P1
P2
=
qx 3
1. P5
/
PROBLEMA DE MOODLE
y
5
Th 5580)
-
1
=
P4 8dar
=
M m &
↑
SS
5s
*
P2 14MPa =
2.
4-
T2 6880C =
T1 T2 >i
41 14MPa = 3, P3 8bar
=
112 0.9
=
157 0.9 = -
5
1
Pz P6 =
7B 0.75
=
B ma
b Mn
-
ligsaturado 40
2
2. -
5 4 S 6
P2 14MPa=
P5 0,8MPa
= Ph 8,8MPa
=
P5 7,584kPa = P0 7,584kPa
=
12 6800 = T5 215,8650
=
T 5500 =
T5 56,950= T0 480 =
v 2,684m3/k9
= v 0.271m3/k9
=
V 0.475m3/k9
=
v 28,589m3/kg
=
v 0,00188m3/kg
=
u 5.216,121k5/k9 u 2657,298k5/k9 5211,595k5/k9 u 2.455,981k5/49 167.56 k5/kg
=
u u
=
= = =
h 5.591,948k5/kg
=
h 2875,961k5/k9
=
h 5589,984k5/k9
=
n 2685,935k5/kg
=
h 167,57k5/k9
=
5 6.519k5/kg.k
=
5 6,888k5/kg.k
= s 8,81k5/kg.K
= 5 8,35645/kg.K
= s
0.5725k5/kg.K
=
hos -no2 h05 2.875,961k5/k9
7+ >05
74y(4035 402) hoe -> =
= =
-
+
1
hoss -ho 2
h055 2.794, 186k5/kg IMPORTANTE:Gando esteros estado de liquida
=
en
40s-how 40s = 2.605,955k5/k9 saturado hay quever las magnitudes la tabla
hoshonythoss-houl;
=
en 5.
74 hoss-hou
hoss 2,496, 605
K5//g
=
1
R;h00:186,226ke
P1 14MPa
(pop.Pod/S; hos
(P07
=
-B:
=
F1 =
41,5280
v 1802m3/kg
=
u 152,195k5/k9
=
n 186,226k5/kg
=
5 0.587125/kg.k
=
(w 11 1(403 h02)) 1/h0s-hul w 1.685k5/Rg
Inc) (w+2)
5.- =
+
+
= -
+
=
4 j m(h0z 40s haa
=
=
-
+
- 4055;0 4.128k5/kg =
, 8 RDINARIO 2821
5
5
1.Wix m(ho =
- hs) WT1= -
=
3.946,575kW
5 6
I I
P5 = 8MPa P0 = SMPa
85 5583C
=
88 598,549°
=
h5 = 3.45 3,185k5/kg hos = 5.163,572ks/kg no 3.270.08 K5/kg
=
U5 = 5.122,490 k5/kg
u6 =
2916,785k5/kg
55 6,819k5/kg.k
= 56 = 6,891k5/kg.K
2.wiz m(hs h +) W+2 4.556,565kW
-
= =
-
=
# 8 9
I I I I
↑F: SMPa P8 = 1MPa Pa: 28kPa
*
= 5883 8
8: 354,68932 8
9:68,60C
n +=5.457,848K5/kg h8s 5.116,955k5/kg
=
48 =
5.167,949K5/kg has: 2.412, +60K5/kg h9 2.689, 151kg
=
48: 2.885,22815/kg u9=2.456, 7(5/kg
u
3.188,461k5/kg
=
.256(5/kg.k 58 7,518k5/kg-k 59 =
5 =
=
7.989k5/kg.k
Mis-hand the
=>
"Mis-o,th
18.103.26)kPa,he 261,55511
u.
y=(Pz Px) he h1 +
P2=h2 257,45k5/kg
-
+
-
+
= =
=
Sw(he- h1) 8,8.1.888
Balance
5.
energetical cambiador decalor: Ym(hs-hel +(1 xlm(n10-hs) 0
-
= = i(h3-hz) + (1 x)(h10.hs) 0
-
=
5 010 179,13C
8 = =
x y(+ 62,852 261,5+5) + (1 Y)(762,87 516+,949) 0
- -
-
=
=
10,85
=
> S
P3 8MPa
-
= 10 P18-1MPa in1 = 12,45kg/s
85 179,132
=
10 179,13C
8 =
↓ 5 62,852k5/kg
= 410 = +62.81k5/kg m2 2,55kgls
=
us + =
53,869k5/kg & 10 =
61,68k5/kg
P2
=
qx 3
1. P5
/
PROBLEMA DE MOODLE
y
5
Th 5580)
-
1
=
P4 8dar
=
M m &
↑
SS
5s
*
P2 14MPa =
2.
4-
T2 6880C =
T1 T2 >i
41 14MPa = 3, P3 8bar
=
112 0.9
=
157 0.9 = -
5
1
Pz P6 =
7B 0.75
=
B ma
b Mn
-
ligsaturado 40
2
2. -
5 4 S 6
P2 14MPa=
P5 0,8MPa
= Ph 8,8MPa
=
P5 7,584kPa = P0 7,584kPa
=
12 6800 = T5 215,8650
=
T 5500 =
T5 56,950= T0 480 =
v 2,684m3/k9
= v 0.271m3/k9
=
V 0.475m3/k9
=
v 28,589m3/kg
=
v 0,00188m3/kg
=
u 5.216,121k5/k9 u 2657,298k5/k9 5211,595k5/k9 u 2.455,981k5/49 167.56 k5/kg
=
u u
=
= = =
h 5.591,948k5/kg
=
h 2875,961k5/k9
=
h 5589,984k5/k9
=
n 2685,935k5/kg
=
h 167,57k5/k9
=
5 6.519k5/kg.k
=
5 6,888k5/kg.k
= s 8,81k5/kg.K
= 5 8,35645/kg.K
= s
0.5725k5/kg.K
=
hos -no2 h05 2.875,961k5/k9
7+ >05
74y(4035 402) hoe -> =
= =
-
+
1
hoss -ho 2
h055 2.794, 186k5/kg IMPORTANTE:Gando esteros estado de liquida
=
en
40s-how 40s = 2.605,955k5/k9 saturado hay quever las magnitudes la tabla
hoshonythoss-houl;
=
en 5.
74 hoss-hou
hoss 2,496, 605
K5//g
=
1
R;h00:186,226ke
P1 14MPa
(pop.Pod/S; hos
(P07
=
-B:
=
F1 =
41,5280
v 1802m3/kg
=
u 152,195k5/k9
=
n 186,226k5/kg
=
5 0.587125/kg.k
=
(w 11 1(403 h02)) 1/h0s-hul w 1.685k5/Rg
Inc) (w+2)
5.- =
+
+
= -
+
=
4 j m(h0z 40s haa
=
=
-
+
- 4055;0 4.128k5/kg =
, 8 RDINARIO 2821
5
5
1.Wix m(ho =
- hs) WT1= -
=
3.946,575kW
5 6
I I
P5 = 8MPa P0 = SMPa
85 5583C
=
88 598,549°
=
h5 = 3.45 3,185k5/kg hos = 5.163,572ks/kg no 3.270.08 K5/kg
=
U5 = 5.122,490 k5/kg
u6 =
2916,785k5/kg
55 6,819k5/kg.k
= 56 = 6,891k5/kg.K
2.wiz m(hs h +) W+2 4.556,565kW
-
= =
-
=
# 8 9
I I I I
↑F: SMPa P8 = 1MPa Pa: 28kPa
*
= 5883 8
8: 354,68932 8
9:68,60C
n +=5.457,848K5/kg h8s 5.116,955k5/kg
=
48 =
5.167,949K5/kg has: 2.412, +60K5/kg h9 2.689, 151kg
=
48: 2.885,22815/kg u9=2.456, 7(5/kg
u
3.188,461k5/kg
=
.256(5/kg.k 58 7,518k5/kg-k 59 =
5 =
=
7.989k5/kg.k
Mis-hand the
=>
"Mis-o,th
18.103.26)kPa,he 261,55511
u.
y=(Pz Px) he h1 +
P2=h2 257,45k5/kg
-
+
-
+
= =
=
Sw(he- h1) 8,8.1.888
Balance
5.
energetical cambiador decalor: Ym(hs-hel +(1 xlm(n10-hs) 0
-
= = i(h3-hz) + (1 x)(h10.hs) 0
-
=
5 010 179,13C
8 = =
x y(+ 62,852 261,5+5) + (1 Y)(762,87 516+,949) 0
- -
-
=
=
10,85
=
> S
P3 8MPa
-
= 10 P18-1MPa in1 = 12,45kg/s
85 179,132
=
10 179,13C
8 =
↓ 5 62,852k5/kg
= 410 = +62.81k5/kg m2 2,55kgls
=
us + =
53,869k5/kg & 10 =
61,68k5/kg
