Adiabatic Flash of a Preheated Crude Oil: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2019, Q3

Question A3: Adiabatic Flash of a Preheated Crude Oil (Part A — equal value)

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. Crude preheated to 237 °C is charged at $\dot V=0.167$ L/min to a flash zone held at 483 K (209.85 °C) and 1.1 atm; vapour and liquid leave in equilibrium at the flash temperature. Flashing is adiabatic (no external heat).

Stream	Density (kg/L)	$C_p$ (kJ/kg·K)
Feed crude	0.85	2.85
Vaporised crude	0.75	2.89
Unvaporised crude	0.892	2.68

Latent heat of vaporisation $\lambda=291$ kJ/kg.

Find. (a) the weight-percent of the feed that vaporises; (b) the overhead (vapour) and bottoms (liquid) flow rates.

Figure A3 — The preheated feed drops into the flash drum; its sensible heat above the 483 K flash temperature vaporises part of the charge, giving an equilibrium overhead vapour and a bottoms liquid, both at 483 K.

Approach. Treat the flash as an adiabatic (isenthalpic) operation: equate the feed’s sensible heat above the flash temperature to the latent heat of the fraction vaporised, then convert the mass split to volumetric flows with the product densities.

Feed mass flow. $$\dot m_F=\rho_F\dot V=0.85(0.167)=0.1420\ \text{kg/min}.$$
Sensible driving force. The feed enters 27.15 °C above the flash temperature ($237-209.85$); this stored sensible heat is what boils off the vapour, since no heat is supplied.
Adiabatic energy balance. Referencing both products to liquid at the flash temperature (so the liquid product carries zero excess enthalpy and the vapour product carries $\lambda$), the feed sensible heat equals the vaporisation duty: $$\dot m_F C_{p,F}(T_F-T_{flash})=f\,\dot m_F\,\lambda \;\Rightarrow\; f=\frac{C_{p,F}\,\Delta T}{\lambda}=\frac{2.85(27.15)}{291}=\boxed{0.266}.$$ So 26.6 wt% of the feed flashes. (The vapour/liquid product $C_p$ values are not needed here because both products leave at the reference temperature.)
Overhead and bottoms amounts (part b). By mass, $\dot m_V=0.266(0.1420)=0.0377$ and $\dot m_L=0.1420-0.0377=0.1042$ kg/min. Converting with the product densities, $$\dot V_V=\frac{0.0377}{0.75}=0.0503\ \text{L/min},\qquad \dot V_L=\frac{0.1042}{0.892}=0.117\ \text{L/min}.$$

Quantity	Result
(a) Fraction vaporised	26.6 wt %
(b) Overhead (vapour)	0.0377 kg/min (0.0503 L/min)
(b) Bottoms (liquid)	0.1042 kg/min (0.117 L/min)

Concept & Theory

Topic. Adiabatic (isenthalpic) equilibrium flash of a preheated liquid.

Key relations. Overall energy balance $\Delta H=0$ at constant $P$; sensible heat $\dot m C_p\Delta T$; latent heat $f\dot m\lambda$; density conversions $\dot V=\dot m/\rho$.

Why this works. In an adiabatic flash the enthalpy of the outlet two-phase mixture equals that of the feed, so the feed’s excess sensible enthalpy above the flash temperature is exactly the energy available to vaporise part of the charge. Choosing the reference state as saturated liquid at the flash temperature collapses the balance to a single sensible-vs-latent equality and makes the vaporised fraction fall out directly.

Common pitfall. Measuring $\Delta T$ from the wrong end (using 483 K as if it were °C, or forgetting the 273.15 offset) mis-sizes the flash; and adding the vapour-product sensible term double-counts, since the vapour already leaves at the reference temperature.

Source. Felder, Rousseau & Bullard, Elementary Principles §8.4 (single-stage flash / isothermal and adiabatic vaporisation energy balances).

Question 3 of 6: Adiabatic Flash of a Preheated Crude Oil

Part A — Process Mass & Energy Balances

Question A3: Adiabatic Flash of a Preheated Crude Oil (Part A — equal value)