Question

A power plant operates on a regenerative vapor power cycle with one open feedwater heater. Steam enters the first turbine stage at 12 \mathrm{MPa}, 560^{\circ} \mathrm{C}12MPa,560 ∘ C and expands to 1 MPa, where some of the steam is extracted and diverted to the open feedwater heater operating at 1 MPa. The remaining steam expands through the second turbine stage to the condenser pressure of 6 kPa. Saturated liquid exits the open feedwater heater at 1 MPa. The net power output for the cycle is 330 MW. For isentropic processes in the turbines and pumps,

determine:
a. the cycle thermal efficiency.
b. the mass flow rate into the first turbine stage, in kg/s.
c. the rate of entropy production in the open feedwater heater, in kW/K.

Answer 1

a. 46.15%

b. 261.73 kg/s

c. 54.79 kW/K

Step-by-step explanation:

a. State 1

The parameters given are;

T₁ = 560°C

P₁ = 12 MPa = 120 bar

Therefore;

h₁ = 3507.41 kJ/kg, s₁ = 6.6864 kJ/(kg·K)

State 2

p₂ = 1 MPa = 10 bar

s₂ = s₁ = 6.6864 kJ/(kg·K)

h₂ = (6.6864 - 6.6426)÷(6.6955 - 6.6426)×(2828.27 - 2803.52) + 2803.52

= (0.0438 ÷ 0.0529) × 24.75 = 2824.01 kJ/kg

State 3

p₃ = 6 kPa = 0.06 bar

s₃ = s₁ = 6.6864 kJ/(kg·K)

sg = 8.3291 kJ/(kg·K)

sf = 0.52087 kJ/(kg·K)

x = s₃/sfg = (6.6864- 0.52087)/(8.3291 - 0.52087) = 0.7896

(h₃ - 151.494)/2415.17 = 0.7896

∴ h₃ = 2058.56 kJ/kg

State 4

Saturated liquid state

p₄ = 0.06 bar= 6000 Pa, h₄ = 151.494 kJ/kg, s₄ = 0.52087 kJ/(kg·K)

State 5

Open feed-water heater

p₅ = p₂ = 1 MPa = 10 bar = 1000000 Pa

s₄ = s₅ = 0.52087 kJ/(kg·K)

h₅ = h₄ + work done by the pump on the saturated liquid

∴ h₅ = h₄ + v₄ × (p₅ - p₄)

h₅ = 151.494 + 0.00100645 × (1000000 - 6000)/1000 = 152.4944113 kJ/kg

Step 6

Saturated liquid state

p₆ = 1 MPa = 10 bar

h₆ = 762.683 kJ/kg

s₆ = 2.1384 kJ/(kg·K)

v₆ = 0.00112723 m³/kg

Step 7

p₇ = p₁ = 12 MPa = 120 bar

s₇ = s₆ = 2.1384 kJ/(kg·K)

h₇ = h₆ + v₆ × (p₇ - p₆)

h₇ = 762.683 + 0.00112723 * (12 - 1) * 1000 = 775.08253 kJ/kg

The fraction of flow extracted at the second stage, y, is given as follows

`y = (762.683 - 152.4944113 )/(2824.01 - 152.4944113 ) = 0.2284`

The turbine control volume is given as follows;

`\frac{\dot{W_t}}{\dot{m_(1)}} = \left (h_(1) - h_(2) \right ) + \left (1 - y \right )\left (h_(2) - h_(3) \right )`

= (3507.41 - 2824.01) + (1 - 0.22840)*(2824.01 - 2058.56) = 1274.02122 kJ/kg

For the pumps, we have;

`\frac{\dot{W_p}}{\dot{m_(1)}} = \left (h_(7) - h_(6) \right ) + \left (1 - y \right )\left (h_(5) - h_(4) \right )`

= (775.08253 - 762.683) + (1 - 0.22840)*(152.4944113 - 151.494)

= 13.17 kJ/kg

For the working fluid that flows through the steam generator, we have;

`\frac{\dot{Q_(in)}}{\dot{m_(1)}} = \left (h_(1) - h_(7) \right )`

= 3507.41 - 775.08253 = 2732.32747 kJ/kg

The thermal efficiency, η, is given as follows;

`\eta = \frac{\frac{\dot{W_t}}{\dot{m_(1)}} -\frac{\dot{W_p}}{\dot{m_(1)}}}{\frac{\dot{Q_(in)}}{\dot{m_(1)}}}`

η = (1274.02122 - 13.17)/2732.32747 = 0.4615 which is 46.15%

(762.683 - 152.4944113)/(2824.01 - 152.4944113)

b. The mass flow rate,
`\dot{m_(1)}`, into the first turbine stage is given as follows;

`\dot{m_(1)} = \frac{\dot{W_(cycle)}}{\frac{\dot{W_t}}{\dot{m_(1)}} -\frac{\dot{W_p}}{\dot{m_(1)}}}`

`\dot{m_(1)}` = 330 *1000/(1274.02122 - 13.17) = 261.73 kg/s

c. From the entropy rate balance of the steady state form, we have;

`\dot{\sigma }_(cv) = \sum_(e)^{}\dot{m}_(e)s_(e) - \sum_(i)^{}\dot{m}_(i)s_(i) = \dot{m}_(6)s_(6) - \dot{m}_(2)s_(2) - \dot{m}_(5)s_(5)`

`\dot{\sigma }_(cv) = \dot{m}_(6) \left [s_(6) - ys_(2) - (1 - y)s_(5) \right ]`

= 261.73 * (2.1384 - 0.2284*6.6864 - (1 - 0.2284)*0.52087 = 54.79 kW/K