A sewer pipe contains 1 mm sand particles of specific gravity 2.65 and 5 mm organic particles of specific gravity 1.2, the minimum velocity required for removing the sewerage, is

ANS:D - 0.45 m/sec

To determine the minimum velocity required for removing the sewerage (i.e., preventing sedimentation and buildup) in the sewer pipe, we need to calculate the critical velocity for both types of particles separately and then take the maximum of the two critical velocities. The critical velocity for particles is given by the formula: Vc​= Gravity×(Specific Gravityparticle​−Specific Gravityfluid​)×Particle Size​​ / 18×Viscosity  Given:

• Gravity (g) ≈ 9.81 m/s²
• Specific gravity of sand particles = 2.65
• Specific gravity of organic particles = 1.2
• Diameter of sand particles = 0.001 m (1 mm)
• Diameter of organic particles = 0.005 m (5 mm)
• Viscosity of water (approximately) = 10−310−3 Pa·s

Let's calculate the critical velocities for both types of particles: For sand particles: Vcsand​​=9.81×(2.65−1)×0.001​​  / 18×10^−3 Vcsand​​=9.81×1.65×0.001​​ / 18×10^−3 Vcsand​​=0.016215/0.018 ​​ Vcsand​​≈0.901​  Vcsand​​≈0.949 m/s For organic particles: Vcorganic​​=9.81×(1.2−1)×0.005​​/18×10^−3 Vcorganic​​=9.81×0.2×0.005​​/18×10−3 Vcorganic​​=0.0981​​/0.018 Vcorganic​​≈5.45 Vcorganic​​≈2.33 m/s The critical velocity for transporting the sand particles is approximately 0.949 m/s, and for organic particles, it is approximately 2.33 m/s. The minimum velocity required to remove the sewerage will be the maximum of the critical velocities of the two types of particles. Therefore, the minimum velocity required is approximately 2.33 m/s2.33m/s. The option provided, 0.45 m/sec, does not match the calculation. It's possible there's a typographical error in the given answer, or there might be a misunderstanding in the problem statement.