Hydraulics and Fluid Mechanics in ME The coefficient of discharge for an external mouthpiece depends upon Velocity of liquid Length of mouthpiece Pressure of liquid Area of mouthpiece Velocity of liquid Length of mouthpiece Pressure of liquid Area of mouthpiece ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME In a centrifugal pump casing, the flow of water leaving the impeller, is Rectilinear flow Radial flow Free vortex motion Forced vortex Rectilinear flow Radial flow Free vortex motion Forced vortex ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The discharge through a wholly drowned orifice is given by (where H1 = Height of water (on the upstream side) above the top of the orifice, H2 = Height of water (on the downstream side) above the bottom of the orifice, and H = Difference between two water levels on either side of the orifice) Q = Cd × bH2 × √(2gh) Q = Cd × b (H2 - H1) × √(2gh) Q = Cd × bH × √(2gh) Q = Cd × bH₁ × √(2gh) Q = Cd × bH2 × √(2gh) Q = Cd × b (H2 - H1) × √(2gh) Q = Cd × bH × √(2gh) Q = Cd × bH₁ × √(2gh) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME A right circular cylinder open at the top is filled with liquid and rotated about its vertical axis at such a speed that half the liquid spills out, then the pressure intensity at the center of bottom is one-fourth its value when cylinder was full one-half its value when cylinder was full cannot be predicted from the given data zero one-fourth its value when cylinder was full one-half its value when cylinder was full cannot be predicted from the given data zero ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The upper surface of a weir over which water flows is known is weir top crest nappe sill weir top crest nappe sill 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 × √H₂/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) 2A × √H₂/Cd × a × √(2g) 2A × (√H₁ - √H₂)/Cd × a × √(2g) 2A × (√H3/2 - √H3/2)/Cd × a × √(2g) 2A × √H₁/Cd × a × √(2g) ANSWER DOWNLOAD EXAMIANS APP