Question 6-M05A A sphere has a diameter da density ?, and constant, uniform surf

ID: 702437 • Letter: Q

Question

Question 6-M05A A sphere has a diameter da density ?, and constant, uniform surface temperature T. A stationary fluid has density p, viscosity H, surface tension y, thermal conductivity k, and temperature T2. The sphere is moving through the fluid at velocity v over a distance L The variable controlling the motion of a sphere through a fluid are the drag force F, the velocity of the sphere relative to the fluid, the diameter of the sphere, the density and dynamic viscosity of the fluid, and gravitational acceleration g. Show the drag force is dependent on the particle's Reynolds number Rep by dimensional analysis using the indicial method Rep pvda

Explanation / Answer

Let's define quantities, symbol and dimensions:

Quantity

Symbol

Dimensions

Drag Force

MLT–2

Sphere velocity

LT–1

Sphere diameter

Dynamic viscosity

ML–1T–1

Gravitational acceleration

LT–2

Density

ML–3

As mentioned in the problem the drag froce is dependent on the sphere velocity, sphere diameter, the density and dynamic visocity of fluid anf gravitational acceleration, so the can be written as:

FD = Constant va dpb ?c gd ?e ... (A) [* Here constant is dimensionless]

Now inserting the dimensions of each symbol in above equation (A):

MLT–2 = Constant (LT–1)a (L)b (ML–1T–1)c (LT–2)d (ML–3)e ... (B)

Equating the indices of M, L, and T on both sides, resultsin following equations:

From M : 1 = c + e ... (1)

From L: 1 = a + b – c + d – 3e ... (2)

From T: – 2 = – a – c – 2d ... (3)

There are three equations but five unknowns. Three of the unknowns may be found in terms of the remaining two. The two letters to be retained can be found arbitrarily. Solving these equations in terms of the unknown d and e, we have:

c = 1 – e ... (4)

a = 1 + e – 2d ...(5)

b = 1 + e + d ...(6)

By substitutine values from (4), (50 and (6) for letters a, b and c, equation (A) becomes:

FD = Constant (v) 1 + e – 2d (dp) 1 + e + d (?) 1 – e (g)d (?)e .. (C)

Collecting all the terms having integral exponent in one group, all terms having exponent e in other group and those having exponent d in third group gives:

FD = constant [v dp ?] [g dp /v2]d [? v dp / ?]e .. (D)

The dimensions of the three brackets groups in eq (D) are zero, and all are dimensionless.

The third group with exponent 'e', is Reynolds number, hence drag force depends upon it and it is proven by above indicial method.

Note: The dimensional analysis determine only the relevant independent dimensionless parameters of a problem, but not the exact relationship between them.

Hope this will help.