8.441. Evaporación de jugo de tomate al vacío. Se concentra el jugo de tomate con 12% de sólidos en peso hasta 25% de solido en un evaporador de tipo película. La temperatura máxima permisible para el jugo de tomate es 135°F, que será la temperatura del producto. (La alimentación entra a 100°F) se usa vapor de agua saturado a 25 lb/ pul2 abs como medio de calentamiento. El coeficiente total de transferencia de valor V es 600btu/h pie°F y el area A es 50 pies2 se estima que la capacidad calorífica de la alimentación ali mentación CP es 0. 95btu/lbm°F. Desprecie cualquier elevación del punto de ebullición que exista calcule la velocidad de alimentación del jugo de tomate al evaporador.
CPF = 95btu/lbm°F
v=
F=
Xv = 0
Xf = 0.12 Tf = 100°F
T1=135°F
5 = 4,410.56 lbm/h
P= Xp = 0.25
Ps = 25lm/pulg 2 Ts = 240°F Sc V = 600btu/h pie°F A = 50 pies pies2
Q absorb = Q cedido Fcp (T1 – T2) + V (H v - hv) = S (H 5 – h5) Sλ = VA∆t
S = VA∆t = (600but/hpie°F) (50pies 2) (240°F – 100°F) ∆ (1.160.7 btu/lbm – 208.44 btu/lbm) 5 = 4,410.56 lbm/h
Q absorbido = Q cedido Fcpf (T1 – Tf ) + V (H v – hv) = 5 (H5 – h5) F (0.95btu/lbm°F) (135°F – 100°F) + V (1,119.85 btu/lbm-102.97 btu/lbm) = (4,410.56 lbm/h) (1160.7 – 208.44btu/lbm) 33.25 F + 1,016.88 V = 4,199, 999.866 (66.25) (2.083 P) + (1016.88) (1,083P) = 4,199,999.866 69.26 P + 1,101.28104 P = 4,199.999.866 P = 4,199,999.866 = 3,588.08 lbm/h 1,170.54104
Interpolar para H v y hv con 135°F X1 130 y1 1117.8 135 X2 140 y2 1121.9
y1 97.98
hv = 1,119.85
y2 107.96
Hv = 102.97
Balance F = v+p fxf = pxp V = 3,885.89 lbm/h F – P = V F (0.12) = P(0.25) 2.083P-P = V F = P(0.25) = 2.083 P F = 7,473 97 lbm/h 1.083 P = V 0.12 EPE1 = 1.87x + 6.22x2 EPE1 = 1.78 (0.133) + 6.22 (0.133) 2 EPE1 = 0.347°F Cp1 = 4.19 – 2.35 (x) Cp1 = 4.19 – 2.35 (0.133) Cp1 = 4.502 Efecto 2 L1 c L1 (28,049.44) (0.133) = (10,979.98) c L 2 0.207 = xL3 EPE2 = 1.78 (0.207) + 6.22 (0.207) 2 EPE2 = 0.635°C Cp2 = 4.19 – 2.35 (0.207)
Cp2 = 5.39 Efecto 3 L2 x L2 = L3 x L3 (10,979.98) (0.207) = (4,536) x L 3 0.5 = x L3
EPE3 = 1.78 (0.5) 7 6.22 (0.5) 2 EPE3 = 2.445 Cp3 = 4.19 – 2.35 (0.5) Cp3 = 3.015 ∆Ŧ T = Ts – T3 – (∑ EPE) ∆TT = 121.1 – 51.54 – (0.347 + 0.635 + 2.445) ∆TT = 66.1 ∆T1 =
1
/u1 /u1 + 1/u1 + 1/u3
1
∆T2 = 66.1
1
/3123 /3123 + 1/1987 + 1/1136
1
∆T2 = 66.1
1
/1987 = 19.53 1 1 /3123 + /1987 + /1136
1
∆T2 = 66.1
1
/1136 /3123 + 1/1987 + 1/1136
1
∆T1 = Ts – T1 T1 = Ts - ∆T1 T1 = 121.1 – 1242 = 108.68 ∆T2 = T1 – T2 T2 = Ts - ∆T1 T2 = 108.68 – 19.53 = 89.15 ∆T3 = T2 – T3 T3 = T2 – ∆T3 T3 = 89.15 – 34.15 = 55
= 34.15
Flujo
Relación
x
cp
T
it
h
>
F
22,680
0.1
3.955
26.7°C
V1
22,680 – L1
108.68
2,689.56
455.72
2,233.84
V2
L1 – L2
89.15
2,658.623
372.84
2,285.78
V3
L2 – 4,536
51.54°C
2,599.12
226.01
2,373.11
2,707.88
508.19
2,199.69
L1
0.133
4.502
L2
0.207
5.38
0.5
3.015
L3
4,536
S
54 121.1°C
Determine las áreas de calefacción, las temperaturas de ebullición y la economía en un evaporador de triple efecto que concentra una disolución de NaOH desde 10% hasta el 50% en peso. El vapor de calefacción del primer efecto es vapor saturado a 6 atm, y el vapor procedente del tercer efecto condensa a 30℃, han de considerarse los siguientes valores para los coeficientes globales de transferencia: U1 1200, U2 1200, U3 1200 Kcal/hm²℃. En alimentación contra corriente con regímenes de alimentación de 5000Kg/h a una temperatura de 30℃. Problema No. 1.
L1= ¿ XL1= 0.5 S=¿ Ps = 6 atm = 607.8 KPa Ts = 159 ℃ Cps = 1.84 KJ/Kg K T3 = 30 ℃
F = 5000 Kg/h Xf = 0.10 Tf = 30 ℃
6 ∗
.∗5
∗
Y1 155 Ts 159 Y2 160
= 607.8
X1 543.1 Ym= 67.8 X2 617.8
Balance de calor Q ENTRA=QSALE Efecto 1: + ( − ℎ ) = + 1 Efecto 2: 3 + 1( − ℎ ) = 2 + 2 Efecto 3: + 2( − ℎ ) = 3 + 3 6 atm * 1.013x105 Pa * 1kPa = 607.8 kPa 1 atm 1000 Pa °C Y1 155 x1 543.1 Ts = 159°C 607.8 Y2 160 X2 617.8 Balance F=V1+V2+V3+L1 E x F = L1 x L1 (5000) (0.10) = L1 (0.5) 500 = L1 0.5 1,000kg / h L1 Flujo Relación
F – L1 = V1 + V2 + V3 5000 – 10000 = V1 + V2 + V3 4,000 = 1,333.33 de V e 3 Cada evaporador
x
T
it
30
h
>
F
5,000
V1
L2 – 1,000
116
2,700.46
486.726 2,213.734
V2
L2 – 1,000
73
2,631.9
305.55
2,326.35
V3
L3 – L2
30
2556.3
125.79
2,430.51
L1
5,000 – L3
0.5
585.6
L2
1,000
1.5
437.5
0.29
135
L3
0.1
cp
65
S
159
2756.96
671.208 2,085.752
Efecto 1 L2 = V1 + L1 L2 = 1,333.33 + 1,000 L2 = 333.33
L2 x L2 = L1 x L1 (333.33) x L 2 = (0.5) (1,000) xL2 = 1.5
Efecto 2 L3 = L2 + V2 L3 = 333.33 + 1,333.33 L3 = 1,666.7
L3 x L3 = L2 x L2 (1,666.7) xL 3 = (333.33) (1.5) xL3 = 0.29
∆TT ¿ T S – T3 ∆TT = 159°C = 129 ∆T1 = ∆TT
1
/u1 1 1 /u1 + /u2 + 1/u3
= 129
1,200 Kcal * 1000 cal * 1h * 1m 2 Hm2°C 1kcal 3,600s 10,000cm 2 0.033cal/5 cm 2°C
2,431.66 BTU/hpie2°F
∆T1 = 43 ∆T2 = 43 ∆T3 = 43 ∆TT T1 = Ts – T1 T1 = Ts - ∆T1 T1 = 159 – 43 = 116 ∆T2 T1 = T1 – T2 T2 = T1 - ∆T2 T2 = 116 – 43 = 73
1
/13,807.7 w/m1K 1 /13,807.7 + 1/13,807. + 1/13,807.7 = 0.033cal 5cm2°C
* 1BTU/hpie2°F = 2,431.66 BTU/hpie°F 1.3571x10 -4cal/scm2°C *
5.6783 w/m2k 1BTU/hpie 2°F
= 13,807.69 w/m2R
∆T3 T2 = T3 T3 = T2 - ∆T3 T3 = 76 – 43 = 30
Estalpias TV1 = 116 X1 115 XM 116 X2 120
H y1 2699.0 2,700.46 y2 2706.3
H y1 482.48 486.726 y2 503.71
Tv2 = 73 X1 10 73 X2 75
H y1 2626.8 2,631.9 y2 2635.3
h y1 292.98 305.55 y2 313.93
TS = X1 X2
y1 2.752.4 y2 2758.1
y1 671.208 y2 675.55
159 155 160
1) L2 = L1 + V1 V1 = L2 – 1,000
2) L3 = L2 + V2
3) F = L3 + V3
2) L3hL3 + V1 (HV2 – hv2) = V2 Hv2 + L2 hL2 L3 (135) + (L 2 – 1,000) (2,213.734) = (L 2 – L2) (2631.9) + L 2 (437.5) 135 L3 + 2,213.734 L 2 – 2,213.734 = 2631.9 L 3 – 2631.9 L2 + 437.5 L 2 - 2,496.7 L 3 + 4,408.134 L2 = 2,213.734 3) FhF + V2 (HV2 – hV2) = V3 HV3 + L3 HL3 (5,000) (65) + (L 3 – L2) (2,326.35) = (5,000 – L3) (2556.3) + L 3 (135) 325,000 + 2,326.35 L 3 – 2,326.35 L 2 = 12,781,500 – 2556.3 L 3 + 135 L 3 4,747365 L 3 – 2,326.35 L 2 = 12,456.500 - 2,496.9 L 3 + 4,408.134 L 2 = 2,213.734 4,747.64 L 3 – 2,326.35 L 2 = 12,456.500 1 5,119.614.17 L 2 = 3.1113 X 10 10 L2 = 2,057.79
4,7473.65 2,496.9
- 2,496.9 L 3 + (4,408.134) (2,057.79) = 2,213.734 - 2,496 L 3 + 9,071,014.064 = 2,213.734 L3 = 3,632.02
1) L2 HL2 + S(HS – hS) = V1 HV1 + L1 hL1 (2,057.79) (437.5) + S(2,085.752) = (2,057.79-1000) (2,700.46) + (1,000) (585.6)
900,283.125 + S(2,085.752) = 2,856,519.583 + 585.600 S (2,085.752) = 2,541,836.458 S = 1,218.67 Q1 = S>S Q1 = (1,218.67) (2,085.752) = 2,541,843.39 RJ/h 2,541,843.39 RJ * 1000j * 1h = 706,067.61 W H 1RJ 3600S
A1 =
Q V1 ∆T1 A1 = 706,067.41 W = 13.807.69 W/m2k °43
1,189 M2
Q2 = V1 > v1 Q2 = (1,057.79) (2,213.734) = 2, 341,665.688 KJ/h 2, 341,665.688 KJ * 1000J * 1b = 650,462.59 w H 1KJ 3600s A3 = 1,017,280,545w 13,807.69 w/m2k °43
=
1.71 m 2
E = V1 + V2 + V3 =1,057.79 + 1,574.23 +,1367.98 = 5 1,218.67
3.28
