Question

Water ideally flows through a pipe of radius 6 centimeters at a rate of 5 meters per second. The pipe then narrows to a radius of 2 centimeters. What is the new velocity of the water?

(a) 50 m/s

(b) 55 m/s

(c) 60 m/s

(d) 65 m/s

I have been asked this question in semester exam.

Question is taken from Fluids in Motion topic in section Fluid Flow and Mixing of Bioprocess Engineering

Answer 1

To solve this, use the principle of continuity, which states that the flow rate remains constant in an incompressible fluid:

A1⋅v1=A2⋅v2A_1 \cdot v_1 = A_2 \cdot v_2A1​⋅v1​=A2​⋅v2​

where:

• A1A_1A1​ and A2A_2A2​ are the cross-sectional areas of the pipe at the two points,
• v1v_1v1​ and v2v_2v2​ are the velocities of the fluid at these points.

Step 1: Find A1A_1A1​ and A2A_2A2​

1. Initial radius r1=6 cmr_1 = 6 \, \text{cm}r1​=6cm (or 0.06 m)

   A1=π(0.06)2=0.0036π m2A_1 = \pi (0.06)^2 = 0.0036\pi \, \text{m}^2A1​=π(0.06)2=0.0036πm2

2. New radius r2=2 cmr_2 = 2 \, \text{cm}r2​=2cm (or 0.02 m)

   A2=π(0.02)2=0.0004π m2A_2 = \pi (0.02)^2 = 0.0004\pi \, \text{m}^2A2​=π(0.02)2=0.0004πm2

Step 2: Apply the Continuity Equation

Given:

• Initial velocity v1=5 m/sv_1 = 5 \, \text{m/s}v1​=5m/s.

0.0036π⋅5=0.0004π⋅v20.0036\pi \cdot 5 = 0.0004\pi \cdot v_20.0036π⋅5=0.0004π⋅v2​

Solving for v2v_2v2​:

v2=0.0036⋅50.0004=0.0180.0004=45 m/sv_2 = \frac{0.0036 \cdot 5}{0.0004} = \frac{0.018}{0.0004} = 45 \, \text{m/s}v2​=0.00040.0036⋅5​=0.00040.018​=45m/s

So, the correct answer is:

None of the options provided (the correct answer should be 45 m/s).