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julia-pushkina [17]

3 years ago

5

Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea

m is then reheated at constant pressure to a temperature of 5008C before it is routed to the second stage, where it exits at 30 kPa and a quality of 97 percent. The work output of the turbine is 5 MW. Assuming the surroundings to be at 258C, determine the reversible power output and the rate of exergy destruction within this turbine.

1 answer:

Nataly [62]3 years ago

6 0

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = $h_{f4}$ + x₄× $h_{fg}$ = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ = $s_{f4}$ + x₄× $s_{fg}$ = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, $\dot X_{dest}$ , is given as follows;

$\dot X_{dest}$ = T₀ × $\dot S_{gen}$ = T₀ × $\dot m$ × (s₄ + s₂ - s₁ - s₃)

$\dot X_{dest}$ = T₀ × $\dot W$ ×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ $\dot X_{dest}$ = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, $\dot W_{rev}$ = $\dot W_{}$ + $\dot X_{dest}$ ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

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