Math Problem Statement
Steam enters, then leaves a turbine operating at steady state with the following properties:
Inlet: Pi=3 MPa, Ti=400°C, vi=10 m/s, mass flow rate=2 kg/s
Outlet: Pi=100 kPa, quality: 0.9, vo=50 m/s
Assume that Heat transfer from the turbine and changes in potential energy are negligible.
What is the power produced by the turbine?
Solution
To calculate the power produced by the turbine, we use the First Law of Thermodynamics for a control volume operating at steady state:
Where:
- = Heat transfer rate (assumed negligible, )
- = Power produced by the turbine
- = Mass flow rate =
- = Specific enthalpies at the inlet and outlet
- = Velocities at the inlet and outlet
- Neglecting potential energy changes.
The equation simplifies to:
Step 1: Determine specific enthalpies ( and )
-
Inlet properties: From the steam tables, at and :
-
Outlet properties: At and quality , the specific enthalpy is: From the steam tables at :
Step 2: Kinetic energy terms
The change in kinetic energy is: Substituting and : Convert to by dividing by 1000:
Step 3: Power calculation
The total power output is: Substitute values:
Final Answer:
The power produced by the turbine is:
Would you like further clarification or an explanation on any specific part?
Here are 5 related questions you might consider:
- How is specific enthalpy determined for different states of steam?
- What are the assumptions in applying the First Law of Thermodynamics in turbines?
- How would the calculation change if heat transfer were not negligible?
- How does steam quality impact turbine performance?
- Why is the velocity change considered, and how significant is its effect?
Tip: Always refer to accurate steam tables for enthalpy and other thermodynamic property data at given conditions.
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Math Problem Analysis
Mathematical Concepts
Thermodynamics
Energy Balance
Kinetic Energy
Steam Tables
Formulas
First Law of Thermodynamics for steady flow: \( \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{v_2^2}{2} - \frac{v_1^2}{2} \right) \)
Specific enthalpy at outlet: \( h_2 = h_f + x \cdot h_{fg} \)
Change in kinetic energy: \( \Delta KE = \frac{v_1^2}{2} - \frac{v_2^2}{2} \)
Theorems
First Law of Thermodynamics
Suitable Grade Level
Undergraduate Engineering
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