 m (see figure). The tank is connected to a short section of pipe that discharges the water into the atmosphere as a vertical jet that is being used to suspend a small metallic plate of mass m. At the discharge from the pipe, the water jet has a diameter of 2 cm. If viscous effects are negligible, and the jet reaches a height h = 1.5 m, estimate the mass (m) of the suspended plate in units of kilograms. There is no heat exchange or temperature change in the fluid at any point. State a if other assumptions if needed

Explanation / Answer

let the velocity of the jet at the origin of jet be v, and the area of the opening be A

for time dt, length of jet coming out of hole = vdt
Volume of water = Avdt
Mass of Water = (rho)Avdt

Assume the atmosphere to have no air resistance

Velocity of this mass of water at hieght h : v'

2*g*h = v^2 - v' ^2
v' = sqroot(v^2 - 2gh)

momentum of the mass at this point = (rho)Av*sqroot(v^2 - 2gh)*dt
Force = momentum / time = (rho)Av*sqroot(v^2 - 2gh)

let this force balance the weight of the plate
then mg = (rho)Av*sqroot(v^2 - 2gh) -- (1)

Now, we have to find the velocity at the jet

Pressure at the base of tank in the outlet = (rho)gH

Assume the outside pressure in the jet be atmospheric pressure (hence all the water that comes in contact with the plate comes to rest immidiately and falls of obliquely without effecting the pressure in the jet stream)

Hence, pressure difference = (rho)gH
From bernoulli's theorem: (rho)gH = 0.5(rho)v^2
or, velocity at base of tank, u = sqroot(2gH) -- (2)

From 1 and 2:

m = (rho)A*sqroot(2gH - 2gh)*sqroot(2gH)/g
m = 1000*3.14*(0.02)^2 *sqroot(2*9.8(2-1.5)) *sqroot(2*9.8*2) / 9.8 = 1000*3.14*2*(0.02)^2 = 2.512 kg

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