Chapter 5Mass and Energy Analysis of Control

Volumes (Open Systems)

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2A. Conservation of Mass Mass Balance (Continuity Equation) for Open System CV:

1. Control volumes-CV: Mass as well as energy can cross the

boundaries (control surface-CS).

So we must keep track of the amount of mass entering and leaving

the control volume.

Q W

CS

2. Closed systems: The mass of

the system remains constant

during a process.

3. Mass flow rate: kg/s

The amount of mass flowing through a cross section per unit time.

= A = V =

kg/s

= fluid density

3, = velocity

, A = cross section area 2

V = A= = volumetric flow rate 3

, = specific volume

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4. Conservation of mass (continuity Eq.): Mass, like energy, is a

conserved property, and it cannot be created or destroyed during a

process.

5. Mass Balance (Conservation of mass) gives:

a.

Total massflow rate

entering CVduring t

Total massflow rateleaving CVduring t

=

Net rate ofchange

of mass in CVduring t

Q W

CS

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General Continuity Equation

Total Mass Flow Rate entering CV during t =

Total Mass Flow Rate leaving CV during t =

Net rate of change of mass in CV during t =

c. Also, = kg

d. These equations are often referred to as the mass balance

(continuity Eq.) and are applicable to any control volume

undergoing any kind of process.

= mfinal cv = (m2 m1)cv OR mi me =( m2 - m1)cv

. =

kg/s

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5. Flow work and the energy of a flowing fluid:

a. Flow work, or flow energy: The work (or energy) required

to push the mass into or out of the control volume. This work

is necessary for maintaining a continuous flow through a

control volume.

F = PA

Wflow = FL = PAL = PV kJ

wflow = Pn kJ/kg

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b. Total Energy of a Flowing Fluid :

= h + V2

2+ gZ kJ/kg

= Pn + e

Recall: h = u + Pv

Energy due to flow ( wflow)

Total Energy per unit mass e = u + ke + pe

= Pv +u +V2

2+ gZ

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d. The total energy consists of three parts for a nonflowing fluid

and four parts for a flowing fluid.

c. The flow energy is automatically taken care of by enthalpy. In

fact, this is the main reason for defining the property enthalpy.

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1. Net Energy Transfer Heat , Work

= Rate of Change in the Total

Energy inside the CV

in out =

,- ,+ i - ee =

is the total energy of a flowing fluid in kJ/kg

= h +2

2+ gZ

B. Conservation of Energy Energy Balance (1st Law ) for Open System CV:

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net,CV - net,CV + (hi +

+ gZi ) - e(he +

+ gZe ) =

in = in + in + flow,in

out = out + out + Eflow,e

flow,in = (Pini + ei ) = i( hi +

+ gZi ) = i

flow,e = (Pene + ee ) = e( he +

+ gZe ) = e

in out =

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net = in - out net = in - out

net,CV - , + (hi +

+ gZi ) - e(he +

+ gZe )=

2. General 1st Law for CV:

The total energy of a flowing fluid entering the CV (i)

The total energy of a flowing fluid Leaving the CV (e)

3. Note the followings:

a. if : = e = 0 closed system net - =

b. Vi is the velocity of the flow entering the CV (i.e., KE of the flow entering the CV.

c. Ve is the velocity of the flow leaving the CV (i.e., KE of the flow leaving the CV.)

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d. Entering flow brings in energy.

e. Exiting flow takes out energy.

b. The rate of heat and work are constants.

a. The properties in the CV Does not change with time.

c. The CV is stationary (not Moving).

C. Steady-flow processes SFP:

= 0 &

= 0

1. The SFP is valid for long term operationdevices such as

Compressors, Turbines, Heat Exchangers, Nozzles, Pipes and Channels,

etc. and it implies the followings:

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3. Conservation of Energy Energy Balance (1st Law ) for SFP:

2. Conservation of Mass Mass Balance (Continuity Equation) for SFP:

Recall: continuity equation: - =

kg/s

= 0

= OR = kg/s

net,CV - , + (hi +

+ gZi ) - e(he +

+ gZe ) =

= 0

CV + (hi +

+ gZi ) = + e(he +

+ gZe ) kW

qCV + (hi +

+ gZi ) = w + (he +

+ gZe ) for single /exit kJ/kg

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4. Some steady-flow SF engineering devices:

Many engineering devices operate essentially under the same

conditions for long periods of time. The components of a steam

power plant (turbines, compressors, heat exchangers, and pumps),

for example, operate nonstop for months before the system is shut

down for maintenance. Therefore, these devices can be analyzed as

steady-flow devices.

Nozzles and diffusers are commonly

utilized in jet engines, rockets,

spacecraft, and even garden hoses.

a. Nozzles and Diffusers:

A nozzle is a device that increases the velocity of a fluid at the expense of pressure.

A diffuser is a device that increases the pressure of a fluid by slowing it down.

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Energy balance for a nozzle or diffuser:

0 , = 0 and PE = 0

(hi +

) = (he +

) kW

= = kg/s- Continuity Eq.

- 1st Law

(hi +

) = (he +

) kJ/kg

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b. Turbines and Compressors:

Turbine drives the electric generator in steam, gas, or hydroelectric power plants. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work.

Compressors, as well as pumps , are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft.

Do work (drive generators).Decrease pressure.

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A compressor is capable of compressing the gas to very high

pressures.

Pumps work very much like compressors except that they

handle liquids instead of gases.

Other examples of SF devices:

Requires work and increase pressure.

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Example: P5-32 Steam at 3 MPa and 400oC enters an adiabaticnozzle steadily with a velocity of 40 m/s and leaves at 2.5 MPa and300 m/s. Determine (a) the exit temperature and (b) the ratio ofthe inlet to exit areas (A1/A2).

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Example: P5-36: Nitrogen gas at 60 kPa and 7oC enters anadiabatic diffuser steadily with a velocity of 200 m/s and leavers at85 kPa and 22oC. Determine (a) the exit velocity of Nitrogen and(b) A1/A2.

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Example: P5-51: Argon gas enters an adiabatic turbine steadily at1600 kpa and 450oC with a velocity of 55 m/s and leaves at 150kpa with a velocity of 150 m/s. The inlet area of the turbine is 60cm2. If the power output of the turbine is 190 kW, determine theexit temperature of argon.

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Example P5-52: Helium is to be compressed from 120 kPa , 310 Kto 700 kPa and 430 K. A heat loss of 20 kJ/kg occurs during thecompression process. Neglecting KE & PE changes determine thepower input required for a mass flow rate of 90kg/min.