wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The exergy of 0.4 kg of wet steam at 0. 1 bar and 0.85 quality is ____ kJ .
(Assume dead state of 1 bar and 300 K)

Data :
At 1 bar, 300 K
h0=113.2kJ/kg;s0=0.395kJ/kgK

At 0.1 bar, 0.85 quality

h1=192kJ/kg;hg=2582kJ/kg

sf=0.649kJ/kgK;sg=8.148kJ/kgK

A
48.8 kJ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
57.9 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
11 2.6 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
93.8 kJ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 48.8 kJ
At dead state: T0=300K,h0=113.2kJ/kg,s0=0.395kJ/kgK,m=0.4kg

Exergy of system, ϕ=m[(hh0)T0(ss0)]

Now, h0T0s0=113.2(300×0.395)

=5.3kJ/kg

At 0.1 bar, 0.85 steam quality,

h=hf+x[hghf]

=192+0.85[2582192)

h=2223.5kJ/kg

s=sf+0.85[sgsf]

=0.649+0.85[8.1480.649]

=7.023kJ/kgK

Now, hT0s=2223.5300×7.023

=116.6kJ/kg

Now, Exergy, ϕ=m[116.6(5.3)]

=0.4×121.9=48.76kJ

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Change in State_Tackle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon