Transferencia de calor, cálculos de coeficiente de convección, distribución de temperaturas, flujo de calor, eficiencia de aletas

Coeficiente global de Convección

T_f=(T_b+T_∞)/2

T_f=((100+22.25) )/2

T_f=61.125 °C=334,125 K

Para T_f=334,125 K usando propiedades del aire de Cengel.

ν=19,071375*〖10〗^(-6) [m^2/s]

k=28,162125*〖10〗^(-3) [W/mK ]

P_r= 0.719919

Ra_D=(gβ(T_s-T_∞)D^3)/ν^2 P_r

Ra_D=(9,81 [m/s^2 ]*(1/334,125)[1/K]*(100-22.25)[K] 〖*(0,02)〗^3 [m^3])/((19,071375*〖10〗^(-6) )^2 [m^4/s^2 ] )*0.719919

Ra_D=36164,8

(Nu) ̅_D={0.6+(0.387〖Ra〗_D^(1/6))/[1+(0.559/Pr)^(9/16) ]^(8/27) }^2

(Nu) ̅_D={0.6+(0.387(36164,8)^(1/6))/[1+(0.559/0.719919)^(9/16) ]^(8/27) }^2

(Nu) ̅_D=5,99945

h ̅=((Nu) ̅_D k)/D

h ̅=(5,99945*28,162125*〖10〗^(-3) [W/mK])/(0,02 [m])

h ̅=8,44786 [W/m^2 K]

Cálculos

Aleta de sección rectangular uniforme de aluminio

Distribución de temperaturas

Ejemplo de cálculo para la posición de 10 cm desde la base (T3)

θ/θ_b =(〖cosh(〗⁡〖m(L-x))〗+(h/mk) sinh⁡(m(L-x)))/(cosh⁡(mL)+(h/mk) sinh⁡〖(mL) 〗 )

T=T_∞+〖(T〗_b-T_∞)*(cosh⁡(m(L-x))+(h/mk) sinh⁡(m(L-x)))/(cosh⁡(mL)+(h/mk) sinh⁡〖(mL) 〗 )

Donde:

m=√((h*P)/(k*A_c ))

m=√((8,44786*[2*(0.03+0.0125)])/(237*[0.03*0.0125]))

m=2.84245

T3=22.25+(102.75-22.25)*(cosh⁡〖(2.84245(0.3-0.1))〗+(8.44786/(2.84245*237)) sinh⁡〖(2.84245(0.3-0.1))〗)/(cosh⁡(2.84245*0.3)+(8.44786/(2.84245*237)) sinh⁡〖(2.84245*0.3) 〗 )

T3= 89.8142 [°C]

T1 [C°] T2 [C°] T3 [C°] T4 [C°]

79.8245 82.3741 89.8142 102.75

Eficiencia

n_f=tanh⁡(mL_c )/(mL_c )

L_c=L+t/2=0.3+0.0125/2=0.30625

n_f=tanh⁡(2.84245*0.30625)/(2.84245*0.30625)

n_f=80.6 %

Transferencia de calor

q=M ( sinh⁡〖(mL) 〗+(h/mk) cosh⁡(mL))/(cosh⁡(mL)+(h/mk) sinh⁡〖(mL) 〗 )

Donde:

M=√(hPkA_c ) θ_b

M=√(8,44786*(2*(0.0125+0.03))*237*(0.0125*0.03) ) *(102.75-22.25)

M=20.33614

q= 20.33614*(sinh⁡〖(2.84245*0.3) 〗+(8.44786/(2.84245*237)) cosh⁡(2.84245*0.3))/(cosh⁡(2.84245*0.3)+(8.44786/(2.84245*237)) sinh⁡〖(2.84245*0.3) 〗 )

q=14.2143 [W]

Aleta cónica (no sé cómo llamarle) de bronce

Distribución

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