An incompressible Newtonian fluid is brought to flow in a pipe by pulling a wire in the center at a velocity of 18 cm/s. The wire has a diameter of 5 cm, while the stationary pipe has a diameter of 10 cm. Calculate the shear stress(tre) in the fluid using Newtonian's law of viscosity. The flow is isothermal at °25 and the fluid has a viscosity of 3.0 cp. Provide the sketch of the velocity profile.

The shear stress is 2.16 Pa

Explanation:

Newton's law of viscosity can be expressed as follows;

Where:

τ = Shear stress in the fluid

Given that the diameter of the wire = 5 cm

The velocity of the wire = 18 cm/s

The diameter of the pipe = 10 cm

The fluid viscosity, μ = 3.0 cp = 1×10⁻³· Pa·s

The change in velocity from the surface of the wire to the internal surface of the pipe = dv = 18 cm/s

The change in the y (perpendicular) direction of motion of the fluid from the surface of the wire to the interior surface of the tube = dy = 10/2 - 5/2 = 2.5 cm

By putting in the values, we have;

The shear stress = 2.16 Pa

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.