SME
Process Design · Problem type
Pump power
Use this for “A pump moves 0.02 m³/s of water against 40 m of head at 70% efficiency. What power does it draw?”
This is an example. Type your own values to start a new problem.
P (kW)
=
ρ (kg/m³)
×
9.81
×
Q (m³/s)
×
H (m)
÷
η
Formulas
P
=
ρ
×
9.81
×
Q
×
H
η
11,211.4286
=
1,000
×
9.81
×
0.02
×
40
0.7
Formulas are worked in W. The step-by-step shows the conversions.
Assumptions
•
H is the total head the pump adds: the lift plus the pipe’s friction losses.
•
P is the shaft power; the efficiency η covers the pump’s own losses.
•
An incompressible liquid; g = 9.81 m/s².
Step-by-step
WE KNOW
Q = 0.02 m³/s, H = 40 m, ρ = 1,000 kg/m³, η = 0.7
FIND
shaft power (P)
Step 1 · Find shaft power (P)
Use P = ρ × 9.81 × Q × H ÷ η
ρgQH
is
the
power
the
liquid
gains;
the
shaft
gives
more,
÷
η
.
P
=
1,000
×
9.81
×
0.02
×
40
0.7
P
=
9,810
×
0.02
×
40
0.7
P
=
196.2
×
40
0.7
P
=
7,848
0.7
P
=
11,211.4286
W
Step 2 · Convert the answers
Change each answer back to the unit you chose.
P = 11,211.4286 W = 11.2114 kW (1 kW = 1000 W)
Check
Put the numbers back in. Both sides should match.
11,211.43
=
1,000
×
9.81
×
0.02
×
40
0.7
✓
Topic check
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