Solve equation from dimensional analysis: 3 eq., 6 unknowns

In summary, the task involves using dimensional analysis to solve a system of three equations with six unknowns. This process requires identifying the dimensions of each variable, establishing relationships among them, and utilizing the given equations to derive a solution. The challenge lies in the underdetermined nature of the system, necessitating additional information or assumptions to fully resolve the unknowns.
  • #1
phillip_at_work
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TL;DR Summary
Text book conducts dimensional analysis to derive an equation. Analysis involves 6 variables and 3 basic units (mass M, length L, and time T). This results in three equations and six unknowns. The final equation can be solved using three of those unknowns. However, with only three equations, I don't see how to find those three unknowns to actually get a solution for the derived equation. Text: Carlton 2007, Marine Propellers and Propulsion p89.
Carlton writes on page 89:
"The thrust of a marine propeller... may be expected to depend upon the following parameters:
(a) The diameter (D)
(b) the speed of advance (Va)
(c) The rotational speed (n)
(d) The density of the fluid (ρ)
(e) The viscosity of the fluid (μ)
(f) The static pressure of the fluid at the propeller station (p0-e)"

What follows is Carlton's derivation:

T ∝ ρaDbVacndμf(p0-e)g

And by dimensional analysis, we get:

MLT-2 = (ML-3)aLb(LT-1)c(T-1)d(ML-1T-1)f(ML-1T-2)g

which results in the following equations:

for mass M: 1 = a + f + g
for length L: 1 = -3a + b + c - f - g
for time T: -2 = -c - d - f - 2g

and hence:

a = 1 - f - g
b = 4 - c - 2f -g
d = 2 - c - f - 2g

And so that proportion can be updated to be:

T ∝ ρ (1-f-g) D (4-c-2f-g)Vacn(2-c-f-2g)μf(po-e)g

For the final equation as:

T = ρn2D4(Va/ nD)c* (μ / ρnD2)f* ( (p0-e) / pn2D2)g

I can follow this derivation without issue. What is confusing is how I solve for T. How can I know the values of `c`, `f`, and `g` as I have three equations and six unknowns? What am I missing?
 
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  • #2
I think I solved my own question. Happy for feedback.

Constant exponents are derived for ρa, nd and Db. These are ρ1, n2, and D4.

That is, a=1, d=2, and b=4. If I apply these to the system of three equations:

a = 1 = 1 - f - g
b = 4 = 4 - c - 2f -g
d = 2 = 2 - c - f - 2g

I compute,

g = f = c = 0

Which feels incorrect, but perhaps I'm over thinking. If this was correct, I'm not sure why Carlton wouldn't write this solution in the chapter, which s/he does not. Any feedback is much appreciated.
 

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