550 ENGINEERING THERMODYNAMICS

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\M-therm\Th12-1.pm5

(iv) Diameter of the pipe connecting turbine with condenser, d :

π

4 d

(^2) × C = ms x 2 vg 2 ...(i)
Here, d = Diameter of the pipe (m),
C = Velocity of steam = 200 m/s (given),
ms = Mass of steam in kg/s,
x 2 = Dryness fraction at ‘2’, and
vg 2 = Specific volume at pressure 0.1 bar (= 14.67 m^3 /kg).
Substituting the various values in eqn. (i), we get
π
4 d
(^2) × 200 =^10000
3600
× 0.9 × 14.67
d =
10000 0 9 14 67 4
3600 200
×× ×^12
××
F
HG
I
KJ
.. /
π
= 0.483 m or 483 mm. (Ans.)
Example 12.2. In a steam power cycle, the steam supply is at 15 bar and dry and satu-
rated. The condenser pressure is 0.4 bar. Calculate the Carnot and Rankine efficiencies of the
cycle. Neglect pump work.
Solution. Steam supply pressure, p 1 = 15 bar, x 1 = 1
Condenser pressure, p 2 = 0.4 bar
Carnot and Rankine efficiencies :
From steam tables :
At 15 bar : ts = 198.3°C, hg = 2789.9 kJ/kg, sg = 6.4406 kJ/kg K
At 0.4 bar :ts = 75.9°C, hf = 317.7 kJ/kg, hfg = 2319.2 kJ/kg,
sf = 1.0261 kJ/kg K, sfg = 6.6448 kJ/kg K
T 1 = 198.3 + 273 = 471.3 K
T 2 = 75.9 + 273 = 348.9 K
∴ ηcarnot = TT
T
12
1
− = 471.3 348.9
471.3
−
= 0.259 or 25.9%. (Ans.)
ηRankine =
Adiabatic or isentropic heat drop
Heat supplied

hh
hhf
12
(^12)
−
−
where h 2 = hf 2 + x 2 hfg 2 = 317.7 + x 2 × 2319.2 ...(i)
Value of x 2 :
As the steam expands isentropically,
∴ s 1 = s 2
6.4406 = sf 2 + x 2 sfg 2 = 1.0261 + x 2 × 6.6448
∴ x 2 =

4406 0261

6448

−

= 0.815

∴ h 2 = 317.7 + 0.815 × 2319.2 = 2207.8 kJ/kg [From eqn. (i)]

Hence, ηRankine = 2789.9 2207.8

2789.9 317.7

−

−

= 0.2354 or 23.54%. (Ans.)