Worked Examples: Preparations

J. Gordon Leishman

74 Worked Examples: Preparations

Many of these worked examples on mathematics, physics, units and conversion factors, and professional responsibilities have been fielded as quizzes, homework, or exam questions. Engineering students often struggle with prerequisite mathematics and physics, as well as units, and so can lose unnecessary points on engineering exams. Use these short worked examples to maintain and consolidate understanding.

Worked Example #1 – Algebra

Solve for the unknown $x$ in the equation $3x + 2 = 5(1 - x).$

Expand the right-hand side to get

$3x + 2 = 5 - 5x$

Collect like terms to give

$3x + 5x = 5 - 2$

and so

$8x = 3$

Solving for $x$ gives

$x = \frac{3}{8} = 0.375$

Worked Example #2 – Differentiation

The position of a particle moving along a line is $x(t) = 4t^3 - 3t^2 + 2t$ , where $x$ is in meters and $t$ is in seconds. Determine:

(a) The velocity as a function of time.
(b) Acceleration as a function of time.
(c) Velocity at $t$ = 2 s.

(a) Velocity is the first derivative of position, i.e.,

$v(t) = \frac{dx}{dt} = 12t^2 - 6t + 2$

(b) Acceleration is the derivative of velocity, i.e.,

$a(t) = \frac{dv}{dt} = 24t - 6$

(c) At $t$ = 2 s, then

$v(2) = 12(2)^2 - 6(2) + 2 = 48 - 12 + 2 = 38~\text{m/s}$

Worked Example #3 – Integration

The velocity of a vehicle is given by $v(t) = 6t^2 + 2t$ , where $v$ is in m/s and $t$ is in seconds. Determine the displacement between $t = 0$ and $t = 3~\mathrm{s}$ .

The displacement is obtained by integrating velocity, i.e.,

$s = \int_0^3 (6t^2 + 2t)\,dt$

Integrate term by term to get

$\int (6t^2 + 2t)\,dt = 2t^3 + t^2$

Evaluating between limits gives

$s = \left[2t^3 + t^2\right]_0^3= (2\times 27 + 9) - 0 = 54 + 9 = 63~\text{m}$

Worked Example #4 – First-order differential equation

Solve the differential equation $\dfrac{dy}{dt} = -3y$ with initial condition $y(0)=5$ .

Separating the variables gives

$\frac{dy}{y} = -3\, dt$

Integrating gives

$\int \frac{1}{y}\,dy = \int -3\,dt$

and so

$\ln y = -3t + C$

Exponentiate to get

$y = Ce^{-3t}$

Applying the initial condition, which gives

$5 = Ce^{0} \Rightarrow C = 5$

Therefore, the final solution is

$y(t) = 5e^{-3t}$

Worked Example #5 – Newton’s second law

A constant 50 N force acts on a 10 kg mass that is initially at rest. Determine:

(a) The acceleration.
(b) The velocity after 4 s.
(c) The distance traveled in 4 s.

(a) From Newton’s second law, then

$F = m \, a$

so that

$a = \frac{F}{m} = \frac{50}{10} = 5~\mathrm{m/s^2}$

(b) Velocity after 4 s is

$v = a \, t = 5 \times 4 = 20~\mathrm{m/s}$

(c) The distance traveled is

$s = \frac{1}{2}a \, t^2 = \frac{1}{2}\times 5 \times 4^2 = 40~\mathrm{m}$

Worked Example #6 – Work and energy

A body of mass 2 kg is lifted vertically by 5 m. Determine:

(a) The work done against gravity.
(b) The change in potential energy.

(a) The work done is

$W = m \, g \, h = 2 \times 9.81 \times 5 = 98.1~\mathrm{J}$

(b) The increase in gravitational potential energy equals the work done, i.e.,

$\Delta PE = m \, g \, h = 98.1~\mathrm{J}$

Worked Example #7 – Pressure in a fluid

Determine the gauge pressure at a depth of 8 m below the water surface. Take the water density as $\varrho$ = 1,000 kg/m³.

The hydrostatic pressure is given by

$p = \varrho \, g \, h$

Substituting the values gives

$p = (1000)(9.81)(8) = 78{,}480~\mathrm{Pa} \approx 7.85\times10^4~\mathrm{Pa} = 78.5~\mathrm{kPa}$

Worked Example #8 – Simple harmonic motion

A mass–spring system has a mass of 0.5 kg and spring constant $k$ = 200 N/m. Determine:

(a) The natural circular frequency.
(b) The oscillation frequency in Hz.
(c) The period of oscillation.

(a) The natural circular frequency is

$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20~\mathrm{rad/s}$

(b) The frequency is

$f = \frac{\omega_n}{2\pi} = \frac{20}{2\pi} = 3.18~\mathrm{Hz}$

(c) The corresponding period is

$T = \frac{1}{f} = 0.314~\mathrm{s}$

Worked Example #9 – Finding the dimensions of quantities

Determine the base dimensions in terms of M, L, and T of each of the following variables:

(a) Plane angle
(b) Specific volume
(c) Force
(d) Stress
(e) Angular velocity

(a) Plane angle: A plane angle $\theta$ is defined in terms of the lines from two points meeting at a vertex and is defined by the arc length of a circle subtended by the lines and the circle’s radius. The unit of plane angle is the radian. Because it is the ratio of an arc length to the radius, the plane angle is dimensionless, i.e., a radian is one measurement unit that is already dimensionless, i.e.,

$\left[ \theta \right] = 1$

(b) Specific volume: The specific volume $SV$ is defined as the ratio of volume to mass, i.e., it is the reciprocal of the density. Therefore,

$\left[ SV \right] = \frac{\rm L^3}{\rm M} = \rm L^3 M^{-1} = \rm M^{-1} L^3$

(c) Force: Force $F$ is the product of mass times acceleration, so

$\left[ F \right] = \rm M L T^{-2}$

(d) Stress: Stress $\sigma$ has dimensions of force per area, but a force is the product of mass times acceleration, so

$\left[ \sigma \right] = \frac{\rm M L T^{-2}}{ \rm L^2} = \rm M L^{-1} T^{-2}$

(e) Angular velocity: Angular velocity $\omega$ is measured in terms of radians per second. A radian is unitless, so

$\left[ \omega \right] = T^{-1}$

Worked Example #10 – Determining the units of surface tension

A group of students becomes curious about why some insects can walk on water. They discover that a fluid property of importance in this problem is called surface tension, which is given the symbol $\sigma_s$ . Surface tension is defined as a force per unit length along an interface. Write the dimensions of surface tension in terms of the base dimensions of mass, length, and time.

We are told that surface tension, $\sigma_s$ , is defined as force per unit length. A force is equivalent to mass times acceleration, which is MLT^-2 in base dimensions. Therefore, the base dimensions of surface tension are

$[\sigma_s] = \frac{\text{Force}}{\text{Length}} = \frac{\rm M L T^{-2}}{L} = \rm M T^{-2}$

Alternatively, surface tension can also be expressed as energy per unit area, i.e.,

$[\sigma_s] = \frac{\text{Energy}}{\text{Area}} = \frac{\rm M L^{2} T^{-2}}{\rm L^{2}} = M T^{-2}$

Further research shows that in SI units, surface tension is expressed as

$\sigma_s = \frac{\mathrm{N}}{\mathrm{m}} = \frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m}} = \mathrm{kg\,s^{-2}}$

while in USC units it is expressed as

$\sigma_s = \frac{\mathrm{lb}}{\mathrm{ft}}$

For example, the surface tension of water at $20^{\circ}$ C is approximately

$\sigma_s \approx 0.072~\mathrm{N/m} = 0.072~\mathrm{N m}^{-1} \ \text{(in SI)} \qquad \text{or} \qquad \sigma_s \approx 0.0049~\mathrm{lb/ft} = 0.0049~\mathrm{lb ft}^{-1} \ \text{(in USC)}$

Worked Example #11 – Determining base dimensions

Write the primary (base) dimensions of each of the following variables from the field of thermodynamics:

(a) Energy, $E$
(b) Specific energy, $e=E/m$
(c) Power, $P$

(a) Energy has units of force times distance, i.e.,

$\left[ E \right] = \left( \rm MLT^{-2} \right) \rm L = \rm ML^{2} T^{-2}$

(b) Specific energy has units of energy per unit mass, i.e.,

$\left[ e \right] = \rm ML^{2} T^{-2} M^{-1} = \rm L^{2} T^{-2}$

(c) Power is the rate of doing work, so a force times distance per unit time, i.e.,

$\left[ P \right] = \left( \rm MLT^{-2} \right) \rm L T^{-1} = \rm M L^{2} T^{-3}$

Worked Example #11 – Determining base dimensions

Write the primary (base) dimensions of each of the following intensive or specific thermodynamic quantities:

(a) Density, $\varrho$
(b) Specific volume, $v$
(c) Pressure, $p$
(d) Specific energy, $e$

(a) Density is mass per unit volume, so

$\left[ \varrho \right] = \frac{\rm M}{\rm L^3} = \rm M L^{-3}$

(b) Specific volume is volume per unit mass, i.e., the reciprocal of density, so

$\left[ v \right] = \frac{\rm L^3}{\rm M} = \rm M^{-1} L^3$

(c) Pressure is force per unit area. Because force has dimensions of mass times acceleration, then

$\left[ p \right] = \frac{\rm M L T^{-2}}{\rm L^2} = \rm M L^{-1} T^{-2}$

(d) Specific energy is energy per unit mass. Because energy has dimensions of force times distance, then

$\left[ e \right] = \frac{\rm M L^2 T^{-2}}{\rm M} = \rm L^2 T^{-2}$

Worked Example #13 – Converting units

Given that jet fuel’s average specific energy is 43 to 45 MJ/kg (megajoules per kilogram), calculate the equivalent value in kilowatt-hours per kilogram (kWh/kg).

A watt is a joule per second. One watt-hour is equivalent to 3,600 joules, so one kilowatt-hour (kWh) is equivalent to 3.6 megajoules (MJ). Therefore, the specific energy in kWh/kg is obtained by dividing the specific energy in MJ/kg by 3.6, i.e.,

$\frac{43}{3.6} = 11.9~\mathrm{kWh/kg}$

and

$\frac{45}{3.6} = 12.5~\mathrm{kWh/kg}$

Therefore, the specific energy of jet fuel is approximately

$11.9 \text{ to } 12.5~\mathrm{kWh/kg}$

This quantity is also sometimes called the gravimetric energy density, but because it is expressed per unit mass, the more precise term is specific energy.

Worked Example #14 – Verifying dimensional homogeneity

Write down the Bernoulli equation and explain the meaning of each term. Verify that each of the terms in the Bernoulli equation has the same base or fundamental dimensions.

The Bernoulli equation can be written as

$p + \frac{1}{2} \varrho \, V^2 + \varrho \, g \, z = \mbox{constant}$

The first term is the local static pressure, the second is the dynamic pressure, and the third is the elevation (head) or hydrostatic pressure term. The sum of the three terms is constant along a streamline under the assumptions of Bernoulli’s equation. If the elevation is unchanged, then $p + \frac{1}{2}\varrho V^2$ is the stagnation or total pressure. The Bernoulli equation is a simplified form of the energy equation for steady, incompressible flow with no losses or energy addition.

Each term in the Bernoulli equation has units of pressure. In terms of fundamental dimensions, then

$\left[ p \right] = ( \rm M L T^{-2} ) ( \rm L^{-2}) = \rm M L^{-1} T^{-2}$

$\left[ \varrho \, V^2 \right] = ( \rm M L^{-3}) \rm (L^2 T^{-2}) = \rm M L^{-1} T^{-2}$

$\left[ \varrho \, g \, z \right] = ( \rm M L^{-3}) ( \rm L T^{-2}) L = \rm M L^{-1} T^{-2}$

So, all terms have the same fundamental dimensions of M L $^{-1}$ T $^{-2}$ .

Worked Example #15 – Converting to base units

In each case, convert the given units into base units (mass, length, and time). Show all the steps.

N (Newton) $\equiv$ mass times acceleration $\equiv \mathrm{M\,L\,T^{-2}} \equiv \mathrm{kg\,m\,s^{-2}}$
Pa (Pascal) or $\mathrm{N\,m^{-2}} \equiv (\mathrm{M\,L\,T^{-2}})(\mathrm{L^{-2}}) = \mathrm{M\,L^{-1}\,T^{-2}} \equiv \mathrm{kg\,m^{-1}\,s^{-2}}$
lb (pound force) $\equiv \mathrm{M\,L\,T^{-2}} \equiv \mathrm{slug\,ft\,s^{-2}}$
J (Joule) or $\mathrm{N\,m} \equiv (\mathrm{M\,L\,T^{-2}})(\mathrm{L}) = \mathrm{M\,L^{2}\,T^{-2}} \equiv \mathrm{kg\,m^{2}\,s^{-2}}$
W (Watt) or $\mathrm{J\,s^{-1}} \equiv (\mathrm{M\,L^{2}\,T^{-2}})(\mathrm{T^{-1}}) = \mathrm{M\,L^{2}\,T^{-3}} \equiv \mathrm{kg\,m^{2}\,s^{-3}}$

Notes: A Newton (N) is an SI unit of force, but not a base unit. However, force equals mass times acceleration, so in base dimensions a Newton has dimensions $\mathrm{M\, L\, T^{-2}}$ and SI units of $\mathrm{kg\,m\,s^{-2}}$ . A Watt (W) is the SI unit of power, which is the rate at which work is done. Work is force times distance, so power has dimensions $\mathrm{M\,L\,T^{-2}}\,\mathrm{L}\,\mathrm{T^{-1}} = \mathrm{M\,L^{2}\,T^{-3}}$ and SI units of $\mathrm{kg\,m^{2}\,s^{-3}}$ .

Worked Example #16 – Converting numerical values to base units

In each case, convert the numerical values in the units given into equivalent numerical values in base units (mass, length, and time). Show all the steps and state the conversion factor(s) you used.

3.3 L (liters) = 3.3/1,000 = 0.0033 m ${^{3}}$ . There are 1,000 liters in a cubic meter.
1.2 hrs (hours) = 3,600 $\times$ 1.2 = 4,320 s. There are 3,600 seconds in an hour.
15.6 hp (horsepower) = 15.6 $\times$ 550 = 8,580 ft-lb s $^{-1}$ . Notice: 550 ft-lb s $^{-1}$ is equivalent to one horsepower. In USC engineering units, ft-lb s $^{-1}$ is the customary unit of power. If expressed strictly in terms of base mass, length, and time units, then because $1~\mathrm{lb} = 1~\mathrm{slug\,ft\,s^{-2}}$ , this result may also be written as $8{,}580~\mathrm{slug\,ft^2\,s^{-3}}$ .
12.8 gals (US gallons) = 12.8 $\times$ 0.13368 = 1.711 ft ${^{3}}$ . One gallon equals 0.13368 ft ${^{3}}$ .
12.4 kN cm $^{-2}$ = 12.4 $\times 10^3 \times 10^{4} = 12.4 \times 10^7$ Pa = 12.4 $\times 10^7$ kg m $^{-1}$ s $^{-2}$ . One kN = 1,000 N, and there are 10,000 square centimeters in a square meter.

Notes: Converting units requires careful attention, and it is essential to do so correctly. The first step is to recognize that many quantities measured daily are not base quantities but must be converted to base quantities for engineering calculations. Volume is often measured in liters, but it is not a base unit. There are 1,000 liters in a cubic meter; a meter is a base SI unit. James Watt figured out from experiments with Scottish farm horses that one horsepower (hp) was equivalent to a rate of doing work of 550 foot-pounds per second or 550 ft-lb s $^{-1}$ . The horsepower was standardized as a rate of doing work equal to 550 foot-pounds per second, or 550 ft-lb s $^{-1}$ . Therefore, 550 is the conversion factor from horsepower to base USC engineering units of power.

Worked Example #17 – Units can become a source of error

On an exam, a student uses the equation of state to compute the density of air using

$\varrho = \frac{p}{R T}$

The question gives the conditions as $p$ = 760 mmHg, $T = 15^{\circ}\mathrm{C}$ , and $R = 287.0~\mathrm{J\,kg^{-1}\,K^{-1}}$ . The student substitutes numbers directly and writes

$\varrho = \frac{760}{287 \times 15} = 0.177~\mathrm{kg/m^3}$

and then attaches “kg/m³ at the end because the density ought to be in those units.” The answer was marked incorrect by the instructor. What went wrong?

Two things went wrong here. The pressure was not converted from mmHg to Pascals, and the temperature was reported in degrees Celsius rather than Kelvin. By treating the units as labels to be pasted on afterward, the student lost the dimensional consistency check and obtained an incorrect value.

The correct procedure is to first convert all given values to consistent SI units, i.e.,

$760~\mathrm{mmHg} = 1.01325 \times 10^{5}~\mathrm{Pa} \quad \text{and} \quad 15^{\circ}\mathrm{C} = 288.15~\mathrm{K}$

While a Pascal (Pa) is not a base unit, it can still be expressed in terms of base units, i.e.

$\text{Pa} = \text{N m}^{-2} = \text{kg m s}^{-2} \text{m}^{-2} = \text{kg m}^{-1} \text{s}^{-2}$

Now substitute only the numbers into the equation, i.e.,

$\varrho = \frac{p}{R \, T} = \frac{1.01325 \times 10^{5}}{287 \times 288.15} = 1.225$

Tracking the units in parallel gives

$\frac{\mathrm{Pa}}{\mathrm{J\,kg^{-1}\,K^{-1}} \, \mathrm{K}} = \frac{\mathrm{N\,m^{-2}}}{\mathrm{N\,m\,kg^{-1}\,K^{-1}} \,\mathrm{K}} = \frac{\mathrm{kg\,m^{-1}\,s^{-2}}}{\mathrm{m^{2}\,s^{-2}}} = \mathrm{kg\,m^{-3}}$

The pitfall was attaching the expected units at the end of the calculation instead of carrying the actual units through the process. The correct method is to convert all inputs to consistent SI units at the start, perform the numerical calculation separately, and check units in parallel. In this way, dimensional consistency serves as an error check.

Worked Example #18 – Units confused between energy and power

On a thermodynamics exam, a student is asked: A heater rated at 2 kW operates for 30 minutes. How much energy is delivered? The student substitutes the numbers directly and writes

$Q = P \, t = 2000 \times 30 = 60,000 \, \mathrm{J}$

and then attaches “J” at the end. The answer was marked as being incorrect by the instructor. What went wrong?

The mistakes are twofold. First, the student used 30 minutes as if it were 30 seconds. Second, units were treated as labels to be applied afterward rather than carried through the calculation. As written, the student used 30 as though it were a time in seconds. The product $2{,}000~\mathrm{W} \times 30~\mathrm{min}$ is still an energy, but it must be converted using $1~\mathrm{min}=60~\mathrm{s}$ before the result can be expressed in joules.

The correct procedure is to first convert all quantities to coherent SI units, i.e., 30 min = 1,800 s. Now substitute only the numbers, i.e.,

$Q = P \, t = 2,000 \times 1,800 = 3.6 \times 10^{6}$

Also, track the units in parallel, i.e.,

$(\mathrm{W})(\mathrm{s}) = (\mathrm{J\,s^{-1}})(\mathrm{s}) = \mathrm{J}$

Therefore, the correct result is

$Q = 3.6 \times 10^{6}~\mathrm{J}$

The pitfall was attaching units at the end of the calculation, rather than converting to coherent units first and tracking them in parallel. The correct method ensures dimensional consistency and avoids missing out factors of 60 or other conversion errors.

Worked Example #19 – Converting weight to mass in USC units

On an engineering exam, a student is asked to compute the mass corresponding to a weight of 50 lb on Earth, where the gravitational acceleration is given as $g$ = 32.17 ft/s². The student writes

$m = \frac{W}{g} = \frac{50}{32.17} = 1.55 \quad \mathrm{lb_m}$

and so claims that the mass is 1.55 lb_m. The instructor marked the units as incorrect. Do you know what went wrong?

The error is the use of “lb_m” (pound-mass). Historically, engineers distinguished between pound-force (lb_f) and pound-mass (lb_m), but this convention can easily lead to confusion in engineering calculations. In the slug-based U.S. customary engineering system, the pound is used as a unit of force, and the slug is used as the corresponding unit of mass. The calculation then uses the actual gravitational acceleration $g$ , with $g = 32.17~\mathrm{ft/s^2}$ near sea level. The correct procedure is

$m = \frac{W}{g} = \frac{50}{32.17} = 1.55~\mathrm{slug}$

Remember, do not use “lb_m” in these engineering calculations. The pound is a unit of force, identical to the old “lb_f,” which is now redundant in this context. The consistent unit of mass in USC engineering units is the slug. Using slugs ensures that $F = m a$ is dimensionally consistent without requiring an additional conversion factor.

Remember! The unit of mass in USC engineering units is the slug. The name comes from “sluggish,” meaning inertia or resistance to acceleration. By definition, one slug accelerated at 1 ft/s² requires a force of 1 lb. This choice of unit ensures that Newton’s second law remains dimensionally consistent.

Worked Example #20 – Hydrostatic pressure in SI

A vertical tank is filled with water of density $\varrho$ = 1,000 kg/m³ to a depth of $h = 2.5~\mathrm{m}$ . Compute the gauge pressure at the base of the tank.

The hydrostatic pressure on the bottom of the tank will be

$p = \varrho \, g \, h = 1,000 \times 9.81 \times 2.5 = 24{,}525~\mathrm{N/m}^{2} = 24{,}525~\mathrm{Pa}$

In SI, using consistent base units (kg, m, s) directly produces the correct pressure in pascals.

Worked Example #21 – Hydrostatic pressure in USC

A reservoir contains water of specific weight $\gamma = 62.4~\mathrm{lb/ft^3}$ . Compute the gauge pressure at a depth of $h$ = 12 ft.

The hydrostatic pressure will be

$p = \gamma \, h = 62.4 \times 12 = 749~\mathrm{lb/ft^2}$

At USC, the specific weight (or weight density) is often given directly in lb/ft³, making the hydrostatic calculation straightforward.

Worked Example #22 – Power from torque (SI)

A shaft transmits a torque of 200 N m at a rotational speed of 1,200 rpm. Compute the shaft power in kilowatts.

The angular frequency is given by

$\Omega = 1,200 \times \frac{2\pi}{60} = 125.7~\mathrm{rad/s}$

and so the power is

$P = \tau \, \Omega = 200 \times 125.7 = 25{,}140~\mathrm{W} = 25.1~\mathrm{kW}$

Always convert rpm to rad/s before applying $P = \tau \, \Omega$ , otherwise the answer will be wrong.

Worked Example #23 – Power from torque (USC)

A shaft delivers a torque of 150 ft lb at 1,800 rpm. Compute the shaft power in horsepower.

Convert the rotational speed to angular frequency, i.e.,

$\Omega = 1,800 \times \frac{2\pi}{60} = 188.5~\mathrm{rad/s}$

Because radians are dimensionless, the power can be computed directly in USC units from

$P = \tau \, \Omega$

so

$P = 150 \times 188.5 = 28{,}275~\mathrm{ft\,lb/s}$

Using $1~\mathrm{hp} = 550~\mathrm{ft\,lb/s}$ , then

$P = \frac{28{,}275}{550} = 51.4~\mathrm{hp}$

As a check, the same result can be obtained by converting the torque to SI units using $150~\mathrm{ft\,lb} = 203~\mathrm{N\,m}$ , computing the power in watts, and then converting to horsepower using $1~\mathrm{hp} = 745.7~\mathrm{W}$ .

Worked Example #24 – Pressure from force and area (USC)

A load of 800 lb is applied to a plate with an area of 16 in². Compute the average pressure over the plate in units of psi.

The pressure is force per unit area, so

$p = \frac{F}{A} = \frac{800}{16} = 50~\mathrm{lb/in^2} = 50~\mathrm{psi}$

Using psi directly avoids unnecessary conversions if the force is in lb and the area is in in².

Worked Example #25 – Heating energy in SI

An immersion heater supplies 1.5 kW of power to water for 45 minutes. Compute the energy delivered in MJ.

The time must be converted to seconds, i.e.,

$t = 45 \times 60 = 2,700~\mathrm{s}$

so that the heat added is

$Q = P \, t = 1,500 \times 2,700 = 4.05 \times 10^6~\mathrm{J} = 4.05~\mathrm{MJ}$

Notice that the time in minutes must be converted to seconds when power is given in watts.

Worked Example #26 – Air density from the gas law (SI)

Air at sea level has a pressure $p = 101{,}325~\mathrm{Pa}$ and a temperature $T = 293~\mathrm{K}$ . Use $R = 287~\mathrm{J\,kg^{-1}\,K^{-1}}$ . Compute its density.

The equation of state gives

$\varrho = \frac{p}{R T} = \frac{101,325}{287 \times 293} = 1.20~\mathrm{kg/m^3}$

Using coherent SI units gives density directly in kilograms per cubic meter (kg/m³).

Worked Example #27 – Air density from the gas law (USC)

Air at sea level has a pressure of $p$ = 2,116 lb/ft² and a temperature $T$ = 518.7 ^oR. The gas constant for air is $R$ = 1,716 ft lb slug^-1 ^oR^-1. Use this information to compute the air density.

Using the equation of state, the solution is

$\varrho = \frac{p}{R T} = \frac{2,116}{1,716 \times 518.7} = 0.00237~\mathrm{slugs/ft^3}$

In USC, always use the slug as the unit of mass; the result here is the standard air density in slugs per cubic foot.

Remember! The unit of mass in USC engineering units is the slug. The name comes from “sluggish,” meaning inertia or resistance to acceleration. By definition, one slug accelerated at 1 ft/s² requires a force of 1 lb. This choice of unit ensures that Newton’s second law remains dimensionally consistent.

Worked Example #28 – Compressed air energy calculation (mixed units)

A storage tank with a volume of $\mathcal{V}$ = 250 L contains compressed air at an absolute pressure of $p$ = 8.0 atm and a temperature of 20°C. Estimate the mass of air in the tank and its internal energy, assuming ideal gas behavior with $R$ = 287 J kg^-1 K^-1 and ${c_v}$ = 718 J kg^-1 K^-1.

Convert to consistent SI units using

$p = 8.0 \times 101{,}325 = 810{,}600~\mathrm{Pa}, \qquad \mathcal{V} = 250 \times 10^{-3} = 0.250~\mathrm{m^3}, \quad \text{and} \quad T = 20 + 273.15 = 293.15~\mathrm{K}$

Substituting the numerical values in the equation of state gives

$m = \frac{p \,\mathcal{V}}{R \, T} = \frac{810{,}600 \times 0.250}{287 \times 293.15} = 2.40~\mathrm{kg}$

The internal energy is then

$U = m \, c_v \, T = 2.40 \times 718 \times 293.15 = 5.05 \times 10^{5}~\mathrm{J}$

Now the units can be checked. Recall that

$\mathrm{Pa} = \mathrm{N\,m^{-2}} = \mathrm{kg\,m^{-1}\,s^{-2}} \quad \text{and} \quad \mathrm{J} = \mathrm{N\,m} = \mathrm{kg\,m^{2}\,s^{-2}}$

So

$\frac{\mathrm{Pa}\, \mathrm{m^3}}{\mathrm{J\,kg^{-1}\,K^{-1}}\, \mathrm{K}} = \frac{(\mathrm{kg\,m^{-1}\,s^{-2}})\, \mathrm{m^3}} {(\mathrm{kg\,m^{2}\,s^{-2}})\,\mathrm{kg^{-1}} \, \mathrm{K^{-1}}\, \mathrm{K}} = \mathrm{kg}$

and

$\mathrm{kg}\times \mathrm{J\,kg^{-1}\,K^{-1}}\times \mathrm{K} = \mathrm{J}$

Notice: Converting (atm, L, °C) to coherent SI (Pa, m³, K) before substitution and carrying a parallel unit check guarantees $m$ in kg and $U$ in J.

Worked Example #29 – Dimensional homogeneity

A student is analyzing the physics of centripetal acceleration, given by the formula $a_c = \Omega^2 \, r$ where $a_c$ is the centripetal acceleration, $\Omega$ is the angular velocity, and $r$ is the radius. The student assigns the dimension of length per unit time to angular velocity. What did the student do wrong?

Dimensionally then

$\left[ r \right] = \rm L \quad \text{and} \quad \left[ \Omega \right] = T^{-1}$

so that

$\left[ \Omega^2 r \right] = (\rm T^{-2})(L) = L\,T^{-2}$

which matches the required dimensions of acceleration. If the student assumes that $\left[ \Omega \right] = \rm L\,T^{-1}$ (confusing angular velocity with speed), then

$\left[ \Omega^2 r \right] = (\rm L^2 T^{-2})(L) = L^3 T^{-2}$

which is not an acceleration because it contains an additional factor of L². Angular displacement (radians) is dimensionless, so angular velocity carries units of T^-1. A dimensional check is a good practice because it often reveals mistakes in the use of physical variables and units.

Worked Example #30 – Approval under pressure

An engineer is asked to approve a design for release even though a known technical deficiency has not been fully resolved. Management argues that the risk is small and that delivery deadlines must be met. What is the engineer’s professional obligation?

An engineer’s primary obligation is to public safety and to the integrity of the engineering process. Approval of a design signifies that it meets applicable technical, safety, and regulatory requirements. Schedule pressure or commercial considerations do not justify approving work that is known to be incomplete or potentially unsafe. The engineer should not approve or sign the design release until the deficiency has either been corrected or formally dispositioned through an appropriate technical and safety review process. The appropriate course of action is to document the deficiency, communicate it clearly through technical and management channels, and ensure that the risks are properly evaluated, accepted by the responsible authority if applicable, and resolved before release. Professional responsibility requires independent technical judgment and accountability for decisions affecting safety and performance.

Worked Example #31 – Work outside expertise or competence

An engineer is asked to sign off on structural calculations for a system outside their area of expertise because no specialist is available and the project is behind schedule. Should the engineer proceed?

Engineers are expected to perform and approve work only within their areas of competence. Signing off on technical work without sufficient expertise increases the risk of error and may violate professional standards and regulatory expectations. The engineer should request review by a qualified specialist or obtain sufficient technical support before approval. If such a review is not possible, the engineer should decline to sign off and formally document the need for appropriate technical evaluation. Professional integrity requires recognizing and respecting the limits of one’s competence.

Worked Example #32 – Accurate reporting of results

During testing, an engineer obtains results showing that a system does not fully meet a required performance specification. A supervisor suggests reporting only the favorable results until further testing can be performed. How should the engineer respond?

Engineering results must be reported honestly, completely, and without intentional omission. Selective reporting of favorable data can misrepresent system performance and may lead to unsafe or noncompliant decisions. The engineer should ensure that all relevant test results are documented and communicated accurately. Any uncertainties or limitations should be clearly stated. Professional responsibility requires transparency in technical reporting and adherence to established standards of accuracy and integrity.

Worked Example #33 – Use of unauthorized software

An engineering team is under time pressure to complete an analysis. A team member proposes using unauthorized or pirated software because licensed tools are not immediately available. Is this acceptable professional practice?

Use of unauthorized software violates legal requirements, licensing agreements, and professional standards. Such actions can expose both the engineer and the organization to legal and financial consequences and undermine professional credibility. Engineers must comply with applicable laws, contractual obligations, and organizational policies. Only properly licensed and validated tools should be used for engineering analysis. Professional responsibility includes maintaining ethical and lawful practices even under schedule pressure.

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2026 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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