Chapter 16, Problem 16.1PP

Calculate the force required to hold a flat plate in equilibrium perpendicular to the flow of water at $25$m/s issuing from a $75$ -mm-diameter nozzle.

To determine

The force required to hold a flat plate in equilibrium perpendicular to the flow of water at 25m/s issuing from a 75-mm-diameter nozzle.

Answer to Problem 16.1PP

The force required, F= $\text{2}\text{.760}\text{kN}$.

Explanation of Solution

Given information:

Flow velocity of water $\text{V=}$ 25m/s and diameter of nozzle $\text{d =}75mm.$

Calculation:

Calculate force using $\text{F = ρ Q V}$

Where, $\text{ρ}$ is the density of water

   $\text{Q}$ is the discharge

   $\text{V}$ is the velocity

   $\text{F}$ is the force

   $\text{Q = AV}$

Where, A is the area of the cross − section.

V is the velocity of flow.

   $\begin{array}{l}\text{A =}\frac{{\text{πd}}^{\text{2}}}{\text{4}}\text{ = }\frac{\text{3}\text{.14×0}{\text{.075}}^{\text{2}}}{\text{4}}\\ \text{A = 0}\text{.00441}{\text{m}}^{\text{2}}\text{.}\end{array}$

   $\begin{array}{l}\text{Q = AV}\\ \text{Q = 0}\text{.00441}{\text{m}}^{\text{2}}\text{ × 25}\frac{\text{m}}{\text{s}}\\ \text{Q = 0}\text{.1104}\frac{{\text{m}}^{\text{3}}}{\text{s}}.\end{array}$

   $\begin{array}{l}\text{F = ρQV}\\ \text{F = 1000}\frac{\text{kg}}{{\text{m}}^{\text{3}}}\text{×0}\text{.1104}\frac{{\text{m}}^{\text{3}}}{\text{s}}\text{×25}\frac{\text{m}}{\text{s}}\\ \text{F = 110}\text{.4}\frac{\text{kg}}{\text{s}}\text{×25}\frac{\text{m}}{\text{s}}\\ \text{F = 2760}\frac{\text{kg}\text{m}}{{\text{s}}^{\text{2}}}\text{ = 2760}\text{N}\\ \text{F = 2}\text{.760}\text{kN}\text{.}\end{array}$