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 empty the tank completely will be (2AH₁3/2)/(Cd × a√2g) (2AH₁²)/(Cd × a√2g) (2A√H₁)/(Cd × a√2g) (2AH₁)/(Cd × a√2g) (2AH₁3/2)/(Cd × a√2g) (2AH₁²)/(Cd × a√2g) (2A√H₁)/(Cd × a√2g) (2AH₁)/(Cd × a√2g) ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The flow in which the particles of a fluid attain such velocities that vary from point to point in magnitude and direction as well as from instant to instant, is known as steady flow turbulent flow one dimensional flow uniform flow steady flow turbulent flow one dimensional flow uniform flow ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME When the Mach number is between ___ the flow is called super-sonic flow. 2.5 and 4 4 and 6 1 and 6 1 and 2.5 2.5 and 4 4 and 6 1 and 6 1 and 2.5 ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The value of mass density in kg-sec-V-m⁴ for water at 0°C is 1 1000 101.9 100.9 1 1000 101.9 100.9 ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME If Hg is the gross or total head and hf is the head lost due to friction, then net or effective head (H) is given by H = Hg + hf H = Hg - hf H = Hg/hf H = Hg × hf H = Hg + hf H = Hg - hf H = Hg/hf H = Hg × hf ANSWER DOWNLOAD EXAMIANS APP
Hydraulics and Fluid Mechanics in ME The total head of a liquid particle in motion is equal to Pressure head - (kinetic head + potential head) Kinetic head - (pressure head + potential head) Potential head - (pressure head + kinetic head) Pressure head + kinetic head + potential head Pressure head - (kinetic head + potential head) Kinetic head - (pressure head + potential head) Potential head - (pressure head + kinetic head) Pressure head + kinetic head + potential head ANSWER DOWNLOAD EXAMIANS APP
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