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23I18 IJAET0118687 v6 iss6 2514 2523.pdf


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International Journal of Advances in Engineering & Technology, Jan. 2014.
©IJAET
ISSN: 22311963



1  dV

V  dT


 , on solving we get


dr  r

dT
3
As V  r 3 , now from Equation (7) and Equation (8), we get
dr  0  r 3T 1 



dT
3  r03T 
Integration of Equation (9) gives

(8)

(9)

1


 3T
1
(10)
r (T )  r0 

 1  T  0 (T  T0 ) 
Now from Equation (10) and Equation (7) we get the expression for thermal expansion coefficient as


1
(11)
 (T )   0 

1



(
T

T
)
T 0
0 

From Equation (6) and Equation (11), one can develops
1


 T
1
V  V0 

 1  T  0 (T  T0 ) 

(12)

Eq. (12) is developed taking the definition of Anderson Gruneisen parameter, where

 T is temperature

independent. But this Eq. (12) only works above the Debye temperature  D [23]. Inspired with this
condition, model should be applicable for the entire range of temperature starting from the room
temperature to melting temperature; we have made an effort to modify Eq. (12) assuming that
not the temperature independent parameter. It has been noted [14] that the value of
temperature. The empirical temperature dependence of

T

T

 T is

changes with

is deliberated as

k

T  T 
 
(13)
T0  T0 
 T0 is the value of Anderson Gruneisen parameter at reference T  T0 . The new dimensionless
parameter k can be calculated from the slope of the graph plotted between log T and log(T / T0 ) .
Under the effect of temperature the product of thermal expansion coefficient and bulk modulus
remains constant [19] that is
(14)
 BT  Constant
On differentiating Equation (14) with respect to temperature and at constant pressure, we get

BT  d 
 dBT 

  
  dT  P
 dT  P

(15)

From Equation (15) and Equation (1)

T 

1  d 
 2  dT  P

(16)

Equation (13) and equation (16) give
k

T 
1  d 
    2 
 T0    dT  P
0
T

(17)

On integrating Equation (18), we get

2516

Vol. 6, Issue 6, pp. 2514-2523