Adiabatic Flash of a Heated Crude Oil: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2015, Q2

Question 2: Adiabatic Flash of a Heated Crude Oil (Part A — 25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A crude-oil feed enters a flash drum hot and partially vaporises; the drum is adiabatic, so the sensible heat released as the feed cools to the flash temperature supplies the latent heat of the vapour formed.

Quantity	Value
Feed rate / density	10 L/h × 0.85 kg/L = 8.5 kg/h
Feed temperature → flash temperature	510 K → 483 K (ΔT = 27 K)
Feed heat capacity $C_{p,F}$	2.85 kJ/kg·K
Latent heat $\lambda$	291 kJ/kg
Overhead / bottoms density	0.75 / 0.892 kg/L

Find. (a) the mass fraction vaporised $f$; (b) the overhead ($V$) and bottoms ($L$) flow rates.

Figure 2 — Heated crude (510 K) flashed to 483 K, 110 kPa: overhead vapour and bottoms liquid leave in equilibrium.

Approach. Take the flash temperature (483 K) as the datum so both exit streams carry zero sensible enthalpy; the feed’s sensible heat above the datum is exactly the energy available to vaporise the overhead fraction (adiabatic energy balance).

Adiabatic energy balance for the vaporised fraction. Per kg of feed, the heat released on cooling 510 → 483 K equals the latent heat of the fraction $f$ vaporised: $$C_{p,F}(T_F-T_{flash})=f\,\lambda\;\Rightarrow\;f=\frac{2.85(510-483)}{291}=\frac{76.95}{291}=\boxed{0.264\;(26.4\%).}$$ The separate vapour/liquid heat capacities are not needed here because both products leave at the datum temperature; they would enter only if the streams left at different temperatures.
Overhead and bottoms rates. Apply $f$ to the 8.5 kg/h feed: $$V=f\,\dot m_F=0.264(8.5)=2.25\ \text{kg/h},\qquad L=\dot m_F-V=\boxed{2.25\ \text{kg/h vapour, }6.25\ \text{kg/h liquid}.}$$
Volumetric flows (using the stream densities). Converting each mass rate with its density gives the laboratory-scale volumetric split: $$\dot V_{OH}=\frac{2.25}{0.75}=3.0\ \text{L/h},\qquad \dot V_{BTM}=\frac{6.25}{0.892}=7.0\ \text{L/h}.$$ These sum to ≈ 10 L/h, consistent with the 10 L/h charged — a useful closure check. Independently, assuming volume additivity, $1/0.85=f/0.75+(1-f)/0.892$ gives $f\approx0.261$, confirming the 26.4% energy-balance result.

Quantity	Result
(a) Percent vaporised	26.4%
(b) Overhead (vapour)	2.25 kg/h (≈ 3.0 L/h)
(b) Bottoms (liquid)	6.25 kg/h (≈ 7.0 L/h)

Concept & Theory

Topic. Single-stage adiabatic (equilibrium) flash — coupled mass and energy balance.

Key relations. Adiabatic energy balance $C_{p,F}\Delta T=f\lambda$; total mass balance $\dot m_F=V+L$; volumetric conversion $\dot V=\dot m/\rho$.

Why this works. With the drum adiabatic and isobaric, the enthalpy of the entering feed must equal the combined enthalpy of the two exit streams. Referencing all enthalpies to the flash temperature makes the exit streams contribute only their phase-change term (latent heat for the vapour, zero for the liquid), so the feed’s sensible heat above 483 K sets the vaporised fraction directly.

Common pitfall. Trying to use every heat-capacity value supplied — the vapour and liquid $C_p$’s are red herrings for this datum choice — or forgetting that the feed enters above the flash temperature, which is what drives the vaporisation in an otherwise adiabatic drum.

Question 2 of 6: Adiabatic Flash of a Heated Crude Oil

Question 2: Adiabatic Flash of a Heated Crude Oil (Part A — 25 marks)