Chapter 5, Problem 5.1P

Interpretation Introduction

Interpretation:

The value of downstream velocity for the given system is to be calculated.

Concept Introduction:

The general continuity equation to be used for a controlled volume is:

  ${\displaystyle {\iint }_{C.S}\rho \left(v\cdot n\right)dA}+\frac{\partial }{\partial t}{\displaystyle {\iiint }_{C.V}\rho dV}=0$ ...... (1)

Here, C.S is the control surface over which the integral dA is taken, C.V is the control volume over which the integral dV is taken, $\rho$ is the density of the fluid, $v$ is the velocity vector, $n$ is the direction of the vector $v$ , The product $v\cdot n$ is scalar defined as:

  $v\cdot n=\left|v\right|\left|n\right|\cos \theta$

Explanation of Solution

Given information:

The figure given for the flow of water through a 4 ft wide tunnel in which a two-dimensional object is placed is given as:

  

Across the cross-section, the upstream velocity $v_1$ is uniform.

The control surface taken for the given system is the entire tunnel given in the figure. The accumulation term will be zero as the system is assumed to be at steady state.

According to the continuity equation, the mass of water entering the tunnel is equal to the mass of water exiting it. Since, the density is constant for the entire system, the reduced form of the continuity equation is written below.

For the two-dimensional object, the trajectory of the outflow velocity is divided into two uniform sections. From the middle of the tunnel to 1 ft towards the edge, the velocity profile is linear while for 1 ft to 2 ft of the tunnel towards the edge, the velocity profile is constant.

  $\begin{array}{c}{\displaystyle {\iint }_{C.S}\rho \left(v\cdot n\right)dA}=0\\ \rho \cdot v_1\cdot w-\left(2v_2\rho {\displaystyle \underset{0}{\overset{1}{\int }}ydy}+2v_2\rho {\displaystyle \underset{1}{\overset{2}{\int }}dy}\right)=0\\ v_1\left(4\right)-\left(2v_2{\left(\frac{y^2}{2}\right)}_0^1+2v_2{\left(y\right)}_1^2\right)=0\\ \left(20\right)\left(4\right)-\left(2v_2\left(\frac{1}{2}\right)+2v_2\left(1\right)\right)=0\\ v_2=26.7\text{ fps}\end{array}$