Download - Chapter 3-First Law of Thermodynamics
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
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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:
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with,
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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)
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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
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outleteach for
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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
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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
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2
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2
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:balance Mass
VhhV
Vhm
Vhm
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mmmmm
outin
outin
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
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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
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TT
dTTC
hh
hh
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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
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:balance Mass
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mmmmm
outin
outin
MIXER
1
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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.
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3
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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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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
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PvPvmW
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mmm
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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.