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Bulk Modulus Formula is bulk modulus = - ( pressure applied / fractional change in volume). Bulk Modulus is defined as the proportion of volumetric stress related to a volumetric strain of the material to which pressure is applied. It is a numerical constant with which product quality is determined. Determining the strength of materials is prior to using them for making required products. Elasticity is another one that increases the value of the product.
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Bulk Modulus Definition
When an object is compressed under a certain pressure the resistance of the substance to withstand changes in its volume is called Bulk Modulus. Each component such as solid, liquid, gas has its bulk modulus value which defines the least value up to which it can be compressed by applying external pressures. It is denoted with K or B.
Bulk Modulus
Read Also: Shearing Stress
Bulk Modulus Formula and Units
As we know that bulk modulus is based on the pressure and volume of the object it is calculated as follows:
K = v(?p)/ ?v
Where K = Bulk Modulus of the certain object in pascal
V = Initial volume of the selected object
?V = change in volume due to application on external pressure
?p = External Pressure applied on the object
As most of the time it is calculated in pascal pressure and volumes are taken in cubic meters.
The SI Unit of Bulk Modulus is Pascal.
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Bulk Modulus of Different Substances
Here are the values of Bulk Modulus for different substances
| Substance | Bulk Modulus Value in GPa |
|---|---|
| Water | 2.2 |
| Air | 0.0004 |
| Methanol | 8.23 |
| Diamond | 443 |
| Alumina | 176 |
| Steel | 160 |
| Limestone | 65 |
| Granite | 50 |
| Glass | 35-55 (based on thickness) |
| Graphite | 34 |
| Sodium Chloride | 24.42 |
| shale | 10 |
| Chalk | 9 |
| Rubber | 1.5 to 2 |
| Sand Stone | 0.7 |
| Glycerine | 4.76 |
| Mercury | 25 |
| copper | 140 |
| Iron | 100 |
Read Also: Stress and Strain
In this bulk modulus also for different states such as isothermal or isentropic states and these values change corresponding to them.
Out of all substances, the bulk modulus of the diamond is higher as it is the hardest one and more pressure is required to compress it.
Things to Remember
- Change in units of pa, MPa are important for the calculation of problems on bulk modulus
- External pressure may be varied due to atmospheric conditions.
- The bulk modulus of any liquid is the measure of its compressibility factor.
- Bulk modulus in negative sign represents an increase in pressure followed by a decrease in volume.
- Bulk modulus helps us to identify the stretch and deform states of an object thereby we can calculate tensile stress and tensile strain of substances.
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Sample Question
Ques. What is an Adiabatic and Isothermal bulk modulus? (2 Marks)
Ans. Adiabatic process means there is no heat exchange with the surroundings in the case of an ideal gas it is defined as the change in pressure to the fractional change in volume of the substance. Isothermal Bulk Modulus is carried out at a constant temperature.
Ques. A pressure of 200pa is exerted upon an iron piece whose bulk modulus is 100*109. Calculate the change in volume of the Iron piece. (2 Marks)
Ans. Given,
Pressure exerted on object(?p) = 200 Mpa
Bulk modulus of Iron = 100*109
Now by using Bulk modulus Formula,
K = v(?p)/ ?v
?v/v = 200*106/100*109
= 0.002
Therefore for the given values of the applied pressure and bulk modulus the change in volume of a piece of iron is 0.002.
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Ques. A rubber ball had a change in volume to 13% due to uniform stress of 1.3*104 N/m2 Find its bulk modulus due to the above pressure. (3 Marks)
Ans. From the above given,
The volumetric strain of object = 13% = 13*10-2
Volumetric stress of object = 1.3*104 N/m2
Then to find the bulk modulus of elasticity
K = volumetric stress/volumetric strain
= 1.3*104/13*10-2
= 105 n/m2
Therefore the Bulk modulus of elasticity of rubber substance is 105 n/m2
Ques. Find the pressure required to compress glycerine to a volume of 0.0005% where the bulk modulus of glycerine 4.35*109 n/m2. (3 Marks)
Ans. From the above given that
Volumetric strain of the body = 0.0005% = 0.0005*10-2
= 10-6
We know that k = volumetric stress/volumetric strain
Pressure = k* volumetric strain
= 4.35*109 *10-6
= 4.35*104
Therefore the pressure required to compress glycerine to a volume of 0.0005% is 4.35*104 N/m2.
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Ques. 4 liters of water when subjected to a pressure of 10pa showed compression of 1.2*10-6m3 find the compressibility of water. (3 Marks)
Ans. Given,
Quantity of water = 4 litres = v = 4*10-3
Pressure = 10pa
Compression by 1.2*10-6m3
Form above calculate the bulk modulus of elasticity
K = dp*v/dv
= 10*4*10-3/1.2*10-6m3
= 3.3 * 104
Compressibility = 1/k
= 1/3.3 * 104
= 0.303* 104
Therefore the compressibility of water for the above conditions is = 0.303* 104 m2/n
Ques. Bulk modulus of rubber is 1.5*109N/m2. What is the pressure required to compress its volume by 1.5%? (3 Marks)
Ans. From the above given that
Bulk modulus of rubber = 1.5*109N/m2
Change in volume of rubber = 1.5%
For calculating the pressure applied on it
k = change in pressure/volumetric strain
dp = k* volumetric strain
dp = 1.5*109*1.5*10-3
dp = 2.25*106
The change in pressure for the compression of rubber to a given volume is 2.25*106N/m2.
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