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
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`
To find the velocity of the water when a nozzle is used, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water remains constant when flowing through a system.
The mass flow rate can be calculated using the equation:
Mass flow rate = Density × Cross-sectional area × Velocity
For the garden hose, we have the following information:
Diameter = 4 cm = 0.04 m
Velocity = 2 m/s
To find the cross-sectional area of the garden hose, we can use the formula for the area of a circle:
Cross-sectional area = π × (Diameter/2)^2
Substituting the values:
Cross-sectional area = π × (0.04/2)^2 = π × 0.02^2 = 0.001256 m^2
We can assume that the density of water remains constant.
Now, let's find the velocity when a nozzle with a diameter of 0.5 cm is used. Using the same formulas:
Diameter = 0.5 cm = 0.005 m
Cross-sectional area = π × (0.005/2)^2 = π × 0.0025^2 = 0.00001963 m^2
We know that the mass flow rate is constant. So, we can equate the mass flow rate for the garden hose and the nozzle:
Density × Cross-sectional area (garden hose) × Velocity (garden hose) = Density × Cross-sectional area (nozzle) × Velocity (nozzle)
Now we can rearrange the formula to solve for the velocity (nozzle):
Velocity (nozzle) = (Density × Cross-sectional area (garden hose) × Velocity (garden hose)) / (Density × Cross-sectional area (nozzle))
The density of water cancels out on both sides:
Velocity (nozzle) = (Cross-sectional area (garden hose) × Velocity (garden hose)) / Cross-sectional area (nozzle)
Substituting the values:
Velocity (nozzle) = (0.001256 m^2 × 2 m/s) / 0.00001963 m^2
Calculating the result:
Velocity (nozzle) ≈ 63.87 m/s
Therefore, the velocity of the water when a nozzle with a diameter of 0.5 cm is used is approximately 63.87 m/s.