A constricted horizontal tube of radius r(1)=4.00 cm tapers to a tube of radius r(2)=2.50 cm. If water flows at speed 2.00 m/s in the larger tube, (a) find the speed of the water v(2) in the smaller tube. (b) Find the flow rate in this tube.
To find the speed of the water (v2) in the smaller tube and the flow rate, we can apply the principle of continuity.
According to the principle of continuity, the product of the cross-sectional area of a fluid moving through a tube and its speed remains constant as long as the flow is steady.
Given:
Radius of the larger tube, r1 = 4.00 cm = 0.04 m
Radius of the smaller tube, r2 = 2.50 cm = 0.025 m
Speed of water in the larger tube, v1 = 2.00 m/s
a) To find the speed of the water (v2) in the smaller tube, we can use the principle of continuity equation:
A1 * v1 = A2 * v2
Where A1 and A2 are the cross-sectional areas of the larger and smaller tubes, respectively.
To find A1 and A2:
A1 = π * (r1)^2
A2 = π * (r2)^2
Substituting the given values:
A1 = π * (0.04 m)^2
A2 = π * (0.025 m)^2
Now, we can rearrange the equation to solve for v2:
v2 = (A1 * v1) / A2
Substituting the values:
v2 = (π * (0.04 m)^2 * 2.00 m/s) / (π * (0.025 m)^2)
Simplifying:
v2 = (0.0016 m^2 * 2.00 m/s) / 0.000625 m^2
v2 = 2.56 m/s
Therefore, the speed of the water in the smaller tube is 2.56 m/s.
b) To find the flow rate in the smaller tube, we can use the equation:
Flow Rate = A * v
Where A is the cross-sectional area and v is the velocity.
Flow Rate = A2 * v2
Substituting the values:
Flow Rate = π * (0.025 m)^2 * 2.56 m/s
Simplifying:
Flow Rate = 0.001963495 m^2 * 2.56 m/s
Flow Rate = 0.005034 m^3/s
Therefore, the flow rate in the smaller tube is 0.005034 m^3/s.