A horizontal pipe narrows from radius of 0.250 m to 0.100m.If the speed of the water in the pipe is 1.00m/s in the larger radius pipe,what is the speed in the smaller pipe
the speed is inversely proportional to the cross sectional area
To find the speed of the water in the smaller pipe, we can use the principle of continuity, which states that the mass flow rate of fluid is constant in a horizontal pipe.
The mass flow rate is given by the equation:
m1 * v1 = m2 * v2
where m1 and v1 are the mass flow rate and velocity of the water in the larger pipe (with radius r1 = 0.250 m), and m2 and v2 are the mass flow rate and velocity of the water in the smaller pipe (with radius r2 = 0.100 m).
The mass flow rate is given by:
m = ρ * A * v
where ρ is the density of water (approximately 1000 kg/m³), A is the cross-sectional area of the pipe, and v is the velocity of the water.
Since we're comparing the two pipes, we can set up the equation:
m1 * v1 = m2 * v2
ρ * A1 * v1 = ρ * A2 * v2
Since the density and velocity of water are the same in both pipes, we can simplify the equation to:
A1 * v1 = A2 * v2
Now, let's calculate the areas of the two pipes:
A1 = π * r1² = π * (0.250 m)²
A2 = π * r2² = π * (0.100 m)²
Plugging in the values:
π * (0.250 m)² * 1.00 m/s = π * (0.100 m)² * v2
Simplifying:
(0.250 m)² * 1.00 m/s = (0.100 m)² * v2
0.0625 m² * 1.00 m/s = 0.01 m² * v2
0.0625 m²/s = 0.01 m²/s * v2
Dividing both sides by 0.01 m²/s:
v2 = 0.0625 m²/s / 0.01 m²/s
v2 = 6.25 m/s
Therefore, the speed of the water in the smaller pipe is 6.25 m/s.