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Mason equation

Mason equation is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Mason equation rather than just read about it. In short: The Mason equation is an approximate analytical expression for the growth (due to condensation) or evaporation of a water droplet—it is due to the meteorologist B. J.

Key takeaways

  • Mason equation belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Mason equation to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Mason equation from memory before moving on to harder problems.

Reference excerpt

The Mason equation is an approximate analytical expression for the growth (due to condensation) or evaporation of a water droplet—it is due to the meteorologist B. J. Mason. The expression is found by recognising that mass diffusion towards the water drop in a supersaturated environment transports energy as latent heat, and this has to be balanced by the diffusion of sensible heat back across the boundary layer, (and the energy of heatup of the drop, but for a cloud-sized drop this last term is usually small).

Equation In Mason's formulation the changes in temperature across the boundary layer can be related to the changes in saturated vapour pressure by the Clausius–Clapeyron relation; the two energy transport terms must be nearly equal but opposite in sign and so this sets the interface temperature of the drop. The resulting expression for the growth rate is significantly lower than that expected if the drop were not warmed by the latent heat. Thus if the drop has a size r, the inward mass flow rate is given by

d M d t = 4 π r p D v ( ρ 0 − ρ w ) {\displaystyle {\frac {dM}{dt}}=4\pi r_{p}D_{v}(\rho _{0}-\rho _{w})\,}

and the sensible heat flux by

d Q d t = 4 π r p K ( T 0 − T w ) {\displaystyle {\frac {dQ}{dt}}=4\pi r_{p}K(T_{0}-T_{w})\,}

and the final expression for the growth rate is

r d r d t = ( S − 1 ) [ ( L / R T − 1 ) ⋅ L ρ l / K T 0 + ( ρ l R T 0 ) / ( D ρ v ) ] {\displaystyle r{\frac {dr}{dt}}={\frac {(S-1)}{[(L/RT-1)\cdot L\rho _{l}/KT_{0}+(\rho _{l}RT_{0})/(D\rho _{v})]}}}

where

S is the supersaturation far from the drop L is the latent heat K is the vapour thermal conductivity D is the binary diffusion coefficient R is the gas constant

References

Worked examples

Example 1 — a first encounter with Mason equation

Start with the simplest possible case. Write down what Mason equation claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Mason equation before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Mason equation ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Mason equation

In research
Mason equation appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Mason equation in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Mason equation is common in secondary-school and first-year university syllabi. It links to neighbouring topics Atmospheric thermodynamics, Equations, Meteorology stubs, so understanding it makes those chapters shorter.
In everyday life
Look for Mason equation outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

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How to study Mason equation in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Mason equation means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Mason equation out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Mason equation in simple terms?

The Mason equation is an approximate analytical expression for the growth (due to condensation) or evaporation of a water droplet—it is due to the meteorologist B. J.

Why does Mason equation matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Mason equation?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Mason equation.

Tags

  • Atmospheric thermodynamics
  • Equations
  • Meteorology stubs
  • Thermodynamic equations
  • Thermodynamics stubs

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