SME
Engineering Thermodynamics · Problem type
Turbine power by the steady-flow energy balance
Use this for “2 kg/s of steam enters a turbine at 3230 kJ/kg and leaves at 2600 kJ/kg, losing 20 kW of heat. Find the power.”
Diagram
Energy in 6,460 = energy out 6,460 kW.
This is an example. Type your own values to start a new problem.
Ẇ (kW)
=
ṁ (kg/s)
×
(
h₁ (kJ/kg)
−
h₂ (kJ/kg)
)
+
(
Q̇ (kW)
)
Formulas
Ẇ
=
ṁ
×
(
h₁
−
h₂
)
+
Q̇
+1,240 = 2 × (3,230 − 2,600) + (−20)
Assumptions
•
Steady flow, one inlet and one outlet; kinetic and potential energy changes neglected.
•
Q̇ is heat into the device (negative when lost); Ẇ is power out (negative when put in).
•
Enthalpies are read from your steam table at the inlet and outlet states.
Step-by-step
WE KNOW
ṁ = 2 kg/s, h₁ = 3,230 kJ/kg, h₂ = 2,600 kJ/kg, Q̇ = −20 kW
FIND
power out (Ẇ)
Step 1 · Find power out (Ẇ)
Use Ẇ = ṁ × (h₁ − h₂) + Q̇
The steady-flow energy balance: the enthalpy the flow gives up, plus the heat in.
Ẇ
=
2
×
(3,230
−
2,600)
+
(−20)
Ẇ
=
2
×
630
+
(−20)
Ẇ
=
1,260
+
(−20)
Ẇ
=
+1,240
kW
Check
Put the numbers back in. Both sides should match.
+1,240 = 2 × (3,230 − 2,600) + (−20) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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