ENGR 2213 Thermodynamics

ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and Mechanical

Engineering

University of Oklahoma

Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles

T

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Turbine

Boiler

Condenser

Pump

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Boiler

Condenser

Pump

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Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles

net out

in in

w q1

q q 6 1

3 2 5 4

h h1

(h h ) (h h )

qin = (h3 – h2) + (h5 – h4)

qout = h6 – h1

p 2 1

t 3 4 5 6

w h hback work ratio

w (h h ) (h h )

T

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6wt = (h3 – h4) + (h5 – h6)

wp = h2 – h1 = v(p2 – p1)

Ideal Reheat Rankine CyclesIdeal Reheat Rankine Cycles

The reheat process in general does not significantlychange the cycle efficiency.

The sole purpose of the reheat cycle is to reducethe moisture content of the steam at the final stagesof the expansion process.

Example 1Example 1

Consider a steam power plant operating on the ideal reheat Rankine cycle. The steam entersthe turbine at 15 MPa and 600 ºC and is condensedin the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4%, determine(a) the pressure at which the steam should be reheated.(b) the thermal efficiency of this cycle.

Example 1 (continued)Example 1 (continued)

State 6: saturated mixture at p1 = 10 kPa, x6 = 0.896

State 5: superheated vapor at T5 = 600 ºC

s5 = s6 = 7.370 kJ/kg·K

h6 = hf + x6hfg = 191.83 + 0.896(2392.8) = 2335.8 kJ/kgs6 = sf + x6sfg = 0.6493 + 0.896(7.5009) = 7.370 kJ/kg·K

Table A-6, p5 = 4 MPa

Example 1 (continued)Example 1 (continued)

State 2: compressed liquid at p2 = 15 MPa

wp = v(p2 – p1) = (0.001008)(15000-10) = 15.11 kJ/kgh2 = h1 + wp = 191.83 + 15.11 = 206.94 kJ/kg

Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg

State 1: saturated liquid at p1 = 10 kPa

Example 1 (continued)Example 1 (continued)

State 3: superheated vapor at p3 = 15 MPa and T3 = 600 ºC

Table A-6 h3 = 3582.3 kJ/kg s3 = 6.6776 kJ/kg·K

State 4: superheated vapor at p4 = 4 MPa

s4 = s3 = 6.6776 kJ/kg·KTable A-6 h4 = 3154.3 kJ/kg T4 = 375.5 ºC

Example 1 (continued)Example 1 (continued)

qin = (h3 – h2) +(h5 – h4) = (3582.3 – 206.94) + (3674.4 – 3154.3) = 3895.46 kJ/kg

qout = h6 – h1 = 2335.8 – 191.83 = 2143.97 kJ/kg

out

in

q 2143.971 1 0.45

q 3895.46

0.43 (without reheat)

Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles

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Open Feedwater Heater

Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles

net

in

w

q 5 6 6 7 2 1 4 3

5 4

(h h ) (1 y)(h h ) (1 y)(h h ) (h h )

(h h )

qin = h5 – h4

qout = (1 – y)(h7 – h1)

wt = (h5 – h6) + (1 – y)(h6 – h7)

wp1 = h2 – h1 = v1(p2 – p1) T

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6wp2 = h4 – h3 = v3(p4 – p3)

1-yywp = (1 – y)wp1 + wp2

Ideal Regenerative Rankine CyclesIdeal Regenerative Rankine Cycles

net out

in in

w q1

q q

7 1

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(1 y)(h h )1

(h h )

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61-yy

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P 2 FWHy

1-y

yh6 + (1-y)h2 = h3

3 2

6 2

h hy

h h

Example 2Example 2

Consider a steam power plant operating on the ideal regenerative Rankine cycle using open feedwater heater. The steam enters the turbineat 15 MPa and 600 ºC and is condensed in the condenser at a pressure of 10 kPa. Some steam leaves the turbine at a pressure of 1.2 MPa and enters the feedwater heater. Determine(a) the fraction of steam extracted from the turbine.(b) the thermal efficiency of this cycle.

Example 2 (continued)Example 2 (continued)

State 2: compressed liquid at p2 = 1.2 MPa

wp1 = v(p2 – p1) = (0.001008)(1200-10) = 1.20 kJ/kgh2 = h1 + wp1 = 191.83 + 1.2 = 193.03 kJ/kg

Table A-5 h1 = hf = 191.83 kJ/kg v1 = vf = 0.001008 m3/kg

State 1: saturated liquid at p1 = 10 kPa

Example 2 (continued)Example 2 (continued)

State 3: saturated liquid at p3 = 1.2 MPa

Table A-5 h3 = 798.65 kJ/kg v3 = 0.001139 m3/kg

State 4: compressed liquid at p4 = 15 MPawp2 = v3(p4 – p3) = (0.001139)(15000-1200) = 15.72 kJ/kg

h4 = h3 + wp2 = 798.65 + 15.72 = 814.37 kJ/kg

Example 2 (continued)Example 2 (continued)

State 5: superheated vapor at p5 = 15 MPa and T5 = 600 ºC

State 6: p6 = 1.2 MPa

s6 = s5 = 6.6776 kJ/kg·KTable A-6, h6 = 2859.5 kJ/kg

Table A-6 h5 = 3582.3 kJ/kg s5 = 6.6776 kJ/kg·K

State 7: p7 = 10 kPas7 = s6 = s5 = 6.6776 kJ/kg·K

Example 2 (continued)Example 2 (continued)

7 f7

fg

s s 6.6776 0.6493x 0.804

s 7.5009

State 7: saturated mixture at p4 = 10 kPa

h7 = hf + x7hfg

= 191.83 + 0.804(2392.8) = 2115.6 kJ/kg3 2

6 2

h h 798.65 193.03y 0.227

h h 2859.5 193.03

Example 2 (continued)Example 2 (continued)

qin = h5 – h4 = 3582.3 – 814.37 = 2767.93 kJ/kg

qout = (1 – y)(h7 – h1) = (1 – 0.227)(2115.6 – 191.83) = 1487.1 kJ/kg

out

in

q 1487.11 1 0.463

q 2767.93

0.43 (without reheat)