Benoulli's Equaiton Pipe Pressure Problem

[mm874]

Homework Statement

Water flows down a pipe which tapers from 200 mm diameter to 125 mm diameter at the lower end. The difference in height between the two ends of the pipe is 4.0 m and the rate of flow is 1.50 m3 min-1. Calculate the pressure difference between the upper and lower ends of the pipe. Which end has the higher pressure?

Homework Equations

P+ρgh+0.5ρv^2=Constant
Volumetric Flow rate = A*v
Mass Flow Rate = ρ*Volumetric Flow rate

The Attempt at a Solution

Velocity at top = 0.025/0.0314 = 0.796m/s
Velocity at bottom = 0.025/0.0122 = 2.49m/s

Then I get lost at I put all the results into benoulli's equation but am not sure what
to do from there?

 

[SteamKing]

Write your Bernoulli equation at the top of the pipe.
Write another Bernoulli equation for the bottom of the pipe.
The two Bernoulli equations equal the same constant, and therefore equal each other.
The only unknowns are the pressures at each end of the pipe.

 

[mm874]

So how do you calculate the unknowns if there is one either side. The pressure cannot be equal at both ends. As it state the faster the speed the lower the pressure.

 

[SteamKing]

You are only asked to find which end of the pipe has the higher pressure. By equating the two Bernoulli equations, you should be able to determine the ratio of the pressures, and depending on the magnitude of the ratio, which is the higher pressure.

 

[DeeNos]

As a scientist, it is important to approach problems like this in a systematic and logical manner. Let's break down the problem and use the given information to solve for the pressure difference between the upper and lower ends of the pipe.

First, we can use Bernoulli's equation to relate the pressure at the two ends of the pipe:

P1 + ρgh1 + 0.5ρv1^2 = P2 + ρgh2 + 0.5ρv2^2

Where P1 and P2 are the pressures at the upper and lower ends of the pipe, respectively; ρ is the density of water; g is the acceleration due to gravity; h1 and h2 are the heights of the two ends of the pipe; and v1 and v2 are the velocities at the two ends of the pipe.

Next, we can use the given information to solve for the velocities at the two ends of the pipe. We know that the volumetric flow rate is 1.50 m^3/min, so we can use the equation Volumetric Flow rate = A*v to solve for v1 and v2. The cross-sectional area of the pipe at the top is A1 = π(0.1m)^2 = 0.0314 m^2 and at the bottom is A2 = π(0.0625m)^2 = 0.0122 m^2. Therefore, v1 = 0.796 m/s and v2 = 2.49 m/s.

Now, we can plug these values into Bernoulli's equation:

P1 + ρgh1 + 0.5ρv1^2 = P2 + ρgh2 + 0.5ρv2^2

We know that the difference in height between the two ends of the pipe is 4.0 m, so h2 = 4.0 m and h1 = 0 m. Also, we can assume that the density of water is constant, so we can cancel it out on both sides of the equation. This leaves us with:

P1 + 0 + 0.5v1^2 = P2 + 4.0 + 0.5v2^2

Substituting in our previously calculated values, we get:

P1 + 0 + 0.5(0.796)^2 = P2 +