thermodynamics t2
TRANSCRIPT
Question 1a) When p=1.5 bar, from the saturated water and
steam table, hf=467 kJ/kg, hg=2693 kJ/kg
h=2226 kJ/kg
Hence:hf<h<hg
So, it’s wet steam
f g
Question 1b) When p=5.5 bar, from the saturated water and steam table, vg=0.3427 m3/kg
vf=0.001m3/kg
v=0.3748 m3/kg
Hence:v>vg
So, it’s superheated steam
f g
Question 1c) When p=9 bar, from the saturated water and steam table, uf=742 kJ/kg, ug=2581 kJ/kg
u=2581 kJ/kg
Hence:u=ug
So, it’s dry saturated steam
f g
Question 1d) When p=13 bar, from the saturated water and steam table, ts=191.6˚C
t=171.6˚C
Hence:t<ts
So, it’s compressed water
Question 1e) When p=20 bar, from the saturated water and steam table, uf=907 kJ/kg, ug=2600 kJ/kg
u=907 kJ/kg
Hence:u=uf
So, it’s saturated water
f g
Question 1f) When p=24 bar, from the saturated water and steam table, hf=952 kJ/kg, hg=2802 kJ/kg
h=932 kJ/kg
Hence:h<hf
So, it’s compressed water
f g
Question 23
3
1) 12 , 8 5
188.03.22) 0.1208 /26.5
3) 12 , 8 5, 0.1632 /
0.1208 0.74020.1632
4) 12 , 8 5, 2784 / , 798 /
(1 ) 0.7402 27
s
x
g
x g
x
g
g f
x g f
p bar from Table
t CVv m gm
p bar from Table v m kg
v x v
vxv
p bar from Table h kJ kg h kJ kg
h xh x h
84 (1 0.7402) 798 2268.0372 /
26.5 2268.0372 601035) 12 , 8 5, 2588 / , 797 /
(1 ) 0.7402 2588 (1 0.7402) 797 2122.6982 /
26.5 2122.6982 56251.5
x x
g f
x g f
x x
kJ kg
H m h kJp bar from Table u kJ kg u kJ kg
u xu x u kJ kg
H m h kJ
26.5 2122.6982 5625.5x xU m u kJ
Question 3• When p=9 bar, from Table8-8
• Hence, it can be calculated that• When T=225°C• v=0.2451• u=2671• h=2891.5
T=200°C T=250°Cv 0.2305 0.2597u 2628 2714h 2835 2948
Question 3(cont’d)• When p=10 bar, from Table8-8
• Hence, it can be calculated that• When T=225°C• v=0.21945• u=2667• h=2886.5
T=200°C T=250°Cv 0.2061 0.2328u 2623 2711h 2829 2944
Question 3(cont’d)• When p=9.2 bar and T=225°C
310 9 9.2 9 0.23997 /0.21945 0.2451 0.245110 9 9.2 9 2670.2 /
2667 2671 267110 9 9.2 9 2890.5 /
2886.5 2891.5 2891.5
v m kgv
u kJ kgu
h kJ kgh
Question 4Solution:(a)Dry saturated steam, p=15 bar, so
vg=0.1317m3/kg
(b)p=6 bar, vg=0.3156 m3/kg
(c) p=15 bar, ug=2595 kJ/kg p=6 bar, uf=669 kJ/kg, ug=2568 kJ/kg
0.1317 0.41730.3156g
g
vv xv xv
Question 42 (1 ) 0.4173 2568 (1 0.4173) 669
1461.4 /g fu xu x u
kJ kg
2 1( ) 19.74 (1461.4 2595) 22377.3U m u u kJ
d) Rigid vessel, so W12=0
e) U2-U1=Q12-W12Q12=U2-U1=-22377.3 kJ
15 bar
6 bar
1
2
Question 5Solution:Given: V1=1.25m3 Wet steam, p=8 bar, x=0.75(a)p=8 bar, wet steam so vg=0.2403m3/kg
(b)p=8 bar, uf=720 kJ/kg, ug=2577 kJ/kg
p=8 bar, vg=0.2403 m3/kg
1 (1 ) 0.75 2577 (1 0.75) 720
2112.75 /g fu xu x u
kJ kg
1( ) 6.936 (2577 2112.75) 3220.038gU m u u kJ
Question 5
U2-U1=Q12-W12Q12=U2-U1+W12
=3220.038+333.3=3553.338 kJ
8 bar1 2
Question 6Solution:
5 210 8314 (32 273) 30.19 10 / 30.1928 0.3
o
o
o
pV mRTmRT
pV
R MRRRMRm TMp N m barV
Question 7Solution:
1 1 2 2
1 2
531 1 2
2 51 2
1.02 10 0.1 115 273 0.0194540 273 6.5 10
pV p VT T
pV TV mT P
Question 8Solution:
1 1 2 2
1 2
1 2
1 2
1 2
52
2 1 51
2.6 10 (28 273) 391.32 10
pV p VT TV Vp pT T
pT T Kp
Question 9Solution:
a)
b)
21 1
1
1.4 10 0.1 0.1620.29 (25 273)
pV mRT
mpV kgRT
1 1 2 2
1 2
231 1 2
2 21 2
1.4 10 0.1 60 273 0.0223525 273 7 10
pV p VT T
pV TV mT p
Question 10Solution:(a)
(b)
215 10 0.05 0.9570.263 (25 273)
pV mRTpVm kgRT
1 1 2 2
1 2
1 2
1 2
1 2
2 12
1
0.08 (25 273) 476.80.05
pV p VT Tp pV VT T
V TT KV
0.92 0.657 0.263 /p vR c c kJ kg K
Question 10Solution:(c)
(d)
(e)
2 1 2 1( )0.957 0.657[476.8 (273 25)]112.4
vU U mc T T
kJ
212 2 1( ) 15 10 (0.08 0.05) 45W p V V kJ
2 1 12 12
12 2 1 12 112.4 45 157.4U U Q WQ U U W kJ
Question 11Solution:(a)
(b)
0.92 0.657 0.263 /p vR c c kJ kg K
21 1
115 10 0.08 476.80.957 0.263
pV mRTpVT KmR
1 1 2 2
1 2
1 2
1 2
1 2
1 22
1
15 (60 273) 10.48476.8
pV p VT TV Vp pT T
pTp barT
Question 11Solution:(c)
(d) Since this is a constant volume process
(e)
2 1 2 1( )0.957 0.657[(273 60) 476.8]90.4
vU U mc T T
kJ
12 0W
2 1 12 12
12 2 1 12 90.4 0 90.4U U Q WQ U U W kJ
Thank youQ & A