Water is flowing through a rectangular channel that is 12 m wide with a speed of 0.75 m/s. The water then flows into four identical rectangular channels that have a width of 3.0 m. The depth of the water does not change as it flows into the four channels. What is the speed of the water in one of the smaller channels?
We can use the principle of conservation of flow rate in this problem. The flow rate in the main channel is equal to the sum of the flow rates in the four smaller channels. Flow rate is given by the product of the cross-sectional area of the channel and the speed of the water flow.
Let's denote the width of the main channel as W1 (12 m), the width of each smaller channel as W2 (3 m), the speed of water in the main channel as V1 (0.75 m/s), and the speed of water in the smaller channel as V2.
The depth of the water remains the same, so we'll just use a generic term, D, to represent the depth of the water.
We can calculate the flow rate in the main channel:
Flow_rate_main_channel = Area_main_channel × V1
Flow_rate_main_channel = (W1 × D) × V1
Flow_rate_main_channel = (12 × D) × 0.75
Now let's calculate the flow rate in one of the smaller channels:
Flow_rate_one_small_channel = Area_one_small_channel × V2
Flow_rate_one_small_channel = (W2 × D) × V2
Flow_rate_one_small_channel = (3 × D) × V2
There are 4 smaller channels, and the flow rate in the main channel is equal to the total flow rate in the smaller channels. So, we can express this as:
Flow_rate_main_channel = 4 × Flow_rate_one_small_channel
Now we can plug in the expressions for the flow rates that we found earlier:
(12 × D) × 0.75 = 4 × (3 × D) × V2
We can see that D is present on both sides of the equation, so we can divide both sides by it:
(12 × 0.75) = 4 × (3) × V2
Next, we can solve for V2:
9 = 12 × V2
V2 = 9 / 12
V2 = 0.75 m/s
So, the speed of the water in one of the smaller channels will be 3 m/s.
To find the speed of the water in the smaller channel, we can use the principle of conservation of mass. According to this principle, the mass of water entering the smaller channels should be equal to the mass of water in the larger channel.
First, let's find the volume flow rate of water in the larger channel. Volume flow rate (Q) is the product of the cross-sectional area (A) and the speed of flow (v):
Q = A * v
In this case, the cross-sectional area of the larger channel (A1) is given by the product of the width (w1) and the depth (d1):
A1 = w1 * d1 = 12 m * d1
Given that the speed of flow (v1) in the larger channel is 0.75 m/s, we can substitute these values into the equation to find the volume flow rate in the larger channel:
Q1 = A1 * v1 = (12 m * d1) * 0.75 m/s = 9 d1 m^3/s
Since the water flows into four identical channels, the volume flow rate in each smaller channel (Q2) would be equal to one-fourth of the volume flow rate in the larger channel (Q1):
Q2 = Q1 / 4 = (9 d1 m^3/s) / 4 = 2.25 d1 m^3/s
Now, let's find the speed of the water in one of the smaller channels (v2). The cross-sectional area of the smaller channel (A2) is given by the product of the width (w2) and the depth (d1) since the depth does not change:
A2 = w2 * d1 = 3.0 m * d1
We can substitute the volume flow rate in the smaller channel (Q2) and the cross-sectional area of the smaller channel (A2) into the volume flow rate equation to find the speed of the water in the smaller channel:
Q2 = A2 * v2
2.25 d1 m^3/s = (3.0 m * d1) * v2
Now we can solve for v2:
v2 = (2.25 d1 m^3/s) / (3.0 m * d1) = 0.75 m/s
Therefore, the speed of the water in one of the smaller channels is 0.75 m/s.
To determine the speed of water in one of the smaller channels, we can apply the principle of continuity, which states that the product of cross-sectional area and velocity remains constant in an incompressible fluid flowing through a channel.
The cross-sectional area of the flow in the wide channel is equal to the product of width and depth. Given that the width of the wide channel is 12 m and the depth remains constant, we can compute the cross-sectional area as follows:
Cross-sectional area in the wide channel = Width × Depth = 12 m × Y m
Next, we need to consider the conservation of mass. Since the mass of water flowing into the small channels is the same as the mass flowing in the wide channel, we can apply the principle of continuity:
Cross-sectional area in the wide channel × velocity in the wide channel = Cross-sectional area in a small channel × velocity in a small channel
Substituting the known values, we can write the equation as:
12 m × Y m × 0.75 m/s = 3.0 m × Y m × V m/s
Simplifying the equation:
9.0 m²/s = 3.0 m²/s × V m/s
Now, let's solve for V:
V = 9.0 m²/s / 3.0 m²/s
V = 3.0 m/s
Therefore, the speed of the water in one of the smaller channels is 3.0 m/s.