The fluid is incompressible and has no sources inside. This means that the continuity (mass conservation) equation is
$$
\text{div}\,\vec{v} = 0. \qquad (1)
$$
Now we follow the standard procedure and represent $\vec{v}$ as follows:
$$
\vec{v} = \text{rot}\,\vec{A}.
$$
The divergence of any curl is zero so equation (1) is satisfied by any smooth vector field $\vec{A}(\vec{r})$.
For 2-dimensional flow we can assume
$$
\vec{A} = \Bigl(0, 0, \psi(x,y)\Bigr)
$$
so that
$$
v_x = \frac{\partial \psi}{\partial y}; \quad v_y = -\frac{\partial \psi}{\partial x}. \qquad (2)
$$
In fluid dynamics $\psi(x,y)$ is called stream function because the lines of constant $\psi$ are the streamlines.
We have two known stream lines:
$$
y = h(x)
$$
and
$$
y = 0.
$$
Let's select the stream function as follows:
$$
\psi(x,y) = C\frac{y}{h(x)}. (3)
$$
For the upper streamline we have $\psi=C$ and for the lower line $\psi=0$. This is a strong assumption and the main point of the solution. The selection of $\psi$ is not definite here. Formula (3) is intuitive, it gives streamlines that are similar to $h(x)$ but coming more straight while approaching to the bottom.
Now we can use (2) and (3) to find $\vec{v}$:
$$
\vec{v}(x,y) = \left(\frac{C}{h(x)},\; Cy\frac{h'(x)}{h^2(x)}\right) \qquad (4)
$$
where $C$ is some constant determined by the boundary conditions.
The velocity field depends on the unknown function $h(x)$.
Finding $h(x)$
Function $h(x)$ can be found by applying the Bernoulli equation to the top streamline. Bernoulli equation for incompressible fluid is
$$
\frac{v^2\bigl(x, y(x)\bigr)}{2} + \frac{p\bigl(x, y(x)\bigr)}{\rho} + gy(x) = \text{const}
$$
where
$y(x)$ is the streamline,
$p(x,y)$ is the pressure,
$\rho$ is the density of the fluid,
$g$ is the gravitational acceleration.
The upper streamline $y(x)=h(x)$ is in the equilibrium with the atmosphere air. This means that the pressure of the fluid is equal to the atmosphere pressure:
$$
p\bigl(x, y(x)\bigr) = p_0.
$$
So
$$
\frac{v^2\bigl(x, h(x)\bigr)}{2} + gh(x) = \text{const} - \frac{p_0}{\rho} = D \qquad (5)
$$
Substitution of (4) into (5) gives the differential equation for $h(x)$:
$$
\frac{C^2}{2h^2}\left(1 + h'^2\right) + gh = D
$$
or
$$
\frac{dh}{dx} = \sqrt{\frac{2h^2}{C^2}(D-gh) - 1}
$$
$$
h(x_1) = h_1
$$
This can be solved numerically if we know $C$ and $D$.
Finding $C$ and $D$
The parameters $C$ and $D$ are determined by the boundary conditions. If we know the velocity at the point $(x_1, h_1)$ then
from (4):
$$
C = h_1 v_x(x_1, h_1)
$$
and from (5):
$$
D = \frac{v^2(x_1, h_1)}{2} + gh_1
$$
Conclusion
There are two weak points in this solution:
- the intuitive assumption (3);
- the undefined constants $C$ and $D$.
Some boundary conditions can violate (3) and/or make calculation of $C$ and $D$ very difficult.
Alternative
There is another way to select the stream function.
If we suppose the flow to be potential the velocity field will have the following form:
$$
\vec{v} = \nabla \varphi
$$
where $\varphi(x,y)$ is the potential of the velocity vector field.
Then in addition to (2) we will have:
$$
v_x = \frac{\partial \varphi}{\partial x}; \quad v_y = \frac{\partial \varphi}{\partial y}. \qquad (6)
$$
Now we can introduce the complex potential of the flow:
$$
W(x+iy) = \varphi(x,y) + i \psi(x,y)
$$
The formulas (2) and (6) together are exactly the Cauchy-Riemann conditions for the function $W(z)$. This means that $W(z)$ describes some conformal map.
If we find a conformal map $W(z)$ that turns some rectangle into the blue area in the picture in the question for any $h(x)$, then we find a potential flow (flow with zero vorticity) that solves the problem. Some manipulations will still be required to find $h(x)$.
In fact any $W(z)$ always turns 2-dimensional potential flow with
$$
\varphi(x,y) = x
$$
$$
\psi(x,y) = y
$$
and
$$
\vec{v} = (v_x, 0)
$$
into something more interesting and still fitting the hydrodynamics equations. This works only for potential flows that are not always a good approximation.
Finding of $W(z)$ in this case is a mathematical problem and perhaps should be discussed somewhere else.
Starting from the conservation of mass:
$$ \dot m_{1}=\dot m_{2} $$
This translates to
$$ \rho_{1} S_{1} V_{1}=\rho_{2} S_{2} V_{2} $$
Assuming incompressible flow, thus $\rho_{1}=\rho_{2}$
gives:
$$S_{1} V_{1}= S_{2} V_{2} $$
With $S_{1} V_{1} = Q_{1}$ , the formula you are using.
This formula follows directly from the mass balance, with only the assumption of incompressible flow. There is no assumption on turbulent or laminar flow, thus this equation holds for both flow types.
With the balance given above you can calculate the speed in at location 2 for the given value of $V_1$ and the ratio of Areas, just as you did. There is no need to account for the flow type.
However, it should be noted here that these are average speeds. If you want to go into further detail, you could include friction forces in the pipe. These friction forces depend on the flow type, and determine the shape in velocity profile, and the resulting velocity.
However, this is significantly more difficult to do than just solving a couple equations.
I know there are some rules of thumb to estimate losses in pipes, but you have to check if their assumptions are valid for your case.
Perhaps you can have a look at Pipe Flow Fluid Mechanics Course
Best Answer
The problem can be addressed using macroscopic thermodynamics without referring to the molecular picture of the fluid. We however need to know something about the chemical composition of the fluid after the reaction. I suppose that not all of your fluid undergoes the chemical reaction, so the resulting fluid is going to be multi-component. Below, I sketch a solution of the problem under the assumption that all the components move with the same macroscopic velocity, so one can treat the fluid after the reaction as a single-component matter with some average thermodynamic properties. A more detailed solution would depend on the exact nature of your fluid, e.g. on whether the ideal-gas approximation is appropriate.
The state of the fluid is completely described by its temperature $T$, pressure $P$, density $\rho$ and velocity $v$. I assume, as you indicated, that the geometry of the problem is one-dimensional so that the velocity can be treated as a scalar. I will use indices 1 and 2 to refer to the fluid before and after the reaction. There is one relation between $T$, $P$ and $\rho$ supplied by the equation of state, so we need three additional relations to fully determine the unknown values of $T_2$, $P_2$, $\rho_2$ and $v_2$.
Mass conservation. Given that the cross-section area of the flow does not change, this amounts to the simple condition $$ \rho_1v_1=\rho_2v_2, $$ equivalent to your equation.
Momentum balance. The flow of momentum per unit area and per second in a fluid of velocity $v$ is $\rho v^2$, which is just the mass flow times velocity. Then the second law of Newton requires that $$ P_1-P_2=\rho_2v_2^2-\rho_1v_1^2. $$
Energy balance. The above two constraints are purely mechanical. Here is the only place where we need some input about the chemical reaction in the system. I will model the reaction as an instantaneous supply of heat per kilogram equal to $q$. Then the energy balance condition can be expressed in terms of increase of specific enthalpy $h$ (enthalpy per kilogram) plus specific kinetic energy (which is just $\frac12v^2$), $$ h_2+\frac12v_2^2=h_1+\frac12v_1^2+q. $$ The enthalpy itself is in turn determined by the temperature and pressure. For an ideal gas, it can be extracted for instance from the knowledge of specific heat at constant pressure.
The above set of equations together with the equation of state is sufficient to determine the final state of your fluid. As said, a more concrete solution can only be given in case you know the equation of state of your fluid. Hope this helps anyway.
EDIT. In classical thermodynamics, thermodynamic potentials such as enthalpy are only defined up to an arbitrary additive constant, which can be chosen independently for different materials. In order that the comparison of $h_1$ and $h_2$ makes sense, we therefore have to fix the reference values of the enthalpy of the fluid before and after the reaction appropriately. In my answer, I assume that the specific enthalpies of these two different fluids are equal at temperature $T_1$ and pressure $P_1$.