Fountain physics problem

Question

Water from a large cylinder is coming to the fountain. The cylinder has a hole of diameter d=2 cm through which the water comes up. The water is coming at a speed of 12 m/s.

I need solutions for these problems:

a) speed of change of water level v1 in the cylinder with diameter D=2 m

b) water pressure p1 of water that is coming to the fountain

c) mass of water in the cylinder

d) how high the water comes up from the fountain

Answer 1

Solution: I am assuming from the problem statement that the water velocity in the 2 cm hole is 12 m/s, and that the required speed V1 =? is that of the Large 2 m diameter cylinder

Let V1 be the speed of water coming from the 2 m diameter Cylinder, V1=?

Let V be the speed of water in the 2 cm diameter hole given as 12 m/s

Let A1 be the area of the 2 m diameter cylinder with Diameter, D1 = 2 m

Let A be the area in the 2 cm diameter hole with Diameter, D = 2 cm

Q = the flow rate in the cylinder

A = π D²/ 4 = π ( 0.02 m)²/ 4 = π ( 10^-4) m²

A1 = π D1²/ 4 = π ( 2m)²/4 = π m²

From The continuity equation Q = AV = A1V1,

and V1 = A V / A1 = π ( 10^-4) m² ( 12m/s) / π m² = 12 x 10^-4 m/s = 0.0012 m/s

2) I am assuming that the pressure required is that in the 2 m diameter cylinder P1=?

The pressure head is equal to the the velocity head , p1 / γw = V1² / 2g

the density of water, γw = 1000 kg/m³ x 9.81 N/kg = 9810 N /m³

The pressure generated in the 2 m diameter cylinder P1 can be calculated as shown below:

P1/ γw = V1² / 2g = (0.0012 m/s)²/ 2 ( 9.81 m/s²) = 7.34 x 10^-8 m , and P1 = 7.34 X 10^-8 m ( 9810 N /m³) =

7.2 x 10^-4 N /m² = 7.2 x 10^-4 pa ( pascals)

3) To calculate the mass of water in the cylinder use, ∑ F = m g , the force F = pressure x area = P1 x A1 =

7.2 x 10^-4 N / m² ( π m² ) = 2.26 x 10^-3 N = 2.26 x 10 ^-3 kg -m /s²

The mass m = F/g = 2.26 x 10 ^-3 kg -m /s²/ 9.81 m/s² = 2.3 x 10^-4 kg

4) The general Kinematic equation for a body shot up directly vertically into space

y = -1/2 gt² + Vo t + yo , taking derivative wrt to to t

dy/dt = V = -gt + Vo , starting from intial velocity of Vo = 12 m/s, the final velocity at maximum height is V = 0

substituting 0 = -gt + 12 m/s, -gt = -12 m/s, and solving for t = 12 m/s / g = 12 m/s / 9.81 m/s² = 1.225 s say 1.23 s

The maximum height is y = -1/2(9.81 m/s²)(1.23 s)² + 12 m/s ( 1.23 s) + yo, where initial height yo = 0

The maximum height attained is: y = -7.42 m + 14.76 m = 7.34 m