SME
Fluid Mechanics · Problem type
Operating point: pump curve meets system curve
Use this for “A pump gives H = 40 − 2000Q² m; the system needs H = 12 + 1500Q² m. Where does it run?”
Chart
Hₚᵤₘₚ(Q) = Hₛᵧₛ(Q)
The curves cross at Q ≈ 0.08944.
This is an example. Type your own values to start a new problem.
H (m)
=
H₀ (m)
−
a (s²/m⁵)
×
(
Q (m³/s)
)
2
H (m)
=
Hₛ (m)
+
b (s²/m⁵)
×
(
Q (m³/s)
)
2
Qₘₐₓ
Free-delivery flow
calculated · m³/s
Formulas
Q
=
√(
H₀
− Hₛ
a
+
b
)
0.08944
=
√(
40 − 12
2,000 + 1,500
)
Qₘₐₓ
=
√(
H₀
a
)
0.1414
=
√(
40
2,000
)
H
=
H₀
−
a
×
Q
2
24
=
40
−
2,000
×
0.08944
2
H
=
Hₛ
+
b
×
Q
2
24
=
12
+
1,500
×
0.08944
2
Assumptions
•
Both curves are fitted as parabolas, Q in m³/s and H in metres.
•
Turbulent pipe losses grow as Q², so the system needs Hₛ + bQ².
•
One pump in steady flow: it runs where its head meets the head the system needs.
Step-by-step
WE KNOW
H₀ = 40 m, a = 2,000 s²/m⁵, Hₛ = 12 m, b = 1,500 s²/m⁵
FIND
flow rate (Q), free-delivery flow (Qₘₐₓ), operating head (H)
Step 1 · Find flow rate (Q)
Use Q = √((H₀ − Hₛ) ÷ (a + b))
Set
the
two
heads
equal:
H₀
−
aQ²
=
Hₛ
+
bQ²,
so
(
a
+
b
)
Q
2
=
H₀
−
Hₛ.
Q
=
√(
40 − 12
2,000 + 1,500
)
Q
=
√(
28
3,500
)
Q
=
√
0.008
Q
=
0.08944
m³/s
Step 2 · Find free-delivery flow (Qₘₐₓ)
Use Qₘₐₓ = √(H₀ ÷ a)
The
pump
curve
reaches
zero
head
at
free
delivery:
H₀
−
aQ²
=
0.
Qₘₐₓ
=
√(
40
2,000
)
Qₘₐₓ
=
√
0.02
Qₘₐₓ = 0.1414 m³/s
Step 3 · Find operating head (H)
Use H = H₀ − a × Q²
Read
the
pump
curve
at
the
operating
flow:
the
head
falls
from
H₀
as
Q
2
grows.
H
=
40
−
2,000
×
0.08944
2
H
=
40
−
2,000
×
0.008
H
=
40
−
16
H
=
24
m
Check
Put the numbers back in. Both sides should match.
0.08944
=
√(
40 − 12
2,000 + 1,500
)
✓
0.1414
=
√(
40
2,000
)
✓
24
=
40
−
2,000
×
0.08944
2
✓
24
=
12
+
1,500
×
0.08944
2
✓
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