↑ROBLEMA 69
h S
va 15m3/min 0,25m3/s
- =
3 1.250kg/m3
=
↑a 283 293,75K;Pa = 1.5.183Pa
Aceite refrigerante
=
=
como
a D (p 2.4585/kg. Tb Ta 10K;Pb 0,83Pa
=
- = =
7
S
v=42m3/min 0,tm/s =
Gas Tc 3t0K; Pc=4,26MPa; M
44kg/Kmo);<p(T,p 0= 94,8K>/Kmol-K
= -
i
=
d
T=2700 = 543, 15K; P1 =P2=5MPa
5 -1001a.0,7
n
3,9kgs PR1 P12 P1/Pc 1,t
1. m f1.v1 P1v1
=
=
=
=
=
= =
=
1Rg.T10.9.188,95-543,15
=
-
R2 T2/Tc 1,25
= =
Rg
Rx 8,314
=
=
188,95
=
TR1 = T1/Tc 1,4 =
M 44.18 3
2. Balance
energetics al aceite:m(hz-hs) G + =
ki
m(p(+b a0 572.52.450(10) (0,85 1)7,510.0 0 0.661,875k
a)
Pb-
-
=
+ +
=
= + =
1.258
1.250kg/m3.0,25m3/s =312.5kgl
-
na-gava -
0 -
=
7.661.875kI
5: Balance energetico al gas: m (h2 h1) 0-
=
he h1 (h2 h2*1 + (he*-hy*) + (hy* h1)
-
=
- -
I
-
h- h RTc = .. . hz hy
-
=
-
h h* RTc
-
+ 94,8(Tz 543,15) + 2.153,326
-
RTc RTc
e e
2.153,326 55%.122.5 o
mhhY
Ric 94,8(42 543,75) +
+
-
+
-
h*hRTc =
-
RTc
1
12 =
462,5K
Proceso de iteracion:
-us = 0cip +
m(52-5x) +
macpintop
TFTi
52 -
51 (52 52*1 + (52 51*) + (5+* 51)
=
- - -
4,98845.kg71oTsmoi-1=
I
-s5R =
-
4,98845.k7kg 62.51 -
=
+ 2,05855kg7.157
-
=
18,147955.4571-
R
2
*- 5R
2.05
=
855.k+kg
7
,9kg/s. 1-412,45);/kg-K +312,5kg/s.2.4505/kg-kIn
=
303. 15 =
Up - Up: 10KW/K
R 293,15
I
IMPORTANTE: Mucho cuidado con las unidades a calcular variaciones de entalpia entropia.
y
Hay que multiplicar sass por msi se usan unidades s.kgfKf
, PROBLEMA 6F
4+ 0,85 =
I T
d
T=
P1
+F3C
= 500 atm
I condiciones alcentrada
P2 = 60 alm
(cp(i)= 5.95 + 15.64.1024.8,344.155+ +77.65. 1891315-kM01-k
1m3/min 1,67 -18m3/s
1
vs = =
-288,15K]
100 = 153 condiciones delambiente
Poo = 1 atm
P1
N
5SIS IMPORTANTE: Aprender sobre los
P2
PP
T
1s 52
1
1. D
diagramas is, P.h. Suelen preguntark
·M
-2
r
enel examen.
81
⑤ i
zgT
2.9 -
Conozco las
↑
condiciones a la entrada.
I
PRx P1(Pc 5,84.10+Pa/5,12.186pa
=
= PR1 5,9575
+ =
It ↑ E1 0,92
=
TR1 π+/c 450,15k/282.4k+ TR1 1,594
=
=
=
504.107 -97 247.5kg/m3
5+
=
91
=
0.92.296.61.650,75
Rx 8,314
Rg =
= = 296,67
M 28,83.18'3
in 7.v1 =
=
24+,5kg/m3.1,66.10m3/s +m 4,72kg/s =
S
P12 P2/pc=
6,88.186Pa/5,12.186PaiPR2 1,1873
=
=
52
TR2= T2/Tc "
Necesito hallar T2
wi m(hoz -nos) m(hz
=
=
-
hs) my, Lhzs.hs]
= * Necesito hallar as para calcular has
hoz h1 h2-hy
4- (z- hx
=
y,(hzs- hy)
-
= =
hols-hot hes- hy
-25 -
51 (525 325)
=
-
+ (525 55) (5y* 51)
-
+ -
I
5
-
sT 1 =...
66512(6)15.95
2
-
15.04.1825.8.344.1851+71.67.
10911d- 8,3791n6.58.100 6.657.1
584.18
+
28,82
34.61 -
-
R
1
process deiteracion para hallarel valor de Tzs.
+ T2S=542,5K
Hay que hacer un
T(zs T2s/xc 342.5k/282,4k
= =
+ TR25:1,213
h S
va 15m3/min 0,25m3/s
- =
3 1.250kg/m3
=
↑a 283 293,75K;Pa = 1.5.183Pa
Aceite refrigerante
=
=
como
a D (p 2.4585/kg. Tb Ta 10K;Pb 0,83Pa
=
- = =
7
S
v=42m3/min 0,tm/s =
Gas Tc 3t0K; Pc=4,26MPa; M
44kg/Kmo);<p(T,p 0= 94,8K>/Kmol-K
= -
i
=
d
T=2700 = 543, 15K; P1 =P2=5MPa
5 -1001a.0,7
n
3,9kgs PR1 P12 P1/Pc 1,t
1. m f1.v1 P1v1
=
=
=
=
=
= =
=
1Rg.T10.9.188,95-543,15
=
-
R2 T2/Tc 1,25
= =
Rg
Rx 8,314
=
=
188,95
=
TR1 = T1/Tc 1,4 =
M 44.18 3
2. Balance
energetics al aceite:m(hz-hs) G + =
ki
m(p(+b a0 572.52.450(10) (0,85 1)7,510.0 0 0.661,875k
a)
Pb-
-
=
+ +
=
= + =
1.258
1.250kg/m3.0,25m3/s =312.5kgl
-
na-gava -
0 -
=
7.661.875kI
5: Balance energetico al gas: m (h2 h1) 0-
=
he h1 (h2 h2*1 + (he*-hy*) + (hy* h1)
-
=
- -
I
-
h- h RTc = .. . hz hy
-
=
-
h h* RTc
-
+ 94,8(Tz 543,15) + 2.153,326
-
RTc RTc
e e
2.153,326 55%.122.5 o
mhhY
Ric 94,8(42 543,75) +
+
-
+
-
h*hRTc =
-
RTc
1
12 =
462,5K
Proceso de iteracion:
-us = 0cip +
m(52-5x) +
macpintop
TFTi
52 -
51 (52 52*1 + (52 51*) + (5+* 51)
=
- - -
4,98845.kg71oTsmoi-1=
I
-s5R =
-
4,98845.k7kg 62.51 -
=
+ 2,05855kg7.157
-
=
18,147955.4571-
R
2
*- 5R
2.05
=
855.k+kg
7
,9kg/s. 1-412,45);/kg-K +312,5kg/s.2.4505/kg-kIn
=
303. 15 =
Up - Up: 10KW/K
R 293,15
I
IMPORTANTE: Mucho cuidado con las unidades a calcular variaciones de entalpia entropia.
y
Hay que multiplicar sass por msi se usan unidades s.kgfKf
, PROBLEMA 6F
4+ 0,85 =
I T
d
T=
P1
+F3C
= 500 atm
I condiciones alcentrada
P2 = 60 alm
(cp(i)= 5.95 + 15.64.1024.8,344.155+ +77.65. 1891315-kM01-k
1m3/min 1,67 -18m3/s
1
vs = =
-288,15K]
100 = 153 condiciones delambiente
Poo = 1 atm
P1
N
5SIS IMPORTANTE: Aprender sobre los
P2
PP
T
1s 52
1
1. D
diagramas is, P.h. Suelen preguntark
·M
-2
r
enel examen.
81
⑤ i
zgT
2.9 -
Conozco las
↑
condiciones a la entrada.
I
PRx P1(Pc 5,84.10+Pa/5,12.186pa
=
= PR1 5,9575
+ =
It ↑ E1 0,92
=
TR1 π+/c 450,15k/282.4k+ TR1 1,594
=
=
=
504.107 -97 247.5kg/m3
5+
=
91
=
0.92.296.61.650,75
Rx 8,314
Rg =
= = 296,67
M 28,83.18'3
in 7.v1 =
=
24+,5kg/m3.1,66.10m3/s +m 4,72kg/s =
S
P12 P2/pc=
6,88.186Pa/5,12.186PaiPR2 1,1873
=
=
52
TR2= T2/Tc "
Necesito hallar T2
wi m(hoz -nos) m(hz
=
=
-
hs) my, Lhzs.hs]
= * Necesito hallar as para calcular has
hoz h1 h2-hy
4- (z- hx
=
y,(hzs- hy)
-
= =
hols-hot hes- hy
-25 -
51 (525 325)
=
-
+ (525 55) (5y* 51)
-
+ -
I
5
-
sT 1 =...
66512(6)15.95
2
-
15.04.1825.8.344.1851+71.67.
10911d- 8,3791n6.58.100 6.657.1
584.18
+
28,82
34.61 -
-
R
1
process deiteracion para hallarel valor de Tzs.
+ T2S=542,5K
Hay que hacer un
T(zs T2s/xc 342.5k/282,4k
= =
+ TR25:1,213
