SME
Engineering Thermodynamics · Problem type
Mixing chamber outlet enthalpy
Use this for “2 kg/s of water at 335 kJ/kg mixes with 3 kg/s at 84 kJ/kg. Find the outlet enthalpy.”
Diagram
Total in: 2 + 3 = 5; out: 5 kg/s.
Energy in 922 = energy out 922 kW.
This is an example. Type your own values to start a new problem.
ṁ₃ (kg/s)
=
ṁ₁ (kg/s)
+
ṁ₂ (kg/s)
ṁ₃ (kg/s)
×
h₃ (kJ/kg)
=
ṁ₁ (kg/s)
×
h₁ (kJ/kg)
+
ṁ₂ (kg/s)
×
h₂ (kJ/kg)
Formulas
ṁ₃
=
ṁ₁
+
ṁ₂
5 = 2 + 3
ṁ₃
×
h₃
=
ṁ₁
×
h₁
+
ṁ₂
×
h₂
5 × 184.4 = 2 × 335 + 3 × 84
Assumptions
•
Steady, adiabatic and no work; kinetic and potential energy changes neglected.
•
Mass in equals mass out, and so does the energy carried, ṁh.
Step-by-step
WE KNOW
ṁ₁ = 2 kg/s, h₁ = 335 kJ/kg, ṁ₂ = 3 kg/s, h₂ = 84 kJ/kg
FIND
outlet flow (ṁ₃), outlet enthalpy (h₃)
Step 1 · Find outlet flow (ṁ₃)
Use ṁ₃ = ṁ₁ + ṁ₂
Mass in equals mass out.
ṁ₃
=
2
+
3
ṁ₃
=
5
kg/s
Step 2 · Find outlet enthalpy (h₃)
Use ṁ₃ × h₃ = ṁ₁ × h₁ + ṁ₂ × h₂
The energy in, shared over the outlet flow: a mass-weighted mean.
h₃
=
ṁ₁
×
h₁
+
ṁ₂
×
h₂
ṁ₃
h₃
=
2 × 335 + 3 × 84
5
h₃
=
670 + 252
5
h₃
=
922
5
h₃
=
184.4
kJ/kg
Check
Put the numbers back in. Both sides should match.
5 = 2 + 3 ✓
5 × 184.4 = 2 × 335 + 3 × 84 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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