CLB 20703

Chemical Engineering

Thermodynamics

Chapter 3:

The First Law of Thermodynamics

Objective of Chapter 3

To discuss ideas about energy for

engineering analysis and develop

equations for applying the principle of

the First Law of Thermodynamics on

conservation of energy in open and

closed systems

Outline

Introduction to First Law of

Thermodynamics.

Energy balance for closed system

Mass and Energy balance for open system

(Steady Flow)

3.1 INTRODUCTION

First Law of Thermodynamics (also known as conservation of energy principle) states that energy can be neither created nor destroyed during a process but can only change forms.

Conservation of Energy Principle the net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process:

This relation is applicable to any kind of system undergoing any kind of process.

systemoutin

system theofenergy

totalin the changeNet

system theleaving

energy Total

system theentering

energy Total

EEE

Energy Balance for any system

system a of changeEnergy

system

system a ofnsfer Energy tra

outin EEE

• Energy transfer are

recognized at the system

boundary as they cross it

represent the energy gained

or lost by a system during a

process.

• Energy can be transferred to

or from a system in three

forms: heat, work and mass

flow.

• Only two forms of energy

transfer associated with a

closed system are heat

transfer and work.

• Energy change of a system during a

process = difference of the energy of the

system at the beginning and at the end of

the process:

Esystem=Efinal-Einitial =E2-E1

• Energy is a property the value of a

property does not change unless the state

of the system changes the energy change

of a system is zero if the state of the system

does not change during the process.

• The change in the total energy of a system

during a process is the sum of the changes

in its internal, kinetic, and potential

energies:

E system= U + KE + PE

3.2 ENERGY BALANCE FOR CLOSED SYSTEMS

• For closed systems, only Q and

W involved.

• By using the sign convention of

heat and work, heat to be

transferred into the system (heat

input) in the amount of Q and

work to be done by the system

(work output) in the amount of W:

Q-W=E

energies etc. internal,potetial, kinetic,in Change

system

mass &heat work,bynsfer energy traNet

outin EEE

• The change in the total energy =

sum of the changes in its

internal, kinetic and potential

energies:

12

2

1

2

221

12

with,

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vvmKE

uumU

PEKEUE

otherb WWW

PEKEUWQ

with,

Overall 1st Law of thermodynamics for closed

systems

Energy Balance for closed systems

(closed tank, rigid tank, piston-cylinder device)

otherb WWWPEKEUWQ with,

CONSTANT VOLUME PROCESS / ISOCHORIC PROCESS (V=0)

RIGID TANK / PISTON-CYLINDER DEVICE

V=0 Wb=0Q=U

STATIONARY SYSTEMS (KE=PE=0)

otherb WWWUWQ with,

CONSTANT PRESSURE PROCESS / ISOBARIC PROCESS (P=0)

PISTON-CYLINDER DEVICE (W=Wb)

HQ

VPUQ

WUQ

UWQ

b

b

definitionEntalphy

Energy change for a cycle

For a closed system undergoing a cycle, the initial and final states

are identical:

Esystem = E2 - E1 = 0.

Then the energy balance for a cycle simplifies to Ein - Eout = 0 or

Ein = Eout.

A closed system does not involve any mass flow across its

boundaries, so the energy balance for a cycle can be expressed in

terms of heat and work interactions the net work output during a

cycle is equal to net heat input:

Q Wnet net

Example 4.1

A rigid tank contains a hot fluid that is cooled while being stirred by a

paddle wheel. Initially, the internal energy of the fluid is 800kJ. During

the cooling process, the fluid loses 500J of heat, and the paddle wheel

does 100kJ of work on the fluid. Determine the final internal energy of

the fluid. Neglect the energy stored in the paddle wheel.

Example 4.2

A 0.5 m3 rigid tank contains refrigerant-134a initially at 160 kPa and

40% quality. Heat is now transfer to the refrigerant until the final

pressure reaches 700kPa. Determine

a) The mass of the refrigerant in tank

b) The amount of heat transferred

c) Show on the process on PV diagram with respect to saturation line

Example 4.3

A piston cylinder device initially contains steam at 200 kPa, 200 OC and

0.5 m3. at this state, a linear spring (F α x) is touching the piston but

exerts no force on it. Heat is now slowly transferred to the steam,

causing the pressure and the volume to rise to 500 kPa and 0.6 m3,

respectively. Show the process on a PV diagram with respect to

saturation line and determine

a) The final temperature

b) The work done by the steam

c) The total heat transferred

3.3 MASS AND ENERGY BALANCE FOR

OPEN SYSTEM

Open systems are characterized by flowing streams, there are 4 common measures of flow:

Velocity, u

Volumetric flowrate, q = uA

Molar flowrate,

Mass flowrate,

where M = molar mass/molecular weight

A = cross-sectional area

ρ = specific or molar density

uAn uAnMm

An open system or a control volume (CV) = a selected region in spaceand usually encloses a device that involves mass flow in and out of thesystem such as a compressor, turbine or nozzle.

Besides heat transfer and work across the boundary, the mass andenergy content of a control volume can change when the mass flows inand out of the system.

To simplify the energy analysis of CV:

The system should be assumed undergoing steady-flow process, and

Conservation of Mass Principle for CV should be firstly defined

before the 1st Law of Thermodynamics can be applied to CV.

Energy Analysis Of Open System

Steady-flow Process A large number of engineering devices such as turbines, compressors,

and nozzles operate for long periods of time under the same conditions steady-flow devices.

Steady-flow process = a process during which a fluid flows through acontrol volume steadily the fluid properties within the control volumemay change with position but not with time.

Therefore, the volume V, mass m, and total energy content E of the CVremain constant during a steady flow process:

dm

dtm

dE

dtE

CVCV

CVCV

0

0

Conservation of Mass Principle for CV: The net mass transfer to orfrom a control volume during a time interval t is equal to the netchange (increase or decrease) in the total mass within the CVduring t:

For CV, mass and volume normally expressed in the rate forms –mass flow rate and volume flow rate. The mass and volume flowrates are related by:

Conservation of Mass Principle

skg or v

AVAVm

v

VVm av

av

outin

CVCVoutin

CVoutin

mm

dtdmmmm

mmm

ttt

0

during CVwithin

massin changeNet

during CV

leaving mass Total

during CV

entering mass Total

Also known as mass balance and applicable to any CV undergoing any kind of process

For control volumes undergoing steady-flow

process

out

outout

in

inin

out h

outout

in h

inin

out

outout

in

inin

outin

systemoutin

gzV

hmWQgzV

hmWQ

gzV

uPvmWQgzV

uPvmWQ

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EE

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outleteach for

2

inleteach for

2

22

mass and work heat,by outnsfer energy tranet of Rate

mass and work heat,by innsfer energy tranet of Rate

CV of energies etc potentialkinetic, internal,in change of Rate

mass and work heat,by CVacrossnsfer energy tranet of Rate

22

22

(kW)

Conservation of Energy Principle The conservation of energy principle (1st Law of Thermodynamics)

for control volumes has the similar definition with that of closed systems:

For steady-flow

process,Ė=0

= energy per

unit mass

flowing in and

out of CV

EPEKHWQ

gzV

hmgzV

hmWQ

gzV

hmWQgzV

hmWQ

gzV

hmWQgzV

hmWQ

ii

ee

netnet

ee

outoutii

inin

out

outout

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outleteach for

2

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2

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outleteach for

2

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inleteach for

2

i

outleteach for

2

inleteach for

2

22

22

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outinnet QQQ

inoutnet WWW Overall 1st Law of

thermodynamics for

CV undergoing

steady-flow process

Energy Balance for control volumes

Steady-flow Engineering Devices

Nozzles and Diffusers

2

1212

2

22

2

11

21

2

22

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:balance Mass

VhhV

Vhm

Vhm

EE

mmmmm

outin

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Nozzle = device that increases the velocity of a fluid at the

expense of pressureDiffuser = device that

increases the pressure of a fluid by slowing it down

The rate of heat transfer between the fluid flowing through a nozzle or a diffuser andthe surroundings is usually very small (Q0), involve no work (W=0), any change inpotential energy is negligible (pe0) but involve very high velocities the kineticenergy changes must be taken into account (ke0).

Nozzle and diffuser are commonly utilized in jet engines, rockets, spacecraft and

even garden hoses.

Turbines

In steam, gas, or hydroelectric power plants, the device that drives

the electric generator TURBINE. As the fluid passes through the

turbine, work is done against the blades, which are attached to the

shaft shaft rotates, and the turbine produces work produce

power output.

By ignoring the change in KE and PE energies (ke=pe=0) through

an adiabatic turbine (Q=0) with a single stream (one inlet-one outlet)

that undergoes a steady flow process:

21

21

21

:balanceEnergy

:balance Mass

hhmW

Whmhm

EE

mmmmm

outin

outin

W

2

1

ControlSurface

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 involve work inputs require power input.

The differences between the three devices:

A fan increases the pressure of a gas slightly and is mainly used to

mobilize a gas at low pressure.

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.

Heat transfer, kinetic and potential energies are also negligible for

compressors (Q=0, pe=0, ke=0):

Compressors and Fans

12

21

21

:balanceEnergy

:balance Mass

hhmW

hmWhm

EE

mmmmm

outin

outin

2

1 W

ControlSurface

Throttling Valves Throttling valves are any kind of flow-restricting devices that

cause a significant pressure drop in fluid.

Unlike turbines, they produce a pressure drop without involvingany work but often accompanied by a large drop in temperature devices are commonly used in refrigeration and air-conditioning applications.

Throttling valves are usually small devices, and the flow throughthem may be assumed to be adiabatic (q=0), no work done (w=0),the change in potential energy is very small (pe=0), the increasein kinetic energy is insignificant (ke=0):

ei

eeii

ei

hh

hmhm

mm

:balanceEnergy

:balance Mass

Enthalpy values at the inlet and exit of a throttling

valve are the same throttling process =

isenthalpic process

Throttling process of an ideal gas

The temperature of an ideal gas remains constant during a throttling process since h=h(T)

ie

T

TP

ie

ie

TT

dTTC

hh

hh

e

i

0)(

0

In engineering applications, mixing two streams of fluids isnot a rare occurrence. The section where the mixing processtakes place mixing chamber.

The mixing chamber does not have to be a distinct“chamber.” An ordinary T-elbow or a Y-elbow in a shower =mixing chamber for the cold and hot water streams.

The conservation of mass principle for a mixing chamberrequires that the sum of the incoming mass flow rates equalthe mass flow rate of the outgoing mixture.

Mixing chambers are usually well insulated (q=0), usually donot involve any kind of work (w=0), the kinetic and potentialenergies of the fluid streams are usually negligible (ke=0,pe=0):

Mixing Chambers

332211

321

:balanceEnergy

:balance Mass

hmhmhm

EE

mmmmm

outin

outin

MIXER

1

23

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3

2

Heat ExchangersHeat exchangers are devices where two moving fluid streams exchange

heat without mixing.

widely used in various industries, and they come in various designs.

The simplest form of a heat exchanger is a double-tube (also called

tube and-shell) heat exchanger.

Heat is transferred from the hot fluid to the cold one through the wall

separating them and the outer shell is usually well insulated to prevent

any heat loss to the surrounding medium.

Heat exchangers typically involve no work interactions (w=0) and

negligible kinetic and potential energy changes (ke=0, pe=0) for each

fluid stream.

Q=0

Heat transfer rate associated with heat exchangers depends on how the control

If only one of the fluids is selected asthe control volume, then heat willcross this boundary as it flows fromone fluid to the other and will not bezero the rate of heat transferbetween the two fluids.

12

3

4

12

3

4

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mmmmmm

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When the entire heat exchangeris selected as the control volume,Q becomes zero, since theboundary for this case lies justbeneath the insulation no heatcrosses the boundary.

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4321

or

or

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and :MB

hhmQhhmQ

hmQhmhmhmQ

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mmmmmm

outin

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The transport of liquids or gases in pipes and ducts is of greatimportance in many engineering applications.

Flow through a pipe or a duct usually satisfies the steady-flowconditions.

Sometimes heat transfer is desirable and is the sole purpose ofthe flow. Water flow through the pipes in the furnace of a powerplant, the flow of refrigerant in a freezer, and the flow in heatexchangers are some examples of this case.

At other times, heat transfer is undesirable, and the pipes orducts are insulated to prevent any heat loss or gain, particularlywhen the temperature difference between the flowing fluid andthe surroundings is large. Heat transfer in this case isnegligible.

Pipes and Duct Flows

Liquid Pumps Work is required to pump a compressed liquid in an adiabatic (q=0)

and steady flow process.

For compressed liquid, the density and specific volumes are

constant (v2=v1=v) and the process of pumping compressed liquid

is isothermal (u=cvdT=0). By neglecting KE and PE:

Pump

Fluid inlet, 1

Fluid exit, 2

h

Liquid flow through a pump

W

12

12

12

0

12

0

12

0

2

1

2

212

2

2

221

2

11

2

22 :EB

:MB

1212

PPvmW

PvPvmW

PvPvuumW

zzgVV

hhmW

gzV

hmgzV

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mmm

u

pe

ke

PvPvuu

outin

Example 4.4

Steam enter a nozzle at 400 OC and 800 kPa with a velocity

of 10m/s, and leaves at 300 OC and 200 kPa while losing

heat at a rate of 25 kW. For an inlet area of 800 m2,

determine the velocity and the volume flow rate of the steam

at the nozzle exit.

Example 4.5

Steam enters the condenser of a

steam power plant at 20 kPa and a

quality of 95% with a mass flow rate of

20000 kg/hr. it is to be cooled by water

from a nearby river by circulating the

water through the tubes within the

condenser. To prevent thermal

pollution, the river water is not allowed

to experienced a temperature rise

above 10 OC. If the steam is to leave

the condenser as saturated liquid at 20

kPa, determine the mass flow rate of

the cooling water required.