Bernoulli and Continuity Question

[tere_lai]

Bernoulli and Continuity Question!

A river (100 m wide) flows through its rectangular channel at a depth of 2.560 m at a velocity of 2.050 m/s. What is the velocity of the discharge if the channel is narrowed to 90 m?

Continuity equation: Q1 = 100m x 2.560 m x 2.050 m/s
Q2 = 90 m x H2 x V2

Bernoulli equation: V^2/2g + P/pg + z = constant

I can't seem to set up the equations properly in order to find intercept. I'm sure that P1 (ie. pressure1) is zero, but P2 isn't. I actually tried solving the question with P2=0 but it's wrong.

A little help?

 

[anathema]

I understand your confusion and I would be happy to help you solve this problem. First, let's review the equations that you have mentioned. The continuity equation states that the flow rate (Q) at any point in a channel must be equal to the flow rate at any other point in the channel. This means that the product of the cross-sectional area (A) and the velocity (V) must remain constant throughout the channel. Therefore, we can set up the following equation:

Q1 = Q2
100 m x 2.560 m x 2.050 m/s = 90 m x H2 x V2

Next, we can use the Bernoulli equation to determine the relationship between the velocity and pressure at different points in the channel. The Bernoulli equation states that the total energy in a fluid system must remain constant. This means that the sum of the kinetic energy, potential energy, and pressure energy at any point in the channel must be equal to the sum at any other point in the channel. Therefore, we can set up the following equation:

V1^2/2g + P1/pg + z1 = V2^2/2g + P2/pg + z2

Since we are dealing with a horizontal channel, we can ignore the potential energy (z) terms. Additionally, we can assume that the pressure at both points is equal, since the channel is open to the atmosphere. Therefore, we can simplify the Bernoulli equation to:

V1^2/2g = V2^2/2g

Now, we can substitute the values from the continuity equation into the Bernoulli equation to solve for the velocity at point 2:

(2.050 m/s)^2/2g = V2^2/2g
V2 = (2.050 m/s)^2 x 100/90 = 2.278 m/s

Therefore, the velocity of the discharge at point 2 is 2.278 m/s. I hope this helps clarify the equations and their use in solving this problem. If you have any further questions, please don't hesitate to ask. Keep up the good work in your scientific studies!

 

[Moetasim]

I understand your confusion with setting up the equations for this problem. Let me walk you through the steps to solve it.

First, we need to understand the concepts of Bernoulli's equation and the continuity equation. Bernoulli's equation states that the total energy of a fluid remains constant along a streamline. This means that the sum of the kinetic energy, potential energy, and pressure energy at any point in the fluid must be the same along a streamline. The continuity equation, on the other hand, states that the mass flow rate of a fluid is constant, meaning that the product of the cross-sectional area, velocity, and density must remain the same at all points along a streamline.

Now, let's apply these concepts to the problem at hand. We know that the river is flowing at a depth of 2.560 m and a velocity of 2.050 m/s in a channel with a width of 100 m. Using the continuity equation, we can set up the following equation:

Q1 = Q2

Where Q1 is the discharge at the initial channel width (100 m) and Q2 is the discharge at the narrowed channel width (90 m). Since the mass flow rate must remain constant, we can write the equation as:

100 m x 2.560 m x 2.050 m/s = 90 m x H2 x V2

Next, we need to use Bernoulli's equation to solve for the velocity in the narrowed channel. We can rearrange the equation to solve for V2:

V2 = √(2[(P1/pg + z1) - (P2/pg + z2) + V1^2/2g])

Where P1 and P2 are the pressures at the initial and narrowed channel, pg is the density of the fluid, z1 and z2 are the heights at the initial and narrowed channel, and V1 is the initial velocity.

Since the problem does not provide information about the pressures or heights at the two points, we can assume that they are equal. This means that the pressure and height terms will cancel out in the equation, leaving us with:

V2 = √(V1^2/2g)

Plugging in the values for V1 and g, we get:

V2 = √(2.050^2/2 x 9.8) = 2.067 m/s

Therefore,