How can I derivate the Kinetic Energy on a practical context?

Question

Lets assume there is a pump or another device that gives more Kinetic Energy to a gas or fluid that is passing through it on an horizontal line, and lets assume perfect conditions like, all the power of the device is only used to do work, there is not an amount that is transformed to heat.

I can say that the power of the device is equal to the change of energy, per time, that the fluid will gain. So, $$ P=\frac{dE}{dt} $$

I know this change of energy will only be in form of Kinetic energy, since the velocity of exit will be greater than the velocity when the fluind enters the device. So, $$ P=\frac{d(KE)}{dt} $$

Now, I know how kinetic energy is defined, so,

$$ P=\frac{1}{2}\frac{d(mv^2)}{dt} $$

Now, I know the velocities will change, and also, there is a thing called "mass flow", which is the amount of mass passing an area, per time, so,

$$ P=\frac{1}{2}\cdot\ v^2 \cdot\frac{dm}{dt} +\frac{2}{2}\cdot\ m \cdot v\cdot\frac{dv}{dt} $$ $$ =\frac{1}{2}\cdot\ v^2 \cdot\frac{dm}{dt} +\ m \cdot v\cdot\frac{dv}{dt} $$

I am exploring this mathematically, using differentials and derivatives, because we are talking about "changes". So, does this makes sense? If not, why? Is that because I am mixing a derivative (velocity) with a flux(mass flow), which is defined by inexact differentials?

Is correct to say Power is equal to "$\frac{dE}{dt}$"?

Answer 1

You are asking about the derivation of the open system (control volume) version of the first law of thermodynamics. You are aware that the starting point for this is the closed system version of the first law, correct? $$\Delta (U+KE+PE)=Q-W$$If we take the time derivative of this, we get: $$\frac{d(U+KE+PE)}{dt}=\dot{Q}-\dot{W}$$where $\dot{Q}$ and $\dot{W}$ are the rate of heat flow into the system through its boundaries and the rate of doing work at the system boundaries. Now suppose that, rather than the system being closed, there is mass flowing into and out of the system through its boundaries (open system). To keep things simple, we will include here only a single entering stream and a single exit stream. The inlet and exit streams carry with them internal energy, kinetic energy, and potential energy that have to be accounted for in the energy balance on our open system. This is done by writing: $$\left[\frac{d(U+KE+PE)}{dt}\right]_{CV}=\dot{Q}-\dot{W}+\dot{m}_{in}(u+ke+pe)_{in}-\dot{m}_{out}(u+ke+pe)_{out}$$ where the u, ke, and pe are the internal energy per unit mass, the kinetic energy per unit mass, and the potential energy per unit mass in a stream, $\dot{m}$ is the mass flow rate of a stream, and the subscript CV the refers to the contents of the control volume (our open system).

There is only one final step in completing the development of the open system (control volume) version of the first law. This is to split the work into two separate contributions: (a) the work done to push the inlet stream(s) forward into the control volume from behind and the work done to push back on the exit stream(s) leaving the control volume from ahead and (b) all other work (called shaft work). The work of type (a) is analogous to the fluid from behind the entering stream acting like a piston to push the fluid forward or the fluid from ahead of the exit stream acting like a piston to push the fluid backward. The rate at which work of type (a) is done is thus given by: $$\dot{m}_{in}(pv)_{in}-\dot{m}_{out}(pv)_{out}$$ where p is the pressure of the fluid in a stream and v is the specific volume of the fluid in a stream. So, the total rate of doing work is: $$\dot{W}=\dot{m}_{in}(pv)_{in}-\dot{m}_{out}(pv)_{out}+\dot{W}_S$$If we substitute this into our first law open system energy balance equation, we obtain: $$\left[\frac{d(U+KE+PE)}{dt}\right]_{CV}=\dot{Q}-\dot{W}_S+\dot{m}_{in}(h+ke+pe)_{in}-\dot{m}_{out}(h+ke+pe)_{out}$$where h is the enthalpy per unit mass of a stream in or out. The shaft work is typically associated with the work for a pump, a turbine, or compressor.

This completes the development.