In control volume analysis, why kinetic energy and flow work are considered seperately? As per work energy principle, the energy spent in moving an object is stored as kinetic energy.Flow work is done to make the fluid move.Why is it not included in kinetic energy (i.e) why kinetic energy and flow work are considered seperately?
 A: Since you are talking about the Bernoulli equation, we can refer directly to the terms of that equation. 
$P+\frac {\rho v^2} 2 +\rho gh=constant$
We have three terms on each side. The interpretation of the kinetic and potential energy terms, $\frac {\rho v^2} 2$ and $\rho gh$, is pretty straightforward, while the interpretation of the pressure term, $P$, is less obvious.
Sometimes, $P$ is interpreted as a "pressure energy", which is added to the kinetic and potential energy of a unit volume of fluid and this unit volume of fluid is treated as a body moving through space without interacting with any external entities and therefore conserving its total energy, which includes the "pressure energy".
Although the units match, this interpretation of the pressure term does not make much physical sense, since a) the unit volume of fluid does interact with the fluid behind it and in front of it and b) there is no recoverable mechanical energy that could be stored in an incompressible fluid. It appears that the whole thing works because (a) and (b) cancel each other.
Potentially, a more physical interpretation of the pressure term could be obtained, if the Bernoulli equation is slightly modified.
Let's write Bernoulli equation for two points along the length of the flow, $x_1$ and $x_2$, and then track the energy changes of the volume of fluid, $V$, between these two points.
$P_1+\frac {\rho v_1^2} 2 +\rho gh_1 =P_2+\frac {\rho v_2^2} 2 +\rho gh_2 $

After a short time interval, the volume of fluid $V$ will move to the right, filling a small volume $\Delta V$ in front of it and leaving the same volume $\Delta V$ behind it. 
$\Delta V=A_1d_1=A_2d_2$
Let's multiply both sides of the Bernoulli equation by $\Delta V$: 
$P_1A_1d_1+\frac {\rho v_1^2} 2 \Delta V+\rho gh_1 \Delta V=P_2A_2d_2+\frac {\rho v_2^2} 2 \Delta V+\rho gh_2 \Delta V$ or
$F_1d_1+\frac {m v_1^2} 2 +mgh_1=F_2d_2+\frac {mv_2^2} 2+mgh_2 $ or
$W_1-W_2=(\frac {mv_2^2} 2 -\frac {m v_1^2} 2) + (mgh_2-mgh_1)$
Here $m=\rho \Delta V$ is the mass of small $\Delta V$ segments on both sides of the volume $V$ and the differences in their kinetic and potential energies reflect the differences in kinetic and potential energies of $V$, since the energy of the rest of the volume $V$ does not change. 
According to the equation, this energy change of the fluid in volume $V$ is the result of the net work, $W_1-W_2$, performed on $V$ by the external forces, $F_1$ from the fluid behind $V$ and $F_2$ from the fluid in front of $V$.
This interpretation of the Bernoulli equation has a clear physical meaning and is in agreement with the work-energy principle. 
Note that the work here was done (or energy spent) by the fluid external to $V$. We can also say that the energy of the external fluid was passed to $V$. 
This work or energy could be positive or negative, i.e., the energy from $V$ could be passed to the rest of the fluid as well.
Let's consider a couple of simple cases to get a feel of how this might work.
First, let's consider a straight uniform pipe with a steady flow. Obviously, all three terms should stay the same everywhere. 
One observation here is that, if the pressure is not zero, we must have some energy flow, which amounts to $PAv$ per unit of time. So, in the absence of friction assumed in the Bernoulli equation, somewhere down the line something still must resist the flow and consume this energy and somewhere upstream there has to be a source of energy maintaining the pressure. 
The energy of the volume $V$ travelling downstream in a uniform pipe will remain constant or we can say that the energy it receives from the upstream is equal to the energy is passes downstream.

Now, let's see what happens, if the pipe has a narrow section and the volume of fluid $V$ has entered it (left section on the diagram below). 

Since, due to the continuity principle, the fluid in the narrow section has to accelerate and, therefore, the kinetic energy of $V$ has to increase, we have to conclude that the pressure force behind $V$, $F_1$, has to be greater than the pressure force in the front of $V$, $F_2$, and that the energy flow into $V$ from the fluid upstream, $W_1$, has to be greater than the energy flow, $W_2$, passed by $V$ downstream. In other words, the external fluid will perform a net positive work on $V$, increasing its energy.
When the trailing edge of $V$ enters the narrow section of the pipe as well (middle section on the diagram), the pressure forces on both ends of $V$ equalize, the acceleration stops and the kinetic energy of $V$ is maintained again.
As the leading edge of $V$ crosses the end of the narrow section (right section on the diagram), it slows down again. Now the energy flowing out of $V$ downstream exceeds the energy flowing in form the upstream and the energy of $V$ decreases. We can say that $V$ performs some work on the external fluid, spending some of its kinetic energy.     
Similar pressure and energy interactions happen when the pipe goes up and down.
To summarize, as a volume of fluid $V$ is travelling down a pipe, the balance of energy it receives from the fluid upstream and passes to the fluid downstream changes as the pipe goes up and down or its diameter increases and decreases, and, as a result, its energy varies as well. The source of all this energy is located somewhere upstream. 
It seems that you were also interpreting the pressure as a source of energy for the volume of fluid, but you were considering work-energy transitions as internal to that volume - hence a contradiction.
