BULK MODULUS Explained!
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In this video:
Materials are also subjected to compressive stress which causes deformation in two or three dimension of the material. To help you visualize this concept, imagine a sponge being squeezed from all dimensions. We will see that there is a change in volume. Hence compressive stress creates a change in volume of a material. This deformation can be expressed mathematically as,
bulk (volume) strain= ∆V/V_o
It is also true that different materials have diferent reactions to forces that tends to deform its volume. Each material have their bulk modulus. Bulk modulus relates the amount of stress to the amount of deformation of the material. We can write it as.
B=(bulk stress)/(bulk strain) (bulk modulus)
A change in the bulk stress creates a change in volume. A change in stress means a change in pressure. Thus we can argue that a change in pressure ∆p produces a change in volume. As is true for gasses and liquids. We can write an equation for this fact by putting in mind that
stress= F_⊥/A=p=pressure
Writing this in fact in terms of differential calculus, we have
∆stress=∆p
We can now rewrite the formula for the bulk modulus as,
B=(bulk stress)/(bulk strain) =-∆p/(∆V/V_o ) (bulk modulus)
The negative sign indicates that an increase in the pressure always causes the material to shrink. In other words if ∆p is positive then ∆V is negative. So B itself is a positive quantity.
The reciprocal of B is called compressibility and is given by the formula,
k=1/B= (∆V/V_o )/∆p= 1/V_o ∙∆V/∆p (compressibility)
Compressibility is the fractional decrease in volume per fractional increase in volume.
Problem solving
A hydraulic press contains 0.25m^3 (250L) of oil. Find the decrease in volume when it is subjected to an increase in pressure ∆p=1.6×〖10〗^2 Pa (about 160atm or 2300psi). the bulk modulus of the oil is B=5.0×〖10〗^9 Pa(about 5.0×〖10〗^4 atm), and its compressibility is k=1/B=20×〖10〗^(-6) atm^(-1).
Identify the target
We need to find the decrease in volume ∆V .
What do we have in hand?
∆p=1.6×〖10〗^2 Pa
B=5.0×〖10〗^9 Pa
k=1/B=20×〖10〗^(-6) atm^(-1).
V_o=0.25m^3 (250L)
Execute
We derive a formula for ∆V.
B∆V/V_o =-∆p/1
∆V=-∆p/1 V_o/B
We have
∆V=-((1.6×〖10〗^7 Pa))/1 ((0.25m^3 ))/((5.0×〖10〗^9 Pa) )=-8.0×〖10〗^(-4) m^(-3) or-.80L
Alternatively we can use the concept of compressibility which will give us,
k=1/B= (∆V/V_o )/∆p= 1/V_o ∙∆V/∆p
So ∆V=-kV_o ∆p
∆V=-kV_o ∆p
∆V=-(20×〖10〗^(-6) atm^(-1) )(0.25m^3 )(1.6×〖10〗^2 Pa)=-8.0×〖10〗^(-4) m^(-3) or-.80L
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