Chapter 15, Problem 48PQ

A fluid flows through a horizontal pipe that widens, making a 45° angle with the y axis (Fig. P15.48). The thin part of the pipe has radius R, and the fluid’s speed in the thin part of the pipe is v₀. The origin of the coordinate system is at the point where the pipe begins to widen. The pipe’s cross section is circular.

1. a. Find an expression for the speed v(x) of the fluid as a function of position for x > 0
2. b. Plot your result: v(x) versus x.

FIGURE P15.48

(a) The continuity equation (Eq. 15.21) relates the cross-sectional area to the speed of the fluid traveling through the pipe.

A₀v₀ = A(x)v(x)

$v\left(x\right)=\frac{A_0{v}_0}{A(x)}$

The cross sectional area is the area of a circle whose radius is y(x). The widening pan of the pipe is a straight line with slope of 1 and intercept y(0) = R.

y(x) = mx + b = x + R

A(x) = π[y(x)]² = π(x + R)²

Plug this into the formula for the velocity.

Plug this into the formula for the velocity.

$v\left(x\right)=\frac{A_0{v}_0}{\pi {(x+R)}^2}$