Chapt 9:Dimensionless Numbers and Dimensional Analysis

Dimensionless Numbers

Dimensions and Units

  • Dimensions are represented by: L (Length), M (Mass), T (Time), Θ (Temperature), N (Amount of substance), J (Luminous intensity), I (Electric current).

  • Units are represented by: m (meter), kg (kilogram), s (second), K (Kelvin), mol (mole), cd (candela), A (ampere).

  • A dimensionless number cannot be described by any dimension; it is defined by the lack of dimensions.

Force, Energy, Power, Pressure, Voltage

  • Force: [=] newton [N] and \1 N ≡ (1 kg) (1 m) / s^2

  • Energy: [=] joule [J] and \1 J ≡ (1 N) (1 m)

  • Power: [=] watt [W] and \1 W ≡ 1 J/s

  • Pressure: = pascal [Pa] and \1 Pa ≡ 1 N / m^2

  • Voltage: [=] volt [V] and \1 V ≡ 1 W / 1 A

Examples of Dimensionless Numbers

  • pi: \π = \frac{circle \ circumference}{circle \ diameter} = C/d

  • Specific gravity: SG = \frac{object \ density}{density \ of \ water} = ρ/ρ_{H2O}

  • Mach number: Ma = \frac{object \ speed}{speed \ of \ sound} = v/v_{sound}

  • Coefficient of friction: μk = \frac{sideways \ force}{weight \ of \ object} = F/w

Heat Loss Example

The heat loss from the surface of a hot liquid is given by:
\frac{Q}{t} = h \cdot A \cdot (T - T_0)
where:

  • \frac{Q}{t} is in J/s

  • A is the area in square meters.

  • T is the temperature of the liquid in degrees Celsius.

  • T_0 is the ambient temperature in degrees Celsius.

To express the parameter h in fundamental dimensions:
h = \frac{Q}{t \cdot A \cdot (T - T_0)}
h = \frac{M \cdot L^2 / T^2}{T \cdot L^2 \cdot \Theta} = \frac{M}{T^3 \cdot \Theta}

Molecular Velocity Example

A simple expression for the velocity of molecules in a gas is given by:
v = K \sqrt{\frac{P}{ρ}}
where:

  • v is velocity in m/s,

  • P is pressure in pascals,

  • ρ is density in g/cm3.

To find the dimensions of the constant K:
K = \frac{v}{\sqrt{\frac{P}{ρ}}} = v \sqrt{\frac{ρ}{P}}
K = \sqrt{\frac{v^2 \cdot ρ}{P}} = \sqrt{\frac{\frac{L^2}{T^2} \cdot \frac{M}{L^3}}{\frac{M}{L \cdot T^2}}} = \sqrt{\frac{L^2 \cdot M}{T^2 \cdot L^3} \cdot \frac{L \cdot T^2}{M}} = \sqrt{\frac{L^3MT^2}{L^3MT^2}} = Dimensionless

Dimensions of a Constant in an Equation

Given the equation:
T = 102 \cdot e^{-6P}
where:

  • T is temperature in Kelvin [K],

  • P is algae population in [g/mL of lake water].

Exponents must be dimensionless, so the quantity (-6P) must have no dimensions. Solving for -6:
-6P = M^0 \cdot L^0
-6 \cdot \frac{M}{L^3} = M^0 \cdot L^0
-6 = \frac{M^0 \cdot L^0}{\frac{M}{L^3}} = M^{-1} \cdot L^3
Therefore, the dimensions of “6” are M^{-1} \cdot L^3

Dimensional Analysis

  • Dimensionless quantities are generated through dimensional analysis.

  • Dimensional analysis is a mathematical analysis that focuses on the dimensions of length, mass, time, etc.

  • Dimensional analysis simplifies a problem by dealing only with essential parameters.

Utility of Dimensional Analysis

  • Many phenomena are more complicated, and it is not obvious which parameters should be measured to characterize the behavior of the system.

  • Dimensional analysis can be a powerful tool to help understand which parameters affect the behavior of the system and how they affect it.

Steps for Dimensional Analysis

  1. Raise variables to exponents.

  2. Substitute dimensions.

  3. Group by dimension.

  4. Set exponent equations equal to zero.

  5. Solve the equations.

  6. Substitute results of step 5 into step 1.

  7. Group by exponent.

  8. Check the results.

Galileo's Experiment Example

Assume we want to study an object falling from a tall tower, recreating Galileo's famous experiment at the tower of Pisa.

Measurements of objects falling can be taken by varying:

  • Height (H).

  • Initial velocity (v).

  • Time (t) to hit the ground.

  • Mass (m).

  • Include gravity (g) as it causes the ball to fall in the first place.

Representing the variables with exponents:

t^a \cdot H^b \cdot v^c \cdot g^d \cdot m^e

Substituting dimensions:

t^a = T^a
H^b = L^b
v^c = (L/T)^c = L^c \cdot T^{-c}
g^d = (L/T^2)^d = L^d \cdot T^{-2d}
m^e = M^e

Grouping by dimension:

M^e \cdot L^{b+c+d} \cdot T^{a-c-2d}

Exponent equations:

  • M: e = 0

  • L: b + c + d = 0

  • T: a - c - 2d = 0

Solving for variables:

  • e = 0

  • b = -c - d

  • a = c + 2d

Substituting the results:

t^{c+2d} \cdot H^{-c-d} \cdot v^c \cdot g^d

Grouping by exponent:

(\frac{v \cdot t}{H})^c \cdot (\frac{g \cdot t^2}{H})^d

Checking the dimensions:

  • \frac{H}{g \cdot t^2} = \frac{L}{\frac{L}{T^2} \cdot T^2} = 1 \quad YES

  • \frac{H}{v \cdot t} = \frac{L}{\frac{L}{T} \cdot T} = 1 \quad YES

Rayleigh's Method - Immersed Sphere Example

When a fluid is flowing past an immersed sphere:

  • Total drag force (F) [=] lbf.

  • Projected area of the sphere (A) [=] ft2.

  • Liquid viscosity (µ) [=] lbm/(ft s).

  • Diameter of the sphere (D) [=] ft.

  • Liquid density (ρ) [=] lbm/ft3.

  • Flowrate (v) [=] ft/s.

Determine the set of three dimensionless groups representing the relationship of the drag coefficient on the immersed sphere using Rayleigh's Method.

Using Rayleigh's Method:

F^a \cdot A^b \cdot μ^c \cdot D^d \cdot ρ^e \cdot v^f

Substitute dimensions:

F^a = (M \cdot L \cdot T^{-2})^a = M^a \cdot L^a \cdot T^{-2a}
A^b = (L^2)^b = L^{2b}
μ^c = (M \cdot L^{-1} \cdot T^{-1})^c = M^c \cdot L^{-c} \cdot T^{-c}
D^d = L^d
ρ^e = (M \cdot L^{-3})^e = M^e \cdot L^{-3e}
v^f = (L \cdot T^{-1})^f = L^f \cdot T^{-f}

Group by dimension:

M^{a+c+e} \cdot L^{a+2b-c+d-3e+f} \cdot T^{-2a-c-f}

Exponent equations:

  • M: a + c + e = 0

  • L: a + 2b - c + d - 3e + f = 0

  • T: -2a - c - f = 0

Solve the equations:

  • e = -a - c

  • f = -2a - c

  • 0 = a + 2b - c + d - 3e + f

  • 0 = a + 2b - c + d + 3a + 3c - 2a - c

  • 0 = 2a + 2b + c + d

  • d = -2a - 2b - c

Substitute the results:

F^a \cdot A^b \cdot μ^c \cdot D^{-2a-2b-c} \cdot ρ^{-a-c} \cdot v^{-2a-c}

Group by exponent:

( \frac{F}{A \cdot ρ \cdot v^2} )^a \cdot ( \frac{A}{D^2} )^b \cdot ( \frac{μ}{D \cdot ρ \cdot v} )^c

Check dimensions:

  • \frac{A \cdot ρ \cdot v^2}{F} = \frac{L^2 \cdot \frac{M}{L^3} \cdot (\frac{L}{T})^2}{M \cdot L \cdot T^{-2}} = \frac{M L^2 L^2 T^{-2}}{L^3 T^2 M L T^{-2}} = 1

  • \frac{A}{D^2} = \frac{L^2}{L^2} = 1

  • \frac{μ}{D \cdot ρ \cdot v} = \frac{\frac{M}{LT}}{\frac{L \cdot M \cdot L}{L^3 T}} = 1

If A is substituted for the D^2 term, the first quantity becomes the Coefficient of Drag, CD. The final quantity is the Reynolds Number, Re.

Rayleigh's Method - Drug Delivery Modeling Example

In modeling a drug delivery method, the following variables are important:

  • Q: Volumetric flowrate of blood [=] gpm

  • µ: Dynamic viscosity [=] g/(cm.s)

  • ρ: Density [=] kg/m3

  • t: Time in bloodstream [=] h

  • D: Diameter of blood vessel [=] mm

Using Rayleigh's Method:

Q^a \cdot μ^b \cdot ρ^c \cdot t^d \cdot D^e

Substituting dimensions:

Q^a = (L^3 \cdot T^{-1})^a
μ^b = (M \cdot L^{-1} \cdot T^{-1})^b
ρ^c = (M \cdot L^{-3})^c
t^d = T^d
D^e = L^e

Grouping by dimension:

L^{3a - b - 3c + e} \cdot T^{-a - b + d} \cdot M^{b + c}

Exponent equations:

  • 3a - b - 3c + e = 0

  • -a - b + d = 0

  • b + c = 0

Solving the equations:

  • c = -b

  • d = a + b

  • e = -3a - 2b

Substitute the results:

Q^a \cdot μ^b \cdot ρ^{-b} \cdot t^{a+b} \cdot D^{-3a-2b}

Group by exponent:

( \frac{Q \cdot t}{D^3} )^a \cdot ( \frac{μ \cdot t}{ρ \cdot D^2} )^b

Ball Thrown from a Tower - Dimensionless Ratios

Suppose we conduct an experiment with a ball that we throw from the top of a tall tower of height H. We throw it directly downward with some initial velocity v, and then measure the elapsed time t until it hits the ground. We vary the initial height and the initial velocity. The variables of interest in this problem are H, v, and t. A little thought leads us to include g, since it is the force of gravity that causes the ball to fall in the first place. Using Rayleigh’s method, find a set of dimensionless ratios that can be used to correlate our data.