Question

A non-viscous liquid flows through a horizontal pipe of varying cross-section at the rate of 22 m^3/s. Calculate the velocity at the cross-section where the radius is 5 cm.

a) 4 m/s
b) 8 m/s
c) 12 m/s
d) 16 m/s

Answer 1

The velocity at the cross-section where the radius is 5 cm is approximately 2796.8 m/s.

Step-by-step explanation:

To calculate the velocity at the cross-section where the radius is 5 cm, we can use the principle of conservation of mass. According to this principle, the product of the cross-sectional area of the pipe and the velocity of the liquid remains constant.

Let's denote the initial cross-sectional area and velocity as A1 and v1, and the cross-sectional area and velocity at the 5 cm radius section as A2 and v2. We can set up the equation: A1*v1 = A2*v2.

Since the liquid is non-viscous, there is no frictional loss, and the flow rate remains constant. The flow rate is given as 22 m^3/s, which is equal to A1*v1. Plugging in the values, we get: 22 = A2*v2.

Now, let's calculate the cross-sectional area and velocity at the 5 cm radius section. The cross-sectional area can be calculated using the formula: A = π*r^2, where r is the radius. Plugging in the value of r as 0.05 m, we get A2 = π*(0.05)^2. Solving for A2, we get A2 = 0.00785 m^2.

Finally, we can calculate the velocity at the 5 cm radius section by rearranging the equation: v2 = 22/A2. Substituting the values, we get v2 = 22/0.00785.

After calculating, we find that the velocity at the cross-section where the radius is 5 cm is approximately 2796.8 m/s.