Hydraulics and Fluid Mechanics in ME If the depth of water in an open channel is less than the critical depth, the flow is called Tranquil flow Turbulent flow Critical flow Torrential flow Tranquil flow Turbulent flow Critical flow Torrential flow ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME If the velocity is zero over half of the cross-sectional area and is uniform over the remaining half, then the momentum correction factor is 2 1 4 43924 2 1 4 43924 ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME According to Chezy's formula, the discharge through an open channel is (where A = Area of flow, C = Chezy's constant, m = Hydraulic mean depth, and i = Uniform slope in bed) C × √(m × i) AC × √(m × i) mi × √(A × C) A × √(m × i) C × √(m × i) AC × √(m × i) mi × √(A × C) A × √(m × i) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A glass tube of small diameter (d) is dipped in fluid. The height of rise or fall in the tube given by (where w = Specific weight of liquid, α = Angle of contact of the liquid surface, and σ = Surface tension) 4wd/σ cosα σ cosα/4wd wd/4σ cosα 4σ cosα/wd 4wd/σ cosα σ cosα/4wd wd/4σ cosα 4σ cosα/wd ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A tank of uniform cross-sectional area (A) containing liquid upto height (H1) has an orifice of cross-sectional area (a) at its bottom. The time required to bring the liquid level from H1 to H2 will be 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The discharge through a large rectangular orifice is given by (where H1 = Height of the liquid above the top of the orifice, H2 = Height of the liquid above the bottom of the orifice, b = Breadth of the orifice, and Cd = Coefficient of discharge) Q = (2/3) Cd × b × √(2g) × (H22 - H12) Q = (2/3) Cd × b × √(2g) × (H2 - H1) Q = (2/3) Cd × b × √(2g) × (H21/2 - H11/2) Q = (2/3) Cd × b × √(2g) × (H23/2 - H13/2) Q = (2/3) Cd × b × √(2g) × (H22 - H12) Q = (2/3) Cd × b × √(2g) × (H2 - H1) Q = (2/3) Cd × b × √(2g) × (H21/2 - H11/2) Q = (2/3) Cd × b × √(2g) × (H23/2 - H13/2) ANSWER DOWNLOAD EXAMIANS APP
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