Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

372 | Thermodynamics

Analysis A sketch of the system and the T- sdiagram of the process are given

in Fig. 7–50.

(a) The enthalpies at various states are

State 1:

State 2a:

The exit enthalpy of the steam for the isentropic process h 2 sis determined from

the requirement that the entropy of the steam remain constant (s 2 ss 1 ):

State 2s:

Obviously, at the end of the isentropic process steam exists as a saturated

mixture since sfs 2 ssg. Thus we need to find the quality at state 2sfirst:

and

By substituting these enthalpy values into Eq. 7–61, the isentropic efficiency

of this turbine is determined to be

(b) The mass flow rate of steam through this turbine is determined from the

energy balance for steady-flow systems:

m#3.64 kg/s

2 MWa

1000 kJ>s

1 MW

bm# 1 3231.72682.4 2 kJ>kg

W

#

a,outm

# 1 h

1 h 2 a^2

m

#

h 1 W

#

a,out
m

#

h 2 a

E

#

inE

#

out

hT

h 1 h 2 a

h 1 h 2 s



3231.72682.4

3231.72407.9

0.667, or 66.7%

h 2 shf
x 2 shfg340.54
0.897 1 2304.7 2 2407.9 kJ>kg

x 2 s

s 2 ssf

sfg



6.92351.0912

6.5019

0.897

P 2 s50 kPa

1 s 2 ss 12

¬S¬

sf1.0912 kJ>kg#K

sg7.5931 kJ>kg#K

¬¬ 1 Table A–5 2

P 2 a50 kPa

T 2 a100°C

f¬h 2 a2682.4 kJ>kg¬¬ 1 Table A–6 2

P 1 3 MPA

T 1 400°C

f¬

h 1 3231.7 kJ>kg

s 1 6.9235 kJ>kg#K

¬¬ 1 Table A–6 2

T,°C

s

1

2 s

s 2 s = s 1

50 kPa

(^400) 3 MPa
(^1002)
P 1 = 3 MPa
T 1 = 400°C
P 2 = 50 kPa
STEAM
TURBINE
2 MW
T 2 = 100°C
Actual process
Isentropic process
FIGURE 7–50
Schematic and T-sdiagram for
Example 7–14.