steam intially at 15 bar 250 degee celcius expai reversibly and polytroplically within a cylinder enclosed by a piston to 1.5 bar.calculate change in entropy of steam
Answers
Answered by
0
Explanation:
ΔS=2.303n×C
p
×log
T
1
T
2
[where,minkgandC
p
inJ/kg]
Entropy change for heating water from 27
o
C to 100
o
C.
ΔS
1
=2.303×
18
1000
×
1000
4180×18
×log
300
373
=910.55J
Entropy change for heating 1 kg H
2
O
(l)
to 1 kg steam at 100
o
C.
ΔS
2
=∫
373
473
T
nC
p
.dT
=m∫
373
473
T
(1670+0.49T)dt
(n is replaced by m since value of C
p
in 3kg
−1
K
−1
m∫
373
473
T
1670dT
+m∫
373
473
0.49dT
m×1670×2.303×[logT]
473
373
+m×0.49[T]
473
373
=1×1670×2.303×log
373
473
+1×0.49×100
=396.73+49=445.73J
Entropy change for heating 1 kg steam at 100
o
C to 1 kg steam at 200
o
C.
ΔS
3
=
T
q
=
373
23×10
5
=6166.2 J K
−1
∴ Total entropy change=ΔS
1
+ΔS
2
+ΔS
3
=910.55+6166.22+445.73
=7522.50J
Similar questions