Flow control and energy

A pump or a fan is sized for the worst day. Most of the time the process needs less. How you take the flow down, with a valve or with speed, decides how much electricity is wasted in between.

Four ways to deliver less water

The Lab II network is designed for 120 m³/h at 25 m. Demand falls to 96 m³/h (80 %). Four ways to follow it:

Method	What happens	Power at 96 m³/h
Throttling valve	pump at full speed, the valve burns the extra head	9.66 kW
On/off cycling	pump runs at the rated point 80 % of the time, needs a tank	8.72 kW (average)
VFD at constant pressure	speed falls until the pump gives exactly 25 m at 96 m³/h	8.72 kW
VFD on the system curve	speed falls to 83 %; the head falls with the flow	6.21 kW

This is question 1 of Lab II: against throttling, the VFD cuts the electrical power by $1 - 6.21/9.66 = 36$ %, less than the 49 % the pure cube law promises from the rated point, because of the static head.

The valve wastes the most: the pump still produces 27.7 m of head while the pipes need only 17.8 m at this flow, so the valve throws 9.9 m away as heat and noise. The VFD removes both the throttling loss and the extra head. Over 6000 hours a year, the difference between the first and last line is about 21 MWh.

Try it: four ways to deliver less flow

Flow needed80 %

Throttling valve / damper
On/off cycling
VFD at constant pressure
VFD speed control

Throttling valve / damper: 89 % · 9.66 kW
On/off cycling: 80 % · 8.72 kW (−10 %)
VFD at constant pressure: 80 % · 8.72 kW (−10 %)
VFD speed control: 57 % · 6.21 kW (−36 %)

Lab II network: 120 m³/h at 25 m rated, 5 m static head, pump efficiency held at 75 %. The VFD curve follows the system curve; constant pressure holds 25 m at the pump. On/off cycling also delivers every litre at 25 m, so in this model it costs the same as constant pressure (dotted line), but it needs a tank.

Predict first

Demand drops to 60 m³/h (50 %). Compared with throttling, how much power does the VFD on the system curve need?

About the sameAbout halfAbout a third

Why is constant pressure less efficient than following the system curve? Because the pressure sensor sits at the pump, but the head that matters is at the far end of the network. Holding 25 m at the pump when demand is low pushes water harder than needed. Many booster drives therefore lower the set-point as flow falls (proportional pressure control).

Fans

A fan in a duct or a boiler draught system has almost no static head: all its pressure goes into friction, so its system curve passes through the origin and the affinity laws hold. At 80 % flow a VFD fan needs about 51 % of its rated power. The alternatives waste more:

an outlet damper throttles like a valve: the fan stays at full speed;
inlet guide vanes pre-swirl the air and do better than a damper, but not as well as speed control;
two-speed motors (pole-changing) give two points of the cube law and nothing in between.

Worked example: HVAC supply fan (course §2.1.2 and §2.4.2)

A supply-air fan at 1780 rpm delivers 30 000 m³/h at 1.2 kPa and draws 12 kW. Airflow must fall by 30 %. The duct system is purely frictional, so the affinity laws apply directly: 1246 rpm, 21 000 m³/h, $1.2 \times 0.7^2 = 0.588$ kPa and $12 \times 0.7^3 = 4.12$ kW. With a damper instead, the fan stays at full speed and typically still draws about 85 % of its power.

Control method	Power	Saving
Full speed, no control	12.00 kW	0 %
Throttling (damper)	10.20 kW	15 %
VFD speed control	4.12 kW	65.7 %

Compressors

A screw or piston compressor is a positive-displacement machine: every turn pushes a fixed volume against the network pressure. Its torque is roughly constant, so its power is proportional to flow (a constant-torque load, Chapter 1, lesson 1).

The classic control is load/unload: the compressor runs at full speed, loads until the pressure reaches an upper limit, then unloads and idles until it falls to a lower limit. An unloaded compressor still draws a large share of its full power, about a quarter in typical figures. At 50 % demand, load/unload therefore needs about 62 % of full power; a VFD compressor needs about 52 %.

Two more rules for compressed air:

Every extra bar of network pressure costs roughly 7 % more energy. Leaks and oversized set-points are paid for continuously.
At full load a VFD is slightly worse than a fixed-speed machine, because the drive has its own losses (2 to 3 %). The VFD pays where demand varies.

ExplorerGo deeper: derivations and open questions

Payback. A 15 kW VFD costs about the same as the motor it drives. Using the table above, how long does it take to pay back on the Lab II pump at 6000 h/year and 0.12 €/kWh, if the demand spends half its time at 96 m³/h and half at 60 m³/h?

Minimum speed. From lesson 1, the Lab II pump delivers nothing below 39 % speed. What happens to a pump running for hours at zero flow? How would you set the drive's minimum speed, and what should happen when demand falls below what that speed delivers?

Proportional pressure. Suppose the far end of the network needs 20 m at any flow, and the pipe friction is $k Q^2$ with $k$ from Lab II. Write the set-point the pump should follow as a function of flow. How much power does that save at 60 m³/h compared with constant 25 m?