Fluid dynamics (continuity equation), pressure on gate

[xzi86]

Homework Statement

Consider the cross section of a sluice gate, which is a device for controlling the flow of water in channels. Determine the force on the gate per unit width of the gate.

Hint: think of each line as a surface, with the length given above, and 1 foot of depth in the direction normal to the plane of the paper. The hydrostatic gauge pressure field p equals ρgh. That is, the gauge pressure at any point in the flow will linearly vary with the depth of the water, h. Use the average values of the pressure (at half-depth) for the pressure forces.

Homework Equations

Continuity equation integral form

The Attempt at a Solution

The answer is : V =16.667 ft/sec, Force in the x- direction on the gate = 1708 lb per foot of width in the direction normal to the plane of the paper. No idea how this was reached. How do you apply the continuity equation to this problem? And what does the hint mean? Any help would be greatly appreciated. Thanks.

 

[KataKoniK]

Thank you for your post. I am happy to assist you in solving this problem.

Firstly, let's define some variables:
- V = velocity of the water in the channel (ft/sec)
- A = cross-sectional area of the channel (ft^2)
- h = depth of the water (ft)
- ρ = density of the water (lb/ft^3)
- g = acceleration due to gravity (32.2 ft/sec^2)
- p = pressure (lb/ft^2)
- F = force (lb)

Now, let's apply the continuity equation to this problem. The continuity equation states that the mass flow rate (ρAV) is constant throughout the channel. In other words, the amount of water flowing through any cross section of the channel per unit time is the same. Therefore, we can write the following equation:

ρAV = constant

Since we are looking for the force per unit width of the gate, we can rearrange the equation to solve for force:

F = ρAV

Now, let's look at the hint given in the problem. It tells us to think of each line in the cross section as a surface with 1 foot of depth in the direction normal to the plane of the paper. This means that the area of each line is 1 foot times the length given in the problem. Therefore, the total cross-sectional area of the channel can be calculated as follows:

A = 4 * 1 ft * 6 ft = 24 ft^2

Next, we need to find the average pressure at half-depth. The hint also tells us that the gauge pressure at any point in the flow will linearly vary with the depth of the water. This means that the average pressure at half-depth will be half of the maximum pressure at the bottom of the channel. The maximum pressure at the bottom of the channel can be calculated using the hydrostatic gauge pressure equation:

p = ρgh = (62.4 lb/ft^3) * 6 ft * 32.2 ft/sec^2 = 12012.48 lb/ft^2

Therefore, the average pressure at half-depth is:

p = 12012.48 lb/ft^2 / 2 = 6006.24 lb/ft^2

Now, we can plug these values into our equation for force:

F = ρAV =