qin ~J, )~~/ 1 qOUl · h2 = hl + W p.in = 191.83 + 10.09 = 201.92 kJ/kg P3 = 10 MPa s T 3 = 500°C h3 = 3373.7 kJ/kg 53 = 6.5966 kJ/kg .K P6 = 10 kPa S6 = Ss (b) WT,ou! = (h3 -h4

Chapter 9 Vapor and Combined Power Cycles

9-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of thesteam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be

determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) From the steam tables (Tables A-4, A-S, and A-6),

hl = hf@ IOkPa = 191.83 kJ/kg T

VI =Vf@IOkPa =O.OO101m3/kg

3/

~

lkJ ...qin

~J,

)~~/ 1 qOUl ""'

Wp.in = VI(P2 -PI)

= ~.00101 m3/kg}10.000-10kPa{ ~\I kPa.m

= 10.09kJ/kg

h2 = hl + wp.in = 191.83+ 10.09 = 201.92 kJ/kg

PJ = 10 MPa hJ = 3373.7 kJ/kg

TJ =500°C JsJ =6.5966kJ/kg.K

s

P4 = 10 kPa

S4 = S3

~ = 6.5966- 0.6493 = 0.793X4 =

Sfg 7.5009

h4 = h f + x4h fg = 191.83 + (0.793 X2392.8 ) = 2089.3 kJ/kg

(b) qin = h3 -~ = 3373.7- 201.92 = 3171.78 kJ/kg

qout = h4 -hl = 2089.3-191.83 = 1897.47 kJ/kg

wne( = qin -qOU( = 3171.78 -1897.47 = 1274.31 kJ /kg

and

Ulrt -1274.31 kJ /kg = 4Q2>/o'1 th = -q::- -3171. 78 kJ /kg

= 165 kg/s(c) m=

,,t;;,~,

i,!

I~J


vqin

2~//!V IOkPa

4s 4J-qOUl

9-23 A steam power plant operates on a simple non ideal Rankine cycle between the specified pressurelimits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of thecooling water are to be determined.


Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

Thl = hf@ IOkPa = 191.83 kJ/kg

VI =Vf@IOkPa =0.00101 m3/kg

Wp,in =VI(P2 -PI)/1lp

= (0.00101 m3/kg)7,000-10 kPa { -1:!!, 1/(0.87)\ 1 kPa .m )

= 8.11 kJ/kg

h2 =hl +Wp,in =191.83+8.11=199.94kJ/kg

PJ = 7 MPa } hJ = 3410.5 kJ/kgTJ = 500°C $3 = 6.7975 kJ/kg .K

P4 = 10 kPa

S4 = S3

!!1::.!!!-

h3 -h4.,

T]T = ~ h4 = h3 -1lT (h3 -h4., )

= 3410.5- (0.87X3410.5 -2153.93) = 2317.28 kJ/kg

Thus,

and

qin = h3 -h2 = 3410.5 -199.94 = 3210.56 kJ Ikg

qout = h4 -hl = 2317.28 -191.83 = 2125.45 kJ Ikg

Wnet = qin -qout = 3210.56- 2125.45 = 1085.11 kJ Ikg

U1Et 1005.11 kJ /kg1]1h = -= = 3l.8:1/ol1in ~10.ffi kJ /kg

45.(00 kJ /s(b) fn- ~ --Utrl -1005.11 kJ /kg

(c) The rate of heat rejection to the cooling water and its temperature rise are

QOUI = mqoul = (41.5 kg/sX2125.45 kJ/kg)= 88,143 kJ/s

LlT = Qout = 88,143 kJ/s = 1054°Ccoo1lngwater ( . ) ( 00 X kJ oC) . mC cooling water 2 O kg/s 4.18 /kg .

= 41.5 kg /s

9-13


~

5

32

//

9.30 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The quality (or

temperature, if superheated) of the steam at the turbine exit, the thermal efficiency of the cycle, and themass flow rate of the steam are to be determined.


Analysis (a) From the steam tables (Tables A-4, A-S, and A-6),

hl = h I@' 10 kPa = 191.83 kJ/kg T

VI =:vI@ 10kPa = 0.00101 m3/kg

w p.in = VI (P2 -~ ) ,

= ~.00101 m3/kg}10,000-10kPa{ ~-~ -

\lkPa.m= 10.09 kJ/kg

h2 = hl + W p.in = 191.83 + 10.09 = 201.92 kJ/kg

sP3 = 10 MPa

T 3 = 500°C

h3 = 3373.7 kJ/kg

53 = 6.5966 kJ/kg .K

P6 = 10 kPa

S6 = Ss

(b) WT,ou! = (h3 -h4 )+ (h5 -h6 ) = 3373.7 -2782.5 + 3478.5- 2460.2 = 1609.5 kJ/kg

qin = (h3 -h2 )+ (h5 -h4 ) =3373.7 -201.92 + 3478.5- 2782.5 = 3867.78 kJ/kg

Wnet = WT.out -W p,in = 1609.5 -10.09 = 1599.41 kJ/kg

.w:m=~=

UlEt

OO.(XX) kJ /s

IW3.41 kJ /kg= 50.0 kg Is

9-17


5

4

~

l-y

7

MPa

~~j( 20 kPa \~,,-~

II q.:;,

9-38 A steam power plant that operates on an ideal regenerative Rankine cycle with an open feed waterheater is considered. The net work output per kg of steam and the thermal efficiency of the cycle are to bedetermined.


Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

ThJ = hf@ ZOkPa = 251.40 kJ/kg

3VI =vf@ZOkPa =0.001017 m /kg

wpl.in = Vl(PZ -PJ)

= ~.001017 m3/kg}400- 20 kPa { ~! 31

\lkPa.m= 0.39 kJ/kg

hz = hJ + wpl,in = 251.40+0.39 = 251.79 kJ/kg

P3 = 0.4 MPa h3 = h f@ 04 MPa = 604.74 kJ/kg

sat.liquid .V3 = V f@ 04 MPa = 0.001084 m3/kg

Wpl/jn = V3(P4 -P3)

= (0.001084 m3/kg}6000-400 kPa { ~

= 6.07 kJ/kg \ kPa .m

h4 = h3 + w pl/jn = 604.74 + 6.07 = 610.81 kJ/kg

lkJ

P5 =6MPa

T 5 = 450°C

h5 = 3301.8 kJ/kg

S5 = 6.1193 kJ/kg .K

6.7193-1.7766$6 -$ fP6 =0.4 MPalx6 ===0.9655

~ $ fg 5.1193$6 = $S

h6 =hf +x6hfg =604.74+(0.9655X2133.8)=2664.93kJ/kg

6.7193-0.8320S7 -S fP7 =20kPa }X7 =-;;:-= 7.0766S7 = s5

( X )h7 =hf +x7hfg =251.40+ 0.8319 2358.3 =2213.3kJ/kg

The fraction of steam extracted is determined from the steady-flow energy balance equation applied to the

feed water heater. Noting that Q = W = Me = f1pe = 0 ,

.Mh' "

= 0.8319

E E -A 1:' ",v\sleaay) -Oin -out -ilLs)'stem -

Ein = Eout

Lni;hi = Lniehe ~ ni6h6 + ni2h2 = ni3h3 ~ yh6 + (1- y }112 = l(h3)

where y is the fraction of steam extracted from the turbine ( = ni6 / ni3 ). Solving for y,

h3 -hz ---

y-~-hz

604.74 -251.79

2664.93-251.79= 0.1463

Then,

qin = h5 -h4 = 3301.8 -610.81 = 2691.0 kJ/kg

qout = (1- y Xh7 -hi ) = (1- 0. 1463X2213.3 -251.4) = 1674.9 kJ/kg

9-21

1Chapter 9 Vapor and Combined Power Cycles

Wnet = qin -qout = 2691.0 -1674.9 = 1016.1 kJ I kg

(b) The thermal efficiency is determined from

TJth = 1-i2!!L = 1- 1674.9 kJ Ikg = 37.8%

qin 2691.0 kJ Ikg

~

and

!!.

~~


/

1~ "/7l-y

9-43 A steam power plant operates on an ideal reheat-regenerative Rankine cycle with an open feedwater

heater. The mass flow rate of steam through the boiler and the thermal efficiency of the cycle are to be

determined.


Analysis (a) From the steam tables (Tables A-4, A-S, and A-6), T ,

hi = hf@ IOkPa = 191.83 kJ/kg

3vI=vf@IOkPa=0.00101m/kg :

Wpl.in = Vi(P2 -PI)= (0.00101 m3/kg)800-10kPa{ ~

\1 kPa.m= 0.80 kJ/kg

h2 =hl +Wpl,in =191.83+0.80=192.63kJ/kg

8PJ = 0.8 MPa 1 hJ = hf@ O.8MPa = 721.11 kJ/kg

sat.liquid Is

lkJ

3

5

6

'-l.:.

7l-y

I 8

V3 [email protected] =0.001115 m3/kg

Wpl/~n = V3(P4 -P3)= ~.001115 m3/kg)10,000- 800 kPa { -

= 10.26 k1/kg \ 1 kPa .m

h4 = h3 + w pl/ ~n = 721.11 + 10.26 = 731.37 k1/kg

P5 = 10 MPa h5 = 3500.9 k1/kg

T5 = 550°C s5 = 6.7561 kJ/kg. K

P6 = 0.8 MPa } h6 = 2811.9 kJ/kgS6 = S5

P7 = 0.8 MPa 1 -

T 7 = 500°C

6

~~:n ~.

-.::-J.

13

4

2

-0~-::--~Pg = 10 kPa

Sg = S7

h7 = 3480.6 kJ/kg

.S7 = 7.8673 kJ/kg. K

X8 =~=!.8673-Q.6493=0.9623s fg 7.5009

h8 = h f + x8h fg = 191.83 + (0. 9623X2392.8) = 2494.4 kJ/kg


feed water heaters. Noting that Q = W = Me = Llpe = O .

E. E.."O(sle.dy) ..in -OUI = !J.Esyslem = 0 ~ Ein = Eoul

Lniihi = Lniehe --+ ni6h6 + ni2h2 = ni3h3 ---+ yh6 + (1- y }JI2 = l(h3)

where y is the fraction of stearn extracted from the turbine ( = ~ / ~ ). Solving for y,

=~= 721.11-192.63 =0.2018y h6 -~ 2711.9-192.63

qin = (h5 -h4)+ (1- y Xh7 -h6 ) = (3500.9- 731.37)+ (1- 0.2098X3480.6 -2811.9) = 3297.9 kJ/kg

qoul = (1- y Xh8 -h1 )= (1- 0.2098X2494.4 -191.83) = 1819.5 kJ/kg

~'net = qin -qout = 3297.9 -1819.5 = 1478.5 kJ/kg

.Wm 8),00) ~ /sm=-=

lqa

Then,

and 1478.4kJ/kg = 54.1 kg /s

-1478.4 kJ/kgWnet(b) 77lh = -=- "" = 44.8%

3297.9 kJ/kg

9-29


9.84 An ideal reheat-regenerative Rankine cycle with one open feedwater heater is considered. The fraction

of steam extracted for regeneration and the thermal efficiency of the cycle are to be determined.


Analysis

-5 I -J,

T I5

~6

7~

6 l-yI 7

y/iI 0.6 MPa /-

4

2

~

3

-&~ 9 s(a) From the steam tables (Tables A-4, A-5, and A-6),

hi = hf @ 15kPa = 225.94 kJ/kg

VI =Vf@15kPa =0.001014 m3/kg

PJ = 0.6 MPa

sat.liquid

h3 = h f @ 0.6 MPa = 670.56 kJ/kg

V3 =Vf@06MPa =0.001101 m3fkg

P5 = 10 MPa } h5 = 3373.7 kJ/kgT5 = 500°C S5 = 6.5966 kJ/kg .K

P9 = 15 kPa

S9 = S7

9-63



feedwater heaters. Noting that Q = W = Me = ~e = O .

...

Documents

qin ~J, )~~/ 1 qOUl · h2 = hl + W p.in = 191.83 + 10.09 = 201.92 kJ/kg P3 = 10 MPa s T 3 = 500°C h3 = 3373.7 kJ/kg 53 = 6.5966 kJ/kg .K P6 = 10 kPa S6 = Ss (b) WT,ou! = (h3 -h4