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• Maria Luisa

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ALETA CONICA

BALANCE DE ENERGIA: Entrada – Salida = 0 ×

−

×

×

+ 2

−

×

×

(

−

=

) = −2

=

−2

−

−

2

𝐴 𝑥 = 𝜋 × 𝑦2 𝑦 = 𝑚𝑥 + 𝑏 𝐴 𝑥 = 𝜋 × 𝑚𝑥 + 𝑏

2

𝑑𝑇 𝑑𝑥

. .

+

2

2

+2

+

. .

2

. .

−

=

−

2

×

= 2

−

.

−

=

Para: x=0, T= Tw

𝑑𝐴 𝑥 = 2𝜋 𝑚 + 𝑏 . 𝑚 𝑑𝑥 𝑞𝑥 = −𝑘

+

x=L, −

Discretizando: 2

𝑑𝑞𝑥 𝑑2 𝑇 = −𝑘 2 𝑑𝑥 𝑑𝑥 −

+ +

+

×

− −

2

−

×

− ×

= −

=

Para i=1 2

−2

+

+

2

×

+

+

2 2

×

−

−

2

−

−

2

×

−

=

−

=

Para i=2 −2

2

2

×

2

2

Para i=3 −2

+

22

+

2

×

−

2

2

−

×

−

= .

. . Para i=10 −2

+

+

2

×

−

−

2

×

 i=10 −

=

− −

=− −

−

=− −

=

+

EFICIENCIA. =

∫ 2

−

∫ 2

−

=

−

∫ −

∫

−

=

ALETA TIPO DISCO DE AREA VARIABLE

BALANCE DE ENERGIA:

Entrada – Salida = 0 ×

−

𝐴𝐿 = 4𝜋𝑟𝛥𝑟𝑠𝑒𝑐𝜃

× ×

−

+2

−

×

=

−

= −

𝐴 𝑇 = 4𝜋𝑟𝑦 ×

𝑦 = 𝑚𝑟 + 𝑤

= −4

𝐴 𝑇 = 4𝜋𝑟 𝑚𝑟 + 𝑤

=

𝑑𝐴 𝑇 = 8𝜋 𝑚𝑟 + 𝑤 . 𝑚 𝑑𝑥 𝑞𝑟 = −𝑘

𝑑𝑇 𝑑𝑟

.

+

2

.

−

= −4

−

2

4

+

.

2

𝑑𝑞𝑟 𝑑 𝑇 = −𝑘 2 𝑑𝑟 𝑑𝑟

−

2

+

2

+

2

8 2.

+

.

=4 −

=

+

.

Para: r=R1, T= Tw r=R2, −

=

−

−

Discretizando: 2

−

+

+

+

2

×

×

−

−

×

−

−

×

+

Para i=1 2

−2

+

+

2

×

+

+

2

×

+

2

−

−

−

−

×

−

=

×

−

=

×

−

= .

Para i=2

−2

2

2

2

2

Para i=3

−2

+

2

×

−

2

−

. . Para i=10

−2

+

+

2

×

−

−

×

 i=10 −

=

− −

=− −

=

=− −

−

+

−

=

=

−

=

Usando los siguientes datos obtenga el perfil de temperatura de la aleta cónica:

=4 =

,

,

= = 2.

,

,

= 2. =2

,

,

2

= .

=2

DESARROLLO UTILIZANDO POLYMATH # TRANSFERENCIA DE CALOR CON ALETA EN TRONCO DE CONO CON TEMPERATURA MAXIMA f(T1) = T2 - 2 * T1 + T0 + 2 * m * dx * (T1 - T0) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T1 - Tinf) / (k * (m * 1 * dx + b)) f(T2) = T3 - 2 * T2 + T1 + 2 * m * dx * (T2 - T1) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T2 - Tinf) / (k * (m * 2 * dx + b)) f(T3) = T4 - 2 * T3 + T2 + 2 * m * dx * (T3 - T2) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T3 - Tinf) / (k * (m * 3 * dx + b)) f(T4) = T5 - 2 * T4 + T3 + 2 * m * dx * (T4 - T3) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T4 - Tinf) / (k * (m * 4 * dx + b)) f(T5) = T6 - 2 * T5 + T4 + 2 * m * dx * (T5 - T4) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T5 - Tinf) / (k * (m * 5 * dx + b)) f(T6) = T7 - 2 * T6 + T5 + 2 * m * dx * (T6 - T5) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T6 - Tinf) / (k * (m * 6 * dx + b)) f(T7) = T8 - 2 * T7 + T6 + 2 * m * dx * (T7 - T6) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T7 - Tinf) / (k * (m * 7 * dx + b)) f(T8) = T9 - 2 * T8 + T7 + 2 * m * dx * (T8 - T7) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T8 - Tinf) / (k * (m * 8 * dx + b)) f(T9) = T10 - 2 * T9 + T8 + 2 * m * dx * (T9 - T8) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T9 - Tinf) / (k * (m * 9 * dx + b)) f(T10) = T11 - 2 * T10 + T9 + 2 * m * dx * (T10 - T9) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T10 - Tinf) / (k * (m * 10 * dx + b)) # Datos k = 40 # W/m°C h = 60 # W/m^2°C L = 0.1 # m dx = 0.001 # m Tinf = 25 # ° T0 = Tw Tw = 250 # °C R1 = 0.025 # m R2 = 0.015 # m m = -(R1 - R2) / L b = R1 T1(0) = 0 T2(0) = 0 T3(0) = 0 T4(0) = 0 T5(0) = 0 T6(0) = 0 T7(0) = 0 T8(0) = 0 T9(0) = 0 T10(0) = 0 T11 = -(h * dx * (T10 - Tinf) / k) + T10

POLYMATH Report

No Title 20-may-2014

Nonlinear Equations

Calculated values of NLE variables Variable Value

f(x)

Initial Guess

1 T1

249.4338 -2.253E-14 0

2 T2

248.8901 3.584E-14 0

3 T3

248.3692 -3.633E-14 0

4 T4

247.8711 2.648E-14 0

5 T5

247.3963 -1.185E-14 0

6 T6

246.9449 9.402E-16 0

7 T7

246.5171 6.169E-15 0

8 T8

246.1132 -2.376E-14 0

9 T9

245.7335 3.893E-14 0

10 T10

245.3783 -4.364E-14 0

Variable Value 1 k

40.

2 h

60.

3 L

0.1

4 dx

0.001

5 Tinf

25.

6 Tw

250.

7 T0

250.

8 R1

0.025

9 R2

0.015

10 m

-0.1

11 b

0.025

12 T11

245.0477

Nonlinear equations 1

f(T1) = T2 - 2 * T1 + T0 + 2 * m * dx * (T1 - T0) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T1 - Tinf) / (k * (m * 1 * dx + b)) = 0

2

f(T2) = T3 - 2 * T2 + T1 + 2 * m * dx * (T2 - T1) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T2 - Tinf) / (k * (m * 2 * dx + b)) = 0

3

f(T3) = T4 - 2 * T3 + T2 + 2 * m * dx * (T3 - T2) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T3 - Tinf) / (k * (m * 3 * dx + b)) = 0

4

f(T4) = T5 - 2 * T4 + T3 + 2 * m * dx * (T4 - T3) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T4 - Tinf) / (k * (m * 4 * dx + b)) = 0

5

f(T5) = T6 - 2 * T5 + T4 + 2 * m * dx * (T5 - T4) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T5 - Tinf) / (k * (m * 5 * dx + b)) = 0

6

f(T6) = T7 - 2 * T6 + T5 + 2 * m * dx * (T6 - T5) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T6 - Tinf) / (k * (m * 6 * dx + b)) = 0

7

f(T7) = T8 - 2 * T7 + T6 + 2 * m * dx * (T7 - T6) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T7 - Tinf) / (k

* (m * 7 * dx + b)) = 0 8

f(T8) = T9 - 2 * T8 + T7 + 2 * m * dx * (T8 - T7) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T8 - Tinf) / (k * (m * 8 * dx + b)) = 0

9

f(T9) = T10 - 2 * T9 + T8 + 2 * m * dx * (T9 - T8) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T9 - Tinf) / (k * (m * 9 * dx + b)) = 0

10

f(T10) = T11 - 2 * T10 + T9 + 2 * m * dx * (T10 - T9) / (m * 1 * dx + b) - 2 * h * dx ^ 2 * (T10 Tinf) / (k * (m * 10 * dx + b)) = 0

Explicit equations 1

k = 40 W/m°C

2

h = 60 W/m^2°C

3

L = 0.1 m

4

dx = 0.001 m

5

Tinf = 25 °

6

Tw = 250 °C

7

T0 = Tw

8

R1 = 0.025 m

9

R2 = 0.015 m

10 m = -(R1 - R2) / L 11 b = R1 12 T11 = -(h * dx * (T10 - Tinf) / k) + T10 General Settings Total number of equations

22

Number of implicit equations 10 Number of explicit equations 12 Elapsed time

0.0000 sec

Solution method

SAFENEWT

Max iterations

150

Tolerance F

0.0000001

Tolerance X

0.0000001

Tolerance min

0.0000001

Ahora con estos datos obtenidos de las temperaturas:

r(m)

T°C

(T-Tinf)°C

0

250

225

0.001

249.4338

224.4338

0.002

248.8901

223.8901

0.003

248.3692

223.3692

0.004

247.8711

222.8711

0.005

247.3963

222.3963

0.006

246.9449

221.9449

0.007

246.5171

221.5171

0.008

246.1132

221.1132

0.009

245.7335

220.7335

0.01

245.3783

220.3783

Grafica ( r Vs Temperatura ) 251

250

249

248

T°C

247

246

245 0

0.002

0.004

0.006

0.008

0.01

0.012

HALLANDO LA EFICIENCIA: Δx= 0.001 m Tinf = 25°C TW= 250°C =

r(m) 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01

−

∫

T°C 250

(T-Tinf)°C 225 224.4338 223.8901 223.3692 222.8711 222.3963 221.9449 221.5171 221.1132 220.7335 220.3783

249.4338 248.8901 248.3692 247.8711 247.3963 246.9449 246.5171 246.1132 245.7335 245.3783

Ahora hallando la eficiencia por el método de trapecio: =∫

=

. 2

[22 + 22 .

−

=

8 + 2 224.4

+ 22 . 44 + 22 .

+

2

8 + 22 .8 + 22 .

+ 2∑

+ 22 . 2 + 22 .

= 2.224 =

∫

−

= = .

2.224 . 2 −2

888

2 + 222.8 + 22 .

8 ]

+ 222.