Question

arolyn, living on the fourth floor of an apartment building, opens a faucet tap 1.45 cm in radius. The tap is 12.0 m above the main water supply pipe to the building. The main water pipe is 4.05 cm in radius and has a volume flow rate of 1.35 L per second. (a) What is the speed of the water exiting Carolyn's faucet? m/s (b) If no other taps are open in the building, what is the gauge pressure in the main water supply pipe? Pa

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of water = = 1000 kg/m³

Volume flow rate of water = Q = 1.35 L = 1.35 x 10^-3 m³

Radius of the main water supply pipe = R₁ = 4.05 cm = 0.0405 m

Radius of the faucet tap = R₂ = 1.45 cm = 0.0145 m

Pressure of the water in the main water supply pipe = P₁

Gauge pressure in the main water supply pipe = P_1g = P₁ - P_atm

Pressure of the water in the faucet tap = P₂ = P_atm (Taps are open to the air)

Speed of flow of water in the main water supply pipe = V₁

Speed of flow of water in the faucet tap = V₂

Height of the tap compared to the main supply pipe = h = 12 m

Area of the main water supply pipe = A₁

A₁ = R₁²

A₁ = (0.0405)²

A₁ = 5.153 x 10^-3 m²

Area of the faucet tap = A₂

A₂ = R₂²

A₂ = (0.0145)²

A₂ = 6.605 x 10^-4 m²

Q = A₁V₁

1.35x10^-3 = (5.153x10^-3)V₁

V₁ = 0.262 m/s

Q = A₂V₂

1.35x10^-3 = (6.605x10^-4)V₂

V₂ = 2.044 m/s

By bernoulli's equation,

P₁ - P_atm = 119774.65 Pa

P_1g = 119774.65 Pa

a) Speed of the water exiting Carolyn's faucet = 2.044 m/s

b Gauge pressure in the main water supply pipe = 119774.65 Pa