Control Volume Analysis

Using Energy

ME-201

Introduction

To develop and illustrate the use of the control volume

form of the conservation of mass and conservation of energy

principles. Mass and energy balances are applied to control

volumes at steady state and for transient applications.

Devices such as turbines, pumps and compressors

through which mass flows can be analyzed in principle by

studying a particular quantity of matter (a closed system) as

it passes through the device, it is normally preferable to think

of a region of space through which mass flows ( CV).

As in the case of a closed system, energy transfer

across the boundary of a control volume can occur by means

of work and heat.

Conservation of Mass and the Control Volume

Control volume. A volume in space in which one has

interest for a particular study or analysis. The surface of this

control volume is referred to as a control surface and always

consist of a closed surface.

The size and shape of the control volume are

completely arbitrary.

The surface may be fixed, or it may move so that it

expands or contracts.

Mass as well as heat and work can cross the

control surface.

Mass in the control volume, as well as the properties

of this mass , can change with time.

The term control volume is used in open system.

A control volume differs from a closed system in that

it involves mass transfer. Mass carries energy with it, and

thus the mass and energy content of a system change

when mass enters or leaves.

The control volume approach will be used for many

engineering problems where a mass flow rate is present,

such as: Turbines, pumps and compressors, Heat

exchangers, Nozzles and diffusers, etc.

Conservation of Mass and the Control Volume

Control Volume

Control Mass. A system of fixed mass is called

a closed system, or control mass .

The closed system boundary does not have to

be fixed.

No mass can cross the closed system

boundary.

Energy in the form of heat and work can cross

the closed system boundary

The term control mass is some times used in

place of closed system.

Conservation of Mass and the Control Volume

Conservation of Mass

The physical law concerning mass, says that we

cannot create or destroy mass.

The rate of change of mass inside the control

volume equal to mass enters minus mass exit.

i.e Rate of change = + in out

Conservation of Mass and the Control Volume

Let us consider the conservation

of mass law as it relates to the control

volume. The law says that we cannot

create or destroy mass. We will express

this law in a mathematical statement

about the mass in the control volume.

To do this we must consider all the

mass flows into and out of the control

volume and the net increase of mass

within the control volume. As a control

volume we consider a tank with a

cylinder and piston and two pipes attaches as shown in Fig.

We know from the first law of thermodynamics for a control mass,

dECM / dt = QCM WCM …………………………………..(1) . .

Rate of energy change in Cm = (Energy that enters into the

Cm Energy that exit from the Cm)

Conservation of Mass and the Control Volume

For Control volume we can write

Rate of energy change in CV = Energy that enters into the

CV Energy that exit from the CV .………………….. (2)

The rate of change of mass inside the control volume equal to mass enters minus mass exit.

i.e Rate of change = + in out

Conservation of Mass and the Control Volume

............................... ……………..(3)

Equation 3 is the mass rate balance for control volumes with

several inlets and exits.

For several inlet and outlet

Conservation of Mass and the Control Volume

The First Law of Thermodynamics

for a Control Volume.

A control volume is shown in

Fig. that involves rate of heat

transfer, rates of work, and

mass flows. The fundamental

physical law states that we

cannot create or destroy

energy such that any rate of

change of energy must be

caused by rates of energy into or out of the control volume.

dEcv / dt = Energy in Energy out............................................... (3)

The fluid flowing across the control surface enters or leaves with an amount of energy per unit mass as

e i = u i + ½. V 2 + g Zi ( flow liquid energy)

and e e = u e + ½. V 2 + g Ze

Whenever a fluid mass enters a control volume at state i, or

exits at state e, there is a boundary movement work associated with that process. This is called flow work .

The First Law of Thermodynamics

for a Control Volume.

Conservation of Energy for a

Control Volume

Fig. 4

Accordingly, the conservation of energy principle applied to a

control volume states:

For the one-inlet and one-exit control volume with one-

dimensional flow shown in Fig. 4 the energy rate balance is:

Conservation of Energy for a

Control Volume

.…....(3)

where,

ECV = Energy of the control volume at time t.

Q and W = The net rate of energy transfer by heat and work

across the boundary of the control volume at t respectively.

Underlined Terms = the rates of transfer of internal, kinetic,

and potential energy of the entering and exiting streams.

If there is no mass flow in or out, the respective mass flow

rates vanish and the underlined terms of Eq. 3 drop out. The

equation then reduces to the rate form of the energy balance

for closed systems.

Conservation of Energy for a

Control Volume

Work is always done on or by a control volume where

matter flows across the boundary, it is convenient to

separate the work term into two contributions:

Wf = Work associated with the fluid pr as mass is

introduced at inlets and removed at exits i.e flow work.

= Includes all other work effects, such as those

associated with rotating shafts, displacement of the

boundary, and electrical effects.

Conservation of Energy for a

Control Volume

With these considerations, the work term W of the energy

rate equation, Eqn. 3, can be written as

.

Since A = m v / V ......................…......(4)

Substituting Eq. 4. in Eq. 3 and collecting all terms referring to

the inlet and the exit into separate expressions, the following

form of the control volume energy rate balance results

Conservation of Energy for a

Control Volume

or

Eqn 5 is the energy rate balance for single inlet and single

out let.

....(5)

Conservation of Energy for a

Control Volume

In practice there may be several locations on the boundary

through which mass enters or exits. This can be accounted

for by introducing summations as in the mass balance.

Accordingly, the energy rate balance is

Conservation of Energy for a

Control Volume

Application of First Law of Thermodynamics to flow process

Steady State. At the steady state of a system, any

thermodynamic property will have a fixed value at a particular

location, and will not alter with time. Thermodynamic

properties may vary along space coordinates, but do not vary

with time. `Steady state` means that the state is steady or

invariant with time.

Steady Flow Process. Steady flow` means that the

rates of flow of mass and energy across the control surface

are constant.

Assumptions.

The CV does not move relative to the co ordinate

frame i.e the CV is fixed.

The state of the mass at each point in the control

volume does not vary with time.

i.e dmCV / dt = 0 and dECV / dt = 0.

For the steady state process we can write

Continuity Eqn: = m

Application of First Law of Thermodynamics to flow process

and First Law :

=

Application of First Law of Thermodynamics to flow process

The rates at which heat and work cross the control

surface remain constant. i.e heat transfer and work

transfer is fixed. If there is only one flow stream entering and

one leaving the CV.

For this type of process, we can write

Continuity Eqn: mi = me = m

and First Law :

= +

Application of First Law of Thermodynamics to flow process

Application of First Law of Thermodynamics to flow process

Example of Steady Flow Process.

Nozzle and Diffuser. A

nozzle is a steady-state device

which increase the velocity or K.E

of a fluid at the expense of its pr

drop, whereas a diffuser increase

the pr of a fluid at the expense of

its K.E. Fig Shows a nozzle and a

Diffuser. The steady flow energy eqn of the control surface gives

The steady flow energy eqn of the control surface gives

QCv+ m [ hi +½ Vi2 +gZi] = Wcv+ m [ he+ ½ Ve

2 + gZe] ....... (1)

Assumption.

a. Nozzle and Diffuser are SSSF device.

b. Change in P.E is negligible i.e ∆ P.E =0.

c. Nozzle is insulated i.e no heat enters or leaves the

system (adiabatic). QCv = 0

d. Work transfer is negligible i.e Wcv= 0.

Example of Steady Flow Process

So energy eqn for nozzle and diffuser becomes

Example of Steady Flow Process

m ( hi + ½ Vi2 ) = m ( he + ½ Ve

2)

or, ( hi + ½ Vi

2 ) = ( he + ½ Ve 2)

or, Ve

2 = Vi2 + 2 ( hi he)

or Ve = Vi

2 + 2 ( hi he) [ Vi is very small as compare to Ve]

or Ve = 2 ( hi he)

Throttling Process. A throttling process occurs

when a fluid flows through a constricted passage, like a

partially opened valve, an orifice or a porous plug with a

significant drop in pr. This is a SSSF process with no heat

transfer to or from CV. There is no means for doing work and little or no change in PE.

Here, ∆PE = 0, ∆KE = 0, QCv = 0 and WCv = 0

Example of Steady Flow Process.

Throttling Process

Example of Steady Flow Process.

Turbine

Example of Steady Flow Process.

Compressor and Pump

Example of Steady Flow Process.

Example of Steady Flow Process

Steam enters a converging–diverging nozzle operating at steady state with p1 = 40 bar, T1 = 4000C, and a velocity of 10 m/s. The steam flows through the nozzle with negligible heat transfer and no significant change in potential energy. At the exit, p2 =15 bar, and the velocity is 665 m/s. The mass flow rate is 2 kg/s. Determine the exit area of the nozzle, in m2.

Problem-1

Solution-1

Solution-1

Solution-1

Steam enters a turbine operating at steady state with a mass flow rate of 4600 kg/h. The turbine develops a power output of 1000 kW. At the inlet, the pr is 60 bar, the temperature is 4000C, and the velocity is 10 m/s. At the exit, the pr is 0.1 bar, the quality is 0.9 (90%), and the velocity is 50 m/s. Calculate the rate of heat transfer between the turbine and surroundings, in kW.

Problem-2

Solution-2

Solution-2