Compressibility
COMPRESSIBILITY
Compressibility is the reciprocal of the bulk modulus of elasticity, K which is defined as the ratio of compressive stress to volumetric strain.Let ∀= Volume of a gas enclosed in the cylinder.
p = Pressure of gas when volume is ∀
Let the pressure is increased to p+dp,the volume of gas decreases from ∀
to ∀ - d∀.
Then increase in pressure = dp kgf/㎡
Decrease in volume = d∀
∴ Volumetric strain = - d∀/∀.
- ve sign means the volume decreases with increase of pressure.
∴ Bulk modulus , K = Increase of pressure / Volumetric strain
= dp/(-d∀/∀) = - dp / d∀.∀ .....(1)
Compressibility. = 1/K ..........(2)
Relationship between Bulk Modulus (K) and Pressure (p) for a Gas :-
The relationship between bulk modulus of elasticity (K) and pressure for a gas for. two different processes of compression are as :
(i) For Isothermal Process :
Equation (1) gives the relationship between pressure (p) and density of a gas as
p/ρ = Constant. Or. p∀ = Constant
Differentiating this equation, we get (p and ∀ both are variables )
p/ρ = Constant. Or. p∀ = Constant
Differentiating this equation, we get (p and ∀ both are variables )
Pd∀ + ∀dp = 0 or pd∀ = - ∀dp or p = (-∀dp /d∀)
Substituting this value in equation (1), we get
K = p .... (3)
(ii) For Adiabatic Process :-
p/ρ^k = Constant or p∀^k = Constant
Differentiating, we get pd(∀) + ∀(dp) = 0
or. pkd∀ + ∀dp = 0
pk = - ∀dP / d∀
K = pk
Where K = Bulk modulus and k = Ratio of specific heats
PROBLEM :- Determine the bulk modulus of elasticity of a liquid, if the pressure. of the liquid is increased from 70 N/ ㎡ to 130 N/c㎡ The Volumeof the liquid decreases by 0. 15 per cent
Solution :- Given :
Initial force. =70N/c㎡
Final pressure = 130 N/c㎡
Decrease in volume. = 0.15% (-d∀/∀) = 0.15/100
Decrease in volume. = 0.15% (-d∀/∀) = 0.15/100
dp =Increase in pressure 130-70= 60N/c㎡
Bulk modulus. K = dp /(-d∀/∀)= 60x100/15 = 4x10^4 N/c㎡ ..Ans
.
Bulk modulus. K = dp /(-d∀/∀)= 60x100/15 = 4x10^4 N/c㎡ ..Ans
.
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