Question

Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the garden hose when a nozzle of diameter 0.0050 m) is attached to the end of the hose in units of m/s.

Answer 1

To calculate the exit velocity of water from the garden hose when a nozzle is attached, we can use the principle of conservation of mass and the equation for flow rate.

Step-by-step explanation:

To calculate the exit velocity of water from the garden hose when a nozzle is attached, we can use the principle of conservation of mass and the equation for flow rate. The equation for flow rate is Q = A1 v1 = A2 v2, where Q is the flow rate, A is the area, and v is the velocity. We can rearrange this equation to solve for v2 as follows:

1. First, we calculate the flow rate using the initial conditions: Q = (π * (0.016/2)^2) * 3 = 0.024 m^3/s.
2. Next, we can use the equation for flow rate to solve for v2: (π * (0.0050/2)^2) * v^2 = 0.024 m^3/s. Rearranging the equation, we find that v2 = (0.024 m^3/s) / (π * (0.0050/2)^2) ≈ 15.24 m/s.

Answer 2

v₂ = 306.12 m/s

Step-by-step explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

v₂ = 306.12 m/s