SME
Fluid Mechanics · Problem type
Pump power with losses
Use this for “A pump lifts 0.01 m³/s of water 20 m with 4 m of pipe losses at 75% efficiency. What shaft power does it need?”
Diagram
The pump lifts the water Δz = 20 m and makes up the losses: hₚ = Δz + hL = 20 + 4 = 24 m.
The energy line is to scale (it ends at the upper surface); the pipe is not.
This is an example. Type your own values to start a new problem.
hₚ (m)
=
Δz (m)
+
hL (m)
P (kW)
=
ρ (kg/m³)
×
9.81
×
Q (m³/s)
×
hₚ (m)
÷
η
Formulas
hₚ = Δz + hL
24 = 20 + 4
P
=
ρ
×
9.81
×
Q
×
hₚ
η
3,139
=
1,000
×
9.81
×
0.01
×
24
0.75
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
Steady flow between two still reservoir surfaces, both open to the air.
•
The energy equation from surface to surface: the pump head lifts Δz and makes up hL.
•
η is the pump’s efficiency as a fraction; g = 9.81 m/s².
Step-by-step
WE KNOW
Q = 0.01 m³/s, Δz = 20 m, hL = 4 m, η = 0.75, ρ = 1,000 kg/m³
FIND
pump head (hₚ), shaft power (P)
Step 1 · Find pump head (hₚ)
Use hₚ = Δz + hL
Energy from surface to surface: the pump lifts the water Δz and makes up the losses.
hₚ = 20 + 4
hₚ = 24 m
Step 2 · Find shaft power (P)
Use P = ρ × 9.81 × Q × hₚ ÷ η
The
water
gains
ρgQhₚ
each
second;
the
shaft
must
give
more,
by
1
η
.
P
=
1,000
×
9.81
×
0.01
×
24
0.75
P
=
9,810
×
0.01
×
24
0.75
P
=
98.1
×
24
0.75
P
=
2,354.4
0.75
P
=
3,139.2
W
Step 3 · Convert the answers
Change each answer back to the unit you chose.
P = 3,139 W = 3.139 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
24 = 20 + 4 ✓
3,139
=
1,000
×
9.81
×
0.01
×
24
0.75
✓
Topic check
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