Water enters a house through a pipe with an inside diameter of 2.0cm at an absolute pressure 4.0×10^5 pascal. a 1.0-cm-diameter pipe leads to the second floor bathroom 5.0m above. when the flow speed at the inlet pipe is 1.5m/s. find the flow speed, pressure, and volume flow rate in the bathroom.

6.0 m/s

3.3×10⁵ Pa

4.7×10⁻⁴ m³/s

Explanation:

Mass is conserved, so the mass flow at the inlet equals the mass flow in the bathroom.  Since the density is the same, that means the volume flow rate is the same.

Q₁ = Q₂

v₁ A₁ = v₂ A₂

v₁ (π D₁² / 4) = v₂ (π D₂² / 4)

v₁ D₁² = v₂ D₂²

Given v₁ = 1.5 m/s, D₁ = 2.0 cm, and D₂ = 1.0 cm:

(1.5 m/s) (2.0 cm)² = v (1.0 cm)²

v = 6.0 m/s

To find the pressure, we use the Bernoulli principle:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Given:

P₁ = 4.0×10⁵ Pa

ρ = 1000 kg/m³

v₁ = 1.5 m/s

h₁ = 0 m

v₂ = 6.0 m/s

h₂ = 5.0 m

4.0×10⁵ + ½ (1000) (1.5)² + 0 = P + ½ (1000) (6.0)² + (1000) (9.8) (5.0)

P = 3.3×10⁵ Pa

Finally, the volume flow rate is:

Q = vA

Q = v (π D² / 4)

Q = π/4 (6.0 m/s) (0.01 m)²

Q = 4.7×10⁻⁴ m³/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures