# When is it appropriate to drop pressure terms when applying conservation of momentum to a fluid?

I'm trying to wrap my head around pressure forces in incompressible, irrotational, invicid flow. Applying conservation of momentum to a control volume gives me

\begin{equation} \frac{\partial}{\partial t} \int_{c.v.} \rho \vec{v} \mathrm{d}V + \int_{c.s} \rho \vec{v} (\vec{v} - \vec{v}_{c.v.})\cdot \hat n \mathrm{d}A = -\int_{c.s.} P \hat n \mathrm{d}A + \int_{c.s.} \overline{\overline{\tau}}^{\prime} \cdot \hat n \mathrm{d}A + \int_{c.v.} \rho \vec{g} \mathrm{d}V\,. \end{equation}

When is it it acceptable to let pressure force terms on the right hand side of the equation equal zero? I know that if there is a uniform pressure on a closed surface, the force is zero, but that is the only case I really understand. Does it matter if the surface on which I'm evaluating the force is collinear with a solid body or perhaps a streamline?

For the kind of flow you mention conservation of mass gives you the equation

\begin{equation} \nabla\cdot\boldsymbol{v}=0 \end{equation}

where $\boldsymbol{v}$ is the velocity field, in integral form it would be

\begin{equation} \oint{\boldsymbol{v}\cdot\boldsymbol{n}dA}=0 \end{equation}

and for "conservation" of linear momentum you have

\begin{equation} \frac{D\boldsymbol{v}}{Dt}=-\nabla\frac{P}{\rho} \end{equation}

if you have an homogeneous pressure field, there would not be a gradient, so in that case you will have the equation

\begin{equation} \frac{D\boldsymbol{v}}{Dt}=0 \end{equation}

which means that velocity is constant along stream lines