Page 61 - DJJ20063- Thermodynamics 1

DJJ20063- Thermodynamics 1

At 120 bar,

T

700
2 −
v − . 3 15 9 x 10 2 − . 3 605 x 10 − . 3 15 9 x 10 2 −
650 1 =
650 − 600 700 − 600
600
2 −
3
v 1 = 3.382 x 10 m /kg
v
-2 -
3.159 x 10 v1 3.605 x 10

o
Secondly, find the specific volume (v2) at 130 bar and 650 C;

At 130 bar,
T

700 2 − 2 − 2 −
v − . 2 901 x 10 3 .318 x 10 − . 2 901 x 10
2 =
650 650 − 600 700 − 600

600
3
-2
v2 = 3.1095 x 10 m /kg
v
2.901 x 10 -2 v2 3.318 x 10 -

o
o
Now interpolate between v1 at 120 bar, 650 C, and v2 at 130 bar, 650 C in order
o
to find v at 125 bar and 650 C.

o
At 650 C,
v − v v − v
1 = 2 1
P 125 − 120 130 − 120
130

2 −
125 v − . 3 382 x 10 2 − 3 .1095 x 10 − . 3 382 x 10 2 −
=
120 125 − 120 130 − 120

v
v1 v v2
3
-2
v = 3.246 x 10 m /kg

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