Modified/simplified continuity equation:
S1*v1*ρ1 = S2*v2*ρ2
where
S1 and S2 = entrance and exit cross-sectional area
v1 and v2 = entrance and exit velocity
ρ1 and ρ2 = initial and final density
Assuming density of water is constant, we can cancel ρ1 and ρ2:
S1 * v1 = S2 * v2
The hose has a cross-section in shape of a circle, so its area is πr^2.
Substituting,
(π(2^2)) * 2 = (π(0.25^2)) * v2
Now solve for v2, units in m/s. Hope this helps~ `u`
Water flows through a 4 cm diameter garden hose at a velocity of 2 m/s. What is the velocity of the water if a
nozzle is used with a diameter of 0.5 cm?
please help..thanks in advanced
1 answer