A fluid with density $d$ flows in a turbulent manner through a pipe of cross-sectional area $A$. Find an expression for the pressure difference $(P_x - P_y)$ between points $x$ and $y$ in the pipe, using the symbols given in the diagram.

1. Problem Description

A fluid with density $d$ flows in a turbulent manner through a pipe of cross-sectional area $A$. Find an expression for the pressure difference $(P_x - P_y)$ between points $x$ and $y$ in the pipe, using the symbols given in the diagram.

2. Solution Steps

We can use Bernoulli's equation to relate the pressure, velocity, and height at points $x$ and $y$. However, since the flow is turbulent and viscous forces are negligible, we can assume that the total energy remains constant, and Bernoulli's equation can be applied.

Bernoulli's equation states:

$P + \frac{1}{2}dv^2 + dgh = constant$,

where

$P$ is the pressure,

$d$ is the density of the fluid,

$v$ is the velocity of the fluid,

$g$ is the acceleration due to gravity, and

$h$ is the height of the fluid above a reference point.

Applying Bernoulli's equation to points $x$ and $y$, we have:

$P_x + \frac{1}{2}dv_x^2 + dgh_x = P_y + \frac{1}{2}dv_y^2 + dgh_y$.

From the diagram, the height of point $x$ above the reference is $h$. Let's assume the reference is the bottom line, so $h_x = h$. Since point $y$ is at the same height as the dashed line, $h_y = h$.

Therefore, $h_x = h_y = h$, and $dgh_x = dgh_y$.

Now, let $A_x$ be the cross-sectional area at point $x$, and $A_y$ be the cross-sectional area at point $y$. Let $A_x = A_1$ according to the figure. Then let $v_x$ and $v_y$ be the flow velocities at $x$ and $y$, respectively.

From the equation of continuity, $A_xv_x = A_yv_y$.

$P_x + \frac{1}{2}dv_x^2 + dgh = P_y + \frac{1}{2}dv_y^2 + dgh$.

$P_x - P_y = \frac{1}{2}dv_y^2 - \frac{1}{2}dv_x^2$.

$P_x - P_y = \frac{1}{2}d(v_y^2 - v_x^2)$.

Since the equation of continuity applies, then $A_1 v_x = A_y v_y$ and thus $v_x = (A_y / A_1) v_y$.

Therefore, $P_x - P_y = \frac{1}{2}d[v_y^2 - (\frac{A_y}{A_1} v_y)^2]$.

$P_x - P_y = \frac{1}{2}d[v_y^2 - \frac{A_y^2}{A_1^2}v_y^2] = \frac{1}{2}dv_y^2 (1 - \frac{A_y^2}{A_1^2})$.

From the diagram it seems that the cross-sectional area at point $y$ is simply $A$, so $A_y=A$. Then $P_x - P_y = \frac{1}{2}dv_y^2(1 - \frac{A^2}{A_1^2})$.

If we assume $A_y \approx A_1$ then $v_x \approx v_y$ and $P_x - P_y \approx 0$.

We assume that $v_y$ is the flow rate. Therefore $P_x - P_y = \frac{1}{2}d (v_y^2 - v_x^2)$

3. Final Answer

$P_x - P_y = \frac{1}{2}d(v_y^2 - v_x^2)$