Question

A 10 m long pipe with inlet and outlet diameters of 40 cm and 20 cm respectively, is carrying an incompressible fluid with a flow rate of \( 0.04 \, \text{m}^3/\text{s} \). The ratio of the velocity at the outlet to that at the inlet is ………….. (rounded off to one decimal place).

Hint:

1. For incompressible fluids, the flow rate \( Q \) remains constant throughout the pipe. 2. The velocity ratio can be directly obtained using the area ratio, \( \frac{v_{\text{outlet}}}{v_{\text{inlet}}} = \frac{A_{\text{inlet}}}{A_{\text{outlet}}} \). 3. Always check the units for consistency when solving fluid mechanics problems.

Answer 1

Step 1: Relationship between velocity, flow rate, and area. For incompressible fluids, the flow rate \( Q \) is related to the velocity \( v \) and the cross-sectional area \( A \) as: \[ Q = v \cdot A \] This applies at both the inlet and the outlet: \[ Q = v_{\text{inlet}} \cdot A_{\text{inlet}} = v_{\text{outlet}} \cdot A_{\text{outlet}} \] Step 2: Calculate the cross-sectional areas. The cross-sectional area of a circular pipe is given by: \[ A = \pi \left(\frac{D}{2}\right)^2 \] For the inlet diameter \( D_{\text{inlet}} = 40 \, \text{cm} = 0.4 \, \text{m} \): \[ A_{\text{inlet}} = \pi \left(\frac{0.4}{2}\right)^2 = \pi (0.2)^2 = 0.04\pi \, \text{m}^2 \] For the outlet diameter \( D_{\text{outlet}} = 20 \, \text{cm} = 0.2 \, \text{m} \): \[ A_{\text{outlet}} = \pi \left(\frac{0.2}{2}\right)^2 = \pi (0.1)^2 = 0.01\pi \, \text{m}^2 \] Step 3: Ratio of outlet to inlet velocities. From continuity: \[ v_{\text{inlet}} \cdot A_{\text{inlet}} = v_{\text{outlet}} \cdot A_{\text{outlet}} \] Rearranging: \[ \frac{v_{\text{outlet}}}{v_{\text{inlet}}} = \frac{A_{\text{inlet}}}{A_{\text{outlet}}} \] Substitute the areas: \[ \frac{v_{\text{outlet}}}{v_{\text{inlet}}} = \frac{0.04\pi}{0.01\pi} = \frac{0.04}{0.01} = 4.0 \] Conclusion: The ratio of the velocity at the outlet to that at the inlet is \( \mathbf{4.0} \).