The drag, D, on a sphere located in a pipe flow through which a fluid is flowing to be determined experimentally. Assume that drag is a function of the sphere diameter, d, the pipe dianeter, H, the fluid velocity, V, and the fluid density, p.

a) What dimensionless parameters would you use for this problem?
b)Same experiments using water indicate that for d=0.5cm H=1.27cm V=0.6m/s the drag is 7x10^-3 N. If possible estimate the drag on a sphere located in a 60cm diameter pipe through which water is flowing at V=1.8 m/s. Geometrical similarity is mainteined.

By the way if gravity had been important (waves), we would have also found the Froude number v/sqrt(gL)

d/H had better be the same for model and experiment

If viscosity were important then rho vL/mu, the dimensionless Reynolds number should be the same, but the problem statement says mu is not important
so
we are left with d/H and the ratio of force to other parameters which we can derive as follows

force = f(d, p, v, h, )

d and H are not separate dimensions, ratio the same for similar geometries

= k * p^q v^r d^w

m l/s^2 = (m/l^3)^q * (l/s)^r * l^w

from mass, q = 1
ml/s^2 = m/l^3 *(1/s)^r *l^w

from time s
-2 = -r so r = 2

from length l
1 = -3+r+w
so w = 2

so in the end
Drag = k ( rho^1 v^2 length^2)

so I claim that if d/H is the same then
Drag/( rho v^2 H^2) will be the same for model and sphere
or
7*10^-3/[(10)^3 (.6)^2(1.27)^2]
= D/[ 10^3 (1.8)^2 (60)^2 ]

7*10^-3 = D (.36)(1.61)/[3.24*3600]

D = 140869 *10^-3
= 140.9 Newtons

Thanks alot Damon :)

You are welcome :)

a) In order to determine the dimensionless parameters for this problem, we need to consider the variables involved and their corresponding units.

The variables in this problem are:
1. Sphere diameter, d (in meters)
2. Pipe diameter, H (in meters)
3. Fluid velocity, V (in meters per second)
4. Fluid density, p (in kilograms per cubic meter)

To construct dimensionless parameters, we can use the Buckingham Pi theorem, which states that the number of dimensionless parameters in a problem is equal to the number of variables minus the number of fundamental dimensions. In this case, the fundamental dimensions are length (L), time (T), and mass (M).

Since there are 4 variables and 3 fundamental dimensions, we have 4 - 3 = 1 dimensionless parameter for this problem.

One possible dimensionless parameter for this problem is the Reynolds number, Re, which is defined as the ratio of inertial forces to viscous forces in a flowing fluid. The Reynolds number can be calculated using the formula:

Re = (p * V * d) / μ

Where μ is the dynamic viscosity of the fluid (in kilograms per meter-second).

b) To estimate the drag on a sphere located in a 60 cm diameter pipe through which water is flowing at a velocity of V = 1.8 m/s, we can assume that geometric similarity is maintained. This means that the ratios of corresponding linear dimensions (such as diameters) are equal.

Given the experimental data for d = 0.5 cm, H = 1.27 cm, V = 0.6 m/s, and D = 7x10^-3 N, we can use this information to estimate the drag for the new situation.

First, we need to calculate the Reynolds number for the experimental data using the given formula and the value of μ for water at the given temperature. Once we have the Reynolds number, we can scale it up to the new situation by maintaining the geometric similarity between the sphere and the pipe.

Finally, using the scaled Reynolds number, we can calculate the drag on the sphere in the 60 cm diameter pipe using the formula for drag.