A sump pump (used to drain water from the basement of houses...

A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of $3.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .$ (a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point? (b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem.

Key Concepts

Bernoulli's Principle

Bernoulli's principle is a statement of energy conservation in fluid dynamics, asserting that for a steady, incompressible, and non-viscous fluid flow, the sum of the pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. It enables the analysis of how pressure and fluid speed vary with elevation and cross-sectional changes in a flow system.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases linearly with depth according to the relation p = ?gh, where ? is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column. This concept is critical when assessing pressure changes resulting from the elevation differences in a fluid system.

Continuity Equation

The continuity equation expresses the conservation of mass for a steady, incompressible fluid flow. It states that the product of the cross-sectional area of the flow and the fluid velocity remains constant along the flow path. This relationship is essential for determining how changes in the size of a conduit (such as a hose) affect the fluid velocity, which in turn influences the pressure profile along the streamline.

Transcript

0:00 You can use bernoulli's equation.

00:09 And essentially, this is saying that the pressure, rather plus one -half times the density times the velocity squared, plus the density times the acceleration due to gravity times height, this will all equal a constant.

00:27 And so here we're going to let subscripts 1 and 2 refer to any two points along the path that the bit of fluid flows.

00:35 And so here we are going to consider a situation where the fluid moves, but its depth is constant.

00:45 So essentially, this term here is not going to be accounted for because the height is, of course, the same.

00:52 And so we can say p sub 1 plus 1 half, the density times v sub 1 squared equals pressure sub 2 plus 1⁄2, plus 1⁄2 times row, the density times v sub 2 squared.

01:07 And so for part a, we can say that we're going to consider a situation here where the velocities are the same.

01:24 So now we have pressure sub 1 plus row g .h .1 plus this would be equal to the pressure sub 2 plus row gh sub 2.

01:38 And so now we have that the pressure at point two is equaling the pressure at point one plus the density times the acceleration due to gravity times h sub 1 minus h sub 2.

01:51 And we can then say that p sub 2 the pressure would be equaling 3 .00 times 10 to the 5 newton's per square meter.

02:01 This would be plus 1 ,000 kilograms per cubic meter, multiplied by 9 .80 meters per second squared, multiplied by the height difference of 2 .50 meters, and we have that 4 part a, the pressure sub 2, would be equaling 3 .245 times 10 to the 5th newton's per square meter.

02:33 This would be our answer for part a...

Please give Ace some feedback