Question

A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 5500C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feedwater in an open feedwater heater. The rest of the steam is reheated to 5000C and is expanded in the low-pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine a. The mass flow rate of steam through the boiler. b. The thermal efficiency of the cycle.

Answer 1

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

-  The mass flow rate of steam through the boiler

-  The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

                                        E_6 + E_2 = E_3

               flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

                                    y*h_6 + (1-y)*h_3 = h_3

                                  y*2810 + (1-y)*192.8 = 721

Compute y:                          y = 0.2018

- Heat produced by the boiler q_b:

                             q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

                    q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b:               q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

                                       q_c = (1-y)*(h_8 - h_1)

                                 q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c:               q_c = 1834.26 KJ/ kg

- Net power output w_net:

                                     w_net = q_b - q_c

                                w_net = 3303.58 - 1834.26

                                    w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

                                     flow(m) = P / w_net

compute flow(m)          flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

                                     u = 1 - (q_c / q_b)

                                     u = 1 - (1834.26/3303.58)

                                     u = 44.48 %

Answer 2

The mass flow rate of steam through the boiler is equal to 54.6 kg/s.

Given the following data:

• Condenser pressure = 10 kPa.

• Steam temperature = 550°C

• Steam-in pressure = 10 MPa.

• Steam-out pressure = 0.8 MPa.

How to determine the mass flow rate of steam.

From the steam table, the properties of saturated steam at a pressure of 10 kPa are:

$h_1=h_f=191.81\; kJ/kg\\\\v_1=v_f=0.00101 \;m^3 / kg$

For the enthalpy at state 2:

$h_2=h_1+w_{in}\\\\h_2=h_1+v_1(P_2-P_1)\\\\h_2=191.81+0.00101(800-10)\\\\h_2=191.81+0.00101(790)\\\\h_2=192.61\;kJ/kg$

From the steam table, the properties of saturated steam at a pressure of 0.8 MPa are:

$h_3=h_f=720.87\; kJ/kg\\\\v_3=v_f=0.001115 \;m^3 / kg$

For the enthalpy at state 4:

$h_4=h_3+w_{in}\\\\h_4=h_3+v_3(P_4-P_3)\\\\h_4=720.87+0.001115(10000-800)\\\\h_4=720.87+0.001115(9200)\\\\h_4=720.87+10.26\\\\h_4=731.12\;kJ/kg$

From the steam table, the properties of saturated steam at a pressure of 10 MPa and temperature of 550°C are:

$h_5=3502\; kJ/kg\\\\s_5=6.7585\;kJ/kgK$

At a pressure of 0.8 MPa:

$h_6=2812.1\; kJ/kg\\\\s_5=s_6=6.7585\;kJ/kgK$

From the steam table, the properties of saturated steam at a pressure of 0.8 MPa and temperature of 500°C are:

$h_7=3481.3\; kJ/kg\\\\s_7=7.8692\;kJ/kgK$

At a pressure of 0.8 MPa:

$s_7=s_8=7.8692\;kJ/kgK\\\\x_8=\frac{s_8-s_f}{s_{fg}} \\\\x_8=\frac{7.8692 - 0.6492}{7.4996} =0.9627\\\\\\\\h_8=h_f+x_8x_{fg}\\\\h_8=191.81+0.9627(2392.1)\\\\h_8=2492.7\;kJ/kg$

Next, we would apply the steady-flow energy balance equation to the feedwater heaters:

$\Delta E_{sys}=E_{in}-E_{out}=0\\\\E_{in}=E_{out}\\\\\sum m_ih_i=\sum m_eh_e\\\\m_6h_6+m_2h_2=m_3h_3\\\\yh_6+(1-y)h_2=1(h_3)\\\\y=\frac{h_3-h_2}{h_6-h_2} \\\\y=\frac{720.87-192.61}{2812.1-192.61}$

y = 0.2017.

Note: y is the fraction of steam that is extracted from the turbine.

For the heat input:

$q_{in}=(h_5-h_4)+(1-y)(h_7-h_6)\\\\q_{in}=(3502-731.12)+(1-0.2017)(3481.3-2812.1)\\\\q_{in}=3305.1\;kJ/kg$

For the heat output:

$q_{out}=(1-y)(h_8-h_1)\\\\q_{out}=(1-0.2017)(2494.7-191.81)\\\\q_{out}=1838.5\;kJ/kg$

For the net work done:

$W_{net}=q_{in}-q_{out}\\\\W_{net}=3305.1-1838.5\\\\W_{net}=1466.6\;kJ/kg$

Now, we can calculate the mass flow rate of this steam:

$M=\frac{P}{W_{net}} \\\\M=\frac{80 \times 10^3}{1466.6}$

M = 54.6 kg/s.

How to calculate the thermal efficiency of the cycle.

Mathematically, thermal efficiency is given by this formula;

$\eta = \frac{W{net}}{q_{in}} \\\\\eta = \frac{1466.6}{3305.1}\\\\\eta = 0.4437$

Thermal efficiency = 44.4%.

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