1.3 Units, Dimensional Analysis & Physical Constants

Q: Crude oil with a density of 55 lbm/ft^3 and a dynamic viscosity of 30 centipoise (cP) flows through an 8-inch inside-diameter pipe at an average velocity of 6.0 ft/s. What is the Reynolds number of the flow?

10,920. To compute Reynolds number in US Customary units using dynamic viscosity, ensure all units match the pound-mass system: Re = (rho * v * D) / mu. Given: rho = 55 lbm/ft^3, v = 6.0 ft/s, D = 8 in * (1 ft / 12 in) = 0.6667 ft. Convert viscosity from centipoise: 1 cP = 0.00067197 lbm/(fts), so mu = 30 * 0.00067197 = 0.020159 lbm/(fts). Computing: Re = (55 * 6.0 * 0.6667) / 0.020159 = 220.01 / 0.020159 = 10,914, which rounds to 10,920.

Q: A fluid flow phenomenon is governed by 6 physical variables (fluid velocity v, pipe diameter D, fluid density rho, dynamic viscosity mu, surface roughness height epsilon, and pressure drop Delta P) involving 3 fundamental dimensions (Mass [M], Length [L], and Time [T]). According to the Buckingham Pi Theorem, how many independent dimensionless Pi groups are required to describe this system?

According to the Buckingham Pi Theorem, the number of independent dimensionless groups Pi is given by Pi = n - k, where n is the total number of physical variables and k is the number of independent fundamental physical dimensions. Here, n = 6 variables and k = 3 dimensions ([M], [L], [T]). Thus, Pi = 6 - 3 = 3 dimensionless groups (typically formulated as Euler number, Reynolds number, and relative roughness ratio epsilon/D).

Q: A hydronic water loop operating with standard water (density = 62.4 lbm/ft^3) experiences a pressure drop of 18.0 psi across a heat exchanger. What is this pressure drop expressed in equivalent feet of water head?

41.5 ft H2O. Head is related to pressure by the relationship: head h = P / gamma, where gamma is the specific weight. Because pressure is given in psi (pounds per square inch), it must be converted to psf (pounds per square foot) by multiplying by 144 in^2/ft^2: P = 18.0 psi * 144 in^2/ft^2 = 2,592 lbf/ft^2. Then, h = 2,592 lbf/ft^2 / 62.4 lbf/ft^3 = 41.538 ft of water. Alternatively, using the standard conversion factor 1 psi = 2.307 ft H2O: 18.0 * 2.307 = 41.53 ft H2O.

Q: What is the density of ideal gaseous nitrogen (molar mass M = 28.013 kg/kmol) at an absolute pressure of 350 kPa and a temperature of 45 degrees Celsius?

3.70 kg/m^3. Apply the Ideal Gas Law: P = rho * R * T, where rho = P / (R * T). First, find the specific gas constant R = R_u / M = 8.31446 kJ/(kmolK) / 28.013 kg/kmol = 0.29681 kJ/(kgK) = 296.81 J/(kg*K). Convert temperature to Kelvin: T = 45 + 273.15 = 318.15 K. Convert pressure to Pascals: P = 350 kPa = 350,000 Pa. Then rho = 350,000 / (296.81 * 318.15) = 350,000 / 94,429.98 = 3.706 kg/m^3.

Key Takeaways

The PE Mechanical exam tests both International System (SI) and US Customary / English Engineering units; flawless unit conversion is essential to passing.
In the English Engineering unit system, the gravitational constant $g_c = 32.174 \text{ lbm}\cdot\text{ft}/(\text{lbf}\cdot\text{s}^2)$ is mandatory to reconcile mass ( $\text{lbm}$ ) and force ( $\text{lbf}$ ) in Newton's second law: $F = ma/g_c$ .
The Universal Gas Constant $R_u = 8.3145 \text{ kJ/(kmol}\cdot\text{K)} = 1545.35 \text{ ft}\cdot\text{lbf/(lbmol}\cdot^\circ\text{R)}$ , requiring specific gas constant calculations via $R = R_u / M$ .
The Buckingham Pi Theorem determines the number of independent dimensionless groups ( $\Pi = n - k$ ) governing physical systems, forming parameters like Reynolds, Mach, and Prandtl numbers.
High-frequency unit conversion traps include mixing gauge and absolute pressures, omitting the $144 \text{ in}^2/\text{ft}^2$ factor in hydraulic head formulas, and confusing dynamic versus kinematic viscosity.

Last updated: August 2026

Units, Dimensional Analysis & Physical Constants

Unit conversion errors and dimensional inconsistencies represent the single most common source of unforced calculation errors on the NCEES PE Mechanical exam. Problems are presented in both International System (SI) and US Customary (English Engineering) units, frequently mixing units within a single problem statement (e.g., pipe diameters in inches, lengths in feet, pressures in psi, flow rates in gallons per minute, and power in horsepower).

1. The Mass vs. Force Distinction: $\text{lbm}$ , $\text{lbf}$ , $\text{slug}$ , and $g_c$

In SI units, mass and force are cleanly delineated through base units: $1 \text{ N} = 1 \text{ kg}\cdot\text{m/s}^2$ . In English units, the historical dual use of the word "pound" for both mass and force requires absolute rigor.

+-----------------------------------------------------------------------------------------+
|                        ENGLISH ENGINEERING UNIT SYSTEMS                                 |
|                                                                                         |
|   1. GRAVITATIONAL SYSTEM (Slug - lbf - ft - s)                                         |
|      - Base mass unit: 1 slug = 32.174 lbm                                              |
|      - Newton's Second Law: F = m * a  (where m is in slugs, F is in lbf)               |
|      - 1 lbf accelerates 1 slug at 1 ft/s^2                                             |
|                                                                                         |
|   2. DUAL POUND SYSTEM (lbm - lbf - ft - s with g_c)                                    |
|      - Base mass unit: pound-mass (lbm)                                                 |
|      - Base force unit: pound-force (lbf)                                               |
|      - Dimensional Constant: g_c = 32.174 (lbm * ft) / (lbf * s^2)                      |
|      - Newton's Second Law: F = (m * a) / g_c                                           |
|      - Weight on Earth (where g = 32.174 ft/s^2): W = (m * g) / g_c = m [in lbf]        |
+-----------------------------------------------------------------------------------------+

Application of $g_c$ in Mechanical Engineering Formulas

Whenever a formula derives from fundamental Newtonian mechanics ( $F = ma$ ) or energy balances and uses mass in $\text{lbm}$ and force/energy in $\text{lbf}$ or $\text{ft}\cdot\text{lbf}$ , $g_c$ must be inserted in the denominator:

\text{Kinetic Energy: } E_k = \frac{m v^2}{2 g_c} \quad [\text{ft}\cdot\text{lbf}]

\text{Potential Energy: } E_p = \frac{m g z}{g_c} \quad [\text{ft}\cdot\text{lbf}]

\text{Hydrostatic Pressure: } P = \frac{\rho g h}{g_c} \quad [\text{lbf/ft}^2]

\text{Dynamic Pressure: } q = \frac{\rho v^2}{2 g_c} \quad [\text{lbf/ft}^2]

\text{Reynolds Number: } Re = \frac{\rho v D}{\mu} \quad \left[\text{where } \rho \text{ is } \frac{\text{lbm}}{\text{ft}^3} \text{ and } \mu \text{ is } \frac{\text{lbm}}{\text{ft}\cdot\text{s}}\right]

Caution

If dynamic viscosity is provided in $\frac{\text{lbf}\cdot\text{s}}{\text{ft}^2}$ , you must convert it to $\frac{\text{lbm}}{\text{ft}\cdot\text{s}}$ by multiplying by $g_c = 32.174 \frac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}$ , or express density $\rho$ in $\frac{\text{slugs}}{\text{ft}^3}$ .

2. Master Conversion Reference Tables

+-----------------------------------------------------------------------------------------+
|                         HIGH-YIELD PE CONVERSION MULTIPLIERS                            |
|                                                                                         |
|   POWER:        1 hp = 550 ft*lbf/s = 33,000 ft*lbf/min = 0.7457 kW = 2,544.4 BTU/hr    |
|                 1 Ton of Refrigeration = 12,000 BTU/hr = 200 BTU/min = 3.51685 kW       |
|                 1 kW = 3,412.14 BTU/hr = 737.56 ft*lbf/s = 1.341 hp                     |
|                                                                                         |
|   ENERGY:       1 BTU = 778.169 ft*lbf = 1,055.06 J = 1.05506 kJ                        |
|                 1 kWh = 3,412.14 BTU = 3.6 MJ = 2.655 * 10^6 ft*lbf                     |
|                                                                                         |
|   PRESSURE:     1 atm = 14.696 psia = 101.325 kPa = 1.01325 bar = 29.921 in. Hg        |
|                       = 33.90 ft H2O = 406.8 in. w.g. = 760 mmHg                        |
|                 1 psi = 144 psf = 2.307 ft H2O = 27.70 in. w.g. = 6.89476 kPa           |
|                 1 bar = 100 kPa = 10^5 Pa = 14.5038 psi = 0.9869 atm                    |
|                 1 in. w.g. (at 60 deg F) = 0.03613 psi = 5.204 psf = 249.08 Pa          |
|                                                                                         |
|   VISCOSITY:    Dynamic (mu): 1 Pa*s = 1 N*s/m^2 = 10 Poise = 1,000 cP                  |
|                               = 0.67197 lbm/(ft*s) = 0.020885 lbf*s/ft^2                |
|                 Kinematic (nu): 1 m^2/s = 10^4 Stokes = 10^6 cSt = 10.7639 ft^2/s       |
+-----------------------------------------------------------------------------------------+

Temperature Scales and Temperature Differences

Thermodynamic equations (ideal gas law, Carnot efficiency, radiative heat transfer) require absolute temperature:

T(\text{Kelvin}) = T(^\circ\text{C}) + 273.15

T(\text{Rankine}) = T(^\circ\text{F}) + 459.67

For temperature differences (such as $\Delta T$ in convective heat transfer $q = h A \Delta T$ or thermal expansion $\Delta L = \alpha L \Delta T$ ):

\Delta T (1.0 \text{ K}) = \Delta T (1.0 ^\circ\text{C}) = \Delta T (1.8 ^\circ\text{F}) = \Delta T (1.8 ^\circ\text{R})

3. Fundamental Physical Constants

Constant Name	Symbol	SI Value & Units	US Customary Value & Units
Standard Acceleration of Gravity	$g_0$	$9.80665 \text{ m/s}^2$	$32.17405 \text{ ft/s}^2$
Universal Gas Constant	$R_u$	$8.31446 \text{ kJ/(kmol}\cdot\text{K)}$	$1545.35 \text{ ft}\cdot\text{lbf/(lbmol}\cdot^\circ\text{R)}$ ; $1.98588 \text{ BTU/(lbmol}\cdot^\circ\text{R)}$
Specific Gas Constant for Air	$R_{\text{air}}$	$0.2870 \text{ kJ/(kg}\cdot\text{K)}$	$53.35 \text{ ft}\cdot\text{lbf/(lbm}\cdot^\circ\text{R)}$ ; $0.06855 \text{ BTU/(lbm}\cdot^\circ\text{R)}$
Stefan-Boltzmann Constant	$\sigma$	$5.67037 \times 10^{-8} \frac{\text{W}}{\text{m}^2\cdot\text{K}^4}$	$0.1714 \times 10^{-8} \frac{\text{BTU}}{\text{hr}\cdot\text{ft}^2\cdot^\circ\text{R}^4}$
Standard Atmospheric Pressure	$P_{\text{atm}}$	$101.325 \text{ kPa}$	$14.696 \text{ psia} = 2116.2 \text{ psf}$
Density of Water at $39.2^\circ\text{F} / 4^\circ\text{C}$	$\rho_{\text{water}}$	$1,000 \text{ kg/m}^3$	$62.428 \text{ lbm/ft}^3 = 1.940 \text{ slugs/ft}^3$
Specific Weight of Water at $60^\circ\text{F}$	$\gamma_{\text{water}}$	$9.790 \text{ kN/m}^3$	$62.37 \text{ lbf/ft}^3 = 8.34 \text{ lbm/gal}$
Specific Heat of Liquid Water	$c_p$	$4.184 \text{ kJ/(kg}\cdot^\circ\text{C)}$	$1.000 \text{ BTU/(lbm}\cdot^\circ\text{F)}$

4. Dimensional Analysis & The Buckingham Pi Theorem

Dimensional analysis allows engineers to group physical variables into dimensionless numbers, reducing experimental complexity and enabling fluid/thermal scale-model testing.

+-----------------------------------------------------------------------------------------+
|                        BUCKINGHAM PI THEOREM WORKFLOW                                   |
|                                                                                         |
|   [STEP 1: Identify all n variables influencing the physical phenomenon]                |
|   [STEP 2: Express each variable in fundamental dimensions (M, L, T, Theta)]            |
|   [STEP 3: Determine k = number of independent fundamental dimensions involved]          |
|   [STEP 4: Calculate number of dimensionless Pi groups: Pi_count = n - k]               |
|   [STEP 5: Select k repeating variables (must contain all k dimensions)]                |
|   [STEP 6: Form equations for each Pi group and solve exponents to ensure dimensionless] |
+-----------------------------------------------------------------------------------------+

Primary Dimensionless Parameters in Mechanical Engineering

Parameter	Formula	Physical Interpretation	Key Applications
Reynolds Number ( $Re$ )	$\frac{\rho v D}{\mu} = \frac{v D}{\nu}$	$\frac{\text{Inertial Forces}}{\text{Viscous Forces}}$	Pipe flow, aerodynamics, boundary layer transition
Mach Number ( $Ma$ )	$\frac{v}{c} = \frac{v}{\sqrt{k R T}}$	$\frac{\text{Flow Velocity}}{\text{Speed of Sound}}$	Compressible flow, nozzle design, gas dynamics
Prandtl Number ( $Pr$ )	$\frac{\nu}{\alpha} = \frac{\mu c_p}{k}$	$\frac{\text{Momentum Diffusivity}}{\text{Thermal Diffusivity}}$	Convection boundary layer analysis
Nusselt Number ( $Nu$ )	$\frac{h L}{k_{\text{fluid}}}$	$\frac{\text{Convective Heat Transfer}}{\text{Pure Fluid Conduction}}$	Convective heat transfer coefficient sizing
Biot Number ( $Bi$ )	$\frac{h L_c}{k_{\text{solid}}}$	$\frac{\text{Internal Conductive Resistance}}{\text{External Convective Resistance}}$	Lumped capacitance validity ( $Bi < 0.1$ )
Fourier Number ( $Fo$ )	$\frac{\alpha t}{L_c^2}$	$\frac{\text{Dimensionless Time}}{\text{Thermal Storage Rate}}$	Transient conduction cooling/heating
Euler Number ( $Eu$ )	$\frac{\Delta P}{\rho v^2}$	$\frac{\text{Pressure Forces}}{\text{Inertial Forces}}$	Fluid machinery, cavitation, pipe friction

5. Step-by-Step Worked Problem: Fan Air Power in Dual Units

Problem: A ventilation fan delivers $Q = 8,500 \text{ CFM}$ ( $14,441 \text{ m}^3/\text{h}$ ) of standard air against a total static pressure resistance of $\Delta P = 2.75 \text{ in. w.g.}$ ( $685 \text{ Pa}$ ). The total fan efficiency is $\eta_t = 68\%$ .

Calculate the required fan shaft brake horsepower (BHP) and shaft power in kilowatts (kW).

+-----------------------------------------------------------------------------------------+
|                        FAN POWER STEP-BY-STEP CALCULATION                               |
|                                                                                         |
|   METHOD 1: US CUSTOMARY CALCULATION                                                    |
|   Static Air Power Formula:                                                             |
|   Air HP = (Q [CFM] * Delta P [in. w.g.]) / 6,356                                       |
|                                                                                         |
|   Derivation of Constant 6,356:                                                         |
|   1 in. w.g. = 5.204 lbf/ft^2                                                           |
|   Power = (Q [ft^3/min] * 5.204 [lbf/ft^2]) = 5.204 * Q ft*lbf/min                      |
|   1 HP = 33,000 ft*lbf/min                                                              |
|   Air HP = (5.204 / 33,000) * Q * Delta P = (Q * Delta P) / 6,341.27 (at standard rho)  |
|   (Standard HVAC engineering uses 6,356 for air at 70 deg F, 29.92 in. Hg).             |
|                                                                                         |
|   Air HP = (8,500 * 2.75) / 6,356 = 23,375 / 6,356 = 3.6776 Air HP                      |
|   Brake Horsepower (BHP) = Air HP / eta_t = 3.6776 / 0.68 = 5.408 BHP                   |
|                                                                                         |
|   METHOD 2: SI CALCULATION                                                              |
|   Q = 8,500 ft^3/min * (0.0283168 m^3/ft^3) * (1 min / 60 s) = 4.0115 m^3/s             |
|   Delta P = 2.75 in. w.g. * (249.08 Pa / 1 in. w.g.) = 684.97 Pa (N/m^2)                |
|   Air Power = Q * Delta P = (4.0115 m^3/s) * (684.97 N/m^2) = 2,747.76 W = 2.748 kW     |
|   Shaft Power = Air Power / eta_t = 2.74776 kW / 0.68 = 4.041 kW                        |
|                                                                                         |
|   CROSS-CHECK CONVERSION:                                                               |
|   4.041 kW * (1 hp / 0.7457 kW) = 5.419 BHP (Exact match within rounding).              |
+-----------------------------------------------------------------------------------------+

Test Your Knowledge

Crude oil with a density of 55 lbm/ft^3 and a dynamic viscosity of 30 centipoise (cP) flows through an 8-inch inside-diameter pipe at an average velocity of 6.0 ft/s. What is the Reynolds number of the flow?

1,850

3,240

10,920

351,600

Test Your Knowledge

A fluid flow phenomenon is governed by 6 physical variables (fluid velocity v, pipe diameter D, fluid density rho, dynamic viscosity mu, surface roughness height epsilon, and pressure drop Delta P) involving 3 fundamental dimensions (Mass [M], Length [L], and Time [T]). According to the Buckingham Pi Theorem, how many independent dimensionless Pi groups are required to describe this system?

Test Your Knowledge

A hydronic water loop operating with standard water (density = 62.4 lbm/ft^3) experiences a pressure drop of 18.0 psi across a heat exchanger. What is this pressure drop expressed in equivalent feet of water head?

7.80 ft H2O

25.4 ft H2O

36.0 ft H2O

41.5 ft H2O

Test Your Knowledge

What is the density of ideal gaseous nitrogen (molar mass M = 28.013 kg/kmol) at an absolute pressure of 350 kPa and a temperature of 45 degrees Celsius?

2.68 kg/m^3

3.70 kg/m^3

4.41 kg/m^3

27.0 kg/m^3

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1.3 Units, Dimensional Analysis & Physical Constants

Key Takeaways

Units, Dimensional Analysis & Physical Constants

1. The Mass vs. Force Distinction: lbm\text{lbm}lbm, lbf\text{lbf}lbf, slug\text{slug}slug, and gcg_cgc​

Application of gcg_cgc​ in Mechanical Engineering Formulas

2. Master Conversion Reference Tables

Temperature Scales and Temperature Differences

3. Fundamental Physical Constants

4. Dimensional Analysis & The Buckingham Pi Theorem

Primary Dimensionless Parameters in Mechanical Engineering

5. Step-by-Step Worked Problem: Fan Air Power in Dual Units

1. The Mass vs. Force Distinction: $\text{lbm}$ , $\text{lbf}$ , $\text{slug}$ , and $g_c$

Application of $g_c$ in Mechanical Engineering Formulas