TITLE.PM5

86 ENGINEERING THERMODYNAMICS

Dharm

\M-therm/th3-2.p65

Temperature of superheated steam = 400°C (Tsup = 400 + 273 = 673 K)

Dryness fraction, x = 0.9

Specific heat of superheated steam, cps = 2.3 kJ/kg K

(i)Internal energy of 1 kg of superheated steam :

At 20 bar. From steam tables,

ts = 212.4°C ; hf = 908.6 kJ/kg ; hfg = 1888.6 kJ/kg, vg = 0.0995 m^3 /kg

Now, hsup = hf + hfg + cps (Tsup – Ts)

= 908.6 + 1888.6 + 2.3(400 – 212.4)

= 3228.68 kJ/kg

Also, hsup = u + p. vsup

or u = hsup – p. vsup

The value of vsup can be found out by Charle’s law

v

T

v

T

g

g

= sup

sup

∴ v

vT

sup T

g sup

s

=

×

= ×

+

0 0995 673.

(212.4 273)

= 0.1379 m^3 /kg

Hence internal energy, u = 3228.68 – 20 × 10^5 × 0.1379 × 10–3
= 2952.88 kJ/kg. (Ans.)
(ii)Internal energy of 1 kg of wet steam :
h = hf + xhfg = 908.6 + 0.9 × 1888.6 = 2608.34 kJ/kg
Again h = u + p. x. vg
∴ u = h – p. x. vg = 2608.34 – 20 × 10^5 × 0.9 × 0.0995 × 10–3
= 2429.24 kJ/kg
Hence internal energy = 2429.24 kJ/kg. (Ans.)
Example 3.19. Two boilers one with superheater and other without superheater are de-
livering equal quantities of steam into a common main. The pressure in the boilers and main is
20 bar. The temperature of steam from a boiler with a superheater is 350°C and temperature of
the steam in the main is 250°C.
Determine the quality of steam supplied by the other boiler. Take cps = 2.25 kJ/kg.
Solution. Boiler B 1. 20 bar, 350°C :
Enthalpy, hh 1 = g 1 + cps (Tsup – Ts)
= 2797.2 + 2.25(350 – 212.4)
= 3106.8 kJ/kg ...(i)
Boiler B 2. 20 bar (temperature not known) :
hh xh 22 =+f 22 fg
= (908.6 + x 2 × 1888.6) kJ/kg ...(ii)
Main. 20 bar, 250°C.
Total heat of 2 kg of steam in the steam main
= 2[hg + cps (Tsup – Ts)]
= 2[2797.2 + 2.25 (250 – 212.4)] = 5763.6 kJ ...(iii)