Question

A water pipe narrows from a radius of \(r_{1}=5.00 \mathrm{~cm}\) to a radius of \(r_{2}=2.00 \mathrm{~cm} .\) If the speed of the water in the wider part of the pipe is \(2.00 \mathrm{~m} / \mathrm{s},\) what is the speed of the water in the narrower part?

Short answer

Answer: The water flow velocity in the narrower section of the pipe is 25.00 m/s.

Step-by-step solution

Identify the known information

We are given: - Radius of the wider pipe section, \(r_1=5.00\) cm - Radius of the narrower pipe section, \(r_2=2.00\) cm - Water flow velocity in the wider pipe section, \(v_1=2.00\) m/s

Understand the principle of continuity

Since the water is incompressible, the principle of continuity states that the mass flow rate is constant at any cross-sectional area of the pipe. The mass flow rate is defined as the product of the cross-sectional area, the velocity, and the density (\(\rho\)) of the fluid: \(m_1 = m_2\) (mass flow rate is conserved) Hence, \(A_1v_1 \rho = A_2v_2 \rho\) As the water's density is constant, the expression reduces to: \(A_1v_1 = A_2v_2\)

Calculate the cross-sectional areas of the sections

The cross-sectional area of a pipe is determined by finding the area of the circular section. For a pipe with radius r, this is given by the formula: \(A = \pi r^2\) So, for both sections of the pipe, we find the areas: \(A_1= \pi r_1^2\) and \(A_2= \pi r_2^2\)

Apply the principle of continuity to find the water velocity in the narrow section

Now, we need to find the water flow velocity in the narrower section of the pipe (\(v_2\)). We can do this by applying the principle of continuity. Using the expression \(A_1v_1 = A_2v_2\) and knowing the areas computed in step 3, we get: \(A_1v_1 = A_2v_2\) Substitute the known values into the equation: \((\pi (5.00\mathrm{~cm})^2)(2.00 \mathrm{~m/s}) = (\pi (2.00\mathrm{~cm})^2)(v_2)\) Divide both sides of the equation by \(\pi (2.00\mathrm{~cm})^2\) to solve for \(v_2\) and remember to convert cm to m (1 cm = 0.01 m): \(v_2 = \frac{(\pi (0.050\mathrm{~m})^2)(2.00 \mathrm{~m/s})}{(\pi (0.020\mathrm{~m})^2)}\)

Calculate the water flow velocity in the narrow section

Perform the calculations: \(v_2 = \frac{(\pi (0.050\mathrm{~m})^2)(2.00 \mathrm{~m/s})}{(\pi (0.020\mathrm{~m})^2)} = 25.00 \mathrm{~m/s}\) The water flow velocity in the narrower part of the pipe is 25.00 m/s.