A pump running at 1764 rpm delivers 0.140 m³/s at 16 m while absorbing 36.5 kW. The same impeller is run at 2961 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2961/1764 = 1.6786$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.140(1.6786) = 0.235 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 16(1.6786)^{2} = 45.1 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 36.5(1.6786)^{3} = 172.6 kW$$

Why the other options are there

• H₂ = 26.9 m (linear scaling of head)
• P₂ = 102.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1759 rpm delivers 0.050 m³/s at 21 m while absorbing 4.0 kW. The same impeller is run at 2446 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2446/1759 = 1.3906$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.050(1.3906) = 0.070 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 21(1.3906)^{2} = 40.6 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 4.0(1.3906)^{3} = 10.8 kW$$

Why the other options are there

• H₂ = 29.2 m (linear scaling of head)
• P₂ = 7.7 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1764 rpm delivers 0.030 m³/s at 23 m while absorbing 40.5 kW. The same impeller is run at 2662 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2662/1764 = 1.5091$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.030(1.5091) = 0.045 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 23(1.5091)^{2} = 52.4 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 40.5(1.5091)^{3} = 139.2 kW$$

Why the other options are there

• H₂ = 34.7 m (linear scaling of head)
• P₂ = 92.2 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1153 rpm delivers 0.100 m³/s at 39 m while absorbing 38.0 kW. The same impeller is run at 2094 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2094/1153 = 1.8161$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.100(1.8161) = 0.182 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 39(1.8161)^{2} = 128.6 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 38.0(1.8161)^{3} = 227.6 kW$$

Why the other options are there

• H₂ = 70.8 m (linear scaling of head)
• P₂ = 125.3 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1680 rpm delivers 0.160 m³/s at 15 m while absorbing 36.0 kW. The same impeller is run at 2840 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2840/1680 = 1.6905$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.160(1.6905) = 0.270 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 15(1.6905)^{2} = 42.9 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 36.0(1.6905)^{3} = 173.9 kW$$

Why the other options are there

• H₂ = 25.4 m (linear scaling of head)
• P₂ = 102.9 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1614 rpm delivers 0.080 m³/s at 23 m while absorbing 55.5 kW. The same impeller is run at 2207 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2207/1614 = 1.3674$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.080(1.3674) = 0.109 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 23(1.3674)^{2} = 43.0 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 55.5(1.3674)^{3} = 141.9 kW$$

Why the other options are there

• H₂ = 31.5 m (linear scaling of head)
• P₂ = 103.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1466 rpm delivers 0.020 m³/s at 13 m while absorbing 6.5 kW. The same impeller is run at 2204 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2204/1466 = 1.5034$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.020(1.5034) = 0.030 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 13(1.5034)^{2} = 29.4 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 6.5(1.5034)^{3} = 22.1 kW$$

Why the other options are there

• H₂ = 19.5 m (linear scaling of head)
• P₂ = 14.7 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1555 rpm delivers 0.030 m³/s at 13 m while absorbing 58.5 kW. The same impeller is run at 1990 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 1990/1555 = 1.2797$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.030(1.2797) = 0.038 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 13(1.2797)^{2} = 21.3 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 58.5(1.2797)^{3} = 122.6 kW$$

Why the other options are there

• H₂ = 16.6 m (linear scaling of head)
• P₂ = 95.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1406 rpm delivers 0.180 m³/s at 13 m while absorbing 21.5 kW. The same impeller is run at 2796 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2796/1406 = 1.9886$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.180(1.9886) = 0.358 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 13(1.9886)^{2} = 51.4 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 21.5(1.9886)^{3} = 169.1 kW$$

Why the other options are there

• H₂ = 25.9 m (linear scaling of head)
• P₂ = 85.0 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws

A pump running at 1291 rpm delivers 0.040 m³/s at 17 m while absorbing 5.5 kW. The same impeller is run at 2631 rpm. Use the affinity laws to predict discharge, head and power.

Start with the thinking

• Flow scales linearly, head with the square and power with the cube of the speed ratio.
• A modest 20% speed increase nearly doubles the power draw — this is why VFD savings are large.

Step-by-step solution

1. Speed ratio

   $$N_{2}/N_{1} = 2631/1291 = 2.0380$$

2. Formula

   $$\dfrac{Q_2}{Q_1}=\dfrac{N_2}{N_1}$$

3. Substituting

   $$Q_{2} = 0.040(2.0380) = 0.082 m^{3}/s$$

4. Formula

   $$\dfrac{H_2}{H_1}=\left(\dfrac{N_2}{N_1}\right)^2$$

5. Substituting

   $$H_{2} = 17(2.0380)^{2} = 70.6 m$$

6. Formula

   $$\dfrac{P_2}{P_1}=\left(\dfrac{N_2}{N_1}\right)^3$$

7. Substituting

   $$P_{2} = 5.5(2.0380)^{3} = 46.6 kW$$

Why the other options are there

• H₂ = 34.6 m (linear scaling of head)
• P₂ = 22.8 kW (squared instead of cubed)

Reference: FE Reference Handbook — Fluid Mechanics → Scaling Laws; Affinity Laws