Calculadora de la densidad del agua

Calculadora de la densidad del agua

La densidad del agua varía en función de la temperatura. Para obtener un cálculo preciso en un amplio rango, se utiliza la siguiente ecuación polinomial empírica:

\[\rho(T) = \frac{999.84 + 16.95 T - 7.99\times 10^{-3} T^2 - 46.17\times 10^{-6} T^3 + 105.56\times 10^{-9} T^4 - 280.54\times 10^{-12} T^5}{1 + 16.88\times 10^{-3} T}\]

Donde:

• \(\rho\) es la densidad del agua en kg/m³.
• \(T\) es la temperatura en °C.

Ingresa la temperatura (°C):

Densidad Calculada:

--- kg/m³

A case of conversion - From $lb_m$ to $gpm$

 Imperial units prove to be somtimes hard to use. Here the case of conversion from mass to volumetric flowrate is presented.

As one may imagine a physical property of the fluid is required: the density $\rho$. This can be written as,

$\dot{Q}= Q\, \rho$        Eq. (01)

where $\dot{Q}$ is the mass flow rate, in $lb_m/hr$ for example, and $Q$ is the volumetric flow rate, in $gpm$, for example.

Let us now consider the case of $40,000.00\, lb_m/hr$ flowing liquid water at $190\,^\circ \, C$. In this case, its density would be $\rho=54.70\,lb_m/ft^3$.

Using Eq. (01), it follows,

$Q=\dfrac{\dot{Q}}{\rho}=\dfrac{40,000.00\, lb_m/hr}{54.70\, lb_m/ft^3}$

$Q=924.56\,ft^3/hr=115.27\,gpm$

which is the desired conversion. If you were working with steam, working pressure must be considered to get the proper fluid density.

The density of a mixture

- Two components of equal mass but different volume - 

Knowing the density of such a mixture is a common strange situation in engineering. This demonstration can be extended to more components provided the mass of each component is the same.

The situation

Consider that you have a solution, of a solid with certain density $\rho_S$ dilute in certain solvent of density $\rho_L$, in which both the solute and solvent are mixed in equal quantities. This is, the mass of the solute is equal to the mass of the solvent. What is the density of such a mixture?

An analytical approach

The solution to this case is not knew and can also be found elsewhere. However, in this post, some different flavor shall be given.

The density of the solute is,

$\rho_S=\dfrac{m_S}{V_S}$        Eq. (01)

while the density of the solvent is,

$\rho_L=\dfrac{m_L}{V_L}$        Eq. (02)

but since the mass of solute and solvent are equal, it follows,

$m=m_S=m_L$        Eq. (03)

Also, the density of the mixture should be,

$\rho=\dfrac{m_S+m_L}{V_S+V_L}=\dfrac{2m}{V_S+S_L}$       Eq. (04) 

The volumes, $V_S$ and $V_L$, in Eq .(04) can be determined from Eqs. (01-02) condering Eq. (03) as follows,

$V_S=\dfrac{m}{\rho_S}$        Eq. (05)

$V_L=\dfrac{m}{\rho_L}$        Eq. (06)

Substitution of Eqs. (05-06) into Eq. (04) produces,

$\rho=\dfrac{2m}{\dfrac{m}{\rho_S}+\dfrac{m}{\rho_L}}$

which can be simplified,

$\rho=\dfrac{2m}{\dfrac{\rho_L m+\rho_S m}{\rho_S \rho_L}}=\dfrac{2m}{\dfrac{m\left( \rho_L+\rho_S \right)}{\rho_S \rho_L}}=\dfrac{2\rho_S \rho_L}{\rho_L+\rho_S}$            Eq. (07)

Of course, Eq. (07) applies for the mixture of two liquids as well. Therefore, if you know the density of both substances, you can readily estimate the density of the mixture (provided the mass of the two components is the same).

A case to estimate the density of a mixture

Let us consider a slurry fluid made of the mixture of coal and water. The coal has specific gravity 2.5 while the slurry has composition 50% coal w/w. What is the density of the slurry?

Well, the slurry mixture is made of coal and water mixed in the same mass proportion. This is, the mass of the coal is the same as the mass of the water used to make the slurry. The volume is unknown: it can be the same but, who knows.

Starting with the specific gravity of the coal (solute), its density should be,

$\rho_S=\left(2.5\right) \left( 1000\,kg/m^3 \right)=2500\,kg/m^3$

For the water, let us consider that it is at room temperature (say, 25 °C). Similar temperatures around this one will give a very similar water density, so that you do not have to worry too much. Then,

$\rho_L=997.05\, kg/m^3$

Therefore, the density of such a slurry should be,

$\rho=1425.56\, kg/m3$

This is the end of the post. I hope you find it useful.

Ildebrando.

Some key basic concepts on thermodynamics

Thermodynamicists tend to use words for technical aspects of this subject. This words are used for other subjects but with some restrictions or further details that help to describe what they are talking about.

Not knowing the meaaning of these concepts would be as trying to comunicate with someone from another country using a different language. This case is not that extreme but difficulties could arise and the worst, you will wste time.

Then, consider the following concepts, presented in a colloquial manner.

Frst, body and system come as words referring to physical things that may be the same for certain circumstances. Perhaps, a difference we can make a first difference between body and system: a system may involve more than one body. In a body and in a system physical and chemical changes, that can be measured, may occur. Next, these measurements, also understood as data, help to describe, in detail, the body or system, or we should say characterized it. Also, the features or parameters to be measured in a body or system may be important due to its changes in time or relation to other features are known as properties.

On the other hand, a system has another feature: it denotes a region or space which is restricted by boundaries with different features such as: thermal conductivity, porosity, etcetera. Measurements of the properties of the system allow to define the system in a given set of conditions, which is called a state. Those properties measured to define the state are also called variables of state. 

From the mathematical point of view, there are independent and dependent variables. Next, state properties can be expressed as independent variables.

Main components of a function.

On thermodynamics, it is known, empirically, that in order to estimate the intensive variables of a system, only two intensive variables need to be known. An intensive variable is one that does not depend on the amount of matter in the system. If the value of the variable is proportional to the quantity of matter, then it would be an extensive variable.

The equations inoliving independent and dependent variables of state are called state equations. 

A thermodynamical process implies changes in time of one or more properties. these changes in time can be referred as state changes as well.

Two types of systems can be found. The first ones are closed systems in which mass entrance or leaving is not allowed or never happens. If mass enters or leaves the system, then we would say that the system is open. Similarly, an adiabatic system is a closed system too since no heat is exchanged with the surroundings.

Thermal equilibrium would mean that state variables remain constant boundaries are allowed to change.

Any question? Write in the comments and I shall try to help.

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Ildebrando.