Energy Balance Equation

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					                    Heat Diffusion Equation                    All go to zero
                         dE dE g                  E  dW
Energy balance equation:         qin  qout  E1  2
                         dt   dt                         dt

Apply this equation to a solid undergoing conduction heat transfer:

                                   E=mcpT=(rV)cpT=r(dxdydz)cpT

               dx
          dy                   qx                                     qx+dx


y                                        T                          T
                              q x   KA                K ( dydz )
                                         x       x                  x   x

                                                              q x
      x                       q x  dx    q x  dq x  q x       dx
                                                               x
                   Heat Diffusion Equation (2)


   Energy Storage = Energy Generation + Net Heat Transfer
   
   t
       r c pT  dxdydz  qdxdydz  qx  qxdx
        T                                  qx
   rcp       dxdydz  qdxdydz  qx  (qx       dx )
         t                                  x
                         T
    qdxdydz  (k           )dxdydz
                   x     x
        T              T         T       T
   rcp         q  (k       )  (k     )  (k       )
         t         x x       y y       z z

Note: partial differential operator   Generalized to three-dimensional
is used since T=T(x,y,z,t)
                  Heat Diffusion Equation (3)
     T           T       T        T
rc p     q  (k )  (k )  (k )
          
     t       x x        y y     z z
                              
Special case1: no generation q = 0
Special case 2: constant thermal conductivity k = constant
     T      2T 2T 2T
rc p     k ( 2  2  2 )  q  k 2 T  q,
                                        
     t      x  y  z
           2  2  2
where  2  2  2  2 is the Laplacian operator
           x  y  z
                
Special case 3:     0 and q = 0
                           
                t
       2T 2T 2T
 2 T  2  2  2  0, The famous Laplace' s equation
       x     y     z
                      1-D, Steady Heat Transfer
Assume steady and no generation, 1- D Laplace' s equation
d 2T
   2
      0, T ( x, y, x, t )  T ( x ), function of x coordinate alone
dx
Note: ordinary differential operator is used since T = T(x) only
The general solution of this equation can be determined by integting twice:
                           dT
First integration leads to      cons tan t  C1 .
                           dx
Integrate again T(x) = C 1 x  C2
Second order differential equation: need two boundary conditions to
determine the two constants C 1 and C 2 .
    T
  100                        T(x=0)=100°C=C2
                             T(x=1 m)=20°C=C1+C2, C1=-80°C
                   20        T(x)=100-80x (°C)
                            x
                   1-D Heat Transfer (cont.)
     Recall Fourier' s Law:
             dT
     q = -kA    , If the temperature gradient is a constant
             dx
     q = constent (heat transfer rate is a constant)
     dT T2  T1
                  , where T ( x  0)  T1 , T ( x  L)  T2
     dx       L
                           dT        T1  T2      T1  T2
                  q   kA       kA          
T1        T2               dx           L        ( L / kA)
                      T1  T2                L
                  q          , where R        :thermal resistance
                         R                  kA
                   q (I)                     Electric circuit analogy
            x
                T1 (V1)            T2 (V2)       I = (V1-V2)/R
     L                     R (R)
          T Composite Wall Heat Transfer
   T1   k1 k2
              T2
                      R1=L1/(k1A) R2=L2/(k2A)
                 T1
                                                             T2
                                         T
                         T1  T2 T1  T2             T1  T2
                      q                   
                            R       R1  R2  L1   L2 
        L1   L2                                            
                                                k1 A   k2 A 
                               T1  T                          L1 
                      Also, q=        , T  T1  qR1  T1  q       
                                 R1                            k1 A 

Note: In the US, insulation materials are often specified in terms of
their thermal resistance in (hr ft2 °F)/Btu ----> 1 Btu=1055 J.
R-value = L/k, R-11 for wall, R-19 to R-31 for ceiling.
                                  R value
The thermal resistance of insulation material can be characterized
by its R-value. R is defined as the temperature difference across the
insulation by the heat flux going through it:
                         T      T      x
                    R                
                         q" k T x       k

The typical space inside the residential frame wall is 3.5 in. Find the R-value if
the wall cavity is filled with fiberglass batt. (k=0.046 W/m.K=0.027 Btu/h.ft.R)

            x        0.292 ft
       R                             10.8( R. ft 2 . h / Btu)  R  11
            k    0.027 Btu / h. ft. R

				
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posted:7/20/2011
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