Reheat and Regenerative Rankine Cycle Problems

Group 2: Garcia, Kim, Memije, TañedaCHE122-1/C32

Reheat Rankine Cycle

Steam power plant that operates on a reheat Rankine cycle with 80 MW of net output. Steam enters the high pressure turbine at 10 MPa and 500 °C, and the low pressure turbine at 1 MPa and 500 °C. Steam leaves the condenser as a saturated liquid at 10 kPa. Isentropic efficiencies of the turbine and compressor are 80 percent and 95 percent, respectively. Find:

a) quality or temperature of steam at turbine exit.

b) thermal efficiency of the cycle. c) mass flow rate of the steam.

Source:http://www.tufts.edu/as/tampl/es7/hw/hw8soln.pdf

Solution

5Superheated Steamh5 = 3478.44 kJ/kg s5 = 7.7621 kJ/kg-K

@Low Pressure Turbine: s5=s6=7.7621 kJ/kg-K

P=10kPahl=191.81kJ/kg sl=0.6492 kJ/kg-Khv=2584.63 kJ/kg sv=8.1501 kJ/kg-K

7.762=0.6492 + xvs(8.1501-0.6492)xvs=0.948h6s=191.81+0.948(8.1501-0.6492)h6s=2460.86 kJ/kg

SolutionFLT: WT2s=h6s-h5=(2460.86-3478.44)kJ/kgWT2s=-1017.58 kJ/kgWT2=n(WT2s)=0.8(-1017.58kJ/kg)WT2=-814.07 kJ/kgh6=WT2+h5=(-814.07+3478.44)kJ/kgh6=2664.37 kJ/kg – Superheated Steam @ 10kPaT(0C) h(kJ/kg)

50 2592.56

T6 2664.37

100 2687.46

Interpolate:

T6=87.83ᵒC

Solution

3Superheated Steamh3 = 3373.63 kJ/kg s3 = 6.5965 kJ/kg-K

@High Pressure Turbine: s3=s4=6.5965 kJ/kg-K

@ 1MPa-Superheated SteamT(0C) h(kJ/kg) s(kJ/kg-

K)

179.91 2778.08 6.5864

T4 H4s 6.5965

200 2827.86 6.6939

Interpolate:

=2782.76 kJ/kg

SolutionFLT: WT1s=h4s-h3=(2782.76-3373.63)kJ/kgWT1s=-590.87 kJ/kgWT1=n(WT1s)=0.8(-590.87kJ/kg)WT2=-472.696 kJ/kgh4=WT1+h3=(-472.696+3373.63)kJ/kgh4=2900.934 kJ/kg

@ Reheat:FLT: qrh=h5-h4=(3478.44-2900.934)kJ/kgqrh= 577.506 kJ/kg

Solution

1Sat. Liquid @10kPah1 = 191.81 kJ/kg v1=0.00101 kg/m3

@Pump:Assume fluid is incompressible:Wps=V(P2-P1)= .00101kg/m3(10,000-10)kPaWps =10.09kJ/kgWp=Wps/n = (10.09/095)kJ/kgWp=10.62 kJ/kgFLT: h2=Wp+h1=(10.62+191.81)kJ/kgh2=202.43 kJ/kg@Boiler:FLT: qh =h3-h2=(3373.63-202.43)kJ/kg

qh=3172.2 kJ/kg

Solution

Wnet=WT1+WT2+WP

Wnet=-472.696-814.07+10.62

-Wnet=1276.146 kJ/kg

nth=/(qh+qrh)nth=1276.146/(3172.2+577.506)nth=0.340

(1276.146kJ/kg)

Regenerative Rankine Cycle

Steam leaves the boiler in a steam turbine plant at 2MPa, 300ᵒC is expanded to 3.5 kPa before entering the condenser. Compute the efficiency of a regenerative cycle, with an open feedwater heater operating at the pressure where the steam becomes saturated vapor.

Source: http://www.che.ufl.edu/ziegler/Problem%20set%20solutions.pdf

Pump 2

Boiler

Turb

ine

OFH

Condenser

Pump 1

qh

wT

qc

wP1wP2

1

23

4

5

6sat’d vapor

7

3.5 KPa

2 MPa, 300ᵒC

@ Stream 5 (superheated vapor, 300 0C, 2000 kPa):

h5 = 3023. 50 kJ/kg s5 = 6.7663 kJ/kg-K

@ Turbine: s5 = s6 = s7 = 6.7663 kJ/kg-K

@ P7 = 3.5 kPa, wet steam (sat’d liquid & vapor)hl=111.29kJ/kg sl=0.3887 kJ/kg-Khv=2549.96 kJ/kg sv=8.526 kJ/kg-K

6.7663 = 0.3887 + xv(8.526-0.3887)xv =0.784h7=111.29+0.784(2549.96-111.29)h7=2023.21 kJ/kg

@ s6 = 6.7663 kJ/kg-K saturated vapor:

P(kPa) s(kJ/kg-K)

550 6.7892

P6 6.7663

600 6.7600

h(kJ/kg)

2752.94

h6

2756.80

Interpolating two times:P6 = 588 MPah6 = 2755.87 kJ/kg

Mass and Energy balance @ OFH:h3 = yh6 – (1-y) h2

y = (h3 - h2)/(h6 – h2)

P(kPa) h(kJ/kg)

550 655.91

588 h3

600 670.54

Interpolating: h3 = 667.03 kJ/kg

@ Pump 1:Saturated liquid at 3.5 kPa:v = 0.00100 kg/m3 h1 = 111.29 kJ/kg

Assume incompressible fluid:wP1 = h2 – h1 = v(p2 – p1)wP1 = 0.00100 kg/m3 (2000 – 3.5) kPawP1 = 2 kJ/kg

h2 = 2 + 111.29 = 113.29 kJ/kg

y = (667.03 – 113.29)/(2755.87 – 113.29)y = 0.209

wT = yh6 + (1-y)h7 – h5 wT = (0.209)(2755.87 kJ/kg) + (1-0.209)(2023.21 kJ/kg) – 3023.50 kJ/kg wT = -847.16 kJ/kg

@ Pump 2:Saturated liquid at 588 kPa:v = 0.00110 kg/m3

Assume incompressible fluid:wP2 = h4 – h3 = v(p4 – p3) = 0.00110 kg/m3 (2000 – 588) kPawP2 = 1.55 kJ/kg

h4 = 667.03 + 1.55 = 668.58 kJ/kg

@ Boiler: qH = h5 – h4 = 3023.50 – 668.50 = 2355 kJ/kg