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71 Worked Examples: Units & Conversion Factors

Many of these worked examples have been fielded as homework problems or exam questions. Many students struggle with units and lose points on exams because they obtain incorrect values in the wrong units.  Study these exemplars to become more proficient in understanding the dimensions of quantities and manipulating units.

Worked Example #1 – 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 #2 – 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 the units of surface tension, \sigma_s, are given as dimensions of 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

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

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

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

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

    \[ \sigma_s \approx 0.072~\mathrm{N/m} \quad \text{(in SI)} \qquad \text{or} \qquad \sigma_s \approx 0.0049~\mathrm{lb/ft} \quad \text{(in USC)}. \]

Worked Example #3 – 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 #4 – Determining base dimensions

Determine the primary (base) dimensions of each of the following parameters from thermodynamics:

  1. Energy, E
  2. Work, W
  3. Power, P
  4. Heat, Q

1. Energy, E, is the ability to do work and is measured in Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. Notice that “foot-pounds” is the USC unit and not “pounds-foot” or “pounds-feet.” Energy has the same units of work (force times distance), so \left[ E \right] = \left( \rm MLT^{-2} \right) \rm L = \rm ML^{2} T^{-2}, which are the base dimensions. Notice that the units of force are obtained from the product of mass and acceleration, i.e., [F] = ( M) (LT^{-2}) = M L T^{-2}.

2. Work, W, is also measured in Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. Work is equivalent to force times distance, so \left[ W \right] = \rm ML T^{-2} L =  M L^{2} \rm T^{-2}, which are the base dimensions of work.

3. Power, P, is the rate of doing work and is measured in Watts (W) in the SI system and foot-pounds per second (ft-lb/s or ft-lb s^{-1} in the USC system. Power is equivalent to a force times distance per unit time (or force times velocity), so \left[ P \right] = \left( \rm MLT^{-2} \right) \rm L T^{-1} = \rm M L^{2} T \rm ^{-3} which are the base dimensions of power. In practice, power is measured in terms of kiloWatts (kW) in SI and horsepower (hp) in USC, where one horsepower is equivalent to 550 ft-lb/s.

4. Heat Q has units of energy (the ability to do work), which have the same units as work, i.e., units of Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. In base dimensions then\left[ Q \right] = \left( \rm MLT^{-2} \right) \rm L = \rm ML^{2} \rm T^{-2}, which are the base dimensions of heat.

Worked Example #5 – Converting units

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

A Watt is a Joule per second. So, one Watt-hour is the equivalent of 3,600 Joules per hour. Therefore, one kilo-watt-hour (kWh) = 3.6 Mega-Joules (MJ). The energy density (in units of kWh/kg) = energy density (in units of MJ/kg)/3.6. Therefore, the energy density of jet fuel in kilowatt-hours per kilogram is approximately 11.9 to 12.5 kWh/kg.

Worked Example #6 – 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 exactly 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 hydrostatic pressure. The sum of the three terms is called total pressure. The Bernoulli equation is a surrogate for the energy equation in a steady, incompressible flow without losses or energy addition.

Each of the terms 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 #7 – Converting to base units

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

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

Notes: A Newton (N) is an SI unit of force, but not a base unit. However, a force is equivalent to a mass times an acceleration, so in terms of base units, a Newton is equivalent to dimensions of M L T^{-2}, so SI units of 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 that power will have units of MLT^{-2} L T^{-1} or M L{^{2}} T{^{-3}} so kg m{^{2}} s{^{-3}}.

Worked Example #8 – 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.

  1. 3.3  l (liters) = 3.3/1,000 = 0.0033 m{^{3}}. There are 1,000 liters in a cubic meter.
  2. 1.2 hrs (hours) = 3,600 \times 1.2 = 4,320 s. There are 3,600 seconds in an hour.
  3. 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.
  4. 12.8 gals (US gallons) = 12.8 \times 0.13368 = 1.711 ft{^{3}}. One gallon equals 0.13368 ft{^{3}}.
  5. 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 1,000 square centimeters in a square meter.

Notes: Converting units takes some work, but it is essential to do it 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}, so 550 is the conversion factor from hp to base USC units.

Worked Example #9 – 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~\mathrm{mmHg}, T = 15^{\circ}\mathrm{C}, and R = 287~\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/m^3” 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 never converted from mmHg to pascals, and the temperature was used in 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.,

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

Track the units in parallel:

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

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. This way, dimensional consistency acts as an error check.

Worked Example #10 – Units confused between energy and power

On a thermodynamics exam, a student is asked: A heater rated at 2~\mathrm{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 a label to be pasted on afterward rather than being carried through the calculation. As written, the student actually computed 2,000~\mathrm{W} \times 30~\mathrm{min}, which is not in units of Joules.

The correct procedure is first to convert all quantities into coherent SI units, i.e., 30~\mathrm{min} = 1,800~\mathrm{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 instead of 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 #11 – 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~\mathrm{lb} on Earth, where the gravitational acceleration is given as g = 32.17~\mathrm{ft\,s^{-2}}. 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 units were marked as being incorrect by the instructor. Do you know what went wrong?

The error is the use of “lb_m” (pound-mass). Historically, engineers tried to distinguish between pound-force (lb_f) and pound-mass (lb_m) by introducing the conversion factor g. This practice is now deprecated. In modern engineering usage, “lb” is always taken as a unit of force, and the correct unit of mass in the USC system is the “slug.” The correct procedure is

    \[ m = \frac{W}{g} = \frac{50}{32.17} = 1.55~\mathrm{slugs} \]

Remember, do not use “lb_m” in engineering calculations. The pound is a unit of force, identical to the old “lb_f,” which is now considered redundant. The consistent mass unit in USC is the “slug.” Using the slug (or sl or slugs) ensures that F = m a is dimensionally consistent without the need for a conversion factor.

Remember! The unit of mass in USC engineering units is the slug. The name comes from  “sluggish,” referring to inertia or resistance to acceleration. By definition, one slug  accelerated at 1~\mathrm{ft/s^2} requires a force of 1~\mathrm{lb}. This choice of unit ensures that Newton’s second law remains dimensionally consistent.

Worked Example #12 – Hydrostatic pressure in SI

A vertical tank is filled with water of density \rho = 1000~\mathrm{kg/m^3} to a depth of h = 2.5~\mathrm{m}. Compute the pressure at the base of the tank.

The hydrostatic pressure on the bottom of the tank will be

    \[ p = \rho g h = 1000 \times 9.81 \times 2.5 = 24{,}525~\mathrm{N/m}^{2}= 24{,}525~\mathrm{Pa} \]

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

Worked Example #13 – Hydrostatic pressure in USC

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

The hydrostatic pressure will be

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

In USC, specific weight (or weight density) in units of lb/ft^3 is often given directly, making the hydrostatic calculation straightforward.

Worked Example #14 – Power from torque (SI)

A shaft transmits a torque of 200~\mathrm{N\,m} at a rotational speed of 1,200~\mathrm{rpm}. Compute the 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 #15 – Power from torque (USC)

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

Convert rotational speed using

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

Convert torque using

    \[ 150~\mathrm{ft\,lb} = 150 \times 1.356 = 203~\mathrm{N\,m} \]

The power in SI will be

    \[ P = 203 \times 188.5 = 38{,}300~\mathrm{W} \]

Convert to hp using

    \[ P = \frac{38{,}300}{745.7} = 51.4~\mathrm{hp} \]

When mixing USC and SI, torque must be expressed in N m for consistency; the final answers can then be converted to horsepower if desired by using the conversion factor that 1 hp =746 W = 0.746 kW.

Worked Example #16 – Pressure from force and area (USC)

A load of 800~\mathrm{lb} is applied to a plate of area 16~\mathrm{in^2}. Compute the average pressure over the plate in units of psi.

The pressure is force per unit area, so

    \[ p = \frac{F}{A} = \frac{800}{16} = 50~\mathrm{psi} \]

Using psi directly avoids unnecessary conversions if both force and area are in lb and in².

Worked Example #17 – Heating energy in SI

An immersion heater supplies 1.5~\mathrm{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

    \[ Q = P t = 1500 \times 2700 = 4.05 \times 10^6~\mathrm{J} = 4.05~\mathrm{MJ} \]

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

Worked Example #18 – 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}. Using R = 287~\mathrm{J\,kg^{-1}\,K^{-1}}. Compute its density.

    \[ \rho = \frac{p}{R T} = \frac{101325}{287 \times 293} = 1.20~\mathrm{kg/m^3} \]

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

Worked Example #19 – Air density from the gas law (USC)

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

The solution is

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

In USC, always use the unit of slug for 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,” referring to inertia or resistance to acceleration. By definition, one slug  accelerated at 1~\mathrm{ft/s^2} requires a force of 1~\mathrm{lb}. This choice of unit ensures that Newton’s second law remains dimensionally consistent.

Worked Example #20 – Compressed air energy calculation (mixed units)

A storage tank  with a volume of \mathcal{V} = 250~\mathrm{L} contains compressed air at a pressure of p = 8.0~\mathrm{atm} and a temperature of 20^{\circ}\mathrm{C}. Estimate the mass of air in the tank and its total energy content (internal energy), assuming ideal gas behavior with R = 287~\mathrm{J\,kg^{-1}\,K^{-1}} and c_v = 718~\mathrm{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 ideal gas law 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, m^3, K) before substitution and carrying a parallel unit check guarantees m in kg and U in J.

Worked Example #21 – 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 dimensions of length per unit time to the 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 carries an extra factor of L^2. Angular displacement (radians) is dimensionless, so angular velocity carries units of  T^{-1}. A dimensional check is a good practice because it generally always reveals mistakes.

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Introduction to Aerospace Flight Vehicles Copyright © 2022–2025 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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