Repeat Problem 2.72 if the flow inside the apparatus is...

Repeat Problem 2.72 if the flow inside the apparatus is liquid water $\left(\rho \simeq 1000 \mathrm{kg} / \mathrm{m}^{3}\right)$ instead of air. Find the pressure difference between the two holes flush with the bottom of the channel. You cannot neglect the two unequal water columns.

Key Concepts

Pressure Differential Due to Unequal Fluid Columns

In systems with two regions of fluid at different elevation levels, the pressure difference is influenced by the weight of the fluid columns above each region. Analyzing these differences requires accounting for the hydrostatic pressure from each column, and it often involves combining this with dynamic flow principles to determine the net pressure difference across openings or boundaries.

Fluid Density

Fluid density is the mass per unit volume of a fluid and plays a key role in determining both dynamic and static pressure within a system. It directly affects the magnitude of the hydrostatic pressure; denser fluids such as water generate significantly higher pressures for a given depth compared to less dense fluids like air.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the gravitational force acting on it. It increases linearly with depth and is calculated by the product of the fluid density, gravitational acceleration, and the height of the fluid column. This concept becomes essential when analyzing pressure differences caused by vertical height disparities in fluid columns.

Bernoulli's Principle

This principle states that for an incompressible, non-viscous fluid in steady flow, the sum of its pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline. It is crucial in problems involving moving fluids as it provides a relationship between pressure differences and changes in flow velocity and elevation.

Transcript

00:01 So what we want to do is we want to find the pressure difference between chamber 1 and chamber 2, and we have water flowing through this top pipe here.

00:10 And we have a monometer attached that's reading a height difference of 200 millimeters.

00:19 So in terms of the values that we have in the diagram, we have h1 minus h2 is equal to delta h.

00:29 So what we want to do now is we want to set up an expression that describes the forces of balance when the monometer is reading.

00:36 Pressure difference.

00:38 So we're going to start by looking at this red point here in chamber one.

00:43 So we have the pressure from the water that's flowing through the tube here that we're looking for times cross -sectional area.

00:54 And then we'll have to take into account the pressure due to this water column here with the height of h1.

00:59 So the density of water times g times h1.

01:05 And then for the mercury, i'll take the density of mercury times g and for the height we don't have the direct value but we have that it's the total height of the monometer h minus h1 which is the space that the water takes up so we have h minus h1 for the height and both of these should be multiplied by the cross -sectional area.

01:35 So now for point two here we have the pressure of the water in chamber two times cross -sectional area plus the pressure due to the water column on this side.

01:53 So that would be the density of water times g times h2.

02:00 And then for mercury as well, again, for the height of the mercury column, it would be the total height h minus h2.

02:17 And this would also be multiplied by the cross -sectional area.

02:23 But what we want to find out is delta p, that's what we're looking for...

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