Question

A water pump takes in water through a 10 cm radius tube 1.5 meters below the surface of a pool. At the surface the water is ejected upwards from a tube 4 cm in radius with enough velocity to reach 4 meters height above the pool. A) What is the velocity of the water as it leaves the tube at the pool surface? B) What is the velocity of the water at the intake tube 1.5 meters below the surface of the pool? C) What is the pressure difference between the water in the intake tube and the exit tube?

Answer 1

a)

H = height gained by water = 4 m

v_e = speed of water at the exit tube = ?

m = mass of water

Using conservation of energy

Kinetic energy of water = Potential energy gained by water

(0.5) m v_e² = mgH

(0.5) v_e² = (9.8) (4)

v_e = 8.85 m/s

b)

r_e = radius of exit tube = 4 cm = 0.04 m

r_i = radius of input tube = 10 cm = 0.10 m

v_i = speed at input tube

Using equation of continuity

A_i v_i = A_e v_e

r_i² v_i = r_e² v_e

r_i² v_i = r_e² v_e

(0.10)² v_i = (0.04)² (8.85)

v_i = 1.42 m/s

c)

P = pressure difference

h = height below the exit tube = 1.5 m

= density of water = 1000 kg/m³

Pressure difference is given as

P = gh + (0.5) v_e² - (0.5) v_i²

P = (1000) (9.8) (1.5) + (0.5) (1000) (8.85)² - (0.5) (1000) (1.42)²

P = 52853.05 Pa