Help: Where Does the Energy Enter?

I have always been unsure about the following issue, I was hoping someone could clear it up. Suppose you have a straight section of pipe carrying a viscous flow ( water for instance). I want to avoid assuming the flow is incompressible for the time being, you'll see why near the end of the discussion.

Starting with and energy rate balance across the inlet and outlet incorporating the Reynolds Transport Theorem:

$\displaystyle \dot{Q} - \dot{W_s} = \frac{d}{dt} \int_{cv} \left( u+ \frac{v^2}{2} + gz \right) \rho dV + \int_{cs} \left( u+ \frac{v^2}{2} + gz + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A}$

$\dot{Q}$ is the net efflux of heat crossing the boundary per unit time.

$\dot{W_s}$ is the shaft work between the inlet/outlet.

$\mathbf{v}$ velocity vector of the flow ( it may vary across the pipe ).

$v$ scalar flow velocity

$\mathbf{A}$ vector describing the cross sectional area of the pipe.

$u$ internal energy of the flow

$\rho$ mass density of the flow

$z$ Centerline elevation of the flow from some reference

$p$ Pressure at inlet/outlet respectively

Simplifications between inlet/outlet:

Steady State $\Rightarrow = \frac{d}{dt} \int_{cv} \left( u+ \frac{v^2}{2} + gz \right) \rho dV = 0$

$\dot{W_s} = 0$ No shaft work present.

$\Delta z = 0$ No change in elevation.

We are left with:

$\displaystyle \dot{Q} = \int_{cs} \left( u+ \frac{v^2}{2} + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A}$

The Second Law of Thermodynamics

$\displaystyle \dot{Q} > 0$ Heat is generated in the process of fluid flowing between inlet and outlet.

Which implies:

$\displaystyle \int_{cs} \left( u+ \frac{v^2}{2} + \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > 0$

Here is where my dilemma begins. If we assume incompressible flow and a uniform velocity distribution over $\mathbf{A}$ , the equation further reduces to:

$\displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > 0$

$\displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} = \dot{m}( h_o - h_i ) > 0$

Where, $h = u+ \frac{p}{\rho}$ which is commonly known as the Enthalpy.

I see two things changing as the flow moves between inlet and outlet. The Pressure $p$ is decreasing from inlet to outlet, and the internal energy $u$ is increasing ( the temp of the flow is increasing from inlet to outlet ).

So it seems like these two remaining forms of energy ( $p$ and $u$ ) are just changing hands, but the result implies that the Enthalpy is not constant. It is growing in the direction of flow.

So where is this energy to change the enthalpy being added to the system?

If we take a step back to the incompressibility assumption, no flow is truly incompressible. As such, there exists a tiny velocity gradient between the inlet and outlet. Is the added energy to change the enthalpy in actuality necessarily coming from the expansion of the flow?

I can't see how accounting for the expansion changes anything.

$\displaystyle \int_{cs} \left( u+ \frac{p}{\rho} \right) \rho \mathbf{v} \cdot d \mathbf{A} > -\displaystyle \int_{cs} \left( \frac{v^2}{2} \right) \rho \mathbf{v} \cdot d \mathbf{A}$

It still seems like the equation above is saying an imbalance of energy exists ( albeilt smaller than its incompressible counterpart), so what am I missing here?

Thanks!