SME
Fluid Mechanics · Problem type
Minor losses at fittings
Use this for “Water at 2 m/s passes fittings whose loss coefficients add to 3.5. Find the minor head loss.”
Diagram
The
energy
grade
line
falls
by
hL
=
0.7136
m
along
the
pipe;
the
hydraulic
grade
line
runs
V
2
2g
=
0.2039
m
under
it.
Heads are to the scale at the left; the pipe’s length is not. ΔP = ρghL = 7,000 Pa.
Sum of loss coefficients ΣK = 3.5
This is an example. Type your own values to start a new problem.
hₘ (m)
=
ΣK
×
(
V (m/s)
)
2
2
×
9.81
ΔP (kPa)
=
ρ (kg/m³)
×
9.81
×
hₘ (m)
Formulas
hₘ
=
ΣK
×
V
2
2 × 9.81
0.7136
=
3.5
×
2
2
2 × 9.81
ΔP
=
ρ
×
9.81
×
hₘ
7,000 = 1,000 × 9.81 × 0.7136
Formulas are worked in Pa. The step-by-step shows the conversions.
Assumptions
•
Each fitting loses K velocity heads V² ÷ 2g, all at the pipe’s one mean speed.
•
ΣK adds the fittings’ coefficients from a table (a sharp entrance 0.5, an exit 1).
•
ΔP = ρghₘ, the same as ΣK × ½ρV², is the pressure the fittings cost; g = 9.81 m/s².
Step-by-step
WE KNOW
ΣK = 3.5, V = 2 m/s, ρ = 1,000 kg/m³
FIND
minor head loss (hₘ), pressure drop (ΔP)
Step 1 · Find minor head loss (hₘ)
Use hₘ = ΣK × V² ÷ (2 × 9.81)
Each
fitting
costs
K
velocity
heads,
so
all
of
them
cost
ΣK
times
V
2
2g
.
hₘ
=
3.5
×
2
2
2 × 9.81
hₘ
=
3.5
×
2
2
19.62
hₘ
=
3.5
×
4
19.62
hₘ
=
14
19.62
hₘ = 0.7136 m
Step 2 · Find pressure drop (ΔP)
Use ΔP = ρ × 9.81 × hₘ
A head of hₘ metres of the fluid is a pressure of ρghₘ.
ΔP = 1,000 × 9.81 × 0.7136
ΔP = 9,810 × 0.7136
ΔP = 7,000 Pa
Step 3 · Convert the answers
Change each answer back to the unit you chose.
ΔP = 7,000 Pa = 7 kPa (1 kPa = 1000 Pa)
Check
Put the numbers back in. Both sides should match.
0.7136
=
3.5
×
2
2
2 × 9.81
✓
7,000 = 1,000 × 9.81 × 0.7136 ✓
Topic check
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