Math / Calculus

QuestionW. A non-viscous incompressible liquid (with homogeneous density $796 \mathrm{~kg} / \mathrm{m}^{3}$ ) is flowing in the pipe shown in the figure. The pipe radius is $R=23.5 \mathrm{~cm}$ in the first section (before the funnelshaped mid section) and $r=14.5 \mathrm{~cm}$ in the second section (after the funnel-shaped mid section). We are informed that the pipe axis is perfectly horizontal, after an analysis of the liquid streamlines. From measurements, we also know that a tracer following the liquid flow moves through at point A with a velocity of $0.790 \mathrm{~m} / \mathrm{s}$ , and the pressure at point B is 725 Pa . What's the pressure at point A?

Studdy Solution

STEP 1

1. The liquid is non-viscous and incompressible with a density of $796 \, \text{kg/m}^3$ .
2. The flow is steady and the pipe is horizontal.
3. The radius of the pipe at point A is $R = 23.5 \, \text{cm}$ .
4. The radius of the pipe at point B is $r = 14.5 \, \text{cm}$ .
5. The velocity at point A is $0.790 \, \text{m/s}$ .
6. The pressure at point B is $725 \, \text{Pa}$ .

STEP 2

1. Use the Continuity Equation to find the velocity at point B.
2. Apply Bernoulli's Equation to find the pressure at point A.

STEP 3

Use the Continuity Equation, which states that the product of cross-sectional area and velocity is constant along the pipe:
$A_A \cdot v_A = A_B \cdot v_B$
Calculate the cross-sectional areas:
$A_A = \pi R^2$ $A_B = \pi r^2$
Substitute the known values:
$\pi (0.235 \, \text{m})^2 \cdot 0.790 \, \text{m/s} = \pi (0.145 \, \text{m})^2 \cdot v_B$
Solve for $v_B$ :
$v_B = \frac{\pi (0.235)^2 \cdot 0.790}{\pi (0.145)^2}$
$v_B = \frac{0.235^2 \cdot 0.790}{0.145^2}$
Calculate $v_B$ .

STEP 4

Apply Bernoulli's Equation, which relates the pressures and velocities at two points in a horizontal flow:
$P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2$
Rearrange to solve for $P_A$ :
$P_A = P_B + \frac{1}{2} \rho (v_B^2 - v_A^2)$
Substitute the known values:
$P_A = 725 \, \text{Pa} + \frac{1}{2} \times 796 \, \text{kg/m}^3 \times (v_B^2 - 0.790^2)$
Calculate $P_A$ .
The pressure at point A is:
$\boxed{P_A}$

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