the horizontal pipe shown in fig. has across-sectional area of 40cm^2 at the wider portions and 10cm^2 at the constriction. water is flowing in the pipe, and the discharge from the pipe is 6*10^-3m^3\s
To determine the velocity of water flowing through the pipe, you can use the equation of continuity, which states that the product of cross-sectional area and velocity remains constant at any point in a non-compressible fluid.
The equation of continuity is given by:
A₁v₁ = A₂v₂
Where:
A₁ and A₂ are the cross-sectional areas at points 1 and 2, respectively.
v₁ and v₂ are the velocities at points 1 and 2, respectively.
In this case, we can set point 1 as the wider portion of the pipe (40 cm²) and point 2 as the constriction (10 cm²). We can assume that the discharge is the same as the velocity at point 2 (v₂).
Let's substitute the given values into the equation of continuity:
(40 cm²) v₁ = (10 cm²) (6 x 10⁻³ m³/s)
To convert the cross-sectional areas from cm² to m²:
40 cm² = 40 x 10⁻⁴ m²
10 cm² = 10 x 10⁻⁴ m²
After substitution:
(40 x 10⁻⁴ m²) v₁ = (10 x 10⁻⁴ m²) (6 x 10⁻³ m³/s)
Now you can solve for v₁:
v₁ = (10 x 10⁻⁴ m²) (6 x 10⁻³ m³/s) / (40 x 10⁻⁴ m²)
Simplifying the expression:
v₁ = 0.6 m/s
Therefore, the velocity of water flowing through the wider portion of the pipe is 0.6 m/s.