PROPERTIES OF PURE SUBSTANCES 83

Dharm

\M-therm/th3-1.p65

Heat supplied per kg of steam

= (1 – x 1 ) hfg 1 + cps (300 – 179.9)

= (1 – 0.85)2013.6 + 2.2 × 120.1

Q t

h

s

fg

=°

=

L

NM

O

QP

179.9 C... at 10 bar,

and 2013.6 kJ / kg

= 566.26 kJ/kg

Total heat supplied

= 0.909 × 566.26 = 514.7 kJ. (Ans.)

External work done during this process

= p()vxvsup 21 − 1 g []Qpp p 12 ==

= 10 × 10^5 v

T

g (^1) Ts xvg
2
1 1
× 1
F
HG
I
KJ
−
L
N
M
M
O
Q
P
P
sup
× 10–3 Q
v
T
v
T
ie v
vT
T
g
s
sup
sup
sup
g sup
s
1
1
2
2
2
12
1

L ×
N
M
M
O
Q
P
P
..,
= 10 × 10^5 0 194
300 273
179 9 273

. () 085
(. )

× +.

+

−×

L

N

M

O

Q

0.194P × 10–3

= 10 10 10

5

3

× (0.245 – 0.165) = 80 kJ/kg

∴ Percentage of total heat supplied (per kg) which appears as external work

= 5662680. = 0.141 = 14.1%. (Ans.)
Example 3.13. Find the specific volume, enthalpy and internal energy of wet steam at
18 bar, dryness fraction 0.85.
Solution. Pressure of steam, p = 18 bar
Dryness fraction, x = 0.85
From steam tables corresponding to 18 bar pressure :
hf = 884.6 kJ/kg, hfg = 1910.3 kJ/kg, vg = 0.110 m^3 /kg, uf = 883 kJ/kg, ug = 2598 kJ/kg.
(i)Specific volume of wet steam,
v = xvg = 0.85 × 0.110 = 0.0935 m^3 /kg. (Ans.)
(ii)Specific enthalpy of wet steam,
h = hf + xhfg = 884.6 + 0.85 × 1910.3
= 2508.35 kJ/kg. (Ans.)
(iii)Specific internal energy of wet steam,
u = (1 – x)uf + xug
= (1 – 0.85) × 883 + 0.85 × 2598
= 2340.75 kJ/kg. (Ans.)
Example 3.14. Find the dryness fraction, specific volume and internal energy of steam at
7 bar and enthalpy 2550 kJ/kg.
Solution. Pressure of steam, p = 7 bar
Enthalpy of steam, h = 2550 kJ
From steam tables corresponding to 7 bar pressure :
hf = 697.1 kJ/kg, hfg = 2064.9 kJ/kg, vg = 0.273 m^3 /kg,
uf = 696 kJ/kg, ug = 2573 kJ/kg.