How to apply first law of thermodynamics closed non-stationary incompressible element of fluid passing a nozzle? [closed]

Question

The First Law of Thermodynamics (FLT) applied to a closed non-stationary incompressible element of fluid is given as

$Q-W = \Delta(KE)+\Delta(PE)+\Delta U$ .......(Eq1)

where:

• $\Delta U$ is change in internal energy;

• $\Delta (KE)$ is the change in Kinetic energy;

• $\Delta (PE)$ is the change in Potential energy;

• $Q$ is the amount of heat supplied to the system;

• $W$ is the amount of work done by the system to the environment.

(Let's assume $\Delta (PE)$ is zero)

If this element of fluid is going through a (converging)nozzle, from Bernoulli's law I can say that the pressure energy is transferred to kinetic energy, and at the end of the nozzle, the fluid's flow rate/velocity has increased. But how to deduce it from FLT applied to a closed non-stationary system (i.e $Eq1$) without using the Steady-flow-energy-equation(SFEE)/FLT-applied-to-open system?

The question is: I know how to analyze the flow through the nozzle as an Open system. How to do the same by applying FLW to a closed incompressible element of fluid passing the nozzle? How is the KE changing there are no pressure terms in Eq1 (and the is no PdV boundary work)?

PS: I am also trying to deduce how to apply the FLT to an incompressible element passing a turbine blade passage causing the turbine to rotate. Any help would be much appreciated.

Ref: Questions about flow work/flow energy?

Answer 1

If I understood your question correctly, I think the problem comes from the use of the expression $-PdV$ that is no longer valid here.

If the pressure at the surface of a closed system is not uniform, the work is not written $-PdV$. It is necessary to integrate on the surface of the closed system. One could write the integral and find that the result is non-zero when the pressure is not uniform. But it may be sufficient to consider the limiting case of a slice of incompressible fluid which advances with a velocity $v$ with different pressures $P_1$ on the left and $P_2$ on the right. The work during $dt$ is $(P_1-P_2)Svdt$ : clearly non zero. So the work term is not zero in your situation.

Hope my poor english is OK !

Edit : mathematical complement

We can write all this mathematically. The work of the pressure forces during $dt$ is an integral over the surface of the system : $\delta W=\iint{-P\vec{dS}\vec{v}dt}$

By Green's theorem, we replace it by a volume integral $\delta W/dt=-\iiint{\vec{\nabla}(P\vec{v})d\tau}$ with $\vec{\nabla}(P\vec{v})=P\vec{\nabla}(\vec{v})+\vec{\nabla}(P)\vec{v}$

For an incompressible flow, $\vec{\nabla}(\vec{v})=0$ it remains $\delta W/dt=-\iiint{\vec{\nabla}(P)d\tau\vec{v}}$.

Finally, for an elementary volume, we find : $\delta W/dt=(-\vec{\nabla}(P)d\tau)\vec{v}$

This is simply the power of the pressure force $\vec{dF}=-\vec{\nabla}(P)d\tau$ on the volume element.