A stream moving with a speed of 0.5 m/s reaches a point where the cross-sectional area of the stream decreases to one-fourth of the original area. What is the water speed in this narrowed portion of the stream?
To find the water speed in the narrowed portion of the stream, we can apply the principle of conservation of mass. According to this principle, the mass of the water remains constant along the stream.
The mass of the water can be calculated using the equation:
Mass = Density × Volume
Since the density of water is constant, we can write:
Mass 1 = Mass 2
(Density × Volume 1) = (Density × Volume 2)
Now, let's assume the original cross-sectional area of the stream as A1 and the water speed in that portion as v1. And let's assume the cross-sectional area of the narrowed portion of the stream as A2 and the water speed in that portion as v2.
The flow rate of the stream, which is the volume of water passing through a given point per unit time, is given by:
Flow rate = Velocity × Cross-sectional area
For the original stream (1):
Flow rate 1 = v1 × A1
For the narrowed portion (2):
Flow rate 2 = v2 × A2
Since the mass of the water remains constant, we can equate the flow rates:
Flow rate 1 = Flow rate 2
v1 × A1 = v2 × A2
We are given that the cross-sectional area of the narrowed portion is one-fourth (1/4) of the original area, so A2 = (1/4)A1.
Substituting this value in the equation above, we have:
v1 × A1 = v2 × (1/4)A1
Now, let's plug in the given value for v1 = 0.5 m/s and solve for v2:
0.5 × A1 = v2 × (1/4)A1
Divide both sides by A1:
0.5 = v2 × (1/4)
Multiply both sides by 4:
2 = v2
Therefore, the water speed in the narrowed portion of the stream is 2 m/s.
To find the water speed in the narrowed portion of the stream, we can use the principle of continuity, which states that the mass flow rate of a fluid is constant within an enclosed system (assuming no external forces or changes in elevation). In simpler terms, this means that the product of the cross-sectional area and the velocity of the fluid remains constant.
Let's denote the initial cross-sectional area of the stream as A1 and the initial water speed as v1. Similarly, we denote the cross-sectional area in the narrowed portion as A2 and the water speed in this portion as v2.
According to the principle of continuity, we have the equation:
A1 * v1 = A2 * v2
Given that A2 is one-fourth of A1, we can substitute that into the equation:
A1 * v1 = (1/4)A1 * v2
Simplifying the equation, we get:
v1 = (1/4)v2
We can rearrange the equation to solve for v2:
v2 = 4v1
Since we know v1 is 0.5 m/s, we can plug it in:
v2 = 4(0.5) = 2 m/s
Therefore, the water speed in the narrowed portion of the stream is 2 m/s.
In steady state flow, which this is, the product of velocity and cross sectional area is a constant that is equal to the volume flow rate.
If the cross sectional is reduced to 1/4 the former value, the flow velocity must increase by a factor of 4.
A*V = constant is called the "continuity equation"