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1.4 Topics in Dimensional Analysis
This section introduces dimensionless groups, the concept of dimensional homogeneity of an equation, and a
process for carrying and canceling units in a calculation. This knowledge is useful in all aspects of engineering,
especially for finding and correcting mistakes during calculations and during derivations of equations.
All of topics in this section are part of dimensional analysis, which is the process for applying knowledge and
rules involving dimensions and units. Other aspects of dimensional analysis are presented in Chapter 8 of this
text.
Dimensionless Groups
Engineers often arrange variables so that primary dimensions cancel out. For example, consider a pipe with an
inside diameter D and length L. These variables can be grouped to form a new variable L/D, which is an
example of a dimensionless group. A dimensionless group is any arrangement of variables in which the primary
dimensions cancel. Another example of a dimensionless group is the Mach number M, which relates fluid speed
V to the speed of sound c:
Another common dimensionless group is named the Reynolds number and given the symbol Re. The Reynolds
number involves density, velocity, length, and viscosity µ:
(1.3)
The convention in this text is to use the symbol [-] to indicate that the primary dimensions of a dimensionless
group cancel out. For example,
(1.4)
Dimensional Homogeneity
When the primary dimensions on each term of an equation are the same, the equation is dimensionally
homogeneous. For example, consider the equation for vertical position s of an object moving in a gravitational
field:
In the equation, g is the gravitational constant, t is time, vo is the magnitude of the vertical component of the
initial velocity, and so is the vertical component of the initial position. This equation is dimensionally
homogeneous because the primary dimension of each term is length L. Example 1.1 shows how to find the
primary dimension for a group of variables using a step-by-step approach. Example 1.2 shows how to check an
equation for dimensional homogeneity by comparing the dimensions on each term.
Since fluid mechanics involves many differential and integral equations, it is useful to know how to find
primary dimensions on integral and derivative terms.
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To find primary dimensions on a derivative, recall from calculus that a derivative is defined as a ratio:
Thus, the primary dimensions of a derivative can be found by using a ratio:
EXAMPLE 1.1 PRIMARY DIME SIO S OF THE
REY OLDS UMBER
Show that the Reynolds number, given in Eq. (1.4), is a dimensionless group.
Problem Definition
Situation: The Reynolds number is given by Re = (ÁVL)/µ.
Find: Show that Re is a dimensionless group.
Plan
1. Identify the variables by using Table A.6.
2. Find the primary dimensions by using Table A.6.
3. Show that Re is dimensionless by canceling primary dimensions.
Solution
1. Variables
· mass density, Á
· velocity, V
· Length, L
· viscosity, µ
2. Primary dimensions
3. Cancel primary dimensions:
Since the primary dimensions cancel, the Reynolds number (ÁVL)/µ is a dimensionless group.
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EXAMPLE 1.2 DIME SIO AL HOMOGE EITY OF THE
IDEAL GAS LAW
Show that the ideal gas law is dimensionally homogeneous.
Problem Definition
Situation: The ideal gas law is given by p = ÁRT.
Find: Show that the ideal gas law is dimensionally homogeneous.
Plan
1. Find the primary dimensions of the first term.
2. Find the primary dimensions of the second term.
3. Show dimensional homogeneity by comparing the terms.
Solution
1. Primary dimensions (first term)
· From Table A.6, the primary dimensions are:
2. Primary dimensions (second term).
· From Table A.6, the primary dimensions are
· Thus
3. Conclusion: The ideal gas law is dimensionally homogeneous because the primary dimensions
of each term are the same.
The primary dimensions for a higher-order derivative can also be found by using the basic definition of the
derivative. The resulting formula for a second-order derivative is
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(1.5)
Applying Eq. 1.5 to acceleration shows that
To find primary dimensions of an integral, recall from calculus that an integral is defined as a sum:
Thus
(1.6)
For example, position is given by the integral of velocity with respect to time. Checking primary dimensions for
this integral gives
In summary, one can easily find primary dimensions on derivative and integral terms by applying fundamental
definitions from calculus. This process is illustrated by Example 1.3
EXAMPLE 1.3 PRIMARY DIME SIO S OF A
DERIVATIVE A D I TEGRAL
Find the primary dimensions of , where µ is viscosity, u is fluid velocity, and y is distance.
Repeat for where t is time, V is volume, and Á is density.
Problem Definition
Situation: A derivative and integral term are specified above.
Find: Primary dimensions on the derivative and the integral.
Plan
1. Find the primary dimensions of the first term by applying Eq. 1.5.
2. Find the primary dimensions of the second term by applying Eqs. 1.5 and 1.6.
Solution
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1.
Primary dimensions of
· From Table A.6:
· Apply Eq. 1.5:
· Combine the previous two steps:
2.
Primary dimensions of
· Find primary dimensions from Table A.6:
· Apply Eqs. (1.5) and (1.6) together:
Sometimes constants have primary dimensions. For example, the hydrostatic equation relates pressure p, density
Á, the gravitational constant g, and elevation z:
For dimensional homogeneity, the constant C needs to have the same primary dimensions as either p or Ágz.
Thus the dimensions of C are [C] = M/LT2 . Another example involves expressing fluid velocity V as a function
of distance y using two constants a and b:
For dimensional homogeneity both sides of this equation need to have primary dimensions of [L/T]. Thus,
[b] = L and [a] = L-1 T-1.
The Grid Method
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Because fluid mechanics involves complex equations and traditional units, this section presents the grid method,
which is a systematic way to carry and cancel units. For example, Fig. 1.2 shows an estimate of the power P
required to ride a bicycle at a speed of V = 20 mph. The engineer estimated that the required force to move
against wind drag is F = 4.0 lbf and applied the equation P = FV. As shown, the calculation reveals that the
power is 159 watts.
Figure 1.2 Grid method
As shown in Fig. 1.2, the grid method involves writing an equation, drawing a grid, and carrying and canceling
units. Regarding unit cancellations, the key idea is the use of unity conversion ratios, in which unity (1.0)
appears on one side of the equation. Examples of unity conversion ratios are
Figure 1.2 shows three conversion ratios. Each of these ratios is obtained by using information given in Table
F.1. For example, the row in Table F.1 for power shows that 1 W = (N · m/s). Dividing both sides of this
equation by N · m/s gives
Table 1.3 shows how to apply the grid method. Notice how the same process steps can apply to different
situations.
Table 1.3 APPLYI G THE GRID METHOD (TWO EXAMPLES)
Process Step Example 1.1 Example 1.2
Situation: Convert a pressure of Situation: Find the force in newtons that is
2.00 psi to pascals. needed to accelerate a mass of 10 g at a rate of
15 ft/s2.
Step 1. p = 2.00 psi F = ma
Problem:
Write the
F(N) = (0.01 kg)(15 ft/s2)
term or
equation.
Include
numbers and
units.
Step 2.
Conversion
Ratios: Look
up unit
conversion
formula(s) in
Table F.1 and
represent
these as unity
conversion
ratios.
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Process Step Example 1.1 Example 1.2
Step 3.
Algebra:
Multiply
variables and
cancel units.
Fix any
errors.
Step 4. p = 13.8 kPa F = 0.0457 N
Calculations:
Perform the
indicated
calculations.
Round the
answer to the
correct
number of
significant
figures.
Since fluid mechanics problems often involve mass units of slugs or pounds-mass (lbm), it is easy to make a
mistake when converting units. Thus, it is useful to have a systematic approach. The idea is to relate mass units
to force units through F = ma. For example, a force of 1.0 N is the magnitude of force that will accelerate a mass
of 1.0 kg at a rate of 1.0 m/s2. Thus,
Rewriting this expression gives a conversion ratio
(1.7)
When mass is given using slugs, the corresponding conversion ratio is
(1.8)
A force of 1.0 lbf is defined as the magnitude of force that will accelerate a mass of 1.0 lbm at a rate of 32.2 ft/s2.
Thus,
Thus, the conversion ratio relating force and mass units becomes
(1.9)
Example 1.4 shows how to apply the grid method. Notice that calculations involving traditional units follow the
same process as calculations involving SI units.
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EXAMPLE 1.4 GRID METHOD APPLIED TO A ROCKET
A water rocket is fabricated by attaching fins to a 1-liter plastic bottle. The rocket is partially filled
with water and the air space above the water is pressurized, causing water to jet out of the rocket and
propel the rocket upward. The thrust force T from the water jet is given by , where m is the
rate at which the water flows out of the rocket in units of mass per time and V is the speed of the
water jet. (a) Estimate the thrust force in newtons for a jet velocity of V = 30 m/s (98.4 ft/s) where the
mass flow rate is m = 9 kg/s (19.8 lbm/s). (b) Estimate the thrust force in units of pounds-force (lbf).
Apply the grid method during your calculations.
Problem Definition
Situation:
1. A rocket is propelled by a water jet.
2. The thrust force is given by .
Find: Thrust force supplied by the water jet. Present the answer in N and lbf.
Sketch:
Plan
Find the thrust force by using the process given in Table 1.3. When traditional units are used, apply
Eq. 1.9.
Solution
1. Thrust force (SI units)
· Insert numbers and units:
· Insert conversion ratios and cancel units:
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2. Thrust force (traditional units)
· Insert numbers and units:
· Insert conversion ratios and cancel units:
Review
1. Validation. Since 1.0 lbf = 4.45 N, answer (a) is the same as answer (b).
2. Tip! To validate calculations in traditional units, repeat the calculation in SI units.
Copyright © 2009 John Wiley & Sons, Inc. All rights reserved.
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