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REPORTE DE PRÁCTICA 03
OPERACIONES UNITARIAS II (AI-441)
“CONDUCCIÓN DE CALOR EN ESTADO TRANSITORIO”
Profesor: Ing. Arones Medina, Edgar
Alumnos: HUAMANI RAMOS,
Grupo I:
I. Objetivos:1. Evaluar la transferencia de calor en estado no estacionario en sistemas de
diferentes geometrías. 2. Determinar el coeficiente de conductividad térmica en condiciones no
estacionarias.
II. Datos y observaciones:
PAPA (considerando como una esfera)
V= 80 ml =8*10-5m3
m= 85.80gr = 0.0858Kg
ρ=mv
= 0.0858kg8∗10−5m3=1072.5
kgm3
V= 43∗π r3
r=3√ 34∗V
π=3√ 34∗8∗10−5m3
π=0.0267m
A=4 π r2=4 π∗0.02672=0.0090m2
cp=3640 jkgºC
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k=0.498 wmºC
=0.498 jseg∗mºC
α= Kρ∗cp
= 0.4981072.5∗3640
=1.2756∗10−7 m2
seg
θ(min) θ(seg) T (r=o ) T (w) T (α )
0 21.53 180 31.2 90.4 926 360 44.4 90.5 929 540 57.8 90.6 9212 720 67.8 90.6 9215 900 75.4 90.6 9218 1080 80.4 90.6 9221 1260 83.8 90.6 9224 1440 86.2 90.6 9227 1620 87.6 90.6 9230 1800 88.5 90.6 9233 1980 89.2 90.6 9236 2160 89.6 90.6 92
III. Cálculos:
Haciendo cumplir con la condición:
Q/A<3000wm2 …………………h=537(92−90.6)1/7=563.4427 w
mºC
Entonces:
3000< Q/A<63000wm2
1Bi
= Khr o
= 0.498563.4427∗0.0267
=0.0331
FO=αθro2 =1.2756∗10
−7∗1800.02672
=0.0322
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Y= f (F0 ,1Bi )=
T c−T α
T i−T α
Y=T c−9221.5−92
Y (21.5−92 )+92=T c
F0 1/Bi Y θ(min) T c (teorico) T c (experi)0.0322 0.0331 0.86 3 31.37 31.2
0.0644 0.0331 0.70 6 42.65 44.40.0966 0.0331 0.50 9 56.75 57.80.1288 0.0331 0.35 12 67.325 67.80.1610 0.0331 0.25 15 74.375 75.40.1932 0.0331 0.18 18 79.31 80.40.2255 0.0331 0.15 21 81.425 83.80.2577 0.0331 0.10 24 84.95 86.20.2899 0.0331 0.08 27 86.36 87.60.3220 0.0331 0.06 30 87.77 88.50.3543 0.0331 0.04 33 89.18 89.20.3864 0.0331 0.03 36 89.885 89.6
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3 6 9 12 15 18 21 24 27 30 33 360
102030405060708090
100
t(min)
Tc(t
eoric
o)
3 6 9 12 15 18 21 24 27 30 33 360
102030405060708090
100
t(min)
Tc(e
xper
i)
La temperatura de cocción de la papa es:
Teóricamente: 89.885ºC
Experimentalmente: 89.6
%ERROR=(89.885−89.6 )
89.885×100
%ERROR=0.31%
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Zanahoria:
V= 70 ml =7*10-5m3
m= 70.70gr = 0.0707Kg
2L=6.3cm
L=0.0315m
ρ=mv
= 0.0707 kg7∗10−5m3=1010
kgm3
V=π r2 L
r=√ VπL
=√ 7∗10−5m3
π 0.0315m=0.0266m
cp=3770 jkgºC
k=0.65 wmºC
=0.65 jseg∗mºC
α= Kρ∗cp
= 0.4981010∗3770
=1.3078∗10−7 m2
seg
θ T (r=o ) T (w) T (α)
0 22.53 41.8 91 926 55.2 91.2 929 67.9 91.2 9212 76.5 91.2 9215 82.1 91.2 9218 85.5 91.2 92
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21 87.5 91.2 9224 88.6 91.2 9227 89.4 91.2 9230 89.7 91.3 9233 90.2 91.3 9236 90.6 91.3 92
I. Cálculos:
Haciendo cumplir con la condición:
Q/A<3000wm2 …………………h=537(92−91)1/7=537 w
mºC
Entonces:
3000< Q/A<63000wm2
1Bi
= Khr o
= 0.498563.4427∗0.0267
=0.0331
FO=αθro2 =1.2756∗10
−7∗1800.02672
=0.0322
Y= f (F0 ,1Bi )=
T c−T α
T i−T α
Y=T c−9221.5−92
Y (21.5−92 )+92=T c
F0 1/Bi Y θ(min) T c (teorico) T c (experi)
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