Answer:
pressure of the water = 3.3 × [tex]10^{5}[/tex] pa
Explanation:
given data
velocity v1 = 1.5 m/s
pressure P = 400,000 Pa
inside radius r1 = 1.00 cm
pipe radius r2 = 0.5 cm
h1 = 0 (datum at inlet)
h2 = 5.0 m (datum at inlet)
density of water ρ = 1000 kg/m³
to find out
pressure of the water
solution
we consider here flow speed in bathroom that is = v2 and Pressure in bathroom is = P2
here we will use both continuity and Bernoulli equations
because here we have more than one unknown so that
v1 × A1 = v2 × A2 × P1 + ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²
now we use here first continuity equation for get v2
v2 = [tex]v1 \frac{A1}{A2}[/tex]
v2 = [tex]1.5 \frac{\pi * 0.01^2}{\pi * 0.005^2}[/tex]
v2 = 6 m/s
and now we use here bernoulli eqution for find here p2 that is
P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )
P2 = 400000 - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )
P2 = 3.3 × [tex]10^{5}[/tex] pa
