# 1.6.1: Fluid Density

The density is a property that is simple to analyzed and understand. The density is related to the other state properties such temperature and pressure through the equation of state or similar. Examples to describe the usage of property are provided.

Example 1.9

A steel tank filled with water undergoes heating from $$10^{\circ}C$$ to $$50^{\circ}C$$. The initial pressure can be assumed to atmospheric. Due to the change temperature the tank, (strong steel structure) undergoes linear expansion of $$8\times10^{-6}$$ {per} \0C. Calculate the pressure at the end of the process. $$E$$ denotes the Young's modulus. Assume that the Young modulus of the water is State your assumptions.

Solution

The expansion of the steel tank will be due to two contributions: one due to the thermal expansion and one due to the pressure increase in the tank. For this example, it is assumed that the expansion due to pressure change is negligible. The tank volume change under the assumptions state here but in the same time the tank walls remain straight. The new density is

$\label{rhoSwater:tankExp}\rho_{2} = \dfrac{\rho_1}\,{\underbrace{\left( 1 + \alpha \, \Delta\,T \right)^3}_{\text{thermal expansion}}}$

The more accurate calculations require looking into the steam tables. As estimated value of the density using Young's modulus and $$V_2 \propto \left(L_2\right)^3$$. $\label{rhoSwater:secondSide} \rho_2 \propto \dfrac{1}{\left(L_2\right)^3} \Longrightarrow \rho_2 \cong \dfrac{m}{\left(L_1 \left( 1 - \dfrac{\Delta P}{E} \right) \right)^3 }$ It can be noticed that $$\rho_1 \cong m/{L_1}^3$$ and thus $\label{rhoSwater:pRho} \dfrac{\rho_1}{\left( 1 + \alpha \Delta\,T \right)^3} = \dfrac{\rho_1}{ \left( 1 - \dfrac{\Delta P}{E} \right)^3 }$ The change is then $\label{rhoSwater:tankTp} 1 + \alpha \Delta\,T = 1 - \dfrac{\Delta P}{E}$ Thus the final pressure is $\label{rhoSwater:tankTp1} P_2 = P_1 - E\,\alpha\, \Delta\,T$ In this case, what happen when the value of $$P_1 - E\,\alpha\, \Delta\,T$$ becomes negative or very very small? The basic assumption falls and the water evaporates. If the expansion of the water is taken into account then the change (increase) of water volume has to be taken into account. The tank volume was calculated earlier and since the claim of "strong'' steel the volume of the tank is only effected by the temperature. $\label{rhoSwater:tankT} \left. \dfrac{V_{2} } {V_{1} } \right|_{tank} = \left( 1+ \alpha\, \Delta T\right)^3$ The volume of the water undergoes also a change and is a function of the temperature and pressure. The water pressure at the end of the process is unknown but the volume is known. Thus, the density at end is also known $\label{rhoSwater:rhoTwo} \rho_2 = \dfrac{m_w} {\left. T_2 \right|_{tank}}$ The pressure is a function volume and the temperature $$P=P(v,T)$$ thus $\label{rhoSwater:PvTg} dP = \overbrace{\left( \dfrac{\partial P }{ \partial v }\right)}^{\sim \beta_v} dv + \overbrace{\left( \dfrac{\partial P }{ \partial T }\right)}^{\sim E} dT$ As approximation it can written as $\label{rhoSwater:PvT} \Delta P = \beta_v \, \Delta v + E \, \Delta T$ Substituting the values results for $\label{rhoSwater:PvTn} \Delta P = \dfrac{0.0002} { \Delta \rho} + 2.15\times10^9 \, \Delta T$ Notice that density change, $$\Delta\rho< 0$$.