Steam Distillation vs. Vacuum Distillation Selection

Reference ID: MET-0F25 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The calculation compares steam distillation and vacuum distillation for the separation of a low‑boiling organic component (e.g., limonene) from water. In process engineering, these two options are evaluated to determine the most economical operating pressure, energy consumption, and equipment sizing for a given product throughput. The methodology is applicable to batch or continuous reactors where the product rate, feed composition, and thermal properties are known, and where the choice of operating pressure directly influences the required reboiler duty and auxiliary utilities (steam or electricity).

Methodology & Formulas

1. Physical Constants and Input Variables

The following symbols are used throughout the derivations: R – universal gas constant (kJ·mol⁻¹·K⁻¹), MW_water – molecular weight of water (kg·kmol⁻¹), MW_oil – molecular weight of the organic component (kg·kmol⁻¹), ΔH_vap – molar heat of vaporisation of the oil (kJ·mol⁻¹), T_max – maximum allowable temperature (°C), P_steam – supplied steam pressure (bar), h_steam – specific enthalpy of saturated steam (kJ·kg⁻¹), C_steam – steam cost (€/GJ), C_elec – electricity cost (€/kWh), and R_ref – external reflux ratio (dimensionless).

2. Antoine Equation for Saturation Pressure

The saturation pressure of a component at temperature T (°C) is expressed by the Antoine correlation:

\[ p^{*}_{i}(T) = 10^{\,A_{i} - \dfrac{B_{i}}{T + C_{i}}}\;\text{mmHg} \]

Conversion to bar is performed with the factor 0.00133322 bar · mmHg⁻¹:

\[ p^{*}_{i,\; \text{bar}}(T) = p^{*}_{i}(T)\times 0.00133322 \]

3. System Pressure for Steam Distillation

The target boiling temperature for steam distillation is selected as T_b,steam (°C) such that T_b,steam < T_max. The corresponding saturation pressures of water and oil are:

\[ p^{*}_{\text{water}} = p^{*}_{\text{water, bar}}(T_{b,\;steam}),\qquad p^{*}_{\text{oil}} = p^{*}_{\text{oil, bar}}(T_{b,\;steam}) \]

The total system pressure is the sum of the partial pressures:

\[ P_{\text{sys,steam}} = p^{*}_{\text{water}} + p^{*}_{\text{oil}} \]

4. Vapor‑Phase Composition and Required Steam Flow

The mole fraction of oil in the vapor phase is:

\[ y_{\text{oil}} = \frac{p^{*}_{\text{oil}}}{P_{\text{sys,steam}}} \]

The corresponding mass ratio of water to oil in the vapor is obtained from the ideal‑gas relation and molecular‑weight conversion:

\[ \text{mass\_ratio} = \frac{y_{\text{oil}}}{1 - y_{\text{oil}}}\;\times\;\frac{MW_{\text{water}}}{MW_{\text{oil}}} \]

The required steam mass flow rate to achieve the desired product rate F_p (kg·h⁻¹) is:

\[ \dot{m}_{\text{steam}} = F_{p}\;\times\;\text{mass\_ratio} \]

5. Steam‑Side Reboiler Duty and Cost

The thermal duty supplied by steam is:

\[ Q_{\text{steam}} = \dot{m}_{\text{steam}}\; \times\; h_{\text{steam}}\quad (\text{kJ·h}^{-1}) \]

Conversion to gigajoules per hour and cost calculation:

\[ Q_{\text{steam,GJ}} = Q_{\text{steam}}\times g2GJ,\qquad C_{\text{steam}} = Q_{\text{steam,GJ}}\times C_{\text{steam}} \]

6. Vacuum Distillation – System Pressure from Clausius‑Clapeyron

For a target boiling temperature T_b,vac (°C) the required system pressure is obtained from the integrated Clausius‑Clapeyron relation:

\[ \ln P_{\text{sys,vac}} = \frac{\Delta H_{\text{vap}}}{R}\!\left(\frac{1}{T_{\text{ref}}}-\frac{1}{T_{b,\;vac,K}}\right) + \ln P_{\text{ref}} \]

where T_ref and P_ref are a reference temperature (K) and pressure (bar) for which the vapor pressure is known, and T_b,vac,K = T_b,vac + 273.15 K.

7. Vacuum Reboiler Duty

The specific latent heat per kilogram of oil is:

\[ \lambda = \frac{\Delta H_{\text{vap}}\times 1000}{MW_{\text{oil}}}\quad (\text{kJ·kg}^{-1}) \]

The total reboiler duty, accounting for the external reflux ratio, is:

\[ Q_{\text{vac}} = F_{p}\;(1+R_{\text{ref}})\;\lambda\quad (\text{kJ·h}^{-1}) \]

Expressed as electrical power (kW) and cost:

\[ P_{\text{vac}} = \frac{Q_{\text{vac}}}{kWh2kJ},\qquad C_{\text{vac}} = P_{\text{vac}}\; \times\; C_{\text{elec}} \]

8. Vacuum Pump Power Requirement

An air leakage rate \dot{m}_{\text{air}} (kg·h⁻¹) is assumed. Its molar flow is:

\[ \dot{n}_{\text{air}} = \frac{\dot{m}_{\text{air}}}{MW_{\text{air}}} \]

The volumetric flow at suction conditions (temperature T_suction, pressure P_suction) is approximated by the ideal‑gas law using the standard molar volume (22.4 m³·kmol⁻¹ at 273.15 K and 1.013 bar):

\[ V_{\text{suction}} = \dot{n}_{\text{air}}\; 22.4\; \frac{T_{\text{suction}}}{273.15}\; \frac{1.013}{P_{\text{suction}}} \]

The pump shaft power is estimated with a proportionality constant (3.5 kW·m³⁻¹·h) and the logarithmic pressure ratio:

\[ P_{\text{pump}} = 3.5\; \frac{V_{\text{suction}}\;\ln\!\left(\dfrac{P_{\text{discharge}}}{P_{\text{suction}}}\right)}{1000} \]

where P_discharge = 1 bar. The associated electricity cost is:

\[ C_{\text{pump}} = P_{\text{pump}}\; \times\; C_{\text{elec}} \]

9. Total Vacuum Distillation Cost

\[ C_{\text{total,vac}} = C_{\text{vac}} + C_{\text{pump}} \]

10. Validity Checks

Check	Condition	Action
Steam‑Distillation Temperature Range	\(10 \le T_{b,\;steam} \le 120\) °C	Issue warning if violated
Steam‑Distillation System Pressure Range	\(0.02 \le P_{\text{sys,steam}} \le 2.0\) bar	Issue warning if violated
Vacuum‑Distillation Pressure Range	\(0.001 \le P_{\text{sys,vac}} \le 1.0\) bar	Issue warning if violated

11. Summary of Output Quantities

Steam‑distillation: boiling temperature T_b,steam, system pressure P_sys,steam, required steam flow \(\dot{m}_{\text{steam}}\), and hourly steam cost \(C_{\text{steam}}\).
Vacuum‑distillation: boiling temperature T_b,vac, system pressure P_sys,vac, reboiler power \(P_{\text{vac}}\), pump power \(P_{\text{pump}}\), and total hourly electricity cost \(C_{\text{total,vac}}\).

Choose steam distillation when the target compound is immiscible with water and has a reasonable vapor pressure below 100 °C. It is ideal for:

Natural product recovery (essential oils, terpenes) where water immiscibility is inherent
Cases where a water phase can be easily separated and the product tolerates traces of moisture
Facilities that already have abundant low-pressure steam and limited vacuum infrastructure

Vacuum distillation is preferred when water contact causes hydrolysis, emulsion, or product degradation, or when the boiling-point reduction achievable under modest vacuum is sufficient to prevent thermal damage.

Compare the latent heats and utility requirements:

Steam distillation adds the latent heat of steam (≈2.3 MJ t⁻¹) plus the energy to vaporize water carried overhead; expect 20–40 % more reboiler duty than pure-component vacuum distillation
Vacuum distillation saves on water vapor load but requires electrical energy for the ejector or mechanical vacuum pump (≈0.1–0.3 kWh per kmol of vapor removed)
Overall, if boiler fuel is cheap and electricity is expensive, steam distillation can still be favorable; otherwise vacuum distillation usually wins on utility cost

Vacuum distillation typically delivers higher purity because no water is present to co-distill or form azeotropes; overhead condensate is almost pure product
Steam distillation can achieve high recovery if the product is hydrophobic, but downstream phase separation and drying steps are required, and losses into the aqueous layer can reach 1–5 %
For trace impurities, vacuum columns can be run at deeper vacuum (1–5 kPa) to separate close-boilers, whereas steam distillation is limited by the total pressure of water plus organics

Steam distillation at 101 kPa lowers the boiling point of an immiscible organic to:

T ≈ 100 °C when the organic vapor pressure is negligible
T ≈ 90–95 °C for moderate organic vapor pressures (5–10 kPa)

Achieving the same 90–100 °C boiling point by vacuum alone requires operating the column at ≈20–30 kPa (150–225 mm Hg) absolute pressure for most mid-molecular-weight organics. If your product is stable at 50–60 °C, deeper vacuum (5–10 kPa) is preferable to avoid water contact altogether.

Worked Example – Choosing Between Steam Distillation and Vacuum Distillation

Scenario: A plant must separate 1 000 kg h⁻¹ of a volatile oil (MW = 136 g mol⁻¹, ΔH_vap = 38 kJ mol⁻¹) from water. The maximum allowable temperature is 90 °C. Two options are considered:

Steam distillation using saturated steam at 6 bar (steam temperature 80 °C).
Vacuum distillation at 0.18 bar absolute (corresponding to a boiling point of 90 °C).

Known Input Parameters:

Product rate, \( \dot{m}_{oil} = 1000 \) kg h⁻¹
Molecular weight of oil, \( MW_{oil} = 136 \) g mol⁻¹
Molecular weight of water, \( MW_{water} = 18 \) g mol⁻¹
Heat of vaporisation of oil, \( \Delta H_{vap} = 38 \) kJ mol⁻¹
Steam pressure, \( P_{steam} = 6 \) bar (saturation temperature 80 °C)
Steam enthalpy, \( h_{steam} = 2750 \) kJ kg⁻¹
Cost of steam, \( C_{steam} = 20 \) EUR GJ⁻¹
Electricity cost, \( C_{elec} = 0.10 \) EUR kWh⁻¹
Antoine constants for water (A = 8.14019, B = 1810.94, C = 244.485)
Antoine constants for oil (A = 7.589, B = 2050.0, C = 235.0)
Vacuum system pressure, \( P_{vac} = 0.1797 \) bar
Pump suction temperature, \( T_{suction} = 303 \) K
Pump efficiency assumed 100 % (for simplicity)

Step‑by‑Step Calculation:

Calculate the molar flow of oil: \[ \dot{n}_{oil}= \frac{\dot{m}_{oil}}{MW_{oil}}= \frac{1000\ \text{kg h}^{-1}}{0.136\ \text{kg mol}^{-1}}=7352.941\ \text{mol h}^{-1} \]
Determine the saturation vapour pressures at 80 °C using the Antoine equation \[ \log_{10} P = A-\frac{B}{T+C} \] where \(T\) is in °C and \(P\) in bar.
- Water: \(P_{water,80}=0.483\) bar
- Oil: \(P_{oil,80}=0.016\) bar
Calculate the mole fraction of oil in the vapour phase: \[ y_{oil}= \frac{P_{oil}}{P_{water}+P_{oil}}= \frac{0.016}{0.483+0.016}=0.0322 \]
Find the total moles of vapour that must be generated to carry the oil: \[ \dot{n}_{vap}= \frac{\dot{n}_{oil}}{y_{oil}}= \frac{7352.941}{0.0322}=228\,300\ \text{mol h}^{-1} \]
Convert the vapour flow to a steam mass flow (assuming the vapour is essentially steam): \[ \dot{m}_{steam}= \dot{n}_{vap}\times MW_{water}=228\,300\ \text{mol h}^{-1}\times0.018\ \text{kg mol}^{-1}=4401.228\ \text{kg h}^{-1} \]
Calculate the thermal energy supplied by the steam: \[ Q_{steam}= \dot{m}_{steam}\times h_{steam}=4401.228\ \text{kg h}^{-1}\times2750\ \text{kJ kg}^{-1}=12\,103.376\ \text{MJ h}^{-1}=12.103\ \text{GJ h}^{-1} \]
Determine the hourly cost of steam: \[ \text{Cost}_{steam}= Q_{steam}\times\frac{C_{steam}}{1\ \text{GJ}}=12.103\ \text{GJ h}^{-1}\times20\ \text{EUR GJ}^{-1}=242.068\ \text{EUR h}^{-1} \]
For vacuum distillation, compute the latent heat per kilogram of oil: \[ \lambda =\frac{\Delta H_{vap}}{MW_{oil}}= \frac{38\ \text{kJ mol}^{-1}}{0.136\ \text{kg mol}^{-1}}=279.412\ \text{kJ kg}^{-1} \]
Thermal power required to vaporise the oil: \[ Q_{vac}= \dot{m}_{oil}\times\lambda=1000\ \text{kg h}^{-1}\times279.412\ \text{kJ kg}^{-1}=279\,412\ \text{kJ h}^{-1}=279.412\ \text{MJ h}^{-1} \] (equivalent to 155.229 kW).
Cost of the reboiler duty (electric heating assumed): \[ \text{Cost}_{reb}= Q_{vac}\times\frac{C_{elec}}{3600}=279\,412\ \text{kJ h}^{-1}\times\frac{0.10\ \text{EUR kWh}^{-1}}{3600\ \text{kJ kWh}^{-1}}=15.523\ \text{EUR h}^{-1} \]
Calculate the vacuum pump power (using the ideal gas work expression): \[ \dot{V}= \frac{\dot{m}_{air}}{\rho_{air}} \approx 0.290\ \text{kW} \] (derived from the given air leak of 10 kg h⁻¹). Pump cost: \[ \text{Cost}_{pump}=0.290\ \text{kW}\times0.10\ \text{EUR kWh}^{-1}=0.029\ \text{EUR h}^{-1} \]
Sum the vacuum‑distillation operating cost: \[ \text{Total}_{vac}=15.523+0.029=15.552\ \text{EUR h}^{-1} \]

Final Answer:

Steam‑distillation operating cost ≈ 242 EUR h⁻¹.
Vacuum‑distillation operating cost ≈ 15.6 EUR h⁻¹.
For the given production rate and temperature limit, vacuum distillation is the more economical choice, costing roughly one‑sixth of the steam‑distillation option.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle