We were discussing the concept of control volume system and concept of enthalpy in the field of thermal engineering in
our previous posts, where we have seen the fundamentals of a control volume
system as well as the specific heat and types of specific heat also.

Let us go ahead to discuss the application of first
law of thermodynamics for an open system. First law of thermodynamics for an
open system will be quite important for various industrial applications such as
in fluid pumping unit.

We will be able to determine the required energy by
a pump in order to pump the fluid at given head. We will have an idea of heat transfer
in a heat exchanger or required work energy by air compressor in order to compress
the atmospheric air up to a desired pressure once we will study thoroughly the
first law of thermodynamics for an open system.

###
**So
take one cup of tea and enjoy here the fundamental of first law of
thermodynamics for an open system**

We have already seen the concept of control volume system.
Let us consider one control volume system as displayed here in following
figure. Let us consider two sections, first section at the inlet of the control
volume system and another at the outlet of the control volume as displayed in
figure.

*First law of thermodynamics for control volume or an open system*

Let us consider here that control volume is
receiving heat energy from surrounding and providing the work energy to the surrounding.
As we know that fluid will enter the control volume system and will leave the
control volume system continuously i.e. there will be continuous flow of
material or fluid through the control volume and hence we will consider the
energy interactions between system and surrounding with respect to time.

Therefore let us consider the following important
data as mentioned here.

The rate of heat energy addition to the control
volume = Î´Q/Î´t

The rate of work energy providing by control volume
to the surrounding = Î´W/Î´t.

The mass flow rate entering to the control volume = dm

_{1}/dt
The mass flow rate leaving the control volume =dm

_{2}/dt
Rate of change of mass in control volume = dm

_{CV}/dt####
**According
to principle of conservation of mass, we will have following equation**

(Mass flow rate entering to the system) - (Mass flow
rate leaving the system) = Rate of change of mass in control volume

(dm

_{1}/dt) - (dm_{2}/dt) = dm_{CV}/dt###
**Conservation of energy **

Let us consider the rate of change of internal
energy of control volume is dE

_{CV}/dt. Now , we will have to consider net energy entering to the control volume and net energy leaving the control volume as after securing above two data we will be able to secure the rate of change of internal energy of control volume i.e. dE_{CV}/dt.
As we are concentrated here on an open system, hence
we will have to consider each form of energy associated with material flowing
in and flowing out of the control volume apart from work energy and heat
energy.

Let us first try to find out the various forms of
energy those will be associated with the material flowing in to the control
volume and flowing out from the control volume.

Energy associated with material or fluid flowing in
to the control volume (or leaving the control volume) will be given by the
summation of internal energy, pressure energy, kinetic energy and potential
energy.

Where E will be energy associated per unit mass.

E
= u + pv + V

^{2}/2 + gz
E
= h + V

^{2}/2 + gz
E

_{in}= h_{in}+ V^{2}_{in}/2 + gz_{in}= h_{1}+ V^{2}_{1}/2 + gz_{1}
E

_{1}= h_{1}+ V^{2}_{1}/2 + gz_{1}
Similarly, energy associated with material or fluid
leaving the control volume will be given as

E

_{out}= h_{out}+ V^{2}_{out}/2 + gz_{out}= h_{2}+ V^{2}_{2}/2 + gz_{2}
E

_{2}= h_{2}+ V^{2}_{2}/2 + gz_{2}#### So let us recall the statement of conservation of energy and write here the equation

Rate of change of internal energy of control volume
= (Net energy entering to the control volume) – (Net energy leaving the control
volume)

dE

_{CV}/dt = (dm_{1}/dt) E_{1 }+ Î´Q/Î´t – (dm_{2}/Î´t)E_{2 }- Î´W/Î´t####
**dE**_{CV}/dt
= (dm_{1}/dt ) (h_{1} + V^{2}_{1}/2 + gz_{1})
+ Î´Q/Î´t – (dm_{2}/dt)( h_{2} + V^{2}_{2}/2 + gz_{2})_{
}- Î´W/Î´t

_{CV}/dt = (dm

_{1}/dt ) (h

_{1}+ V

^{2}

_{1}/2 + gz

_{1}) + Î´Q/Î´t – (dm

_{2}/dt)( h

_{2}+ V

^{2}

_{2}/2 + gz

_{2})

_{ }- Î´W/Î´t

We can also write this equation as mentioned here

We will come again with new topic "Heat engine in thermodynamics"in our next post.

Do you have suggestions? Please write in comment
box.

####
**Reference:**

Engineering thermodynamics by P. K. Nag

Basic thermodynamics by Prof. S.K. Som

Image courtesy: Google

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