SME
Engineering Thermodynamics · Problem type
Nozzle exit speed
Use this for “Steam enters a nozzle at 30 m/s and its enthalpy drops 50 kJ/kg. How fast does it leave?”
Diagram
Energy in 3,050 = energy out 3,050 kJ/kg.
This is an example. Type your own values to start a new problem.
(
V₂ (m/s)
)
2
=
(
V₁ (m/s)
)
2
+
2
×
1000
×
(
h₁ (kJ/kg)
−
h₂ (kJ/kg)
)
Formulas
V₂
=
√(
V₁
2
+
2
×
1,000
×
(
h₁
−
h₂
))
317.6
=
√(
30
2
+
2
×
1,000
×
(3,050
−
3,000))
Assumptions
•
Steady, adiabatic and no work; the potential energy change is neglected.
•
The enthalpy drop becomes kinetic energy: h₁ − h₂ = (V₂² − V₁²) ÷ 2.
•
h is in kJ/kg: × 1000 turns it into J/kg (m²/s²), so V comes out in m/s.
Step-by-step
WE KNOW
V₁ = 30 m/s, h₁ = 3,050 kJ/kg, h₂ = 3,000 kJ/kg
FIND
exit speed (V₂)
Step 1 · Find exit speed (V₂)
Use V₂ = √(V₁² + 2 × 1,000 × (h₁ − h₂))
The
energy
balance
h₁
+
V₁
2
2
=
h₂
+
V₂
2
2,
with
h
in
J/kg
(×
1000),
solved
for
V₂
.
V₂
=
√(
30
2
+
2
×
1,000
×
(3,050
−
3,000))
h₁
−
h₂
=
50
kJ/kg
=
50,000
J/kg
V₂
=
√
30
2
+ 2 × 50,000
V₂
=
317.6
m/s
Check
Put the numbers back in. Both sides should match.
317.6
=
√(
30
2
+
2
×
1,000
×
(3,050
−
3,000))
✓
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Mixing chamber outlet enthalpy
Topic check
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