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Chapter 15: Problem 36

A fire hose \(10 \mathrm{cm}\) in diameter delivers water at \(15 \mathrm{kg} / \mathrm{s}\). The hose terminates in a 2.5 -cm-diameter nozzle. What are the flow speeds (a) in the hose and (b) at the nozzle?

Short Answer

Expert verified
(a) The flow speed in the hose is \(v_{hose} = Q/A_{hose}\). (b) The flow speed at the nozzle is \(v_{nozzle} = A_{hose}v_{hose}/A_{nozzle}\).

Step by step solution

01

Understand the continuity equation

The continuity equation in fluid dynamics is given by \(A_1v_1 = A_2v_2\). This equation tells that the mass flow rate must remain constant throughout the fluid and hence, the product of the cross-sectional area (A) and flow speed (v) at any two points in the fluid must be equal.
02

Compute the cross-sectional areas

First, calculate the cross-sectional areas of the hose and the nozzle. The cross-sectional area of a pipe, A, is given by \(\pi r^2\), where r is the radius of the pipe. Convert diameters to radii and to meters before substituting into the formula. So, \(A_{hose} = \pi (0.05 m)^2\), and \(A_{nozzle} = \pi (0.0125 m)^2\).
03

Identify known quantities

The hose delivers water at \(15 kg/s\). Since the density of water (\(ρ\)) is \(1000 kg/m^3\), this mass flow rate corresponds to a volume flow rate of \(Q = 15 kg/s / 1000 kg/m^3 = 0.015 m^3/s\). This volume flow rate is equal to the product of the cross-sectional area and the flow speed, \(Q = A_{hose}v_{hose}\). Hence, we have the flow rate and the cross-sectional area.
04

Compute the flow speeds (a) in the hose and (b) at the nozzle

For the hose, rearrange the formula from step 3 to solve for \(v_{hose}\): \(v_{hose} = Q/A_{hose}\). For the nozzle, apply the continuity equation: \(A_{hose}v_{hose} = A_{nozzle}v_{nozzle}\). Rearrange and solve for \(v_{nozzle}\).

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