Question 1

Steam ; 8 bar ; P1 = P2

S = 6.211

T = 350

Q = h2 - h1

h2 = 2769

h1, find x first

entropy = 6.211

S1 = sf + xsfg

x = S1 - Sf
––––––
sfg
= 6.211 - 2.046
–––––––––––
4.617
x = 0.902

Now, find h1
h1 = hf + xhfg
= 721 + 0.902 (2048)
= 721 + 847.3
= 2568.3 kJ.K

.
. . Q = h2 - h1
= 2769 - 2568.3
= 200.7 kJ.kg
Question 2

Steam at 110 bar (P1)

T = 400

S1 = S2

P2 = 3 bar

W = 165.5

T2 = ?

n = u + pv

u1 = n + pv
= 2705 + (11000)(0.01598)
= 2880.78

W = u2 - u1
165.5 = u2 - 2880.78
u2 - 2880.78 = 165.5
u2 = 2715.28 kJ/kg

Find,

2715.28 = 561 + x(1983)

2154.28 = 1983x
x = 0.92
Question 3

v = 0.85 m3 of CO2

M = 44

P1 = 1.05 ; 105 kPa

P2 = 4.8 ; 480 kPa

T = 17 ; 290K

T1 = T2

a) Mass of CO2

R = RO
–––
M
= 8314.4
–––––
44
= 188.96 m3/kg

b) ∆ in entropy

v = xvy
0.85 = xvy
P vg

1.00 1.694

1.05 vg

1.10 1.549

1.05 - 1.00 v - 1.694

––––––––– = –––––––––––
1.10 - 1.00 1.549 - 1.694

1 v - 1.694
–– = ––––––––
2 -0.145

2v - 3.388 = -0.145
2v = 3.243
vg = 1.622

v = xvg
0.85 = x(1.622)
x = 0.524
S1 ; 1.05 Bar

P S1

1.00 4.476

1.05 S1

1.10 4.474

1.05 - 1.00 S1 - 4.476

––––––––– = –––––––––––
1.10 - 1.00 4.474 - 4.476

1 S1 - 4.476
–– = ––––––––
2 -0.002

-0.002 = 2S1 - 8.952

-2S1 = -8.952 + 0.002
-2S1 = -8.95
S1 = 4.475

S2 ; 4.8 Bar

P S2

4.5 4.459

4.8 S2

5.0 4.460

4.8 - 4.5 S2 - 4.459

––––––– = –––––––––––
5.0 - 4.5 4.460 - 4.459

3 S2 - 4.459
–– = ––––––––
5 0.001

0.003 = 5S2 - 22.295

-5S2 = -22.298
S2 = 4.4596

.
. . ∆ in entropy = 4.475 - 4.4596
= 0.0154 kJ/kg.K
Question 4

m = 1kg

P1 = 1.01

T1 = 27

PV1.3 = constant

P2 = 5

Cp = 1.005

R = 0.287

T2 = ?

∆ of entropy = ?

r-1
––
r
T2 P2
–– = –––
T1 P1

1.3-1
––––
1.3
T2 500
––– = ––––
300 101

T2
––– = 2.34
300

T2 = 702 K

∆ of entropy

Pv = MRi –––––––––––––– (1)

101(v) = 1(0.287)(300)
v1 = 0.852

Pv = MRi –––––––––––––– (2)

500(v) = 1(0.287)(702)
v2 = 0.402