Unit 02 · Lesson 03 of 09

What does vapor quality mean?

A little vapor can occupy most of the space.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do

Quality measures mass.
Derive the mixture relation.
Invert when another property is given.

Before you touch the controls

For 10% vapor by mass, will vapor occupy 10% of the volume?

An experiment you can change

Vapor mass is not vapor volume

Predict the direction of change, move one control, then explain the response using the equation.

Equal total mass · volume fraction shown

Vapor: 99.4% of volume

Layer ordering is illustrative; height proportions represent occupied volume.

Vapor mass fraction0.1

v=0.0010431+0.1(1.694-0.0010431)=0.17034\,\mathrm{m^3/kg}

Vapor occupies 99.449% of volume but only 10% of mass. The two proportions differ because vapor is much less dense.

Liquid volume per kg mixture0.001 m³

Vapor volume per kg mixture0.169 m³

Model assumptions and units

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

Explore the saturation dome

Saturation temperature100 °CVapor mass fraction0.25

p_{sat}=101.42\,\mathrm{kPa},\quad v=0.41875\,\mathrm{m^3/kg}

The tie line is straight in volume, but the position looks nonlinear because the volume axis is logarithmic. Saturation properties are calculated using IAPWS-IF97.

Saturated water at 100 kPa · IAPWS-IF97
Phase	h (kJ/kg)	u (kJ/kg)	v (m³/kg)
Liquid · 99.606°C	417.436	417.332	0.001043
Vapor · same temperature	2,674.95	2,505.547	1.694023

Quality measures mass.

Quality is vapor mass divided by total mixture mass. It ranges from zero for saturated liquid to one for saturated vapor. It is not defined for compressed liquid or superheated vapor.

x=\frac{m_g}{m_f+m_g}

Derive the mixture relation.

Add liquid and vapor volumes, then divide by total mass. The mass fractions are $1-x$ and x. The same weighted sum applies to specific internal energy and enthalpy because their totals are additive.

v=(1-x)v_f+xv_g=v_f+xv_{fg}

Invert when another property is given.

If specific volume and saturation pressure are known, find both endpoint volumes in the saturation table, then solve the linear relation for quality. Check that the result lies between zero and one; otherwise the mixture assumption was inappropriate.

x=\frac{v-v_f}{v_g-v_f},\quad h=h_f+xh_{fg}

Volume fractions tell a different story.

The vapor volume fraction weights mass by the phase-specific volume. Vapor can dominate the occupied space because its density is far lower. That does not make the vapor mass dominant.

\phi_g=\frac{xv_g}{(1-x)v_f+xv_g}

A worked example

Find quality from a measured volume.

At 100 kPa, take $v=0.20\,\mathrm{m^3/kg}$ and read $v_f,v_g$ from the saturation table.
Use $x=(0.20-v_f)/(v_g-v_f)\approx0.1174$ .
Only about 11.7% of the mass is vapor. Its volume fraction is about 99.5%; the liquid mass occupies a small part of the volume.

Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Ten percent vapor by mass

At 100 kPa, saturated water has quality 0.10. Find specific volume and compare mass fraction with vapor volume fraction.

Hint

Use v = vf + x(vg − vf).

Explore the visual solution

Equal total mass · volume fraction shown

Vapor: 99.4% of volume

Layer ordering is illustrative; height proportions represent occupied volume.

Vapor mass fraction0.1

v=0.0010431+0.1(1.694-0.0010431)=0.17034\,\mathrm{m^3/kg}

Vapor occupies 99.449% of volume but only 10% of mass. The two proportions differ because vapor is much less dense.

Liquid volume per kg mixture0.001 m³

Vapor volume per kg mixture0.169 m³

Model assumptions and units

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

Controls explore variations. The steps below retain the original question’s values.

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

2 · Compare

Half vapor by mass

At 100 kPa, find specific volume for $x = 0.50.$ Is this half the saturated-vapor volume plus half the saturated-liquid volume?

Hint

The mixture relation is a mass-weighted sum.

Explore the visual solution

Equal total mass · volume fraction shown

Vapor: 99.9% of volume

Layer ordering is illustrative; height proportions represent occupied volume.

Vapor mass fraction0.5

v=0.0010431+0.5(1.694-0.0010431)=0.84753\,\mathrm{m^3/kg}

Vapor occupies 99.938% of volume but only 50% of mass. The two proportions differ because vapor is much less dense.

Liquid volume per kg mixture0.001 m³

Vapor volume per kg mixture0.847 m³

Model assumptions and units

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

Controls explore variations. The steps below retain the original question’s values.

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

3 · Challenge the model

Trace vapor in mostly liquid mass

At 100 kPa, use quality 0.01. Find specific volume and explain why even this tiny vapor mass expands the occupied volume greatly.

Hint

Compare vg/vf before calculating.

Explore the visual solution

Equal total mass · volume fraction shown

Vapor: 94.3% of volume

Layer ordering is illustrative; height proportions represent occupied volume.

Vapor mass fraction0.01

v=0.0010431+0.01(1.694-0.0010431)=0.017973\,\mathrm{m^3/kg}

Vapor occupies 94.254% of volume but only 1% of mass. The two proportions differ because vapor is much less dense.

Liquid volume per kg mixture0.001 m³

Vapor volume per kg mixture0.017 m³

Model assumptions and units

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

Controls explore variations. The steps below retain the original question’s values.

Equilibrium saturated water at 100 kPa. Quality is defined only inside the two-phase region, including its saturated endpoints.

x=\frac{m_g}{m_f+m_g},\quad v=v_f+x(v_g-v_f)

Your learning path

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