What does EnerTherm Engineering do?

EnerTherm Engineering is a UK-based industrial engineering company that designs and delivers thermal process solutions, energy auditing services, and bespoke thermal equipment for manufacturing and process industries. Services include energy audits, pinch analysis, process modelling, thermal design simulation, and project management.

What thermal products does EnerTherm Engineering manufacture?

EnerTherm Engineering manufactures process heaters, heat exchangers, thermal oxidisers, incinerators, condensers, dehydrators, and the Omni Vision industrial energy intelligence platform. All products are custom-designed for specific industrial applications and process requirements.

Does EnerTherm Engineering provide energy auditing services?

Yes. EnerTherm Engineering provides comprehensive energy audits, heat loss audits, and pinch analysis to identify energy waste, quantify potential savings, and prioritise capital investments in industrial facilities across sectors including chemical, pharmaceutical, food & beverage, and manufacturing.

What industries does EnerTherm Engineering serve?

EnerTherm Engineering serves chemical processing, pharmaceutical, food & beverage, oil & gas, power generation, automotive, textile, paper & pulp, general manufacturing, waste management, agriculture, healthcare, and mining & metals industries.

Where is EnerTherm Engineering located?

EnerTherm Engineering is headquartered in the United Kingdom and provides engineering services and thermal equipment to clients internationally.

Can EnerTherm Engineering help with industrial decarbonisation?

Yes. EnerTherm Engineering specialises in waste heat recovery system design, process optimisation for energy reduction, and net zero carbon roadmapping. They also provide ISO 14001 and ISO 50001 compliant energy management strategies.

does EnerTherm Engineering offer CFD simulation and thermal modelling?

Yes. EnerTherm Engineering provides Computational Fluid Dynamics (CFD) and Finite Element Analysis (FEA) for thermal equipment design. These simulations validate heat transfer performance before fabrication, reducing costly physical prototyping.

What is a thermal oxidiser and does EnerTherm Engineering supply them?

A thermal oxidiser is an industrial system that destroys volatile organic compounds (VOCs) and hazardous air pollutants (HAPs) through controlled high-temperature combustion. EnerTherm Engineering designs and manufactures regenerative, recuperative, and direct-fired thermal oxidisers for chemical, pharmaceutical, and manufacturing industries.

Does EnerTherm Engineering design custom process equipment?

Yes. EnerTherm Engineering provides bespoke engineering design of process vessels, heat exchangers, reactors, and thermal treatment systems from concept through to fabrication-ready drawings, as well as rapid prototyping and pilot plant commissioning.

What is EnerTherm Engineering's Omni Vision platform?

Omni Vision is EnerTherm Engineering's industrial energy intelligence platform. It provides turnkey utility metering, CO2 tracking, and AI-powered production KPI intelligence for industrial compliance monitoring and operational optimisation.

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Louvre Head Loss

Pressure drop across angled louvers (blinds) in a duct.

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Every K = 1 component costs one velocity head — about 60 Pa at 10 m/s in air — and fan power rises with the cube of the flow you push through it.

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Figures assume typical conditions and the stated method. For measured, guaranteed numbers on your plant, our engineers run site surveys, heat loss audits, and full process models.

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Louvre Head Loss

Calculates the pressure-drop loss coefficient and pressure drop across angled louvers (blinds) fitted inside a duct or pipe.

Reynolds Number

R e = \frac{V D _{h}}{ν}

where $V$ is the mean flow velocity, $D_{h}$ is the hydraulic diameter of the duct, and $ν$ is the kinematic viscosity of air. For a square-section duct, $D_{h}$ equals the side width.

Loss Coefficient

The loss coefficient has two parts — a geometric base term and a Reynolds-number-dependent turbulence correction:

K = K_{0} + max (0, 1 - \frac{50}{R e}) \cdot k_{q u}

When Re ≤ 50 the turbulence correction term is taken as zero (the factor is clamped).

The geometric base term is:

K_{0} = (\frac{1 - F A R}{F A R})^{2}

The angle-and-count coefficient $k_{q u}$ increases with louver angle $θ$ and number of louvers n:

k_{q u} = C (n) sin^{2} θ

Pressure Drop

Δ p = K \cdot \frac{1}{2} ρ V^{2}

References

Idel'chik, I.E. and Fried, E., Flow Resistance, a Design Guide for Engineers, Taylor & Francis, Washington D.C., 1989, p. 306.

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Frequently asked questions

Pressure drop across angled louvers (blinds) in a duct. Enter your inputs and press Calculate — the worked solution shows every step of the method with your numbers substituted in.

K expresses a component's pressure drop in velocity heads: Δp = K·½ρV². A K of 1 means the component destroys exactly the kinetic energy of the approaching flow. It is dimensionless, so the same K applies at any flow rate — the pressure drop then scales with velocity squared.

Compute each component's Δp at its local velocity and add them, together with straight-duct friction. Where the duct area is constant you can equivalently sum the K values first. The result is the static pressure the fan must develop at that flow rate.