Basic Mechanical Engineering (3110006)

BE | Semester-1   Summer-2020 | 11-06-2020

Q3) (c)

1.5kg of steam at a pressure of 10bar and temperature of 250°C is expanded until the pressure becomes 2.8bar. The dryness fraction of steam is then 0.9. Calculate change in Internal Energy.

Given data:
m = 1.5  kg
p1 = 10 bar
Tsup,1 = 300°C
p2 = 2.8 bar
x2 = 0.9
Calculate:
du = ?
 
Solution:
 
By using steam table for superheated steam (at state 1): at p1 = 10 bar and Tsup,1 = 300°C
 
hsup,1 = 2943 kJkg
vsup,1 = 0.2328 m3kg
 
By using steam table (for dry saturated steam): at p2 = 2.8 bar
 
hf,2 = 551.4 kJkg
hg,2 = 2721.5  kJkg
hfg,2 = 2170.1 kJkg
vg,2 = 0.646  m3kg
 
The internal energy per kg of superheated steam,
 
usup,1 = hsup,1 - p1 × vsup,1
usup,1 = 2943 - 1000 × 0.2328
usup,1 = 2710.2 kJkg
 
But, the change of internal energy = u2 - usup,1  kJkg
 
Finding unknown u2 :
 
Internal energy of wet steam after expansion, = u2 = h2 - p·v2
 
Finding unknown, h2 :
 
Specific enthalpy of wet steam, 
 
h2 = hf,2  + x2 · hfg,2
h2 = 551.4 + 0.9 · 2170.1
h2 = 2309.181 kJkg
 
Finding unknown, v2 :
 
Specific enthalpy of wet steam,
 
v2 = x2 · vg,2
v2 = 0.9 ×  0.646
v2 = 0.5814 m3kg
 
Therefore,  u2 = h2 - p·v2 = 2309.181 - 1000 × 0.5814 = 1721.781 kJkg
 
Hence, change of internal energy, u2 - usup,1 = 1721.181 - 2710.2 = -988.419  kJkg
 
But, change of internal energy for 1.5 kg steam is, = 1.5 × -988.419 = -1482.628  kJkg
 
Negative sign indicates decrease in internal energy.