Practica Calorimetria

INSTITUTO POLITECNICO NACIONAL

ESCUELA SUPERIOR DE INGENIERIA QUIMICA
E INDUSTRIAS EXTRACTIVAS


DEPARTAMENTO DE INGENIERIA QUIMICA INDUSTRIAL


LABORATORIO DE FENOMENOS DE TRANSPORTE


PRACTICA No. 1
“TRANSFERENCIA DE CALOR”

GRUPO: 3IV3 SALON: 7305

EQUIPO: 1


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PROFESOR : JORGE PALAFOX RAMOS
FECHA DE ENTREGA: 12 DE ABRIL DEL2010









PRACTICA-. “TRANSFERENCIA DE CALOR” (CONDUMETRIA)

OBJETIVO:
Obtener la distribución de temperaturas a lo largo de una barra y comparar los resultados experimentales con los teóricos.








































Ecuaciones :
Energía en estado estacionario
[pic]
Donde:
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Donde:
L (=)longitud de la barra de bronce 350 mm
T1 (=) termo par 1
TA,T2, T3,T4,etc… (=) temperatura en los termopares

Distancia entre termopar 50 mm
TA(=)Temperatura del ambiente
Ɵ (=) temperatura adimensional
m (=) pendiente de la curva
p (=) perímetro de la barra
diámetro de la barra 10mm
A (=) área de transferencia de calor [pic]
KBronce(=) coeficiente de transferencia de calor
H (=)coeficiente global de transferencia de calor por convección y radiación

Ecuación para determinar la pendiente
[pic]
T9(=) Ta

Para calcular el calor total que se pierde en la barra

[pic]

Donde :
As (=) area total de la superficie extendida (=) ΠDL
Ts (=) temperatura de la superficie promedio (T1- T8 + 273)
H (=) Hcm + Hrm
[pic]
[pic]

Donde:
σ (=) es la constante de boltzman (=)56.7x10-9
ε(=) emisividad de la superficie
f (=) factor = 1
Emisividad = 0.95












T1 = 339.95 °k
Ta = 296.65°k
Supuesta m =7.103




TABLA 1
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Ө1= 1
T(x 1)=339.95

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Ɵ2= 0.857142857
T(x 2)=333.6214286

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Ɵ3= 0.714285714
T(x 3)=327.2928571

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Ɵ4= 0.571428571
T(x 4)=320.9642857

[pic]
Ɵ5= 0.428571428
T(x 5)=314.6357143

[pic]
Ɵ6=0.285714285
T(x 6)=308.3071429

[pic]
Ɵ7= 0.14285714
T(x 7)=301.9785714

[pic]
Ɵ3= 0
T(x 3)=295.65



Ts = 317.8


H (=) Hcm + Hrm
H= 15.27


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TABLA 1
  |L (mm) |V |I |T °C |T °K |T. TEORIK °K |Ɵ |H | |X1 |0 |9.5 |0.36 |T1= 66.8 |339.95 |339.95 |1 |15.27 | |X2 |50 |9.5 |0.36 |T2= 55.2 |328.35 |333.6214 |0.857142 |  |  | |X3|100 |9.5 |0.36 |T3= 46.3 |319.45 |327.2928 |0.714285 |Hcm |Hrm | |X4 |150 |9.5 |0.36 |T4=39.9 |313.05 |320.9642 |0.571428 |9.0556 |6.22527134 | |X5 |200 |9.5 |0.36 |T5=34.6 |307.75 |314.6357 |0.428571 |  |  | |X6 |250 |9.5 |0.36 |T6=31.1 |304.25 |308.3071 |0.285714 |K =121.06 | |X7 |300 |9.5 |0.36 |T7=29 |302.15 |301.9785 |0.142857 |  |  | |X8 |350 |9.5 |0.36 |T8=28.1 |301.25 |295.65 |0 |Qp= 3.71 ||X9 |Ta |9.5 |0.36 |T9=22.5 |295.65 |  | | | | | | |Q= 3.42 Watts | | |TS= 317.8 | | | | |[pic]







CALCULOS 2
[pic]
Ɵ1= 1
T(x 1)=338.95

[pic]
Ɵ2= 0.857142857
T(x 2)=332.7642857

[pic]
Ɵ3= 0.714285714
T(x 3)=326.5785714

[pic]
Ɵ4= 0.571428571
T(x 4)=320.3928571

[pic]
Ɵ5= 0.428571428
T(x 5)=314.2071428

[pic]
Ɵ6= 0.285714285
T(x 6)=308.0214285

[pic]
Ɵ7=0.14285714
T(x 7)=301.8357142

[pic]
Ɵ3= 0
T(x 3)=295.65



Ts = 317.3


H (=) Hcm + Hrm
H= 15.214


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TABLA 2
  |L (mm) |V |I |T °C |T °K |T. TEORIK °K |Ɵ |H | |X1 |0 |11.5 |0.44 |T1= 65.8 |338.95 |338.95 |1 |15.214 | |X2 |50 |11.5 |0.44 |T2= 52.8 |325.95 |332.76 |0.85714286 |  |  | |X3 |100 |11.5 |0.44 |T3= 44.1 |317.25|326.58 |0.71428571 |Hcm |Hrm | |X4 |150 |11.5 |0.44 |T4=38.5 |311.65 |320.39 |0.57142857 |9.004 |6.21 | |X5 |200 |11.5 |0.44 |T5=34.1 |307.25 |314.21 |0.42857143 |  |  | |X6 |250 |11.5 |0.44 |T6=31.4 |304.55 |308.02 |0.28571429 |K =120.7 | |X7 |300 |11.5 |0.44 |T7=29.7 |302.85 |301.84 |0.14285714 |  |  | |X8 |350 |11.5 |0.44 |T8=28.9 |302.05 |295.65 |0 |Qp= 3.62 | |X9 |Ta |11.5 |0.44 |T9=22.5...