Evaporator Heat Balance Calculator
Calculate the steam consumption rate (kg/h) for a single-effect evaporator concentrating a dilute feed to a target product concentration. Helps process engineers size steam supply and heat transfer area.
Last updated: September 2026
Formula below · 2 sources (aiche.org, Wikipedia) · Updated Sep 2026
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About this calculator
In a single-effect evaporator, water is evaporated from a feed stream to raise its solute concentration. The water evaporation rate W is found from a solute mass balance: W = F × (x_P − x_F) / x_P, where F is the feed rate (kg/h), x_F is the feed concentration (wt%), and x_P is the product concentration (wt%). The steam consumption S then follows from the energy balance: S × λ_s = F × c_p × (T_b − T_F) + W × λ_v. The calculator assumes: feed at T_F = 25 °C with c_p = 4.18 kJ/(kg·K); boiling at atmospheric pressure, T_b = 100 °C, with no boiling-point elevation; vapor latent heat λ_v = 2,257 kJ/kg at 100 °C; and heating steam saturated at 3 bar abs (133.5 °C) with λ_s = 2,163 kJ/kg. So S = [F × 4.18 × 75 + W × 2,257] / 2,163. Concentrated solutions with boiling-point elevation, a lower specific heat, or vacuum operation will differ; use this for preliminary sizing.
How to use
Suppose F = 5,000 kg/h, x_F = 10 wt%, x_P = 40 wt%. Step 1 — evaporation rate: W = 5,000 × (40 − 10) / 40 = 3,750 kg/h. Step 2 — feed preheat from 25 to 100 °C: 5,000 × 4.18 × 75 = 1,567,500 kJ/h. Step 3 — evaporation: 3,750 × 2,257 = 8,463,750 kJ/h. Step 4 — steam: S = (1,567,500 + 8,463,750) / 2,163 = 4,637.6 kg/h. That is about 1.24 kg of steam per kg of water evaporated (economy 0.81), typical for a single effect with cold feed.
Frequently asked questions
What is the steam economy of a single-effect evaporator and how can it be improved?
Steam economy is the kg of water evaporated per kg of steam consumed. For a single-effect evaporator, economy is typically 0.8–0.9 because some steam energy goes toward heating the feed from its inlet temperature to the boiling point. Multiple-effect evaporators reuse vapor from one effect as the heating steam for the next, achieving economies of 2–4 for triple-effect systems. Mechanical vapor recompression (MVR) can push economy above 10 by mechanically compressing the evaporated vapor back to a higher pressure for reuse. Selecting multi-effect or MVR systems significantly reduces operating costs for large-scale processes.
How does feed concentration affect steam consumption in an evaporator?
A lower feed concentration means more water must be evaporated to reach the target product concentration, directly increasing steam consumption proportionally. For example, doubling the feed concentration from 5 wt% to 10 wt% (while keeping product at 40 wt%) reduces the evaporation rate by roughly half. Boiling point elevation (BPE) in concentrated solutions also increases steam consumption because a higher-temperature steam is needed to maintain the driving force. Engineers account for BPE using Dühring's rule when designing evaporators for sugar, salt, or caustic solutions.
Why is the latent heat of vaporization used in evaporator heat balance calculations?
Evaporation is a phase-change process: water transitions from liquid to vapor, absorbing a large amount of energy (λ_v ≈ 2,260 kJ/kg at 100 °C) without changing temperature. This latent heat dominates the energy balance, far exceeding the sensible heat needed to raise the feed temperature. The heating steam condenses on the other side of the heat exchanger surface, releasing its own latent heat λ_s to drive the process. The ratio λ_v / λ_s is close to 1, which is why single-effect steam economy approaches (but never reaches) 1 kg evaporated per kg of steam.