Question

ان المتفانوس a) Explain the methods that can be adopted to increase the thermal efficiency of a vapor power cycle. You may use a T-s diagram to explain your reasons. b) A steam regenerative cycle, shown below, is delivering 100 MW of power. Steam from the boiler is supplied to the turbine at 8 MPa, 400° C. After expanding isentropically in the 1st stage turbine to a pressure of 1 MPa, steam is taken to the reheater where its temperature is raised back to 400°C and supplied to the 2nd stage turbine. Expansion in the 2nd stage turbine is also isentropic to a pressure of 400 kPa, where a fraction (y = 10%) of the steam is extracted and fed to the closed FWH as shown. Remaining steam continues to expand isentropically to the condenser pressure of 50 kPa. Steam is saturated liquid at the exit of the condenser. The extracted steam is also saturated liquid at the exit of the FWH. Assume work of both pumps are negligible (i.e., Wpump I = 0, Wpump II = 0 ). Assume there is no pressure drop in the system. Do the following: i) Draw the corresponding T- s diagram. Your plot must represent correct state corresponding to the steam at each point. Determine: ii) The required steam mass flow rate, kg/s iii) The amount of heat supplied at the boiler, kJ/s iv) The thermal efficiency of the cycle, % FOR PROPERTY VALUES USE THE CLOSEST MATCH ONLY IF THE DIFFERENCE IS NOT SIGNIFICANT. OTHERWISE INTERPOLATE. 8 MP, 400C 1 Turbine 1 MPa () 4 2. 400 kPa- Qpotter Boiler SO

Answer 1

Given data 1

P1 8 MPa

Ti 400 °C

P2 = 1 MPa

P3 = 8 MPa

T3 400 °C

P4 = 400 kPa

P5 50 kPa

T-s diagram →

T A 3 1 kg 11 у 9,10 2,4 7,8 1-y 5 6 S

Solution →

from superheated steam table, properties at 8 MPa and 400 °C

ni = 3139.4 kJ/kg

$1 = 6.3658 kJ/kgK

Properties at 1 MPa

h = 762.52 kJ/kg

hfg = 2014.6 kJ/kg

SA 2.1381 kJ/kgK

Sfg 4.4470 kJ/kgK

from T-s diagram, → $1 = S2

S1 = sf + 2 * 89

6.3658 -2.1381+T*4.4470

r= 0.9506

h2 = hy +2*hfa

h2 = 762.52 +0.9506 * 2014.6

h2 2677.7717 kJ/kg

Properties at 400 kPa

h = 604.65 kJ/kg

hfg = 2133.4 kJ/kg

SA 1.7765 kJ/kgK

Sfg 5.1190 kJ/kgK

from T-s diagram, → $1 = 84

S1 = sf + 2 * 89

6.3658 = 1.77654*5.1190

r= 0.8965

14 = hy +r*hta

h4 604.65 +0.8965 * 2133.4

h4 = 2517.291 kJ/kg

from superheated steam table, properties at 1 MPa and 400 °C

13 3264.5 kJ/kg

S3 7.4669 kJ/kgK

Properties at 50 kPa

h = 340.54 kJ/kg

hfg = 2304.7 kJ/kg

SA 1.0912 kJ/kgK

Sfg 6.5018 kJ/kgK

from T-s diagram, → S3 = 85

S3 = Sf +*819

$7.4669=1.0912+x*6.5018$

$h_{5}=h_{f}+x*h_{fg}$

$h_{5}=340.54+0.9806*2304.7$

$h_{5}=2600.5411\textup{ kJ/kg}$

$h_{11}=s_{f}\textup{ at 400 kPa}=604.65\textup{ kJ/kg}$

$w_{T}=\left ( h_{1}-h_{2} \right )+\left ( 1-y \right )*\left ( h_{2}-h_{4} \right )+\left ( 1-y \right )*\left ( h_{3}-h_{5} \right )$

$w_{T}=\left ( 3139.4-2677.77 \right )+\left ( 1-0.1 \right )*\left ( 2677.77-2517.291 \right )+\left ( 1-0.1 \right )*\left ( 3264.5-2600.541 \right )$

WT = 672.456 kJ/kg

$\textbf{steam mass flow rate}\left ( \dot{m} \right )\rightarrow$

$\dot{m}=\frac{100*10^{3}}{672.456}$

m 148.7084 kg/s

Heat supplied in boiler (3) ►

$q_{s}=\dot{m}*\left [ \left ( h_{1}-h_{11} \right )+\left ( 1-y \right )*\left ( h_{2}-h_{3} \right ) \right ]$

$q_{s}=148.7084*\left [ \left ( 3139.4-604.65 \right )+\left ( 1-0.1 \right )*\left ( 3264.5-2677.77 \right ) \right ]$

95 = 455.465 * 10kJ/s

$\eta =0.2195$