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Units and Dimensions — Comprehensive Study Note

Why units and dimensions matter

Units and dimensions are foundational in bioprocess and biochemical engineering. They are used in calculations involving concentration, volume, mass, length, reaction rate, and bioreactor operation. Dimensional analysis helps you:

convert quantities safely;

derive the dimensions of unfamiliar quantities;

check whether an equation is physically sensible;

identify errors in powers, denominators, and unit conversions.

The safest conversion method is to convert both quantities through a common base unit, such as metre for length or gram/kilogram for mass.

Length units and conversions

The metric length scale is:

1\,\mathrm{m}=10\,\mathrm{dm}=100\,\mathrm{cm}=1000\,\mathrm{mm}

Continuing toward smaller units:

1\,\mathrm{m}=10^6\,\mu\mathrm{m}=10^9\,\mathrm{nm}=10^{10}\,\mathrm{\mathring{A}}=10^{12}\,\mathrm{pm}=10^{15}\,\mathrm{fm}

Therefore:

\(1\,\mu\mathrm{m}=10^{-6}\,\mathrm{m}\)

\(1\,\mathrm{nm}=10^{-9}\,\mathrm{m}\)

\(1\,\mathrm{nm}=10^{-3}\,\mu\mathrm{m}\)

\(1\,\mathrm{\mathring{A}}=10^{-10}\,\mathrm{m}\)

\(1\,\mathrm{\mathring{A}}=10^{-4}\,\mu\mathrm{m}\)

Example: convert nanometres to micrometres

Because \(1\,\mathrm{m}=10^9\,\mathrm{nm}=10^6\,\mu\mathrm{m}\),

1\,\mathrm{nm}\times\frac{10^6\,\mu\mathrm{m}}{10^9\,\mathrm{nm}}

=10^{-3}\,\mu\mathrm{m}

The \(\mathrm{nm}\) units cancel, leaving the answer in \(\mu\mathrm{m}\). This common-base-unit approach works for any pair of length units.

Metric-prefix pattern

For smaller quantities, the prefixes follow powers of ten:

milli: \(10^{-3}\)

micro: \(10^{-6}\)

nano: \(10^{-9}\)

pico: \(10^{-12}\)

femto: \(10^{-15}\)

For larger quantities:

kilo: \(10^3\)

mega: \(10^6\)

giga: \(10^9\)

tera: \(10^{12}\)

peta: \(10^{15}\)

Mass units and conversions

The mass scale is:

1\,\mathrm{g}=10\,\mathrm{dg}=100\,\mathrm{cg}=1000\,\mathrm{mg}

For still smaller masses:

1\,\mathrm{g}=10^6\,\mu\mathrm{g}=10^9\,\mathrm{ng}=10^{12}\,\mathrm{pg}=10^{15}\,\mathrm{fg}

Important relationships include:

\(1\,\mathrm{mg}=10^{-3}\,\mathrm{g}\)

\(1\,\mu\mathrm{g}=10^{-6}\,\mathrm{g}\)

\(1\,\mathrm{ng}=10^{-9}\,\mathrm{g}\)

\(1\,\mathrm{pg}=10^{-12}\,\mathrm{g}\)

\(1\,\mathrm{fg}=10^{-15}\,\mathrm{g}\)

The ångström is a length unit, so it does not belong on the mass scale.

Area and volume from length

Dimensions change when a length is squared or cubed.

Area

For a square with side length \(l\):

A=l\times l=l^2

Therefore, area has dimensions \(L^2\).

Volume of a cube

For a cube with side length \(l\):

V=l\times l\times l=l^3

Therefore, volume has dimensions \(L^3\).

A key laboratory relationship is:

1\,\mathrm{cm^3}=1\,\mathrm{mL}

This follows from the practical definition that a cube measuring \(1\,\mathrm{cm}\times1\,\mathrm{cm}\times1\,\mathrm{cm}\) contains \(1\,\mathrm{mL}\) of liquid.

Volume of a sphere

For a sphere of radius \(r\):

V=\frac{4}{3}\pi r^3

Since radius is a length, the result has dimensions \(L^3\). For \(r=1\,\mathrm{cm}\):

V=\frac{4}{3}\pi(1\,\mathrm{cm})^3

Important correction: if the video states \(4\pi r^3\) at one point, that is missing the factor \(\frac{1}{3}\). The correct sphere-volume formula is \(\frac{4}{3}\pi r^3\).

Fundamental dimensions and SI base quantities

Dimensions describe the physical nature of a quantity independently of the particular unit used.

A formula can be analyzed by replacing each physical quantity with its fundamental dimension. This is often easier and safer than manipulating named units directly.

Mechanical quantities and their dimensions

Velocity

Velocity is distance travelled per unit time:

v=\frac{\text{distance}}{\text{time}}

Thus,

[v]=\frac{L}{T}=LT^{-1}

Acceleration

Acceleration is change in velocity per unit time:

a=\frac{\Delta v}{\Delta t}

Since \([v]=LT^{-1}\),

[a]=\frac{LT^{-1}}{T}=LT^{-2}

Momentum

Momentum is mass multiplied by velocity:

p=mv

Therefore,

[p]=M\times LT^{-1}=MLT^{-1}

Force

Newton's second law gives:

F=ma

Therefore,

[F]=M\times LT^{-2}=MLT^{-2}

Work, heat, and energy

Work is force multiplied by displacement:

W=F\times l

Therefore,

[W]=MLT^{-2}\times L=ML^2T^{-2}

Work, heat, and energy have the same dimensions:

[W]=[Q]=[E]=ML^2T^{-2}

They are different physical ideas, but dimensionally equivalent.

Power

Power is work done per unit time:

P=\frac{W}{t}

Therefore,

[P]=\frac{ML^2T^{-2}}{T}=ML^2T^{-3}

These derivations illustrate the central method: begin with the defining equation, substitute the dimensions of the variables, and simplify the powers.

Bioprocess and transport quantities

Pressure

Pressure is force per unit area:

P=\frac{F}{A}

Using \([F]=MLT^{-2}\) and \([A]=L^2\):

[P]=\frac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}

Density

Density is mass per unit volume:

\rho=\frac{M}{V}

Because volume has dimensions \(L^3\):

[\rho]=\frac{M}{L^3}=ML^{-3}

Surface tension

Surface tension is force per unit length:

\gamma=\frac{F}{l}

Therefore,

[\gamma]=\frac{MLT^{-2}}{L}=MT^{-2}

Frequency

Frequency is the reciprocal of time:

f=\frac{1}{t}

Thus,

[f]=T^{-1}

Molar concentration

Molar concentration is amount of substance per unit volume:

C=\frac{n}{V}

Therefore,

[C]=\frac{N}{L^3}=NL^{-3}

Electric charge

Electric charge is current multiplied by time:

q=It

Therefore,

[q]=IT

Voltage

Voltage is work done per unit charge:

V=\frac{W}{q}

Using \([W]=ML^2T^{-2}\) and \([q]=IT\):

[V]=\frac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}

Entropy

The video introduces entropy through heat transferred relative to a temperature change:

S=\frac{Q}{\Delta T}

Since \([Q]=ML^2T^{-2}\):

[S]=\frac{ML^2T^{-2}}{\Theta}=ML^2T^{-2}\Theta^{-1}

Entropy is described in the video in connection with temperature-related randomness.

Diffusion coefficient

A diffusion coefficient has dimensions of area per unit time:

D=\frac{\text{area}}{\text{time}}

Therefore,

[D]=L^2T^{-1}

Heat flux

Heat flux is the rate of heat transfer per unit area:

q''=\frac{\text{heat}}{\text{time}\times\text{area}}

Using heat dimensions \(ML^2T^{-2}\), time \(T\), and area \(L^2\):

[q'']=\frac{ML^2T^{-2}}{T\times L^2}=MT^{-3}

Equivalently, this may be written as:

[q'']=ML^0T^{-3}

because \(L^0=1\).

Worked dimensional-analysis strategy

Use this five-step process for almost any question:

Write the defining equation. For example, \(P=F/A\).

Replace each quantity with its fundamental dimension. For example, \([F]=MLT^{-2}\) and \([A]=L^2\).

Keep numerator and denominator powers separate. This prevents sign mistakes.

Subtract exponents when dividing. For example, \(L/L^2=L^{-1}\).

Check whether the final dimensions are physically reasonable.

Example for pressure:

[P]=\frac{[F]}{[A]}

=\frac{MLT^{-2}}{L^2}

=ML^{1-2}T^{-2}

=ML^{-1}T^{-2}

The most common errors are forgetting that area contributes \(L^2\), volume contributes \(L^3\), or changing the sign of an exponent incorrectly when a quantity is in the denominator.

Homework: specific heat capacity

Specific heat capacity is defined as heat energy required per unit mass per unit temperature change:

c=\frac{\text{heat energy}}{\text{mass}\times\Delta T}

Using the dimensions of heat energy:

[Q]=ML^2T^{-2}

mass:

[M]=M

and temperature change:

[\Delta T]=\Theta

we obtain:

[c]=\frac{ML^2T^{-2}}{M\Theta}

Cancel \(M\):

[c]=L^2T^{-2}\Theta^{-1}

So the dimensional formula for specific heat capacity is:

\boxed{[c]=L^2T^{-2}\Theta^{-1}}

This is a useful example of why a quantity can have dimensions different from the quantity in its numerator: dividing by mass removes the mass dimension.

Common mistakes to avoid

Converting directly without a common reference unit: use metre, gram, or another common base unit first.

Forgetting powers: area is \(L^2\), not \(L\); volume is \(L^3\), not \(L\).

Handling denominators incorrectly: dividing by \(L^2\) produces \(L^{-2}\) in the result.

Mixing mass and length prefixes: an ångström belongs to length, not mass.

Dropping units before cancellation: write the conversion fraction so unwanted units visibly cancel.

Memorizing instead of deriving: when uncertain, return to the defining equation.

Using the incorrect sphere formula: the correct expression is \(V=\frac{4}{3}\pi r^3\), not \(4\pi r^3\).

Confusing unit and dimension: metre is a unit of length; \(L\) is the dimension of length.

One-page revision sheet

Core geometry

A=l^2,\qquad V{\text{cube}}=l^3,\qquad V{\text{sphere}}=\frac{4}{3}\pi r^3

1\,\mathrm{cm^3}=1\,\mathrm{mL}

Core mechanical dimensions

[v]=LT^{-1},\quad [a]=LT^{-2},\quad [p]=MLT^{-1}

[F]=MLT^{-2},\quad [W]=[Q]=[E]=ML^2T^{-2},\quad [P]=ML^2T^{-3}

Core bioprocess dimensions

[\text{pressure}]=ML^{-1}T^{-2}

[\rho]=ML^{-3},\quad [\gamma]=MT^{-2},\quad [f]=T^{-1}

[C]=NL^{-3},\quad [q]=IT,\quad [V]=ML^2T^{-3}I^{-1}

[S]=ML^2T^{-2}\Theta^{-1},\quad [D]=L^2T^{-1},\quad [q'']=MT^{-3}

Final study rule

Whenever you forget a dimensional formula, do not guess. Write the physical definition, substitute the fundamental dimensions, and simplify the exponents carefully.