Dimostrazioni Fisica tecnica - parte 4

dT

dT = (9_gr_g²) * (c_B)

dr 2λ r₁

r₁ < r < r₂ => non ho flusso conv.

c_B = 0. Parete con q_B

c₂ = 0

T(r)T(||) == == > 9_gr²   + c₂ + t₅

(||) |r = r_g 2λ r

(r²)  

c₂ = (|_S| + 9_S{ 5² }

 ^r2

(|9_g|)_4λ   (9_gr_g²) |9_gr_g²| |r²| = ( |2|   + t₅)

(r)   (|)⁵²

(|){ ( 4λ ) } (4λ) (  4λ

Fig =  |9_g| Fig. = |““””  (  R₁ ) |

(9_g) 9_gr_g

Q_conv = h ( | "2πr/%K| ) (|T_s('100U| )

( 4λ ) ( r ) )

{senza generazione}

-==--

d -== x dT

dr -== | r² dT

dr -== r² | dr

t  =  0 . |A₁| = |A₂

|c₃| = A₂tA₁

==

--==--r( x_B| + |C₄| == --2t +1 |&₄ +1|&sub>4

==

|c₂| = B₂c₂

( |r| )(|r m)

l(x) = |c₄| + c₂

T_sg = T(r) == > ( |c₄| + c₂

( |2| ) + c₂

c₄| + B₄ + ( &endash;__/|&endash; )

(|2₃| - 2₁ )/ c₃ = (T_sg| - T_sg|

==

|(&redsub;_2) )/

(|c_{2t|_c)2)}

|==

(r) == ,

{___}T(r) = T{__} + F&sub> + T(>_{SS2 - TS2}} + ( |Xs1 - (1cs1{{) ) ) + 1/(t|)}

53. Aletta a sezione circolare di lunghezza infinita

9_QX = 9_X + dX + dq∞w

9_X = λ dT/dx

9_X+dX = 9_X + d/dx (9_X) dX = (λ dT/dx) + (dλ/dx) Ac dX

dQ = h(1-s∞ dA) = h(Q(1-s∞) dx

dq = dq(T-∞,0) = dT/dx - D∞o e∞s

θ