A fireman's water hose has a diameter of about 4.0 inches. The water is flowing through it at about 14.0 m/s. The nozzle's diameter drops to about 0.80 inches. If there is no friction, what would the exit velocity of the water be?
area proportional to diameter^2
V2 A2 = V1A1
V2 / V1 = A1/A2 = 4^2/.8^2 = 16 / .64
V2 = 14 * 16/.64 = 350, agree
I got it 350 m/s
To determine the exit velocity of the water, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water remains constant throughout the hose.
The mass flow rate (ṁ) is given by the equation:
ṁ = ρ * A * V
where:
ṁ is the mass flow rate,
ρ is the density of water,
A is the cross-sectional area of the hose, and
V is the velocity of the water.
First, let's convert the diameters to radius and the velocity to metric units:
The initial diameter of 4.0 inches is equivalent to a radius of 2.0 inches or 0.0508 meters (since 1 inch = 0.0254 meters).
The reduced diameter of 0.80 inches is equivalent to a radius of 0.40 inches or 0.0102 meters.
Now let's calculate the cross-sectional areas:
The initial cross-sectional area (A1) = π * R1^2
= π * (0.0508)^2
The reduced cross-sectional area (A2) = π * R2^2
= π * (0.0102)^2
Next, we'll set up the equation for the conservation of mass:
ṁ1 = ṁ2
ρ * A1 * V1 = ρ * A2 * V2
Since the densities cancel out, and we are solving for V2, we can rearrange the equation:
V2 = (A1 * V1) / A2
Now we can substitute the values:
V2 = (π * (0.0508)^2 * 14.0 m/s) / (π * (0.0102)^2)
Simplifying further:
V2 = (0.0012916 m^2 * 14.0 m/s) / (0.00010404 m^2)
V2 = 16.716 m/s
Therefore, the exit velocity of the water would be approximately 16.716 m/s.
To find the exit velocity of the water, we can use the principle of continuity, which states that the volume flow rate, represented by Q, remains constant in an incompressible fluid.
The equation for the principle of continuity is: Q = A1 * v1 = A2 * v2, where A1 and v1 are the cross-sectional area and velocity at the first point, and A2 and v2 are the cross-sectional area and velocity at the second point.
In this case, the water is flowing through a fireman's water hose, which has a larger diameter of 4.0 inches at the inlet (point 1), and it narrows down to a smaller diameter of 0.80 inches at the nozzle (point 2).
First, let's convert the diameters to meters:
Diameter at point 1 (inlet): 4.0 inches = 0.1016 meters (1 inch = 0.0254 meters)
Diameter at point 2 (nozzle): 0.80 inches = 0.02032 meters
Next, we'll find the cross-sectional areas at each point.
Area at point 1 (A1):
A1 = π * (diameter at point 1/2)^2
= π * (0.1016 / 2)^2
Area at point 2 (A2):
A2 = π * (diameter at point 2/2)^2
= π * (0.02032 / 2)^2
The velocity at point 1 (v1) is given as 14.0 m/s.
Now, we can use the principle of continuity to find the velocity at point 2 (v2).
Q = A1 * v1 = A2 * v2
Since Q is constant and we're assuming there is no friction, we can solve for v2:
v2 = (A1 * v1) / A2
Substituting the values we found:
v2 = (A1 * 14.0) / A2
Now, let's calculate the values of A1 and A2 and find the exit velocity:
A1 = π * (0.1016 / 2)^2
A2 = π * (0.02032 / 2)^2
v2 = (π * (0.1016 / 2)^2 * 14.0) / (π * (0.02032 / 2)^2)
Simplifying the equation:
v2 = (0.0254^2 * 14.0) / (0.00508^2)
Calculating the values:
v2 = (0.00064516 * 14.0) / 0.0000258016
v2 ≈ 13.59 m/s
Therefore, the exit velocity of the water would be approximately 13.59 m/s if there is no friction.