Bulk modulus

The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its base unit is Pascal.

As an example, suppose an iron cannon ball with Bulk Modulus 160 GPa is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPA = 0.8 GPA.

THE RATIO OF HYDRAULIC STRESS TO THE CORRESPONDING HYDRAULIC STRAIN IS CALLED AS BULK MODULUS. IT DONATED BY SYMBOL ‘B’. B=-p/(∆V/V)

THE RECIPROCAL OF THE BULK MODULUS IS CALLED COMPRESSIBILITY. IT IS DENOTED BY ‘k’. IT IS DEFINED AS THE FRACTIONAL CHANGE IN VOLUME PER UNIT INCREASE IN PRESSURE.

k=(1/B)=-(1/∆p) * (∆V/V)

STATE MATERIAL B(10^9 N/m^2 OR GPa)

SOLIDS

ALUMINIUM 72

BRASS 61

COPPER 140

GLASS 37

IRON 100

NICKEL 260

STEEL 160

LIQUIDS

WATER 2.2

ETHANOL 0.9

CARBON DISULPHIDE 1.56

GLYCERIN 4.76

MERCURY 25

GASES AIR (at STP) 1.0 x 10^-4

IT CAN BE SEEN FROM THE DATA THAT BULK MODULI FOR SOLIDS ARE MUCH LARGER THAN FOR LIQUIDS, WHICH ARE AGAIN LARGER THAT BULK MODULUS OF GASES.THUS SOLIDS ARE LEAST COMPRESSIBLE AND GASES ARE THE MOST COMPRESSIBLE.

Q: THE AVERAGE DEPTH OF INDIAN OCEAN IS ABOUT 3000M. CALCULATE THE FRACTIONAL COMPRESSION. ∆V/V, OF WATER AT THE BOTTOM OF THE OCEAN .GIVEN: BULK MODULUS OF THE OCEAN IS 2.2 x 10^9 N/m^2, g= 10 m/S^2

ANS : THE PRESSURE EXERTED BY THE COLUMN OF WATER ON THE BOTTOM LAYER IS P=ρgh

= 3000m x 1000kg/m^3 x 10m/s^2= 3 x 10^7 kg/ms^2

= 3 x 10^7 N/m^2

FRACTIONAL COMPRESSION ∆V/V, IS ∆V/V =STRESS/B

=(3 x10^7 N/m^2)/(2.2 x 10^9 N/m^2) = 1.36 x 10^-2 or 1.36%

Q1.WHAT IS THE DENSITY OF OCEAN WATER AT A DEPTH,WHERE THE PRESSURE IS 80.0 ATM,GIVEN THAT IT’S DENSITY AT THE SURFACE IS 1.03 x 10^3 Kg/m^3 ? COMPRESSIBILITY OF WATER =45.8 x 10^-11 /Pa. GIVEN 1 ATM =1.013 x 10^5.

ANS: COMPRESSIBILITY =1/K=45.8 x 10^11/PACHANGE IN PRESSURE,P=80-1=79 ATM

=79x1.013 x 10^5 PADENSITY AT SURFACE,ρ= 1.03 x 10^3 Kg/m^3 K=P/ (∆v/v) ∆v/v= P/K = 3.665 x 10^-3 ∆v/v=(V-V’)/V=(M/ ρ -M/ ρ’)/(M/ ρ)

= 1-(ρ /ρ’)ρ /ρ’=1-(∆v/v)

ρ’= 1.034 x 10^3

Q2. CALCULATE THE PRESSURE REQUIRED TO STOP THE INCREASE IN VOLUME OF A COPPER BLOCK WHEN IT IS HEATED FROM 50˚ TO 70˚C. COEFFICIENT OF LINEAR EXPANSION OF COPPER = 8.0 x 10^-6/˚C AND BULK MODULUS OF ELASTICITY = 3.6 x 10^11 N/m^2.

ANS: WHEN A BLOCK OF VOLUME V IS HEATED THROUGH A TEMPERATURE OF ∆T,THE CHANGE IN VOLUME IS ∆v/v =γ ∆TWHERE γ(=3α=24 x 10^-6) IS THE COEFFICIENT OF CUBICAL EXPANSION.:. VOLUME STRAIN = ∆v/v=γ∆T

BULK MODULUS = K=P/(∆v/v)=P/(γ ∆T)

PRESSURE,P = K γ ∆T

P= 1.728 x 10^8 N/m^2

Type Of STRESS

TENSILE OR COMPRESSIVE

SHEARING HYDRAULIC

STRESS Two equal & opp. forces _|_ to opp. faces (σ=F/A)

Two equal & opp. Forces ||

Forces _|_ everywhere to the surface, per unit area same everywhere

STRAIN Elongation or compression || to force direction(∆v/v)(longitudinal strain)

Pure Shear, θ Volume Change (compression or elongation) ∆v/v

Change In

Shape YES YES NO

Volume NO NO YES

Elastic modulus

Y=(FxL)/(Ax∆L) G=(Fxθ)/A B=-P/(∆V/V)

Name Of Modulus

Young’s Modulus Shear Modulus Bulk Modulus

State Of Matter

Solid Solid Solid, Liquid & Gas