Pdf Finite Diff.pdf

29/11/2018

PROBLEMA

DIFERENCIAS FINITAS REGIMEN TRANSITORIO

A great plane plate of steel has 2 cm of thickness. If the initial temperatures inside the plate are function from the distance to a face according to the equations: T=100x for 0≤x≤1 ; T=100*(2-x) for 1≤x≤2 determine the temperatures as a function of x and t when both faces stay to 0 oC. to use Δx=0.4 cm. and kΔt /ρ CpΔx2 =1/4. (to apply 5 increments of time).

To use Δx=0.4 cm. and kΔt /ρ CpΔx2 =1/4. (to apply 5 increments of time).

∆𝑥 = 0.4𝑐𝑚

∆𝑥 = 0.4𝑐𝑚

Propiedades del material

∆𝑡 =

Steel, AISI 1010 (0.1% carbon) 1.88 x 10−5 m2/s Steel, 1% carbon

1.172 × 10−5 m2/s

Steel, stainless 304A at 27 °C

4.2 × 10−6 m2/s

Steel, stainless 310 at 25 °C

3.352 × 10−6 m2/s

∆𝑡 = 0.2128𝑠𝑒𝑔

∆𝑥 4 𝛼

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Difusividad térmica y lambda (Fourier:Fo)

T=100x for 0≤x≤1 ; T=100*(2-x) for 1≤x≤2 0°𝐶

40°𝐶

80°𝐶

80°𝐶

40°𝐶

0°𝐶

2cm y x 0.4cm

0.4cm

0.4cm

0.4cm

0.4cm

1cm

1cm

𝑇 = 100𝑥, 0 ≤ 𝑥 ≤ 1

𝑇 = 100 2 − 𝑥 , 1 ≤ 𝑥 ≤ 2

COMPROBAR SISTEMA DE UNIDADES

clc, clear all ti=0; Nt=5; xi=0/100; xf=2/100; Dx=0.4/100; Nx=(xf-xi)/Dx; alfa=1.88e-5; %m^2/s Dt=Dx^2/4/alfa; t=ti:Dt:Nt*Dt; lambda=1/4; %alfa*Dt/(Dx^2)

0 0 0 0 0 0

40

80

80

40

0 0 0 0 0 0

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T=zeros(Nt+1,Nx+1); con=1; for c=xi:Dx:xf if c<=1/100 T(1,con)=100*c*100; else T(1,con)=100*(2-c*100); end con=con+1; end

T=

𝑻𝒊

T=

0 0 0 0 0 0 𝟏,𝒋0𝟏 =0𝑻𝒊,𝒋 0 0 0 0

for i=1:Nt for j=1:Nx-1 T(i+1,j+1)=T(i,j+1)+lambda*(T(i,j+2)-2*T(i,j+1)+T(i,j)); end end (1,1)

i=1 %fila j=1 %columna

(1,2)

0 0 0

(1,3)

40 (2,2)

80

80

40

0 0 0

y x

0 0 0 0 0 0 0 0 0 0 0 0 0𝟏 + 𝝀0𝑻𝒊,𝒋 0𝟐 − 𝟐𝑻 0 𝒊,𝒋 0 0 0 0 0 0 0 0

𝟏

+ 𝑻𝒊,𝒋

0 40 80 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

80 0 0 0 0 0

40 0 0 0 0 0

0

40.0000 80.0000 80.0000 40.0000 0

0

40.0000 70.0000 70.0000 40.0000 0

0

37.5000 62.5000 62.5000 37.5000 0

0

34.3750 56.2500 56.2500 34.3750 0

0

31.2500 50.7813 50.7813 31.2500 0

0

28.3203 45.8984 45.8984 28.3203 0

0 0 0 0 0 0

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0 0 0 0 0 0

40

80

80

40

40

70

70

40

37.5

62.5

62.5

37.5

34.375

56.25

56.25

34.375

31.25 50.7813 50.7813 31.25 28.3203 45.8984 45.8984 28.3203

0 0 0 0 0 0

T x=(xi:Dx:xf)*100; plot(x,T’) grid on title('Gráfica de Schmidt') xlabel('Distancia') ylabel('Temperaturas')

T([1:end],1)=100 T([1:end],end)=10 T(1,[2:end-1])=50

100 100 100 100 100 100

50

50

50

50

10 10 10 10 10 10

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DIFERENCIAS FINITAS REGIMEN ESTACIONARIO

DATOS 𝒌 = 𝟒𝟓

𝑾 𝒎°𝑪

𝒆̇ = 𝟓 × 𝟏𝟎𝟔

𝒉 = 𝟓𝟓 𝑾 𝒎𝟑

𝑾 𝒎𝟐 °𝑪

𝒒̇ 𝑳 = 𝟖𝟎𝟎𝟎

𝑾 𝒎𝟐

𝑻𝟎 = 𝟏𝟐𝟎°𝑪

#𝒏𝒐𝒅𝒐𝒔 = 𝟏𝟑

𝑻

∆𝒙 = ∆𝒚 = 𝟏.5cm

= 𝟑𝟎°𝑪

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BALANCE DE ENERGIA

ANALISIS NODOS #1, #2 y #3 − 𝑻𝟑 𝑻𝒆̇ 𝟎 𝒍−𝟐 𝑻 −𝟐𝑻−𝟐 𝑻𝟑 +𝒍 𝒌𝑻𝒍𝟑𝒍 𝑻𝑻 −𝟐𝟔𝑻− 𝑳 − 𝑻𝟐𝑳 +𝒍 𝒌𝑻𝟏𝒍 𝑻 𝑻𝟐𝟒𝟎 −𝒆̇ 𝟎𝑻𝒍𝟏𝟐 𝒆̇ 𝟎 𝒍𝟐 𝟐 𝑻𝟏+ 𝟓 𝒍 = 𝒉𝒍 𝒉𝒍 𝑻𝒒̇ 𝑻− 𝑻 + 𝒌 + 𝒌 + =𝟎 𝟐 𝑻 𝟐−𝒍 𝑻𝒍𝟏 + 𝟐𝒌 𝟐 𝒍 𝒍 + 𝒌𝒍+ 𝒌 =𝟎 𝟐𝒍 𝑳 +𝒉 𝟐 𝟐 +

𝟐

𝑇

+𝑇

+𝑇

+𝑇

− 4𝑇

+

𝑒̇

𝑙 𝑘

𝟐

𝟐

𝒍

𝟐

𝒍

𝟒

=0

ANALISIS NODOS #4, #5 y #6 𝒍 𝑻𝟏 −𝒍𝑻𝑻𝟒𝟑 − 𝑻𝟔𝒍 𝟏𝟐𝟎𝒍 𝑻 −𝟓 𝑻−𝟒𝑻𝟔 𝒆̇𝑻𝟎𝟓𝟏𝟐𝟎 𝒍𝟐− 𝑻 −𝟒𝑻𝟔 𝒆̇ 𝟎𝒆̇𝒍𝟎𝟐𝒍𝟐 𝒉𝒍𝒒̇ 𝑳𝑻𝒍 +−𝒌𝑻𝟔 +𝑻𝒌𝟒 + 𝑻𝟐++𝒌𝑻𝟔++𝒌𝟏𝟐𝟎 − 𝟒𝑻 +𝒌𝒍 = 𝟎 + + ++ 𝒌𝒍 ==𝟎𝟎 𝒍 𝟐 𝟐𝒍 𝒍 𝟓 𝒌 𝒍 𝒍 𝟐 𝒍𝟐 𝟒𝟐

ANALISIS NODOS #7 y #8 𝒉

𝒍 𝒍 𝑻 𝟕 − 𝑻𝟖 𝒍 𝑻𝟖 − 𝑻𝟕 𝟏𝟐𝟎 − 𝑻𝟕 𝒆̇ 𝟎 𝒍𝟐 𝑻 − 𝑻𝟖 + 𝒌 +𝒌 + 𝒌𝒍 + =𝟎 𝟐 𝟐 𝒍 𝟐 𝒍 𝒍 𝟐

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SISTEMA DE ECUACIONES LINEALES

𝑳 𝑳 𝒍 𝑻𝟐 − 𝑻𝟏 𝒍 𝑻𝟒 − 𝑻𝟏 𝒆̇ 𝟎 𝒍𝟐 + 𝒉 𝑻 − 𝑻𝟏 + 𝒌 +𝒌 + =𝟎 𝟐 𝟐 𝟐 𝒍 𝟐 𝒍 𝟒 𝒍 𝑻𝟏 − 𝑻𝟐 𝒍 𝑻𝟑 − 𝑻𝟐 𝑻𝟓 − 𝑻𝟐 𝒆̇ 𝟎 𝒍𝟐 𝒉𝒍 𝑻 − 𝑻𝟐 + 𝒌 +𝒌 + 𝒌𝒍 + =𝟎 𝟐 𝒍 𝟐 𝒍 𝒍 𝟐

𝒒̇ 𝑳

𝟏𝟔𝟑. 𝟔

𝟏𝟔𝟎. 𝟓

𝟏𝟓𝟔. 𝟒

𝒉𝒍 𝑻 − 𝑻𝟐 + 𝒌

𝟏𝟓𝟒. 𝟎 𝟏𝟓𝟏. 𝟎 𝟏𝟒𝟒. 𝟒 𝟏𝟑𝟒. 𝟓 𝟏𝟑𝟐. 𝟔

𝒍 𝑻𝟏 − 𝑻𝟒 𝒍 𝟏𝟐𝟎 − 𝑻𝟒 𝑻𝟓 − 𝑻𝟒 𝒆̇ 𝟎 𝒍𝟐 +𝒌 + 𝒌𝒍 + =𝟎 𝟐 𝒍 𝟐 𝒍 𝒍 𝟒 𝒆̇ 𝟎 𝒍𝟐 𝑻𝟒 + 𝑻𝟐 + 𝑻𝟔 + 𝟏𝟐𝟎 − 𝟒𝑻𝟓 + =𝟎 𝒌 𝒍 𝑻𝟑 − 𝑻𝟔 𝒍 𝑻𝟓 − 𝑻𝟔 𝟏𝟐𝟎 − 𝑻𝟔 𝒆̇ 𝟎 𝒍𝟐 𝒉𝒍 𝑻 − 𝑻𝟔 + 𝒌 +𝒌 + 𝒌𝒍 + =𝟎 𝟐 𝒍 𝟐 𝒍 𝒍 𝟐 𝒒̇ 𝑳 𝒍 + 𝒌

𝒉

DIFERENCIAS FINITAS SUPERFICIES EXTENDIDAS

𝒍 𝑻𝟐 − 𝑻𝟑 𝒍 𝑻𝟔 − 𝑻𝟑 𝒆̇ 𝟎 𝒍𝟐 +𝒌 + =𝟎 𝟐 𝒍 𝟐 𝒍 𝟐

𝒍 𝒍 𝑻𝟕 − 𝑻𝟖 𝒍 𝑻𝟖 − 𝑻𝟕 𝟏𝟐𝟎 − 𝑻𝟕 𝒆̇ 𝟎 𝒍𝟐 𝑻 − 𝑻𝟖 + 𝒌 +𝒌 + 𝒌𝒍 + =𝟎 𝟐 𝟐 𝒍 𝟐 𝒍 𝒍 𝟐

𝑻𝟏 𝑻𝟐 𝑻𝟑 𝑻𝟒 𝑻𝟓 𝑻𝟔 𝑻𝟕 𝑻𝟖

= 𝟏𝟔𝟑. 𝟔°𝑪 = 𝟏𝟔𝟎. 𝟓°𝑪 = 𝟏𝟓𝟔. 𝟒°𝑪 = 𝟏𝟓𝟒. 𝟎°𝑪 = 𝟏𝟓𝟏. 𝟎°𝑪 = 𝟏𝟒𝟒. 𝟒°𝑪 = 𝟏𝟑𝟒. 𝟓°𝑪 = 𝟏𝟑𝟐. 𝟔°𝑪

Determine the temperature distribution based to time and distance, a thin copper fin joined by one side to a wall having a constant temperature of 200°C. The air around the fin is 10°C. Consider that initially fin is 30°C and the convective coefficient is 40W/ °C m2 . Consider Δx = 0.5 cm , time of 5 minutes. The answer must be shown in a table and a graph. (Properties of copper: k=332 Kcal/mh°C, ρ=8950 Kg/m3, Cp=0.0915 Kcal/Kg).(L=10 cm; w=3 cm; e=0.5 cm).(Using λ = 1/4 and λ = 1/2 discuss the results corresponding obtained). The equation for the fin is:

∝

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ALETA RECTANGULAR

ALETA CILINDRICA

En nuestro caso: 𝐴 = 𝑤𝑡 𝐴 = 2𝑤 𝐿 + 𝑡⁄2

10°C

Properties of copper: k=332 Kcal/mh°C ρ=8950 Kg/m3 Cp=0.0915 Kcal/Kg

L=10 cm w=3 cm e=0.5 cm

e=0.5 cm

200°C 30°C

𝐴𝑐 w=3 cm

L=10 cm

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𝑘

𝑘

𝜕 𝜕𝑇 𝜕𝐴 𝐴 −ℎ 𝑇−𝑇 𝜕𝑥 𝜕𝑥 𝜕𝑥

= 𝜌𝐶 𝐴

𝜕𝐴 𝜕𝑇 𝜕 𝑇 𝜕𝐴 +𝐴 −ℎ 𝑇−𝑇 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝐴 =0 𝜕𝑥

𝜕𝑇 𝜕𝑡

= 𝜌𝐶 𝐴

𝜕𝐴 = 𝑤𝐿 𝜕𝑥

𝐵𝑖 =

ℎ𝐿 𝑘

𝐹 =

𝛼𝑡 𝐿

𝑘 𝜕 𝑇 ℎ 𝑤𝑒 − 𝑤𝐿 𝑇 − 𝑇 𝜌𝐶 𝜕𝑥 𝜌𝐶 𝜕𝑇 𝜕𝑡

𝛼

𝜕 𝑇 ℎ 𝐿 − 𝑇−𝑇 𝜕𝑥 𝜌𝐶 𝑒

= 𝜌𝐶 𝑒𝑤

=

𝜕𝑇 𝜕𝑡

𝜕𝑇 𝜕𝑡