Question 13.2: Determine all of the dimensionless groups possible to descri...
Determine all of the dimensionless groups possible to describe the drag force on a white blood cell, using Equation 13.23.
h(F, υ, μ, ρ, d) = 0 (13.23)
List the dimensional parameters:
F υ μ ρ D
Select MLT as the fundamental dimension.
List the fundamental dimensions of each parameter of interest:
F υ μ ρ D
\frac{ML}{t^{2}} \frac{L}{t} \frac{M}{Lt} \frac{M}{L3} L
The minimum number of independent parameters is 3 (M, L, and T).
The number of Pi groups is 2 (5 – 3 = 2).
Pick υ, d, and ρ as the sub-set of dimensional parameters and relate this to F (they represent M, L, and T). At this point, any grouping of three parameters that have M, L, and T represented would suffice.
The first Π group will be set up as
\Pi _{1}= F \upsilon^{a} d^{b} \rho^{c}
\left(\frac{ML}{t^{2}}\right) \left(\frac{L}{t}\right)^{a} \left(L\right)^{b} \left(\frac{M}{L^{3}}\right)^{c} = M^{0}L^{0}t^{0}
\left(\frac{ML}{t^{2}}\right) \left(\frac{L^{a}}{t^{a}}\right) \left(L^{b}\right) \left(\frac{M^{c}}{L^{3c}}\right) = M^{0}L^{0}t^{0}
ML(t^{-2})(L^{a})(t^{-a})(L^{b})(M^{c})(L^{-3c}) = M^{0}L^{0}t^{0}
Separate each dimension and relate the exponents:
M : 1 + c = 0
L : 1 + a + b – 3c = 0
t : – 2 – a = 0
From the mass equation,
c = – 1
From the time equation,
a = – 2
From the length equation,
b = 3( – 1) – 1 + 2 = – 2
Therefore, the first Π group is
\Pi _{1}= \frac{F}{\upsilon^{2} d^{2} \rho}
Pick υ, d, and ρ as the sub-set of dimensional parameters and relate this to μ (they represent M, L, and T).
The second Π group will be set up as
\Pi _{2}= \mu \upsilon^{a} d^{b} \rho^{c}
\left(\frac{M}{Lt}\right) \left(\frac{L}{t}\right)^{a} \left(L\right)^{b} \left(\frac{M}{L^{3}}\right)^{c} = M^{0}L^{0}t^{0}
\left(\frac{M}{Lt}\right) \left(\frac{L^{a}}{t^{a}}\right) \left(L^{b}\right) \left(\frac{M^{c}}{L^{3c}}\right) = M^{0}L^{0}t^{0}
M(L^{-1})(t^{-1})(L^{a})(t^{-a})(L^{b})(M^{c})(L^{-3c}) = M^{0}L^{0}t^{0}
Separate each variable and relate the exponents:
M : 1 + c = 0
L : – 1 + a + b – 3c = 0
t : – 1 – a = 0
From the mass equation,
c = – 1
From the time equation,
a = – 1
From the length equation,
b = 3( – 1) + 1 + 1 = – 1
Therefore, the second Π group is
\Pi _{2}= \frac{\mu }{\rho \upsilon d}
From these two Π groups, we can state that
\Pi _{1}= f(\Pi _{2})
\frac{F}{\upsilon^{2} d^{2} \rho} = f\left(\frac{\mu }{\rho \upsilon d}\right)
but the actual relationship between the Π groups would need to be determined experimentally (e.g., are there constants of proportionality?). However, the usefulness of this analysis technique is that we can obtain dimensionless parameters that can relate fluid properties. In our example, drag force was related to the fluid’s velocity, density, and a characteristic length, and the viscosity was related to the same three parameters.