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I am not understanding the origin of the second term on RHS of momentum conservation equation (cf. the Wiki page), $$ \frac{\partial}{\partial t}\int_V\rho\mathbf u\,dV=-\oint_S\left(\rho\mathbf u\cdot d\mathbf{S}\right)\mathbf{u}-\oint_Sp\,d\mathbf{S}+\int_V\rho\mathbf{f}_{body}\,dV+\mathbf{F}_{surf} $$

I see that it is a surface integral of the pressure and it has the correct dimensions of force, but I am confused about the origin of the term.

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This is the pressure-gradient term integrated over all volume, converted to a surface integral and using Gauss' theorem.

Note that physicists prefer the differential form of such equations (see also this Wikipedia article), when the corresponding equation becomes $$ \frac{\text{d}\boldsymbol{u}}{\text{d} t} = \frac{\partial\boldsymbol{u}}{\partial t} + \boldsymbol{u}\cdot\boldsymbol{\nabla}\boldsymbol{u} = -\frac{1}{\rho} \boldsymbol{\nabla}p + \boldsymbol{f}_{\text{body}}. $$ This is for inviscid flow. For viscous flow another term appears on the right (obtaining the Navier-Stokes equation).

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Under the hypotheses of continuum mechanics, one can isolate any subdomain $V$ of the continuum and in general assume it is affected by

  1. a field of surface forces per unit area (tractions), denoted $\mathbf{t}$, which is a surface field defined on its boundary and
  2. a field of body forces (per unit volume), denoted $\mathbf{f}_{\text{body}}$ as in the Wikipedia article, defined in its interior.

Then, one can postulate that the conservation of linear momentum applies to the body (generalizing Newton's Second Law) as $$\frac{\mathrm{d} \mathbf{P}}{\mathrm{d} t} = \mathbf{F}_{\text{body}} + \mathbf{F}_{\text{surface}} \tag{1}$$ where the total linear momentum in V is defined as $$\mathbf{P} = \int_V { \rho \mathbf{u} \, \mathrm{d}V}$$ and where the total surface and body forces can be calculated as $$\mathbf{F}_{\text{body}} = \int_V { \mathbf{f}_{\text{body}} \, \mathrm{d}V}, \quad \mathbf{F}_{\text{surface}} = \oint_S { \mathbf{t} \, \mathrm{d}S}$$ Therefore, the LHS term and the first term on the RHS in your equation correspond, by Reynolds' transport theorem, to the LHS in Eq.(1); while the third term in your equation is the first term on the RHS of Eq. (1).

Furthermore, one can prove (see this paper about the foundations of the theory) that there must exist a tensor field $\boldsymbol{\sigma}$ such that, for any point on the boundary of a body (subdomain) with a well-defined local normal $\mathbf{n}$, the local value of traction vector (taking the normal such that it points towards the outside of the body) is given by $$\boldsymbol{\sigma} \cdot \mathbf{n} \tag{2}$$

The tensor $\boldsymbol{\sigma}$ is the Cauchy stress tensor, and it can be proven that it must be symmetric as a consequence of the local conservation of angular momentum. Furthermore, in fluids one postulates Pascal's Principle: normal stresses (the normal component of the tractions) do not depend on the direction.

Now, note that since according to Eq.(2) the normal stresses are the eigenvalues of the stress tensor, they must all be equal to a single scalar field, which we denote as $-p$, the (minus) pressure. The minus sign results from the compressive nature of the force. According to this we have $$\mathbf{F}_{\text{surface, normal}} = \oint_S { (\mathbf{t} \cdot \mathbf{n}) \mathbf{n} \, \mathrm{d}S} = \oint_S { (\boldsymbol{\sigma}\cdot \mathbf{n}) \cdot \mathbf{n} \, \mathbf{\mathrm{d}S}} = - \oint_S { p \, \mathbf{\mathrm{d}S}}$$

where $\mathbf{\mathrm{d}S} = \mathbf{n}\mathrm{d}S$ and $\mathbf{F}_{\text{surface, normal}}$ is defined such that $\mathbf{F}_{\text{surface, normal}} + \mathbf{F}_{\text{surf}} = \mathbf{F}_{\text{surface}}$ (note that $\mathbf{F}_{\text{surf}}$ are shear forces as stated in the Wikipedia article, i.e., $\mathbf{F}_{\text{surf}} = \mathbf{t} - \mathbf{t}\cdot \mathbf{n} $)

Thus, the term you asked about corresponds to the contribution of the normal surface forces to the rate of change in linear momentum of the volume considered.

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