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VFD turndown by the affinity laws

Pumps · affinity laws and variable-speed savings

A condenser water pump delivers 500 gpm at 80 ft of head, drawing 13.5 kW at its full 1800 rpm. The balancing report shows the loop was oversized, and a new VFD will run the pump at 1200 rpm through the shoulder seasons — about 6,000 hours a year. Electricity is billed at 11 cents per kilowatt-hour.

Given
  • Q₁ = 500 gpm — Flow at full speed
  • H₁ = 80 ft — Head at full speed
  • P₁ = 13.5 kW — Power at full speed
  • N₁ = 1800 rpm — Full speed
  • N₂ = 1200 rpm — Turndown speed
  • t = 6000 h — Hours at turndown
  • p_e = 0.11 $/kWh — Electricity rate
Determine
  1. (a)the flow at 1200 rpm
  2. (b)the head at 1200 rpm
  3. (c)the power at 1200 rpm
  4. (d)the energy and cost of a year at turndown
Step 1 of 5(a) · solve for Flow at speed 2

Flow tracks speed one-for-one: two-thirds speed is two-thirds flow, 333 gpm. This is the law to check FIRST, because it is the one the process feels — if 333 gpm starves the condenser on a design day, the drive program needs a floor, not a debate about energy.

Rearranged for Q₂
Q2=Q1⋅N2N1Q_2 = Q_1 \cdot \tfrac{N_2}{N_1}
Your values, in your units
Q2=(500 gpm)⋅(1,200 rpm)(1,800 rpm)Q_2 = \left( 500\ \text{gpm} \right) \cdot \tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}
Converted to base units
Q2=(1,892.71 L/min)⋅(1,200 rpm)(1,800 rpm)Q_2 = \left( 1{,}892.71\ \text{L/min} \right) \cdot \tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}
Answer
Q2=1.2618 m3/minQ_2 = 1.2618\ \text{m}^{3}\text{/min}

Carried onward at full precision, not this rounded figure.

Open the Pump Affinity Law — Flow vs Speed solver →

Step 2 of 5(b) · solve for Head at speed 2

Head falls with the SQUARE: (2/3)² = 4/9, so 80 ft becomes 35.6 ft. This is the law that bites static-head systems — a loop that must lift water 40 ft simply stops flowing before the cube-law savings arrive. A condenser loop is nearly all friction, which is why this turndown works here.

Rearranged for H₂
H2=H1(N2N1)2H_2 = H_1 \left(\tfrac{N_2}{N_1}\right)^{2}
Your values, in your units
H2=(80 ft)((1,200 rpm)(1,800 rpm))2H_2 = \left( 80\ \text{ft} \right) \left(\tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}\right)^{2}
Converted to base units
H2=(24.384 m)((1,200 rpm)(1,800 rpm))2H_2 = \left( 24.384\ \text{m} \right) \left(\tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}\right)^{2}
Answer
H2=10.837 mH_2 = 10.837\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Pump Affinity Law — Head vs Speed solver →

Step 3 of 5(c) · solve for Power at speed 2

Power falls with the CUBE: (2/3)³ = 8/27, and 13.5 kW collapses to 4.0 kW. One-third of the speed gone, seventy per cent of the power gone — this single line is the business case for every pump VFD ever sold, and the reason oversized pumps are an opportunity rather than just a mistake.

Rearranged for P₂
P2=P1(N2N1)3P_2 = P_1 \left(\tfrac{N_2}{N_1}\right)^{3}
Your values, in your units
P2=(13.5 kW)((1,200 rpm)(1,800 rpm))3P_2 = \left( 13.5\ \text{kW} \right) \left(\tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}\right)^{3}
Converted to base units
P2=(13,500 W)((1,200 rpm)(1,800 rpm))3P_2 = \left( 13{,}500\ \text{W} \right) \left(\tfrac{\left( 1{,}200\ \text{rpm} \right)}{\left( 1{,}800\ \text{rpm} \right)}\right)^{3}
Answer
P2=4 kWP_2 = 4\ \text{kW}

Carried onward at full precision, not this rounded figure.

Open the Pump Affinity Law — Power vs Speed solver →

Step 4 of 5(d) · solve for Energy

6,000 hours at 4.0 kW is 24,000 kWh. The same hours at full speed would have been 81,000 kWh — the drive is not saving a percentage, it is deleting two-thirds of the pump's energy line.

Rearranged for E
E=PtE = P t
4 kWcarried from step 3
Your values, in your units
E=(4,000 W) (6,000 h)E = \left( 4{,}000\ \text{W} \right) \, \left( 6{,}000\ \text{h} \right)
Converted to base units
E=(4,000 W) (21,600,000 s)E = \left( 4{,}000\ \text{W} \right) \, \left( 21{,}600{,}000\ \text{s} \right)
Answer
E=86.4 GJE = 86.4\ \text{GJ}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Step 5 of 5 · solve for Energy cost

24,000 kWh at 11 cents is $2,640 a year at turndown, against $8,910 at full speed — a $6,270 annual saving, which pays for a drive on this size of motor in well under two years.

Rearranged for C_e
Ce=E peC_e = E \, p_e
86.4 GJcarried from step 4
Your values, in your units
Ce=(86,400,000,000 J) (0.11 $/kWh)C_e = \left( 86{,}400{,}000{,}000\ \text{J} \right) \, \left( 0.11\ \text{\$/kWh} \right)
Converted to base units
Ce=(24,000 kWh) (0.11 $/kWh)C_e = \left( 24{,}000\ \text{kWh} \right) \, \left( 0.11\ \text{\$/kWh} \right)
Answer
Ce=2,640 $C_e = 2{,}640\ \text{\$}

Carried onward at full precision, not this rounded figure.

Open the Energy Cost from a Utility Rate solver →

At 1200 rpm the pump delivers about 333 gpm at 35.6 ft and draws 4.0 kW, so 6,000 hours of turndown consumes 24,000 kWh and costs $2,640 — roughly $6,270 a year less than running the same hours at full speed.

Why this order

The three affinity laws are one fact about centrifugal machines wearing three exponents: at a fixed impeller diameter, flow goes with speed, head with speed squared, power with speed cubed. The chain runs them in that order deliberately. Flow first because it is the service — the condenser either gets enough water or it does not, and no energy saving survives a tripped chiller. Head second because it is the feasibility check: the pump can only push water around the loop if its (falling) head still clears the system's static component, and a loop with real lift in it has a hard floor below which turndown means zero flow, not less flow. Power third, because once service and feasibility clear, the cube law is pure profit. The ratios keep the arithmetic honest: 1200/1800 = 2/3 exactly, so the three answers are 2/3, 4/9 and 8/27 of the originals — 333.3 gpm, 35.56 ft, 4.000 kW — no calculator required, which is exactly how to sanity-check a drive vendor's savings claim in the meeting where it is made.

What goes wrong in practice: claiming cube-law savings on a system that cannot take the turndown. A cooling tower needs a minimum flow over the fill for even distribution, condensers have minimum tube velocities to stay clean, and static lift puts a floor under head — the honest study finds the lowest workable speed first and prices that, rather than pricing 30% speed and installing 85%. Second, the laws assume the system curve stays put: throttle a discharge valve while the drive slows and the two fight, with the operating point landing somewhere neither predicted — commission the drive with the valves OPEN and let speed do the throttling. Third, small print on the meter: drive losses (2–3%) and reduced motor efficiency at part load shave the ideal 8/27 to something nearer 0.32 in the wild — real, still spectacular, and worth stating honestly in the proposal instead of being discovered on the first bill.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.