Water flows through a fire hose of diameter 6.35 centimeters at a rate of 0.012 m3/s. the fire hose ends in a nozzle of inner diameter 2.2 centimeters. what is the velocity with which the water exits the nozzle?

Question

Nathan Ashcroft · Accepted Answer

#31.6ms^-1# Since the mass flow rate must always be the same and the density is constant, this solution assumes an incompressible fluid, which means the volumetric flow rate is constant throughout the hose and nozzle. In order to calculate the of velocity of the water in the nozzle we need to know the cross-sectional area of the nozzle.
#A_N=(πD_N^2)/4=(π(0.022)^2)/4=3.8*10^-4 m^2# Now we can use the cross-sectional area with the volumetric flow rate to calculate the velocity of the water in the nozzle:
#(dV)/(dt)=A(ds)/(dt)# Where #(dV)/(dt)# is the volumetric flow rate and #(ds)/(dt)# is the rate of change of displacement, i.e. the velocity. #v=((dV)//(dt))/A=0.012/(3.8*10^-4)=31.6ms^-1# A little over 70 mph, that is!

Adam Arnold · Answer

undefined To find the velocity with which the water exits the nozzle, we can use the principle of continuity, which states that the flow rate remains constant throughout the hose. The formula to calculate velocity (v) using the principle of continuity is:

$$A_1v_1 = A_2v_2$$

Where:
- $A_1$ is the cross-sectional area of the hose.
- $v_1$ is the velocity of water in the hose.
- $A_2$ is the cross-sectional area of the nozzle.
- $v_2$ is the velocity of water exiting the nozzle.

Given:
- Diameter of the hose (D1) = 6.35 cm = 0.0635 m
- Diameter of the nozzle (D2) = 2.2 cm = 0.022 m
- Flow rate (Q) = 0.012 m^3/s

We can calculate the cross-sectional areas of the hose and the nozzle using the formula $A = \pi \times (radius)^2$, where the radius is half of the diameter.

$$A_1 = \pi \times (0.0635/2)^2$$
$$A_2 = \pi \times (0.022/2)^2$$

Then, we can rearrange the equation and solve for $v_2$:

$$v_2 = \frac{A_1}{A_2} \times v_1$$

Substitute the known values and solve for $v_2$.