SME
Fluid Mechanics
Pumps and turbomachinery
Use this for “A pump at 1450 rpm gives 0.05 m³/s at 30 m head using 18 kW. What does it give at 1750 rpm?”
Chart
H(0.06034) ≈ 43.7
Original speed n₁ = 1,450 rpm
New speed n₂ = 1,750 rpm
Original power P₁ = 18 kW
New power P₂ = 31.64 kW
This is an example. Type your own values to start a new problem.
Q₂ (m³/s)
=
Q₁ (m³/s)
×
n₂ (rpm)
n₁ (rpm)
H₂ (m)
=
H₁ (m)
×
n₂ (rpm)
n₁ (rpm)
2
P₂ (kW)
=
P₁ (kW)
×
n₂ (rpm)
n₁ (rpm)
3
Formulas
Q₂
=
Q₁
×
(
n₂
n₁
)
0.06034
=
0.05
×
(
1,750
1,450
)
H₂
=
H₁
×
(
n₂
n₁
)
2
43.7
=
30
×
(
1,750
1,450
)
2
P₂
=
P₁
×
(
n₂
n₁
)
3
31.64
=
18
×
(
1,750
1,450
)
3
H₂
=
H₁
×
(
Q₂
Q₁
)
2
43.7
=
30
×
(
0.06034
0.05
)
2
Assumptions
•
The same pump with the same impeller diameter, so the two points are similar.
•
The efficiency stays the same at both speeds.
•
Flow scales with speed, head with speed², power with speed³.
Step-by-step
WE KNOW
n₂ = 1,750 rpm, n₁ = 1,450 rpm, Q₁ = 0.05 m³/s, H₁ = 30 m, P₁ = 18 kW
FIND
new flow (Q₂), new head (H₂), new power (P₂)
Step 1 · Find new flow (Q₂)
Use Q₂ = Q₁ × (n₂ ÷ n₁)
Flow grows in step with the speed: each turn moves the same volume.
Q₂
=
0.05
×
(
1,750
1,450
)
Q₂
=
0.05
×
1.207
Q₂
=
0.06034
m³/s
Step 2 · Find new head (H₂)
Use H₂ = H₁ × (n₂ ÷ n₁)²
Head grows as speed²: the blade speed is squared in the energy it gives.
H₂
=
30
×
(
1,750
1,450
)
2
H₂
=
30
×
1.2069
2
H₂
=
30
×
1.4566
H₂
=
43.7
m
Step 3 · Find new power (P₂)
Use P₂ = P₁ × (n₂ ÷ n₁)³
Power is flow times head, so it grows as speed³.
P₂
=
18
×
(
1,750
1,450
)
3
P₂
=
18
×
1.2069
3
P₂
=
18
×
1.758
P₂
=
31.64
kW
Check
Put the numbers back in. Both sides should match.
0.06034
=
0.05
×
(
1,750
1,450
)
✓
43.7
=
30
×
(
1,750
1,450
)
2
✓
31.64
=
18
×
(
1,750
1,450
)
3
✓
43.7
=
30
×
(
0.06034
0.05
)
2
✓
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Operating point: pump curve meets system curve
Topic check
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