Question

The town of Hana on the island of Maui maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 8.70 cm. The hose ends with a nozzle of diameter 2.90 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 8.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 1.50 h? (c) Calculate the gauge pressure of the flowing water in the hose just behind the nozzle.

Answer 1

(a) This one is easy, you just need the static pressure of the water at nozzle level times the area of the end of the stopper.
Force = Pressure * Area
Area = pi * d^2 / 4 = pi * (2.9 cm)^2 / 4 = 6.605 m^2
Area = 6.605 cm^3 * (1 m^2/(100 cm)^2)
Area = 0.0004337 m^2
Pressure = density * gravity * height
Pressure = 1000 kg/m^3 * 9.81 m/s^2 * 8.5 m = 1 N/m^2

Pressure = 75375/0.0004337= Pa

Force = P * A = 75375 Pa * 0.0004337 m^2

Force = 0.0004334 N

(b) Bernoulli's equation: Lets call the top of the tank point one and the end of the nozzle point 2

z1 + P1/rho*g + V1^2/2g = z2 + P2/rho*g + V2^2/2g

Let's call our point of reference for elevation as the height of the nozzle

z1 = 7.5 m
z2 = 0 m
V1 = 0 m/s, because the water at the top is not moving.
P1 and P2 are both zero because they are both open to the atmosphere (gauge pressure of course)

The equation reduces to:

z1 = V2^2/2g

solve for V2

V2 = sqrt (2*g*z1)

V2 = 12.13 m/s

And the flow rate at point 2 is the velocity times the area of the nozzle

Q = V * A

And the mass flow rate would equal the density times the flow rate, and the mass flow rate is also the mass of water (m) over a given time (t).

m/t = Q * rho

m/t = V * A * rho

m = V * A * rho * t

m = 12.13 m/s * 0.0004337 m^2 * 1000 kg/m^3 * 2.00 hr * (60 min/1 hr) * (60 sec/1 min)

m = 37,878 kg

This assumes however that the tank is so large that the height of the water in the tank doesn't change during those 2.00 hours. But since they don't give us the diameter of the tank, only that it is a large tank, that is the best that we can do.

(c) Now lets call the point behind the nozzle as point 3. We can either take Bernoulli's equation between point 1 and point 3 or between point 2 and point 3. Lets choose points 2 and 3 first.

z2 + P2/rho*g + V2^2/2g = z3 + P3/rho*g + V3^2/2g

z2 and z3 are both zero.
P2 = zero guage pressure.

V2^2/2g = P3/rho*g + V3^2/2g

Solve for P3

P3 = rho*g ((V2^2/2g) - (V3^2/2g))

P3 = (rho/2)*(V2^2 - V3^2)

Now we know V2, but not V3. However, they are related by the flow rate and the Area's. They both have to have the same flow rate Q

Q = V2 * A2

Q = V3 * A3

V3 * A3 = V2 * A2

V3 = V2 * (A2/A3)

A3 = pi * d3^2 / 4 = pi * (0.0695 m)^2 /4

A3 = 0.0038 m^2

V3 = V2 * (A2/A3) = 12.13 m/s * (0.0004337 m^2 / 0.0038 m^2)

V3 = 1.3844 m/s

And from above we have

P3 = (rho/2)*(V2^2 - V3^2)

P3 = (1000 kg/m^3 / 2) ((12.13 m/s)^2 - (1.3844 m/s)^2)

P3 = 72,600 Pa

P3 = 72.6 kPa

Now lets go from point 1 to point 3

z1 + P1/rho*g + V1^2/2g = z3 + P3/rho*g + V3^2/2g

z1 = 7.5 m
z3 = 0 m
P1 = 0 Pa gauge pressure
V1 = 0 m/s

z1 = P3/rho*g + V3^2/2g

Solve for P3

P3 = rho * g * (z1 - (V3^2/2g)

P3 = 1000 kg/m^3 * 9.81 m/s^2 * (7.5 m - ((1.3844 m/s)^2/2*9.81 m/s^2)

P3 = 72.6 kPa

A fun question, thank you.

Good Luck