Solids and Liquids: Internal Energy and Enthalpy

Both liquids and solids can be considered incompressible substances.  As a result, their specific volume will remain constant during a process.  This means that any energy associated with a volume change is  considered negligible in comparison to other forms of energy.  Due to this fact, the specific heat $c_p$ and $c_v$ will be equal to each other for a liquid and a solid.  Hence, $c_p=c_v=c$.  In turn, this will effect how the internal energy and enthalpy are calculated.

Internal Energy: Solids and Liquids

Similar to an ideal gas, the internal energy of a liquid and solid is dependent on temperature only.  As a result, the partial differential equation in relation to $c_v$ will be as follows.

(Eq 1) $du=c_{v}dT=c(T)dt$

For a process, the change in internal energy from state 1 to state 2 is obtained using the following integration.

(Eq 2) $\Delta u=u_2-u_1=\int {}_1^{2}c(T)dT$  (kJ/kg)

Finally, for small temperature differences, the average specific heat $c$ can be used to determine the change in internal energy.

(Eq 3) $\Delta u\cong c_{avg}(T_2-T_1)$  (kj/kg)

Enthalpy: Solids and Liquids

Enthalpy is defined by $h=u+P\nu$, where $P$ is the absolute pressure and $\nu$ is the specific volume.  For a liquid and solid, the specific volume will be constant.  As a result,  the differential equation for enthalpy change of a liquid and solid is the following.

(Eq 4) $dh=du+\nu \Delta P+Pd\nu =du+\nu dP$

where $Pd\nu \to 0$

After integrating equation 4, the following equation will result.

(Eq 5) $\Delta h=\Delta u+\nu \Delta P\cong c_{avg}\Delta T+\nu \Delta P$ (kJ/kg)

Regarding solids, only the term $\nu \Delta P$ can be considered insignificant.  As a result, $\Delta h=\Delta u\cong c_{avg}\Delta T$.  On the other hand, for liquids, there are two common cases that can be encountered.  The first case is when there is a constant pressure during the process, or $\Delta P=0$.  When this is the case, $\Delta h=\Delta u\cong c_{avg}\Delta T$.  The second case occurs when the temperature is constant during the process, or $\Delta T=0$.  If this is the case than $\Delta h=\nu \Delta P$.

Finally, if we are trying to find the enthalpy of a compressed liquid, than the following equation is used.

(Eq 6) $h_{@P,T}\cong h_{f@T}+{\nu }_{f@T}(P-P_{sat@T})$

This equation is used when state 1 of a process starts as a saturated liquid.  Than as the process occurs the temperature remains the same until the process ends as a compressed liquid at state 2.

 

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