# 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
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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 views: 136 posted: 7/20/2011 language: English pages: 7