Overall Coefficient and Boiling Mechanisms

Reference ID: MET-51F1 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The calculation of the overall heat transfer coefficient (U) is a fundamental requirement in the design and operation of falling-film evaporators. In process engineering, these units are critical for concentrating heat-sensitive liquids, such as tomato juice, where maintaining product quality requires precise control over thermal exposure. The overall coefficient accounts for the series of thermal resistances encountered by heat as it transfers from the condensing steam, through the tube wall, and into the boiling liquid film.

This analysis is essential for determining the required heat transfer area, predicting the impact of fouling on evaporator performance over time, and ensuring that the boiling regime remains within the stable nucleate boiling range to prevent equipment damage or product degradation.

Methodology & Formulas

The system is modeled using a three-resistance approach, augmented by a fouling factor. The driving force for heat transfer is the temperature difference between the steam and the boiling liquid, adjusted for the Boiling Point Elevation (BPE) caused by the concentration of solutes.

Overall Heat Transfer Coefficient

The calculation of the overall heat transfer coefficient (U) is a fundamental requirement in the design and operation of falling‑film evaporators, and a comparison of falling‑film versus flooded configurations for food applications helps engineers select the most suitable mode for concentrating heat‑sensitive liquids such as tomato juice, where maintaining product quality requires precise control over thermal exposure.

\[ \frac{1}{U_{\text{clean}}} = \frac{1}{\alpha_{\text{c}}} + \frac{\varepsilon}{k_{\text{w}}} + \frac{1}{\alpha_{\text{b,clean}}} \]

Under fouled conditions, an additional resistance term is introduced:

\[ \frac{1}{U_{\text{f}}} = \frac{1}{\alpha_{\text{c}}} + \frac{\varepsilon}{k_{\text{w}}} + \frac{1}{\alpha_{\text{b,fouled}}} + R_{\text{f}} \]

Boiling Point Elevation and Driving Force

The effective temperature driving force is reduced by the BPE, which is calculated empirically based on the concentration of the juice:

\[ \Delta T_{\text{useful}} = T_{\text{steam}} - T_{\text{vapor,space}} - BPE_{\text{final}} \] \[ BPE = 0.02 \cdot Brix + 0.0001 \cdot Brix^{2} \]

Fouling and Heat Flux

Fouling resistance is modeled as a function of the cumulative mass evaporated per unit area:

\[ R_{\text{f}} = K_{\text{f}} \cdot \left( \frac{m_{\text{evap,total}}}{A} \right) \]

The resulting heat flux, which must be monitored to ensure it remains below the critical heat flux threshold, is given by:

\[ \dot{q} = U_{\text{f}} \cdot \Delta T_{\text{useful}} \]

Operational Regimes and Criteria

Parameter	Condition/Threshold	Engineering Significance
\(\Delta T_{\text{useful}}\)	≤ 40.0 °C	Exceeding this limit risks transition from nucleate to film boiling.
\(\alpha_{\text{b,fouled}}\)	500 – 5000 W/(m² · K)	Empirical range for viscous food liquids; values outside indicate model deviation.
\(R_{\text{f}}\)	≤ 0.001 m² · K/W	Critical threshold for fouling; indicates a need for system cleaning.

The overall heat transfer coefficient undergoes a dramatic reduction as the boiling regime shifts. This occurs because:

In the nucleate boiling regime, bubble formation promotes high levels of fluid agitation and surface wetting.
As the heat flux reaches the critical heat flux point, a continuous vapor film begins to form on the heating surface.
This vapor film acts as a significant thermal insulator, drastically increasing the thermal resistance and causing the overall coefficient to plummet.

When calculating the overall coefficient, process engineers must account for several resistances in series. Key variables include:

The individual convective heat transfer coefficients for both the tube-side and shell-side fluids.
The thermal conductivity and thickness of the tube wall material.
The fouling factors, which represent the accumulation of deposits on either side of the heat transfer surface.
The geometry of the heat exchanger, specifically the ratio of outer to inner surface areas.

Operating near or above the critical heat flux is dangerous for industrial equipment for the following reasons:

It triggers the Leidenfrost effect, where a stable vapor film prevents liquid contact with the metal surface.
The sudden drop in the heat transfer coefficient causes the surface temperature of the tubes to spike rapidly.
This temperature excursion can lead to mechanical failure, tube burnout, or structural degradation of the heat exchanger materials.

Worked Example: Overall Coefficient and Boiling Mechanisms for a Falling-Film Evaporator

Scenario: A single-effect falling-film evaporator concentrates tomato juice from 30 °Brix to 60 °Brix. Saturated steam at 101.0 °C condenses on the outside of vertical tubes; the vapor space is maintained at 60.0 °C (0.2 bar). The evaporator has 100 m² of heat transfer area, and the feed mass flow is 10,000 kg/h. After 5 hours of continuous operation, fouling has built up and the juice has reached the final concentration. The following knowns and calculated values from the model are used.

Feed: tomato juice, 30 °Brix, 10,000 kg/h
Steam temperature: \(T_{\text{steam}} = 101.0\,^\circ\mathrm{C}\)
Vapor space temperature: \(T_{\text{vapor}} = 60.0\,^\circ\mathrm{C}\)
Heat transfer area: \(A = 100.0\,\mathrm{m^{2}}\)
Wall thickness: \(\varepsilon = 0.002\,\mathrm{m}\); wall conductivity \(k_{\text{w}} = 16.0\,\mathrm{W/(m\cdot K)}\)
Condensing steam coefficient: \(\alpha_{\text{c}} = 10000.0\,\mathrm{W/(m^{2}\cdot K)}\)
Clean boiling coefficient: \(\alpha_{\text{b,clean}} = 4000.0\,\mathrm{W/(m^{2}\cdot K)}\)
Fouled boiling coefficient: \(\alpha_{\text{b,fouled}} = 1500.0\,\mathrm{W/(m^{2}\cdot K)}\)
Fouling proportionality constant: \(K = 2\times10^{-7}\,\mathrm{m^{2}\cdot K/W\;per\;kg/m^{2}}\)

Calculate boiling point elevations (BPE).
Initial (30 °Brix): \(\mathrm{BPE}_{\text{i}} = 0.02\cdot30 + 0.0001\cdot30^{2} = 0.69\,^\circ\mathrm{C}\).
Final (60 °Brix): \(\mathrm{BPE}_{\text{f}} = 0.02\cdot60 + 0.0001\cdot60^{2} = 1.56\,^\circ\mathrm{C}\).
(Matches provided values: bpe_initial = 0.69, bpe_final = 1.56.)
Determine the useful temperature driving force.
\[ \Delta T_{\text{useful}} = T_{\text{steam}} - T_{\text{vapor}} - \mathrm{BPE}_{\text{f}} = 101.0 - 60.0 - 1.56 = 39.44\,^\circ\mathrm{C}. \] (delta_t_useful = 39.44.) Check: 39.44 °C ≤ 40 °C (nucleate boiling limit satisfied).
Compute the total mass evaporated over 5 hours.
Feed mass in 5 h: \(10\,000 \ \mathrm{kg/h} \times 5\ \mathrm{h} = 50\,000\ \mathrm{kg}\). Concentration ratio: \(\mathrm{Brix}_{\text{i}} / \mathrm{Brix}_{\text{f}} = 30/60 = 0.5\). Mass evaporated (water removed): \[ m_{\text{evap,total}} = 50\,000 \times (1 - 0.5) = 25\,000\ \mathrm{kg}. \] (mass_evap_total = 25000.0.)
Calculate the evaporated mass per unit area.
\[ m/A = \frac{25\,000\ \mathrm{kg}}{100\ \mathrm{m^{2}}} = 250\ \mathrm{kg/m^{2}}. \] (m_evap_per_area = 250.0.)
Compute the fouling resistance.
\[ R_{\text{f}} = K \cdot (m/A) = 2.0\times10^{-7} \times 250.0 = 5.0\times10^{-5}\ \mathrm{m^{2}\cdot K/W}. \] (r_f = 5.0e-05.) Check: \(R_{\text{f}} < 0.001\ \mathrm{m^{2}\cdot K/W}\) – acceptable.
Evaluate the clean overall heat transfer coefficient.
Three resistances in series: \[ \frac{1}{U_{\text{clean}}} = \frac{1}{\alpha_{\text{c}}} + \frac{\varepsilon}{k_{\text{w}}} + \frac{1}{\alpha_{\text{b,clean}}}. \] Using the resistances from the data: \[ \frac{1}{U_{\text{clean}}} = 0.000100 + 0.000125 + 0.000250 = 0.000475\ \mathrm{m^{2}\cdot K/W}, \] hence \[ U_{\text{clean}} = \frac{1}{0.000475} \approx 2105.263\ \mathrm{W/(m^{2}\cdot K)}. \] (u_clean = 2105.263.)
Determine the fouled overall heat transfer coefficient.
With fouling resistance added: \[ \frac{1}{U_{\text{f}}} = \frac{1}{\alpha_{\text{c}}} + \frac{\varepsilon}{k_{\text{w}}} + R_{\text{f}} + \frac{1}{\alpha_{\text{b,fouled}}}. \] Insert values: \[ \frac{1}{U_{\text{f}}} = 0.000100 + 0.000125 + 0.000050 + \frac{1}{1500}. \] Since \(1/1500 = 0.000666667\), the sum is \(0.000100 + 0.000125 + 0.000050 + 0.000666667 = 0.000941667\ \mathrm{m^{2}\cdot K/W}\). Thus \[ U_{\text{f}} = \frac{1}{0.000941667} \approx 1061.947\ \mathrm{W/(m^{2}\cdot K)}. \] (u_f = 1061.947.)
Compute the operating heat flux (fouled condition).
\[ q = U_{\text{f}} \cdot \Delta T_{\text{useful}} = 1061.947\ \mathrm{W/(m^{2}\cdot K)} \times 39.44\ \mathrm{K} \approx 41883.186\ \mathrm{W/m^{2}}. \] (q_flux = 41883.186 W/m², which is about 41.9 kW/m².)

Final Answer:
Clean overall coefficient: \(U_{\text{clean}} = 2105.263\ \mathrm{W/(m^{2}\cdot K)}\);
Fouled overall coefficient: \(U_{\text{f}} = 1061.947\ \mathrm{W/(m^{2}\cdot K)}\);
Heat flux under fouled conditions: \(q = 41883.186\ \mathrm{W/m^{2}}\).

"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