Cubical Expansion
Cubical Expansion
VOLUMETRIC EXPANSION
Increase in volume of a body on heating is referred to as
“VOLUMETRIC EXPANSION ” or “CUBICAL EXPANSION“.
“VOLUMETRIC EXPANSION ” or “CUBICAL EXPANSION“.
Consider a metallic body of volume = V1 . Let it’s temperature is raised by T. then experimentally “Increase in volume (V) is directly proportional to its initial volume (V1) and rise in temperature (T).
V V1———(1)
V T———(2)combining (1) and (2)
V V1 T
OR
Where is constant known as “coefficient of volumetric expansion”
V T———(2)combining (1) and (2)
V V1 T
OR
Where is constant known as “coefficient of volumetric expansion”
EXPRESSION FOR FINAL VOLUME
We know that
Final volume = Initial volume + increase in volume
V2 = V1 + V
Putting the value of V
V2 = V1 + V1T
V2 = V1( 1+ T)
COEFFICIENT OF VOLUMETRIC EXPANSION
Coefficient of volumetric expansion () is defined as:
“increase in volume per unit original volume per Kelvin rise in temperature is called coefficient of volumetric expansion.” Unit: .
SHOW THAT:
.
Consider a metallic cube of each side of length of “L”.
Before heating at T1 K……………………
After heating at T2 K…………………..
Volume of cube at T1 K :
V1 = L3 ———- 1
[V = L.B.H]
If it is heated to T2 K, then the length of each side will become (L + D L) and its volume will become
V2 = (L + D L) 3
since V2= V1 + D V
comparing above two equations
V1 + D V = (L + D L)3
as [ D L = a L D T ]
V1 + D V = (L + a L D T)3
V1 + D V = L3 (1 + a D T)3
Using formula [(a+b)3 = a3 + b3 + 3a2b + 3ab2]
V1 + D V = L3 {1 + a 3D T3 + 3(1)( a D T) + 3(1)( a 2 D T2)}
V1 + D V = L3 {1 + a 3D T3 + 3a D T + 3a 2 D T2}
Since “a “ is a fractional value of range 10-5 or 10-6, therefore we
can neglect all the terms containing high powers of “a “ thus.
V1 + D V = L3 (1 + 3a D T)Putting the value of DV
V1 + b V1D T = V1 (1 + 3a D T) Since
[D V = V1b D T and L3 = V1]
1 + b D T = 1 + 3a D T
V1 = L3 ———- 1
[V = L.B.H]
If it is heated to T2 K, then the length of each side will become (L + D L) and its volume will become
V2 = (L + D L) 3
since V2= V1 + D V
comparing above two equations
V1 + D V = (L + D L)3
as [ D L = a L D T ]
V1 + D V = (L + a L D T)3
V1 + D V = L3 (1 + a D T)3
Using formula [(a+b)3 = a3 + b3 + 3a2b + 3ab2]
V1 + D V = L3 {1 + a 3D T3 + 3(1)( a D T) + 3(1)( a 2 D T2)}
V1 + D V = L3 {1 + a 3D T3 + 3a D T + 3a 2 D T2}
Since “a “ is a fractional value of range 10-5 or 10-6, therefore we
can neglect all the terms containing high powers of “a “ thus.
V1 + D V = L3 (1 + 3a D T)Putting the value of DV
V1 + b V1D T = V1 (1 + 3a D T) Since
[D V = V1b D T and L3 = V1]
1 + b D T = 1 + 3a D T
This expression shows that the coefficient of the cubical expansion is three times of the coefficient of the linear expansion.