Transwiki:Relative density

Relative density is a dimensionless ratio of the densities of two materials. The term specific gravity is similar, except that the reference material is water. A relative density can help quantify the buoyancy between two materials, or determine the density of one "unknown" material using the "known" density of another material. Mathematically, relative density is expressed as:

G = \frac{\rho_\mathrm{object}}{\rho_\mathrm{reference}}\, $$

where $$G$$ is the relative density, and $$\rho$$ is the densities of the two materials in the same units (e.g., kg/m³, g/cm³).

Relative density is dimensionless, since it is a ratio between two quantities of same unit. If the ratio is greater than 1, the object will be heavier than the same volume of the reference. If it is less than 1, it will be lighter than the reference. It is important to specify the reference material when reporting a relative density, but when the reference material is not specified it is usually understood to be water at 3.98 ° C.

Specific gravity
Using water as a reference material can make the calculations using SI units convenient, since density of water is (approximately) 1000 kg/m³ or 1 g/cm³. Determining the specific gravity (same as relative density of an object relative to water) entails little effort, as the density of the object only needs to be divided by 1 or 1000, depending on the unit, e.g.:

G = \frac{\rho_\mathrm{object}}{\rho_\mathrm{H_2O}} = \frac{\rho_\mathrm{object}\ \mathrm{kg/m^3}}{1000\ \mathrm{kg/m^3}} = \frac{\rho_\mathrm{object}\ \mathrm{g/cm^3}}{1\ \mathrm{g/cm^3}}\, $$

Unitless properties of specific gravity
While it may seem that this would result in two different specific gravities based upon which unit is used, this is not the case. Consider an object with a density of $$4\ \mathrm{g/cm^3} $$. According to the above formula, the specific gravity of this object would be

G = \frac{\rho_\mathrm{object}}{\rho_\mathrm{H_2O}} = \frac{\rho_\mathrm{object}\ \mathrm{g/cm^3}}{1\ \mathrm{g/cm^3}} = \frac{4\ \mathrm{g/cm^3}}{1\ \mathrm{g/cm^3}}\, = 4$$

Convert this object into $$ \mathrm{kg/m^3} $$. The density of this object is:

\rho_\mathrm{object}=4\frac{\mathrm{g}}{\mathrm{cm^3}}*\frac{1\ \mathrm{kg}}{1000\ \mathrm{g}}*\frac{100^3\ \mathrm{cm^3}}{\mathrm{m^3}} = 4*1000 \frac{\mathrm{kg}}{\mathrm{m^3}} = 4000 \frac{\mathrm{kg}}{\mathrm{m^3}}\, $$

Hence the specific gravity of this object is:

G = \frac{\rho_\mathrm{object}}{\rho_\mathrm{H_2O}} = \frac{\rho_\mathrm{object}\ \mathrm{kg/m^3}}{1000\ \mathrm{kg/m^3}} = \frac{4000\ \mathrm{kg/m^3}}{1000\ \mathrm{kg/m^3}}\, = 4$$

Since the specific gravity is, in fact unitless, as long as the same reference is used and consistent units are used the specific gravity will be the exact same number. Hence specific gravity will also work with gravitational density instead of mass-based density. Calling the gravitational density of an object $$\gamma_\mathrm{object} = \rho_\mathrm{object}g_\mathrm{local}$$; where $$g_\mathrm{local}$$ is the the local gravitational constant...

G = \frac{\rho_\mathrm{object}}{\rho_\mathrm{H_2O}} = \frac{g_\mathrm{local}\ \mathrm{N/kg}\ \rho_\mathrm{object}\ \mathrm{kg/m^3}}{g_\mathrm{local}\ \mathrm{N/kg}\ \rho_\mathrm{H_2O}\ \mathrm{kg/m^3}} = \frac{\gamma_\mathrm{object}\ \mathrm{N/m^3}}{\gamma_\mathrm{H_2O}\ \mathrm{N/m^3}}\, $$

Difference between specific gravity and relative density
Specific gravity is a special case of relative density. While specific gravity has a reference density of water, relative density can have any reference density that is used. It is best to specify the reference material when using relative density, using subscripts:

RD_\mathrm{Object/Reference} \, $$

Which states "the relative density of the object with respect to the reference material".

While relative density will not change as long as consistent units are maintained, the relative density is relative to its reference. The relative density of an object relative to mercury, is diffirent then the relative density with respect to water (specific gravity).

Taking the relative density with respect to alcohol, the water has a specific gravity of 1.2, and the iron has a specific gravity of 10. Taking the relative density relative to water, the numbers would be .78, 1, and 7.9. With respect to iron, the numbers are .1, .12, and 1. These are all correct relative densities, however it is with different reference points. Note that again, units are not needed and hence these numbers are correct no matter what unit system is used.

Practical uses of the unitless properties of relative density
Maintaining a consistent set of units is important. The units do not need to even be in terms of "amount of substance" per "volume" directly. For example, suppose someone wanted to find out the specific gravity of a rock. They could see that the rock deflected a certain spring by 3 inches. Then they put on a reference substance, which deflected the spring by 5 inches. The first rock made the water in a certain graduated cylinder rise by 20 mm, and the reference made it rise by 34 mm. The relative density between these two objects can easily be determined without having to figure out several constants which would be needed to determine the density directly (like the spring constant or the cross sectional area of the cylinder).

Since they know that the density of each object would be:

\rho_\mathrm{object}=\frac{Mass}{Volume} = \frac{Deflection*Spring\ Constant*Gravity}{Displacement_\mathrm{Water Line} * Area_\mathrm{Cylinder}}\, $$

Then the relative density is:

RD=\frac{\rho_\mathrm{object}}{\rho_\mathrm{H_2O}} =\frac{\frac{Deflection_\mathrm{Obj.}*Spring\ Constant*Gravity}{Displacement_\mathrm{Obj.} * Area_\mathrm{Cylinder}}} {\frac{Deflection_\mathrm{Ref.}*Spring\ Constant*Gravity}{Displacement_\mathrm{Ref.} * Area_\mathrm{Cylinder}}}\, $$

RD=\frac{\frac{Deflection_\mathrm{Obj.}}{Displacement_\mathrm{Obj.}}}{\frac{Deflection_\mathrm{Ref.}}{Displacement_\mathrm{Ref.}}} = \frac{\frac{3\ \mathrm{in}}{20\ \mathrm{mm}}}{\frac{5\ \mathrm{in}}{34\ \mathrm{mm}}}=\frac{3\ \mathrm{in} * 34\ \mathrm{mm}}{5\ \mathrm{in} * 20\ \mathrm{mm}} = 1.02\, $$

Now if they knew the actual density of the reference substance, then they can find the actual density of the test object. In this example, the density of the reference substance is 300000 N/inch³ (which would be absurd, but shows the power of the RD system). Then the density of the test substance is 300000 N/in³ * 1.02 = 306000 N/in³.

Temperature dependence

 * See Density for a table of the measured densities of water at various temperatures.

Changes in temperature affect the densities of materials, and the relative densities between two materials need to be corrected at certain temperatures. It is best to record the temperatures of the two materials, expressed here:
 * relative density: $$ 8.15_{4^\circ \mathrm{C}}^{20^\circ \mathrm{C}} \,$$ or specific gravity: $$ 2.432_0^{15} $$

where the superscript indicate the temperature at which the density of the material is measured, and the subscripts indicate the temperature of the water to which it is compared.

Relative density and hydrometers
One important use of relative density is hydrometers. It is relatively easy to show that the displacement of a hydrometer is approximately directly proportional to the change of relative density for small changes with respect to any reference that the hydrometer is placed into.

local gravitational constant.

Since the hydrometer is in static equilibrium, this force is equal to the gravitational force pulling down on the hydrometer, or $$mg \,$$, where $$m \,$$ is the mass of the entire hydrometer. Hence:
 * $$mg = \rho_\mathrm{Ref.} V_\mathrm{red} g \,$$ (1)
 * $$m = \rho_\mathrm{Ref.} V_\mathrm{red} \,$$ (2)
 * $$V_\mathrm{red} = \frac{m}{\rho_\mathrm{Ref.}}$$ (3)

At this line, the specific gravity is 1. When the hydrometer is placed in the new fluid, shown as green, a new mark is placed. Note that the hydrometer has dropped slightly, hence the specific gravity is lighter in this fluid then in the first fluid. Since the mass does not change and it is assumed that the gravitational constant does not change, the force holding up the hydrometer in the second fluid is the exact same as the force holding up the hydrometer in the first fluid, i.e $$mg$$

mg = \rho_\mathrm{New} (V_\mathrm{red}+(-\Delta x)A) g \, $$ (4)

where
 * $$\Delta x$$ is the displacement of the line. Since the line went down, the displacement is negative.
 * $$A$$ is the cross sectional area of the shaft.

m = \rho_\mathrm{New} (V_\mathrm{red}-\Delta x A) \, $$ (5)

m = \rho_\mathrm{New} (V_\mathrm{red}) - (\rho_\mathrm{New}\Delta x A) \, $$ (6)

Substituiting from equation 3

m = m \frac{\rho_\mathrm{New}}{\rho_\mathrm{Ref.}} - (\rho_\mathrm{Ref.}\Delta x A) \, $$ (7)

but

\frac{\rho_\mathrm{New}}{\rho_\mathrm{Ref.}}\, $$

is simply the relative density of the new fluid. Hence...

m = m RD_\mathrm{New/Ref.} - (\rho_\mathrm{New}\Delta x A) \, $$ (8)

1 = RD_\mathrm{New/Ref.} - \rho_\mathrm{New}\frac{\Delta x A}{m} \, $$ (9)

RD_\mathrm{New/Ref.}-1 = \rho_\mathrm{New}\frac{(\Delta x) A}{m} \, $$ (10)

For small differences in RD, $$\rho_\mathrm{New} = \rho_\mathrm{Ref.}$$, and hence

RD_\mathrm{New/Ref.}-1 = \rho_\mathrm{Ref.}\frac{(\Delta x) A}{m} \, $$ (11)

Knowing the original density, the cross-sectional area of the stalk, and the mass of the hydrometer, one can calculate the change of the relative density with respect to the reference fluid by simply measuring the displacement. Note that if the hydrometer falls down, the RD has gone less then 1. And if the hydrometer rises, RD is higher then the reference fluid.

For more percise measurements:

RD_\mathrm{New/Ref.}-1 = RD_\mathrm{New/Ref.}\rho_\mathrm{Ref.}\frac{(\Delta x) A}{m} \, $$ (12)

1=RD_\mathrm{New/Ref.}(1 - \rho_\mathrm{Ref.}\frac{(\Delta x) A}{m}) \, $$ (13)

RD_\mathrm{New/Ref.} = \frac{1}{(1 - \rho_\mathrm{Ref.}\frac{(\Delta x) A}{m})} \, $$ (14)

Expanding this as an infinite geometric series:



RD_\mathrm{New/Ref.} = \frac{1}{(1 - \rho_\mathrm{Ref.}\frac{(\Delta x) A}{m})} \, $$ (15)

RD_\mathrm{New/Ref.} = \sum_{k=0}^\infty ar^{k} = \frac{a}{1-r} = \sum_{k=0}^\infty (\frac{\rho_\mathrm{Ref.}(\Delta x) A}{m})^{k} $$ (16)

RD_\mathrm{New/Ref.} = 1 + (\frac{\rho_\mathrm{Ref.}(\Delta x) A}{m}) + (\frac{\rho_\mathrm{Ref.}(\Delta x) A}{m})^2 + (\frac{\rho_\mathrm{Ref.}(\Delta x) A}{m})^3 +.... $$ (17)

As a first-order of approximation, this reduces to the previous formula. Note that this formula becomes invalid if the hydrometer's density is greater than the density of the reference fluid, i.e. it sinks below the surface in the reference fluid. That is the only way the term $$(\frac{\rho_\mathrm{Ref.}(\Delta x) A}{m})$$ (which can be seen as a relative density comparing the density of the reference fluid and the density of the entire mass of the hydrometer divided by the small displaced volume) can be greater than 1.

Uses of relative density
Relative density is often used by geologists and mineralogists to help determine the mineral content of a rock or other sample. Gemologists use it as an aid in the identification of gemstones. The reason that relative density is measured in terms of the density of water is because that is the easiest way to measure it in the field. Basically, density is defined as the mass of a sample divided by its volume. With an irregularly shaped rock, the volume can be very difficult to accurately measure. One way is to put it in a water-filled graduated cylinder and see how much water it displaces. Relative density is more easily and perhaps more accurately measured without measuring volume. Simply suspend the sample from a spring scale and weigh it under water. The following formula is used to determine the specific gravity:

G = \frac{W}{W - F}\, $$

where
 * $$G$$ is the relative density,
 * $$W$$ is the weight of the sample (measured in pounds-force, newtons, or some other unit of force),
 * $$F$$ is the force, measured in the same units, while the sample was submerged.

Note that with this technique it is difficult to measure relative densities less than one, because in order to do so, the sign of F must change, requiring the measurement of the downward force needed to keep the sample underwater.

Another practical method uses three measurements. The mineral sample is weighed dry. Then a container filled to the brim with water is weighed, and weighed again with the sample immersed, after the displaced water has overflowed and been removed. Subtracting the last reading from the sum of the first two readings gives the weight of the displaced water. The relative density result is the dry sample weight divided by that of the displaced water. This method works with scales that can't easily accommodate a suspended sample, and also allows for measurement of samples that are less dense than water. Surface tension of the water may keep a significant amount of water from overflowing, which is especially problematic for small objects being immersed. A workaround would be to use a water container with as small a mouth as possible.

Relative density is not important physically as much as it is in reporting data to the masses. It is difficult to explain the high density of iridium as simply being 22650 kg/m³. To the masses, this number is just a large number. But if the facts instead detailed that the density of iridium is nearly twice that of lead, people can understand how heavy iridium really is. As a reporter, the relative density tells more information to the general public then the actual number does.

Specific gravity of water
The specific gravity of liquids or solids is defined as the ratio of density of the material to the density of distilled water. (S = density of the material/density of water). This implies that if the specific gravity is approximately equal to 1.000, then the density of the material is close to the density of water. If the specific gravity is large this means that the density of the material is much larger than the density of water and if the specific gravity is small this implies that the density of the material is much smaller than the density of water.

The specific gravity of a gas is generally defined by comparing the density of the gas with the density of air at a temperature of 20 degrees Celsius and a pressure of 101.325 kPa absolute, where the density is 1.205 kg/m3. Specific Gravity is unitless.

External link

 * Weight Per Cubic Foot And Specific Gravity