In what context is enthalpy a convenient concept? Internal energy $U$ is clearly an important concept; the first law of thermodynamics states that for an isolated system internal energy is constant $(\Delta U=0)$ and that for a closed system the change in internal energy is the heat absorbed by the system $Q$ and work done on the system $W$ $(\Delta U=Q+W)$.
Enthalpy $H$ is the sum of the internal energy and the product of the pressure and volume of the system $(H=U+PV)$. I was taught that enthalpy is a preferred quantity to internal energy for constant pressure systems where $\Delta H=Q$, as opposed to constant volume systems where $\Delta U=Q$. But why would anyone care about what quantity is equal to $Q$ under certain conditions instead of simply reporting $Q$? Enthalpy seems redundant in this context.
Is enthalpy a convenient concept in other contexts, such as systems with varying pressure? Is it described by any fundamental laws as internal energy is by the first law?
 A: I mostly encounter enthalpy in discussions of open-air phase changes, such as freezing and boiling.
Take the example of boiling a pot of water.  There is some change in internal energy associated with the change from the liquid to the solid phase.  However, the phase change also includes a dramatic change in the density of the water, from $\rho_\text{liquid} \approx 1000\rm\,kg/m^3$ to $\rho_\text{vapor} \approx 1\rm\,kg/m^3$.  To boil away a kilogram of water at ordinary pressure therefore requires pushing away nearly a cubic meter of air, which requires work
$$
P\Delta V = 100\,\mathrm{kPa} \cdot 1\rm\,m^3 = 100\,kJ
$$
This is a substantial ($\sim5\%$) correction to the heat of vaporization of water (about 2200 kJ/kg), and is present is every boiling-water experiment that doesn't take place inside some kind of constant-volume vessel.
I think that you already know this, and that the core of your question is

But why would anyone care about what quantity is equal to $Q$ under certain conditions instead of simply reporting $Q$?

The answer here is that, if you report the heat required to drive some phase transition under some conditions, you have to also state what those conditions are.  Reporting the enthalpy at a given temperature and pressure is a very efficient way to state that you've boiled the water (or whatever) at constant pressure, like most people do, rather than having done something more elaborate to measure only the internal energy.
A: The quick answer is: flow processes. e.g. fluid flowing into a chemical plant; gas running through a turbine; chemicals reacting at constant pressure. The equation $dU = T dS - p dV$ gives us 
$$ dH = T dS + V dp$$
which you can regard as a 'fundamental law' (equivalent to the familiar one for $dU$).
An important application is the Joule-Kelvin or throttling process, where $H$ is constant. This in turn leads to Bernoulli's equation for laminar flow in the absence of heat transport. In chemistry you get the van't Hoff equation.
A: In addition to what @Bob D and and others said about the use of enthalpy (primarily) in performing energy balances on continuous flow systems, enthalpy is also important in quantifying the temperature dependence of the equilibrium constant for chemical reactions (via the van't Hopf equation) and the temperature dependence of vapor-liquid equilibrium of single- and multicomponent chemical systems.  And, of course, such equilibrium constants are important in designing and operating distillation equipment and chemical reactors.  
So, if you are ever going to be working with industrial scale continuous flow systems, you are going to be working with enthalpy rather than internal energy.  And, if you are ever going to understand phase equilibrium and chemical equilibrium, you better have more than a nodding acquaintance with enthalpy.
A: Enthalpy is a particularly convenient concept when analyzing components, such as turbines, compressors, pumps, condensers, evaporators, and expansion valve, in open steady flow processes. Here are some examples for the case of an ideal Rankine power cycle where the $h$ is enthalpy:
Reversible, adiabatic turbines: $\dot W_{out} =\dot m(h_{in}-h_{out})$
Reversible, constant pressure condensers (heat extraction): $\dot Q_{out}=\dot m(h_{in}-h_{out})$
Reversible, adiabatic compressors, pumps: $\dot W_{in}=\dot m(h_{in}-h_{out})$
Reversible, constant pressure, expansion (boiler heat addition): $\dot Q_{in}=\dot m(h_{out}-h_{in})$
Hope this helps
