6.2 Bubble Point, Dew Point, and Flash Concepts

6.2 Bubble Point, Dew Point, and Flash Concepts

Quick Answer: Bubble point = first vapor from a liquid of known (x); dew point = first liquid from a vapor of known (y). An isothermal flash at fixed T,P splits feed F into vapor V and liquid L using balances plus $K_i = y_i/x_i$. Relative volatility $\alpha_{ij} = K_i/K_j$ drives how hard distillation must work.

Section 6.1 gave equilibrium criteria. This section is the process calculation layer: when does a liquid start to boil, when does a vapor start to condense, and what happens in a single-equilibrium-stage flash drum—the unit that appears in flowsheets from LNG trains to condensate stabilization in Qatar’s upstream and midstream facilities.

Bubble Point and Dew Point Meanings

Consider a mixture of fixed overall composition held at a given pressure (or temperature). As you heat a compressed liquid, you eventually hit the bubble-point temperature $T_{\mathrm{bub}}$: the temperature where the first infinitesimal vapor bubble appears. As you cool a superheated vapor, you hit the dew-point temperature $T_{\mathrm{dew}}$: the temperature where the first liquid droplet appears. Between dew and bubble at fixed P lies the two-phase region (for a mixture, $T_{\mathrm{dew}} > T_{\mathrm{bub}}$ in the usual isobaric diagram sense for a given overall z—careful with which composition is held fixed).

Equivalently at fixed T you speak of bubble pressure and dew pressure.

Ideal binary equations (Raoult)

Bubble pressure at given T, x:

$$P_{\mathrm{bub}} = \sum_i x_i P_i^{\mathrm{sat}}(T)$$

Then $y_i = x_i P_i^{\mathrm{sat}}/P_{\mathrm{bub}}$.

Dew pressure at given T, y:

$$\frac{1}{P_{\mathrm{dew}}} = \sum_i \frac{y_i}{P_i^{\mathrm{sat}}(T)} \quad \Rightarrow \quad P_{\mathrm{dew}} = \left(\sum_i \frac{y_i}{P_i^{\mathrm{sat}}}\right)^{-1}$$

Then $x_i = y_i P_{\mathrm{dew}} / P_i^{\mathrm{sat}}$.

Bubble temperature at fixed P and x: find T such that $\sum x_i P_i^{\mathrm{sat}}(T) = P$ (iterative if vapor pressures are nonlinear in T). Dew temperature at fixed P and y: find T such that $\sum y_i P / P_i^{\mathrm{sat}}(T) = 1$.

Multicomponent memory hooks

• Bubble calculations start from known liquid x.
• Dew calculations start from known vapor y.
• For a pure fluid, bubble and dew coincide: boiling point at that P.
• For mixtures, the first vapor is richer in lights than the liquid; the first liquid is richer in heavies than the vapor.

Simple Isothermal Flash

A classic flash drum receives feed F (mole basis) of composition $z_i$ at conditions that produce two phases at specified T and P. Outlet streams:

• Vapor rate (V), composition $y_i$
• Liquid rate (L), composition $x_i$

With vapor fraction $\psi = V/F$ (moles vapor per mole feed):

Overall: (F = V + L) or $1 = \psi + (L/F)$.

Component balance:

$$z_i F = y_i V + x_i L$$

Equilibrium:

$$K_i(T,P,\text{composition}) = \frac{y_i}{x_i}$$

Combining yields the Rachford–Rice structure (you need the idea, not necessarily the name):

$$\sum_i \frac{z_i (K_i - 1)}{1 + \psi (K_i - 1)} = 0$$

Solve for $\psi$ between 0 and 1 when the feed is truly in the two-phase region, then:

$$x_i = \frac{z_i}{1 + \psi(K_i - 1)}, \quad y_i = K_i x_i$$

Qualitative material-balance framing for MCQs: lights prefer the vapor (high $K_i$); heavies prefer the liquid (low $K_i$). Raising flash T at fixed P increases vapor fraction and tends to enrich vapor in heavies slightly as more material boils—but equilibrium still enforces K-values. Lowering P at fixed T also increases vapor fraction (same qualitative Le Chatelier-style “more volume for vapor”).

Worked numerical flash (binary, constant K)

Feed: F = 100 mol, $z_A = 0.60$, $z_B = 0.40$. At flash T,P: $K_A = 2.0$, $K_B = 0.5$.

Find $\psi$, then x and y.

Rachford–Rice for binary can be solved directly. Try $\psi = 0.40$:

$$\frac{0.60(2-1)}{1+0.40(1)} + \frac{0.40(0.5-1)}{1+0.40(-0.5)} = \frac{0.60}{1.4} + \frac{-0.20}{0.8} = 0.429 - 0.250 = 0.179$$

Still positive; try higher $\psi$ (more vapor). At $\psi = 0.60$:

$$\frac{0.60}{1+0.60} + \frac{-0.20}{1+0.60(-0.5)} = \frac{0.60}{1.6} + \frac{-0.20}{0.70} = 0.375 - 0.286 = 0.089$$

At $\psi = 0.80$:

$$\frac{0.60}{1.8} + \frac{-0.20}{1-0.40} = 0.333 - 0.333 = 0$$

So $\psi = 0.80$: V = 80 mol, L = 20 mol.

$$x_A = \frac{0.60}{1+0.80(1)} = \frac{0.60}{1.8} = 0.333, \quad y_A = 2.0 \times 0.333 = 0.667$$ $$x_B = 1 - 0.333 = 0.667, \quad y_B = 0.5 \times 0.667 = 0.333$$

Check component A: $y_A V + x_A L = 0.667(80) + 0.333(20) = 53.3 + 6.7 = 60 = z_A F$ ✓

Relative Volatility and Separation

Relative volatility of i with respect to j:

$$\alpha_{ij} = \frac{K_i}{K_j} = \frac{y_i/x_i}{y_j/x_j}$$

For ideal binary Raoult at low P:

$$\alpha_{AB} = \frac{P_A^{\mathrm{sat}}}{P_B^{\mathrm{sat}}}$$

(approximately independent of composition if both are ideal).

A single flash is only one equilibrium stage. Distillation (Chapter 9) stacks many stages with reflux so that relative volatility compounds into high-purity products. Exam logic chain:

1. VLE ⇒ K-values
2. $\alpha = K_{\mathrm{light}}/K_{\mathrm{heavy}}$
3. Large $\alpha$ ⇒ fewer stages / easier column
4. Flash alone rarely meets fuel-spec or polymer-grade purity

Linking flash to bubble/dew

• If flash P is above bubble P at that T for composition z, feed is liquid (no vapor).
• If flash P is below dew P, feed is all vapor.
• Two-phase flash requires operating between dew and bubble pressures (at fixed T) for that overall composition.

Worked Bubble vs Dew Contrast

Same ideal binary as Section 6.1 style: at T with $P_A^{\mathrm{sat}} = 120\,\mathrm{kPa}$, $P_B^{\mathrm{sat}} = 60\,\mathrm{kPa}$.

Liquid z = x_A = 0.40 (compressed liquid heated at P = 84 kPa):

• Bubble P at this T is 84 kPa (from earlier). At P = 84 kPa the liquid is at bubble point; first vapor has $y_A = 0.571$.

Vapor with y_A = 0.40 at same T:

$$P_{\mathrm{dew}} = \left(\frac{0.40}{120} + \frac{0.60}{60}\right)^{-1} = (0.003333 + 0.010)^{-1} = (0.013333)^{-1} = 75\,\mathrm{kPa}$$

First liquid: $x_A = y_A P_{\mathrm{dew}}/P_A^{\mathrm{sat}} = 0.40 \times 75 / 120 = 0.25$.

Exam takeaway: same numerical composition 0.40 means different things as liquid x vs vapor y—bubble and dew pressures differ (84 vs 75 kPa here).

UPDA Strategy

1. Underline bubble vs dew vs flash in the stem.
2. Ideal ⇒ Raoult forms; given K-values ⇒ use them directly in flash balances.
3. Always check $\sum x_i = 1$, $\sum y_i = 1$, and component closures.
4. Relate $\alpha$ to distillation difficulty; do not claim one flash equals a full column.
5. Phase rule: binary VLE has F = 2—specifying T and P fixes equilibrium x–y pairs along the isotherm/isobar, which is why flash needs z and energy/material specs to fix $\psi$.

Section 6.3 leaves physical VLE and treats chemical reaction equilibrium—another “equilibrium vs rate” theme that Domain D (reactors) will reuse with kinetics.