Centrifugal Pump Characteristics

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

0 formulas

10 exam-style examples

~45 min

All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

MODEL 3656/3756 S-GROUP 1750 RPM NOTE: NOT RECOMMENDED FOR OPERATION

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Pump brake horsepower — Centrifugal Pump Characteristics

A pump delivers 10.5 cfs against 147 ft of head at 84% efficiency. Find the water and brake horsepower.

Given

$Q = 10.5 cfs$
$H = 147 ft$
$\eta = 0.84$

Find

WHP and BHP

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

Water power — WHP = γQH/550
Substituting
$WHP = 62.4(10.5)(147)/550 = 175.1 hp$
Brake power
$BHP = WHP/\eta$
Substituting
$BHP = 175.1/0.84 = 208.5 hp$

Answer:

WHP ≈ 175.1 hp; BHP ≈ 208.5 hp

Why the other options are there

147.1 hp (efficiency multiplied)
2.81 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 2

Water horsepower, shaft power and pump operating cost — Centrifugal Pump Characteristics

A pump delivers 0.220 m³/s against a total dynamic head of 17 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 4 h/day and $0.17/kWh.

Given

$Q = 0.220 m^{3}/s$
$H = 17 m$
$\eta = 0.84$
4 h/day at $0.17/kWh

Find

Water power, shaft power and annual cost

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

Formula
$P_water = \rho g Q H$
Substituting
$P = 1000(9.81)(0.220)(17) = 36.69 kW$
Formula
$P_shaft = P_water/\eta$
Substituting
$P_shaft = 36.69/0.84 = 43.68 kW$
Annual energy
$43.68 kW \times 4 h/day \times 365 = 63,770 kWh$
Cost — 63,770 kWh × $0.17 = $10,841 per year

Answer:

P_water = 36.7 kW, P_shaft = 43.7 kW, cost ≈ $10,841/yr

Why the other options are there

30.8 kW (efficiency multiplied)
49.2 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 3

Pump brake horsepower — Centrifugal Pump Characteristics (2)

A pump delivers 14.5 cfs against 164 ft of head at 84% efficiency. Find the water and brake horsepower.

Given

$Q = 14.5 cfs$
$H = 164 ft$
$\eta = 0.84$

Find

WHP and BHP

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

Water power — WHP = γQH/550
Substituting
$WHP = 62.4(14.5)(164)/550 = 269.8 hp$
Brake power
$BHP = WHP/\eta$
Substituting
$BHP = 269.8/0.84 = 321.2 hp$

Answer:

WHP ≈ 269.8 hp; BHP ≈ 321.2 hp

Why the other options are there

226.6 hp (efficiency multiplied)
4.32 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 4

Water horsepower, shaft power and pump operating cost — Centrifugal Pump Characteristics (2)

A pump delivers 0.140 m³/s against a total dynamic head of 16 m at 82% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 17 h/day and $0.13/kWh.

Given

$Q = 0.140 m^{3}/s$
$H = 16 m$
$\eta = 0.82$
17 h/day at $0.13/kWh

Find

Water power, shaft power and annual cost

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

Formula
$P_water = \rho g Q H$
Substituting
$P = 1000(9.81)(0.140)(16) = 21.97 kW$
Formula
$P_shaft = P_water/\eta$
Substituting
$P_shaft = 21.97/0.82 = 26.80 kW$
Annual energy
$26.80 kW \times 17 h/day \times 365 = 166,282 kWh$
Cost — 166,282 kWh × $0.13 = $21,617 per year

Answer:

P_water = 22.0 kW, P_shaft = 26.8 kW, cost ≈ $21,617/yr

Why the other options are there

18.0 kW (efficiency multiplied)
29.5 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 5

Pump brake horsepower — Centrifugal Pump Characteristics (3)

A pump delivers 4.0 cfs against 41 ft of head at 76% efficiency. Find the water and brake horsepower.

Given

$Q = 4.0 cfs$
$H = 41 ft$
$\eta = 0.76$

Find

WHP and BHP

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

Water power — WHP = γQH/550
Substituting
$WHP = 62.4(4.0)(41)/550 = 18.61 hp$
Brake power
$BHP = WHP/\eta$
Substituting
$BHP = 18.61/0.76 = 24.48 hp$

Answer:

WHP ≈ 18.6 hp; BHP ≈ 24.5 hp

Why the other options are there

14.1 hp (efficiency multiplied)
0.30 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 6

Water horsepower, shaft power and pump operating cost — Centrifugal Pump Characteristics (3)

A pump delivers 0.240 m³/s against a total dynamic head of 10 m at 80% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 20 h/day and $0.16/kWh.

Given

$Q = 0.240 m^{3}/s$
$H = 10 m$
$\eta = 0.80$
20 h/day at $0.16/kWh

Find

Water power, shaft power and annual cost

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

Formula
$P_water = \rho g Q H$
Substituting
$P = 1000(9.81)(0.240)(10) = 23.54 kW$
Formula
$P_shaft = P_water/\eta$
Substituting
$P_shaft = 23.54/0.80 = 29.43 kW$
Annual energy
$29.43 kW \times 20 h/day \times 365 = 214,839 kWh$
Cost — 214,839 kWh × $0.16 = $34,374 per year

Answer:

P_water = 23.5 kW, P_shaft = 29.4 kW, cost ≈ $34,374/yr

Why the other options are there

18.8 kW (efficiency multiplied)
31.6 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 7

Pump brake horsepower — Centrifugal Pump Characteristics (4)

A pump delivers 13.5 cfs against 71 ft of head at 66% efficiency. Find the water and brake horsepower.

Given

$Q = 13.5 cfs$
$H = 71 ft$
$\eta = 0.66$

Find

WHP and BHP

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

Water power — WHP = γQH/550
Substituting
$WHP = 62.4(13.5)(71)/550 = 108.7 hp$
Brake power
$BHP = WHP/\eta$
Substituting
$BHP = 108.7/0.66 = 164.8 hp$

Answer:

WHP ≈ 108.7 hp; BHP ≈ 164.8 hp

Why the other options are there

71.8 hp (efficiency multiplied)
1.74 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 8

Water horsepower, shaft power and pump operating cost — Centrifugal Pump Characteristics (4)

A pump delivers 0.240 m³/s against a total dynamic head of 57 m at 80% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 12 h/day and $0.14/kWh.

Given

$Q = 0.240 m^{3}/s$
$H = 57 m$
$\eta = 0.80$
12 h/day at $0.14/kWh

Find

Water power, shaft power and annual cost

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

Formula
$P_water = \rho g Q H$
Substituting
$P = 1000(9.81)(0.240)(57) = 134.2 kW$
Formula
$P_shaft = P_water/\eta$
Substituting
$P_shaft = 134.2/0.80 = 167.8 kW$
Annual energy
$167.8 kW \times 12 h/day \times 365 = 734,749 kWh$
Cost — 734,749 kWh × $0.14 = $102,865 per year

Answer:

P_water = 134.2 kW, P_shaft = 167.8 kW, cost ≈ $102,865/yr

Why the other options are there

107.4 kW (efficiency multiplied)
179.9 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 9

Pump brake horsepower — Centrifugal Pump Characteristics (5)

A pump delivers 17.5 cfs against 63 ft of head at 68% efficiency. Find the water and brake horsepower.

Given

$Q = 17.5 cfs$
$H = 63 ft$
$\eta = 0.68$

Find

WHP and BHP

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

Water power — WHP = γQH/550
Substituting
$WHP = 62.4(17.5)(63)/550 = 125.1 hp$
Brake power
$BHP = WHP/\eta$
Substituting
$BHP = 125.1/0.68 = 183.9 hp$

Answer:

WHP ≈ 125.1 hp; BHP ≈ 183.9 hp

Why the other options are there

85.1 hp (efficiency multiplied)
2.00 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics

Example 10

Water horsepower, shaft power and pump operating cost — Centrifugal Pump Characteristics (5)

A pump delivers 0.150 m³/s against a total dynamic head of 18 m at 82% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 15 h/day and $0.09/kWh.

Given

$Q = 0.150 m^{3}/s$
$H = 18 m$
$\eta = 0.82$
15 h/day at $0.09/kWh

Find

Water power, shaft power and annual cost

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

Formula
$P_water = \rho g Q H$
Substituting
$P = 1000(9.81)(0.150)(18) = 26.49 kW$
Formula
$P_shaft = P_water/\eta$
Substituting
$P_shaft = 26.49/0.82 = 32.30 kW$
Annual energy
$32.30 kW \times 15 h/day \times 365 = 176,849 kWh$
Cost — 176,849 kWh × $0.09 = $15,916 per year

Answer:

P_water = 26.5 kW, P_shaft = 32.3 kW, cost ≈ $15,916/yr

Why the other options are there

21.7 kW (efficiency multiplied)
35.5 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Centrifugal Pump Characteristics