Q. The capillary rise in the glass tube is not to exceed 0.2 mm of water. Determine its minimum size, given that surface tension for water in contact with air = 0.0725 N/m. SOLUTION: Capillary rise, h = 0.2mm = 0.2 * 10³ m Surface tension σ = 0.0725N / m Let dia. of tube … Read more
Q. Determine the viscosity of a liquid having kinematic viscosity 6 stokes and specific gravity 1.9. SOLUTION: Kinematic viscosity u = 6 stokes = 6cm²/ s = 6 * 10-⁴m²/ s Sp. Gravity of liquid = Density of liquid/Density of water 1.9=Density of liquid/ 1000 Density of liquid= 1000×1.9=1900kg / (m³) . Using the relations … Read more
Q. Find the kinematic viscosity of an oil having density 981 kg / (m ^ 3) . The shear stress at a point in oil is 0.2452N / (m ^ 2) and velocity gradient at that point is 0.2 per second. SOLUTION: Mass density,rho = 981kg / (m³) Shear stress, τ = 0.2452N / (m²) … Read more
Q. Two horizontal plates are placed 1.25 cm apart, the space between them being filled with oil of viscosity 14 poises. Calculate the shear stress in oil if upper plate is moved with a velocity of 2.5 m/s. SOLUTION: Given: Distance between plates, dy 1.25 cm = 0.0125 m Viscosity μ = 14 poise = … Read more
Q. Determine the intensity of shear of an oil having viscosity = 1 poise. The oil is used for lubricating the clearance between a shaft of diameter 10 cm and its journal bearing. The clearance is 1.5 mm and the shaft rotates at 150 rpm SOLUTION: Given: μ= 1 poise- 10 Ns/m² Dia of shaft … Read more
Q. If the equation of a velocity profile over a plate is V = 2 y2/3, in which ‘v’ is the velocity in mis at a distance of ‘y’ metres above the plate, Determine the shear stress at y = 0 and y = 0.075m (or) 7.5 сm. Given: µ = 0.835 N.S/m² SOLUTION: The … Read more
Q. If liquid pressure is increased from 4000 kN/m² to (b) Ir 6000 kN/m², the volume of liquid decreased by 0.2 percent, Determine the bulk modulus of elasticity of the liquid. SOLUTION: Bulk modulus of elasticity K = – (dp)/((dv)/v) Change in pressure p = 6000 – 4000 = 2000kN / (m ^ 2) =(dv)/v … Read more
Q. The specific gravity and dynamic viscosity of a certain liquid are 2.0 and 1.3N- S/ (m ^ 2) respectively. Calculate the kinematic viscosity of the liquid. SOLUTION: Kinematic viscosity = μ/ρ Dynamic viscosity μ= 1.3 N- S / (m ^ 2) Specific gravity S = 2.0 Specific gravity= specific weight of liquid/ specific weight … Read more
Q. At a point of layer of oil, the shear stress is 0.2 N/m² and velocity gradient is 0.25 m/sec/m. Calculate the coefficient of dynamic. SOLUTION: Shear stress τ= 0.2N / (m ^ 2) Velocity gradient, (dv)/(dy) =0.25 m/sec/m From the relationship, τ=μ * dv/dy 0.2 = μ * 0.25 μ = 0.8N -sec/m^ 2 … Read more
.Q. Find the values of absolute pressure given that (a) Pressure gauge reading as 150 kPa, (b) Vacuum gauge reading as 40 hPa, when the atmospheric pressure is 101.3 kPa. Given data: Atmospheric pressure=101.3 kPa Gauge pressure = 150kPa Vacuum pressure = 40kPa SOLUTION: (a) Absolute Pressure = atmospheric pressure + gauge pressure = 101.3 … Read more