Enthalpy during compression of water What happens to the temperature of water when compressed?
Enthalpy $H = U + PV$.
$H$ is conserved in a closed system.  By which I mean adiabatic and negligible external work applied.
We compress a litre of water to 10 bar (say).  This requires negligible work because water is almost incompressible.  But $P$ goes up, $V$ hardly changes, so $PV$ goes up, so $U$ should go down.  It should get quite a bit colder.
But is that correct?  Because very little work was done, yet $PV$ changes quite a lot, and it would produce a lot of cooling.
 A: Enthalpy certainly isn't conserved in a system you do work on. However, entropy is if you do the work quasistatically and adiabatically. Thus, you seek $\left(\frac{\partial T}{\partial P}\right)_S$, the temperature rise upon isentropic compression. Applying the triple product rule and a Maxwell relation, we have
$$\left(\frac{\partial T}{\partial P}\right)_S=-\left(\frac{\partial S}{\partial T}\right)^{-1}_P\left(\frac{\partial S}{\partial P}\right)^{-1}_T=\left(\frac{\partial S}{\partial T}\right)^{-1}_P\left(\frac{\partial T}{\partial V}\right)^{-1}_P=\frac{\alpha TV}{C_P}=\frac{\alpha T}{\rho c_P},$$
where $\alpha$ is the volumetric thermal expansion coefficient, $c_P$ is the specific heat capacity, and $\rho$ is the density. For water at room temperature, we should therefore expect a temperature increase of about 0.015°C/bar.
A: For control masses at constant volume it is common to analyse them in terms of the internal energy; the enthalpy is commonly used for reversible isobaric control mass problems (in which the change of enthalpy is equal to the heat flux), and for open systems (such as flowing fluids). The reason for the latter can be seen in derivations given elsewhere; the related concept is sometimes termed flow work.
I'm not sure where your notion that the enthalpy is conserved in a closed system comes from (especially as you don't specify for what kind of process).
Returning to your problem: from the fundamental relation for the internal energy:
$$\mathrm{d}U = T\cdot\mathrm{d}S - P\cdot\mathrm{d}V$$
If the fluid were truly incompressible and the system was initially at equilibrium, pressurisation by application of an external force, with no heat input or output, would cause no change of the internal energy (no work has been done and the system's thermodynamic state has not changed; note that $P$ is independent of $u$ and $v$ for an incompressible fluid, as $v$ is a constant, so from the equation above we have that $u$ is solely temperature-dependent). A mechanical analogue is the zero compression work done to a spring of infinite stiffness when the applied force is varied. If the fluid were compressible, the previous answer shows that a temperature rise would be expected.
