A pump delivers 0.06 m³/s against a total head of 32 m at 74% efficiency. What brake power is required?

Start with the thinking

• Water power first, then divide by efficiency.
• Dividing by efficiency increases the power — a common sign check.

Step-by-step solution

1. Water power — P_w = γQH = 9.81(0.06)(32)

2. Evaluate

   $$P_w = 18.8 kW$$

3. Brake power

   $$P = P_w/\eta = 18.8/0.74$$

4. Result

   $$P = 25.5 kW$$

Why the other options are there

• 13.9 kW (multiplied by efficiency)
• 18.8 kW (efficiency ignored)

Reference: FE Reference Handbook — Fluid Mechanics — Pump power

A pump delivers 8.5 cfs against 160 ft of head at 64% efficiency. Find the water and brake horsepower.

Start with the thinking

• Water horsepower is γQH/550; efficiency only affects input power.
• Efficiency divides — the motor always draws more.

Step-by-step solution

1. Water power — WHP = γQH/550

2. Substituting

   $$WHP = 62.4(8.5)(160)/550 = 154.3 hp$$

3. Brake power

   $$BHP = WHP/\eta$$

4. Substituting

   $$BHP = 154.3/0.64 = 241.1 hp$$

Why the other options are there

• 98.8 hp (efficiency multiplied)
• 2.47 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 0.400 m³/s against a total dynamic head of 44 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 19 h/day and $0.20/kWh.

Start with the thinking

• Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
• Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

1. Formula

   $$P_water = \rho g Q H$$

2. Substituting

   $$P = 1000(9.81)(0.400)(44) = 172.7 kW$$

3. Formula

   $$P_shaft = P_water/\eta$$

4. Substituting

   $$P_shaft = 172.7/0.84 = 205.5 kW$$

5. Annual energy

   $$205.5 kW \times 19 h/day \times 365 = 1,425,440 kWh$$

6. Cost — 1,425,440 kWh × $0.20 = $285,088 per year

Why the other options are there

• 145.0 kW (efficiency multiplied)
• 231.4 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 2.5 cfs against 151 ft of head at 84% efficiency. Find the water and brake horsepower.

Start with the thinking

• Water horsepower is γQH/550; efficiency only affects input power.
• Efficiency divides — the motor always draws more.

Step-by-step solution

1. Water power — WHP = γQH/550

2. Substituting

   $$WHP = 62.4(2.5)(151)/550 = 42.83 hp$$

3. Brake power

   $$BHP = WHP/\eta$$

4. Substituting

   $$BHP = 42.83/0.84 = 50.99 hp$$

Why the other options are there

• 36.0 hp (efficiency multiplied)
• 0.69 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 0.040 m³/s against a total dynamic head of 31 m at 82% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 15 h/day and $0.14/kWh.

Start with the thinking

• Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
• Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

1. Formula

   $$P_water = \rho g Q H$$

2. Substituting

   $$P = 1000(9.81)(0.040)(31) = 12.16 kW$$

3. Formula

   $$P_shaft = P_water/\eta$$

4. Substituting

   $$P_shaft = 12.16/0.82 = 14.83 kW$$

5. Annual energy

   $$14.83 kW \times 15 h/day \times 365 = 81,220 kWh$$

6. Cost — 81,220 kWh × $0.14 = $11,371 per year

Why the other options are there

• 10.0 kW (efficiency multiplied)
• 16.3 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 16.0 cfs against 178 ft of head at 88% efficiency. Find the water and brake horsepower.

Start with the thinking

• Water horsepower is γQH/550; efficiency only affects input power.
• Efficiency divides — the motor always draws more.

Step-by-step solution

1. Water power — WHP = γQH/550

2. Substituting

   $$WHP = 62.4(16.0)(178)/550 = 323.1 hp$$

3. Brake power

   $$BHP = WHP/\eta$$

4. Substituting

   $$BHP = 323.1/0.88 = 367.2 hp$$

Why the other options are there

• 284.3 hp (efficiency multiplied)
• 5.18 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 0.150 m³/s against a total dynamic head of 53 m at 80% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 10 h/day and $0.11/kWh.

Start with the thinking

• Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
• Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

1. Formula

   $$P_water = \rho g Q H$$

2. Substituting

   $$P = 1000(9.81)(0.150)(53) = 77.99 kW$$

3. Formula

   $$P_shaft = P_water/\eta$$

4. Substituting

   $$P_shaft = 77.99/0.80 = 97.49 kW$$

5. Annual energy

   $$97.49 kW \times 10 h/day \times 365 = 355,827 kWh$$

6. Cost — 355,827 kWh × $0.11 = $39,141 per year

Why the other options are there

• 62.4 kW (efficiency multiplied)
• 104.5 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 5.0 cfs against 68 ft of head at 84% efficiency. Find the water and brake horsepower.

Start with the thinking

• Water horsepower is γQH/550; efficiency only affects input power.
• Efficiency divides — the motor always draws more.

Step-by-step solution

1. Water power — WHP = γQH/550

2. Substituting

   $$WHP = 62.4(5.0)(68)/550 = 38.57 hp$$

3. Brake power

   $$BHP = WHP/\eta$$

4. Substituting

   $$BHP = 38.57/0.84 = 45.92 hp$$

Why the other options are there

• 32.4 hp (efficiency multiplied)
• 0.62 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 0.130 m³/s against a total dynamic head of 39 m at 74% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 19 h/day and $0.11/kWh.

Start with the thinking

• Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
• Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

1. Formula

   $$P_water = \rho g Q H$$

2. Substituting

   $$P = 1000(9.81)(0.130)(39) = 49.74 kW$$

3. Formula

   $$P_shaft = P_water/\eta$$

4. Substituting

   $$P_shaft = 49.74/0.74 = 67.21 kW$$

5. Annual energy

   $$67.21 kW \times 19 h/day \times 365 = 466,114 kWh$$

6. Cost — 466,114 kWh × $0.11 = $51,272 per year

Why the other options are there

• 36.8 kW (efficiency multiplied)
• 66.7 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation

A pump delivers 12.5 cfs against 191 ft of head at 76% efficiency. Find the water and brake horsepower.

Start with the thinking

• Water horsepower is γQH/550; efficiency only affects input power.
• Efficiency divides — the motor always draws more.

Step-by-step solution

1. Water power — WHP = γQH/550

2. Substituting

   $$WHP = 62.4(12.5)(191)/550 = 270.9 hp$$

3. Brake power

   $$BHP = WHP/\eta$$

4. Substituting

   $$BHP = 270.9/0.76 = 356.4 hp$$

Why the other options are there

• 205.9 hp (efficiency multiplied)
• 4.34 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Pump Power Equation