HEAT AND MASS TRANSFER

Differential equation of heat transfer

In order to find out the rate of heat transfer ,let consider an element from the bulk

  

Heat flowing in x direction  

During the time dt                          =-k­­­­­­­­­­­­­­­­­­­­­­_x dz dy ∂T/∂x dt

Heat flowing in x direction in time dt = Q_xdt

 By expansion of the equation

(f_(x+h)­) =f_x +h f’_x+h²f’’_{x+………..}

Then Q_(x+dx) = Q_x+(∂ Heat flowing in x direction is Qx = -k­­­­­­­­­­­­­­­­­­­­­­_x dz dy ∂T/∂x

Qx/∂x) dx

Then heat accumulation = Q_x+dx– Q_x

Q_X +  ∂Q_X/∂_X - Q_x

(∂Qx/∂x)             = (∂(-kdzdy)/∂x) × ∂T/∂x × dt dx

                            =  -k_x ∂²T/∂x² dx dy dz dt

Similarly

Heat accumulated due to heat flow in y direction

K_y ∂²T/∂y²dxdydzdt

Total heat accumulated = sum of heat accumulation in all direction

=(k_x∂²T/∂x² + k_y∂²T/∂y^{2 +}∂²T/∂z²)dxdydzdt

If q^. is the heat generated in differential element /volume

Then heat generated in time dt is=  q^.dxdydzdt

Net heat accumulation =

(k_x∂²T/∂x²+k_y∂²T/∂y²+k_z∂²T/∂z²+q^.)dxdydzdt                                            6

Due to heat accumulation body temperature changes by dT

Q= mC_pdT                                                                                          

=D×dxdydz×C_pdT                                                                                          7

Equating  6 and 7

D C_pdT = (k_x∂²T/∂x²+k_y∂²T/∂y²+k_z∂²T/∂z²+q^.)dt 

Dividing by dt

(kx∂²T/∂x²+k_y∂²T/∂y²+k_z∂²T/∂z²+q^. )= DC_p∂T/∂t

For isotropic material k= constant

K(∂²T/∂x²+∂²T/∂y²+∂²T/∂z²+q^./k)    = DC_p∂T/∂t

(∂²T/∂x²+∂²T/∂y²+∂²T/∂z²+q^./k)    = 1/(k/DC_p) ×∂T/∂t

1/(k/DC_p) = thermal diffusibility  =1/α