Chapter 5

Mass and Energy Analysis of Open Systems

Course Outcomes

• Ability to acquire and explain the basic

concepts in thermodynamics.

• Ability to comprehend and apply the concept

to the actual conditions and problems; i.e. to the actual conditions and problems; i.e.

closed and open systems.

• Ability to apply and correlate the concept with

the appropriate equations and principles to

analyze and solve engineering problems.

Course Learning Outcomes

The student should be able to :

• Explain the mechanisms of energy transfer for an open system.

• Write down the general energy and mass balances for an open system and simplify the

• Write down the general energy and mass balances for an open system and simplify the energy and mass balances for steady flow systems.

• Solve energy balance problems for common steady-flow devices namely, nozzle, compressors, turbine and throttling valve.

Contents

5.1 Energy Balance for Open Systems

5.2 Conservation of Mass

5.3 Flow Work and Energy of Flowing Fluid

5.4 Energy Transport by Mass5.4 Energy Transport by Mass

5.5 Energy Analysis of Steady-Flow Systems

5.1 Energy Balance for Open Systems

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5.2 CONSERVATION OF MASS

Conservation of mass:

Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process.

Closed systems:

The mass of the system remain constant during a process.

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The mass of the system remain constant during a process.

Control volumes:

Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume.

Mass Flow Rates Volume Flow Rates

( )3m /savg cV V A=&( )kg/s

Vm V

vρ= =

&

&∙

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The average velocity Vavg is defined as

the average speed through a cross

section.

The volume flow rate is the volume of

fluid flowing through a cross section

per unit time.

Conservation of Mass Principle

The conservation of mass principle for a control volume:

The net mass transfer to or from a control volume during a time interval ∆t is

equal to the net change (increase or decrease) in the total mass within the

control volume during ∆t.

Conservation of Mass for General Control Volume

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CV

CV

(kg)

/ (kg/s)

in out

in out

m m m

m m dm dt

− = ∆

− =& &

( ) ( ) ( )Total mass entering Total mass leaving Net change in massthe CV during the CV during within the CV during t t t

− =∆ ∆ ∆

Conservation of Mass for General Control Volume

The conservation of mass principle for the open system or control volume is expressed as

Steady-flow process, mCV = constant ,

Conservation of mass principle � the total amount of mass entering a control volume

equal the total amount of mass leaving it.

Multiple inlets and exits

Single stream

( )kg/sin outm m=∑ ∑& &

1 2 1 1 2 2 2 ( )1 kg/sm m V A V Aρ ρ= → =& &

Mass Balance for Steady-Flow Processes

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Conservation of mass principle for a two-

inlet–one-outlet steady-flow system.

Nozzles, diffusers, turbines,

compressors and pumps involve a

single stream (only one inlet and

one outlet).

Special Case: Incompressible Flow

The conservation of mass relations can be simplified even further when the

fluid is incompressible, which is usually the case for liquids.

Steady,

incompressible

Steady,

incompressible flow

(single stream)

3( )m /sin outV V=∑ ∑& &

1 2 1 1 2 2V V V A V A= → =& &

10During a steady-flow process, volume flow rates are not

necessarily conserved although mass flow rates are.

(single stream)

For steady flow of liquids, the volume flow rates, as well as the

mass flow rates, remain constant since liquids are essentially

incompressible substances.

5.3 FLOW WORK AND THE ENERGY OF A

FLOWING FLUID

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.

11Schematic for flow work.

In the absence of acceleration, the force

applied on a fluid by a piston is equal to the

force applied on the piston by the fluid.

Open System:

1212

The energy content of a

control volume can be

changed by mass flow as

well as heat and work

interactions.

Total Energy of a Flowing Fluid

h = u + Pv

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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The total energy consists of 3 parts for a nonflowing fluid

and 4 parts for a flowing fluid.

2

2

Ve u gz= + +

2

2v

VP u gzθ = + + +

Nonflowing fluid Flowing fluid

Internal

energyKinetic

energy

Potential

energy Internal

energy

Kinetic

energy

Potential

energy

Flow

energy

5.4 Energy Transport by Mass

When the kinetic and potential energies of

a fluid stream are negligible

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The product is the energy

transported into control volume by

mass per unit time.

iimθ&

a fluid stream are negligible

When the properties of the mass at

each inlet or exit change with time as

well as over the cross section

Example 5.3Steam is leaving 4-L pressure cooker

whose operating pressure is 150 kPa. It

is observed that the amount of liquid in

the cooer has decreased by 0.6 L in 40

min after the steady operating conditions

are established, and the cross-sectional

area of the exit operating is 8 mm2. area of the exit operating is 8 mm2.

Determine

(a) The mass flow rate of the steam and the exit velocity

(b) The total and flow energies of the steam per unit mass

(c) The rate at which energy leaves the cooker by steam

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5.5 Energy Analysis of Steady-Flow Systems

Many engineering systems such as power plants operate under steady conditions.

Under steady-flow conditions, the mass and

energy contents of a control volume remain

constant.

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Under steady-flow conditions, the fluid

properties at an inlet or exit remain constant

(do not change with time).

Mass and Energy balances for a

steady-flow process

A water heater in

steady operation.

Mass balance

Energy balance

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Rate of all work

done, excluding

Wflow and Wb

∆pe (m2/s2 or J/kg)

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Rate of heat

transfer

∆h (kJ/kg) =

∆u + P∆v∆ke (m2/s2 or J/kg)

Energy balance relations with sign conventions

(i.e., heat input and work output are positive)

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Under steady operation, shaft work and electrical

work are the only forms of work a simple

compressible system may involve.

when kinetic and potential energy

changes are negligibleSome energy unit equivalents

STEADY-FLOW ENGINEERING DEVICES

Example: Turbines

Compressors

Heat exchangers

Pumps

Conveniently analyzed as

steady-flow devices

NOZZLES & TURBINES &

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NOZZLES &

DIFFUSERS

TURBINES &

COMPRESSORS

THROTTLING

VALVES

NOZZLES &

DIFFUSERS

• Nozzles and diffusers are commonly

utilized in jet engines, rockets,

spacecraft, and even garden hoses.

• 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.

• The cross-sectional area of a nozzle

decreases in the flow direction for

subsonic flows and increases for

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Nozzles and diffusers are shaped so

that they cause large changes in

fluid velocities and thus kinetic

energies.

subsonic flows and increases for

supersonic flows. The reverse is true for

diffusers.

Energy balance for a nozzle or diffuser

Example 5.4

Air at 10°C and 80 kPa enters the diffuser of a

jet engine steadily with a velocity of 200 m/s.

The inlet area of the diffuser is 0.4 m2. The air

leaves the diffuser with a velocity that is very

small compared with the inlet velocity.

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small compared with the inlet velocity.

Determine:

(a) The mass flowrate of the air

(b) The temperature of the air leaving the

diffuser

Example 5.5

Steam at 1.8 MPa and 400°C steadily enters a

nozzle whose inlet area is 0.02 m2. The mass

flowrate of steam through the nozzle is 5 kg/s.

Steam leaves the nozzle at 1.4 MPa with a velocity

of 275 m/s. Heat losses from the nozzle per unit

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of 275 m/s. Heat losses from the nozzle per unit

mass of the steam are estimated to be 2.8 kJ/kg.

Determine:

(a) The inlet velocity

(b) The exit temperature of the steam

TURBINES &

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 and fans, are

devices used to increase the pressure of a

fluid. Work is supplied to these devices from

an external source through a rotating shaft.

• A fan increases the pressure of a gas slightly

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• A fan increases the pressure of a gas slightly

and is mainly used to mobilize a gas.

• 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.

Energy balance for the compressor

Example 5.6

Air at 100 kPa and 280 K is compressed

steadily to 600 kPa and 400 K. The mass

flowrate of the air is 0.02 kg/s and a heat loss

of 16 kJ/kg occurs during the process.

Assuming the changes in kinetic and potential

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Assuming the changes in kinetic and potential

energies are negligible, determine the

necessary power input to the compressor.

Example 5.7

The power output of an adiabatic

steam turbine is 5 MW. The inlet

and the exit conditions of the

steam are as indicated the figure.

(a) Compare the magnitudes of Δh,

Δke and Δpe

P1 = 2 MPa

T1 = 400°C

V1 = 50 m/s

z1 = 10 m

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Δke and Δpe

(b) Determine the work done per

unit mass of the steam flowing

through the turbine

(c) Calculate the mass flowrate of

the steam P2 = 15 MPa

x2 = 90%

V2 = 180 m/s

z2 = 6 m

THROTTLING

VALVES

Throttling valves are any kind of flow-restricting devices

that cause a significant pressure drop in the fluid.

What is the difference between a turbine and a throttling

valve?

The pressure drop in the fluid is often accompanied by a

large drop in temperature, and for that reason throttling

devices are commonly used in refrigeration and air-

conditioning applications.

Energy balance

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The temperature of an ideal gas does not

change during a throttling (h = constant)

process since h = h(T).

During a throttling process, the enthalpy of a

fluid remains constant. But internal and flow

energies may be converted to each other.

Example 5.8

R-134a enters the capillary tube of a

refrigerator as saturated liquid at

0.8 MPa and is throttled to a pressure of 0.12

MPa. Determine the quality of the refrigerant

at the final state and the temperature drop

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at the final state and the temperature drop

during this process.