FORCED CIRCULATIONFOOD CONCENTRATION SYSTEMS

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Circulation pump demand: flow, pressure and efficiency

Estimate pump shaft power from a defined operating point, separately from net evaporation capacity.

Q × ΔP / EFFICIENCY

Illustrative case: 60 m³/h × 180 kPa gives 3 kW hydraulic power; at 60% pump efficiency, the shaft demand is 5 kW.

Pshaft [kW] = Q [m³/h] × ΔP [kPa] ÷ 3,600 ÷ (η [%] / 100). Excludes motor and drive losses, other auxiliaries and transient demand. Efficiency must match the actual operating point. This is not a pump selection or a steam-saving calculation.

Where the inputs come from

Flow

Use circulation volume flow at operating conditions. Feed kg/h and water evaporated kg/h are different quantities. Converting mass to volume requires the liquid density.

Pressure rise

Include pressure losses across the heater, pipes, valves and fittings in the specified circulation route. The separator absolute pressure alone is not the pump differential pressure.

Efficiency

Use pump efficiency at the selected flow, head and product properties. Motor and drive losses require additional data. The illustrative default is not a proposed pump operating point.

Count the full electrical service

Add motor/drive losses, feed and discharge equipment, vacuum and any MVR system when comparing plant demand. Starting requirements and control strategy may matter to available electrical capacity.

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