transferencia de calor agua-aire

7

Click here to load reader

Upload: roycaves

Post on 12-Dec-2015

222 views

Category:

Documents


1 download

DESCRIPTION

Cálculos para la transferencia de calor de agua a aire.

TRANSCRIPT

Page 1: Transferencia de Calor Agua-Aire

PRESENTACIÓN DE RESULTADOS

Lecturas 1 2 3 4 5 6Control de Temperatura

50 50 70 70 80 80

Control del Flujo de Aire

40 40 60 60 70 70

R.P.M 4 000 4 000 6 000 6 000 7 000 7 000Voltaje 100 105 150 150 170 170Amperaje 3.5 3.5 5 5 5.5 5.5Presión 300 mm Manómetro

20 20 57 56 74 74

Presión 200 mm Manómetro

7 7 26 25 33 33

Temperatura de Entrada de Agua

22 23 24 23 23 23

Temperatura Ambiente

23 23 23 23 23 23

Termopar 1 28 29 38 36 39 39Termopar 2 21 23 26 24 24 29Termopar 3 23 24 26 30 32 28Termopar 4 54 57 105 117 132 136Temperatura Salida de Agua

23.5 23.8 25 27 28.5 30

Caudal de Agua por Minuto

1 1 1 1 1 1

Tipo de Flujo Contraflujo Paralelo Paralelo Contraflujo Contraflujo Paralelo

Page 2: Transferencia de Calor Agua-Aire

CÁLCULOS

Primera Lectura: Flujo Paralelo

Q=VI

Q= (100V ) (3.5 A )

Q=350W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=350W

1010J

Kg°C(54 °C−28 ° C )

mA=0.0133Kgs

∆T 1=T 1−T o

∆T 1=28 ° C−23° C

∆T 1=5 °C

∆T 2=T 4−T i

∆T 2=54 ° C−22 °C

∆T 2=32 °C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

5 °C−32 °C

ln( 5 °C32 °C )∆T DMLT=14.5450 ° C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 350W

(0.1144m2 ) (14.5450 °C )

U=210.167 W

m2 °C

Page 3: Transferencia de Calor Agua-Aire

Segunda Lectura: Flujo en Contraflujo

Q=VI

Q= (105V ) (3.5 A )

Q=367.5W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=367.5W

1010J

Kg°C(57 °C−29 °C )

mA=0.0129Kgs

∆T 1=T 4−T i

∆T 1=57 ° C−23° C

∆T 1=34 ° C

∆T 2=T 1−T o

∆T 2=29 ° C−24 °C

∆T 2=5 °C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

34 ° C−5° C

ln( 34 ° C5 ° C )∆T DMLT=15.1284 ° C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 367.5W

(0.1144m2 ) (15.1284 ° C )

U=212.16 W

m2 °C

Page 4: Transferencia de Calor Agua-Aire

Tercera Lectura: Flujo Paralelo

Q=VI

Q= (150V ) (5 A )

Q=750W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=750W

1010J

Kg°C(105° C−38 °C )

mA=0.0110Kgs

∆T 1=T 1−T i

∆T 1=38 °C−24 °C

∆T 1=14 ° C

∆T 2=T 4−To

∆T 2=105 ° C−25° C

∆T 2=80 ° C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

14 ° C−80 °C

ln( 14 ° C80 ° C )∆T DMLT=37.8664 ° C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 750W

(0.1144m2 ) (37.8664 °C )

U=172.98 W

m2° C

Page 5: Transferencia de Calor Agua-Aire

Cuarta Lectura: Flujo en Contraflujo

Q=VI

Q= (150V ) (5 A )

Q=750W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=750W

1010J

Kg°C(117° C−36° C )

mA=0.0091Kgs

∆T 1=T 4−T i

∆T 1=117° C−23 °C

∆T 1=9 4 °C

∆T 2=T 1−T o

∆T 2=36 ° C−24 °C

∆T 2=12 °C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

94 ° C−12° C

ln( 94 ° C12 °C )∆T DMLT=39.8369 ° C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 750W

(0.1144m2 ) (39.8369° C )

U=164.43 W

m2° C

Page 6: Transferencia de Calor Agua-Aire

Quinta Lectura: Flujo en Contraflujo

Q=VI

Q= (170V ) (5 .5 A )

Q=935W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=935W

1010J

Kg°C(132° C−39 °C )

mA=0.0099Kgs

∆T 1=T 4−T i

∆T 1=132 °C−23 ° C

∆T 1=109 °C

∆T 2=T 1−T o

∆T 2=39 ° C−28.5° C

∆T 2=10.5 °C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

109 °C−10.5 °C

ln( 109 °C10.5 °C )∆T DMLT=42.0945 °C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 935W

(0.1144m2 ) (42.0945 ° C )

U=193.99 W

m2° C

Page 7: Transferencia de Calor Agua-Aire

Sexta Lectura: Flujo Paralelo

Q=VI

Q= (170V ) (5 .5 A )

Q=935W

Q=mACpaire (T 4−T 1 )

mA=Q

Cpaire (T4−T 1)

mA=935W

1010J

Kg°C(136 °C−39 °C )

mA=0.0095Kgs

∆T 1=T 1−T i

∆T 1=39 °C−23° C

∆T 1=16 ° C

∆T 2=T 4−To

∆T 2=136 ° C−30° C

∆T 2=106 ° C

∆T DMLT=∆T 1−∆T 2

ln(∆T 1∆T 2 )∆T DMLT=

16 °C−106 ° C

ln( 16 ° C106 ° C )∆T DMLT=47.5976 °C

Para el área del conducto:

A=L∅ π

A=(1.850m ) (0.0197m )π

A=0.1144m2

Q=UA ∆T DMLT

U= QA ∆T DMLT

U= 935W

(0.1144m2 ) (47.5976 ° C )

U=171.56 W

m2 °C