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:

Q˙W˙=m˙(h2h1+v222v122)\dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{v_2^2}{2} - \frac{v_1^2}{2} \right)

Where:

  • Q˙\dot{Q} = Heat transfer rate (assumed negligible, Q˙=0\dot{Q} = 0)
  • W˙\dot{W} = Power produced by the turbine
  • m˙\dot{m} = Mass flow rate = 2 kg/s2\ \text{kg/s}
  • h1,h2h_1, h_2 = Specific enthalpies at the inlet and outlet
  • v1,v2v_1, v_2 = Velocities at the inlet and outlet
  • Neglecting potential energy changes.

The equation simplifies to: W˙=m˙(h1h2)+m˙(v122v222)\dot{W} = \dot{m} \left( h_1 - h_2 \right) + \dot{m} \left( \frac{v_1^2}{2} - \frac{v_2^2}{2} \right)

Step 1: Determine specific enthalpies (h1h_1 and h2h_2)

  • Inlet properties: From the steam tables, at Pi=3 MPaP_i = 3\ \text{MPa} and Ti=400CT_i = 400^\circ\text{C}: h1=3213.6 kJ/kgh_1 = 3213.6\ \text{kJ/kg}

  • Outlet properties: At Po=100 kPaP_o = 100\ \text{kPa} and quality x=0.9x = 0.9, the specific enthalpy is: h2=hf+xhfgh_2 = h_f + x \cdot h_{fg} From the steam tables at Po=100 kPaP_o = 100\ \text{kPa}: hf=417.46 kJ/kg,hfg=2257.92 kJ/kgh_f = 417.46\ \text{kJ/kg}, \quad h_{fg} = 2257.92\ \text{kJ/kg} h2=417.46+0.92257.92=2459.57 kJ/kgh_2 = 417.46 + 0.9 \cdot 2257.92 = 2459.57\ \text{kJ/kg}

Step 2: Kinetic energy terms

The change in kinetic energy is: ΔKE=v122v222\Delta KE = \frac{v_1^2}{2} - \frac{v_2^2}{2} Substituting v1=10 m/sv_1 = 10\ \text{m/s} and v2=50 m/sv_2 = 50\ \text{m/s}: ΔKE=10225022=501250=1200 m2/s2\Delta KE = \frac{10^2}{2} - \frac{50^2}{2} = 50 - 1250 = -1200\ \text{m}^2/\text{s}^2 Convert to kJ/kg\text{kJ/kg} by dividing by 1000: ΔKE=1.2 kJ/kg\Delta KE = -1.2\ \text{kJ/kg}

Step 3: Power calculation

The total power output is: W˙=m˙[(h1h2)+ΔKE]\dot{W} = \dot{m} \left[ (h_1 - h_2) + \Delta KE \right] Substitute values: W˙=2[(3213.62459.57)+(1.2)]\dot{W} = 2 \cdot \left[ (3213.6 - 2459.57) + (-1.2) \right] W˙=2[754.031.2]\dot{W} = 2 \cdot \left[ 754.03 - 1.2 \right] W˙=2752.83=1505.66 kW\dot{W} = 2 \cdot 752.83 = 1505.66\ \text{kW}

Final Answer:

The power produced by the turbine is: 1505.7 kW\boxed{1505.7\ \text{kW}}


Would you like further clarification or an explanation on any specific part?
Here are 5 related questions you might consider:

  1. How is specific enthalpy determined for different states of steam?
  2. What are the assumptions in applying the First Law of Thermodynamics in turbines?
  3. How would the calculation change if heat transfer were not negligible?
  4. How does steam quality impact turbine performance?
  5. 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