Chapter 5 - Energy Analysis for Open Systems Compatibility Mode

7/30/2019 Chapter 5 - Energy Analysis for Open Systems Compatibility Mode

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Energy Analysis of Control Volumes Chapter 5

Conservation of Mass, Flow W ork and Energy of a Flowing Fluid, Energy Analysis of Steady-Flow Systems, Examples (Nozzles, Turbines, Compressors, Throttling Valves, Mixing Chambers, Heat Exchangers), Unsteady-

Flow Systems


2/35

Balance EquationsMost equipments in chemical, petroleum and related industries are designed to

operate fluids that run continuously through them. Therefore, it is importantto understand the fluid flow (mechanics and characteristics). This is donebased on the law of mass conservation, Newtons 2 nd law and first and secondlaw of thermodynamics

Objectives

1. Develop the general balance equation

Department of Chemical Engineering Thermodynamics ChE 340 5-2

2. Develop the conservation of mass principle3. Apply first law of thermodynamics to control volumes

4. Identify energy carried by flowing streams crossing the control surface

5. Solve energy balance problems for common steady-flow systems

6. Apply the energy balance to general unsteady-flow processes


3/35

Measure of flow

Mass flow rate.

m (kg/h)Molar flow rate

.n (kmol/h)

VelocityV (m/s)

These measures are interrelated, i.e.

..)( n MW m = &

Remember Ac is thearea of flowcVAV

=.

Amount of material flowing through a cross section per unit time

Volumetric flow rate

) / (m 3.

sV


Mass flow rate: cVAm =.

Molar flow rate:) /(

. MW VAn

c =

= A

cnc

avg dAV AV

1Average velocity:


4/35

Mass Balance for Open Systems

Control volume: Region of space identified for analysis of open system (C.V.).

Control surface: a surface which separate control volume from its surrounding.

Consider the C.V. shown in the drawing,

where fluid with the C.V. is thethermodynamic system for which mass andenergy balances are written

What is the relationshi between the enterin


and leaving mass as well as the rate of changeof mass within the C.V.?

{Net rate of mass flow into C.V.} = {Rate of mass change within the C.V.}

dt

dm

mmmCV

=+ 3.

2

.

1

. -

or 0in

.

out

. =+ iiCV mmdt dm


5/35

2

.

1

.mm =

Since, = Mass flow rate.

m = VA

Continuity equation

Steady state flow process: condition within the control volume do not change withtime C.V. contains constant mass of fluid

For single entrance and single exit system: = constant

0)()(inout

=+ iiCV VAVAdt dm

0)()(inout

= ii VAVA


222111 AV AV =or

VAV AV

m ===2

22

1

11.or

Read Examples 5-1

For single entrance and single exit involving incompressible fluid ( 1 = 2):

VA AV AV m === 2211.


6/35

Example: A 1.2 high, 0.9 m diameter cylindrical water tank

whose top is open to the atmosphere is initially filled withwater. Now the discharge plug near the bottom of the tank ispulled out, and a water jet whose diameter is 1.3 cm streamsout, as shown. The average velocity of the jet is approximatedas V = (2 gh )1/2 . Determine how long it takes for the water

level in the tank to drop to 0.6 m from the bottom.

dt dm

mm CV out in =..

Mass balance:

0.

=mwhere: VAm =.

= hV m ==


et

dt h Ad

gh A k jet )(

2 tan

=

Separating the variables

dt dh

Dgh D k jet 2tan2 42

4

=

= 6 9 4 s = 11.6 min

hg

dh

D

Ddt

jet

k

22

2tan= =

2

022

2tan

0

h

h jet

k

t

h

dh

g D

Ddt

g

hh

D D

t jet

k

2220

2

tan

=

8.9226.02.1

013.09.0

2

=


7/35

Flow Work and the Energy of a Flowing FluidFlow work or flow energy: work required to push the mass into or out of thecontrol volume when dealing with flowing system.

Consider a fluid element of volume V as shownin the drawing.

Fluid immediately upstream will force thisfluid element to enter the control volume;thus, it can be regarded as imaginary piston.


In such case, the force applied on the fluidelement by the imaginary piston is:

PAF =

PV PALFLW ===flow

To push the entire fluid element into the control volume, this force must actthrough a distance L . Thus the work done in pushing the fluid element across theboundary (i.e., the flow work) is:

(kJ)


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Pw =flow

Per unit mass, this work is:

(kJ/kg)

It is worth mentioning that the flow work relation is the same whether thefluid is pushed into or out of the control volume:


Since it is expressed in terms of properties, flow work is referred as flow energy , convected energy or transport energy .


9/35

Total Energy of a Flowing Fluid


(kJ/kg) But P + u = h For flowing fluid;

= h + ke + pe = h + V 2 /2 + gz (kJ/kg)

Note that the reason for defining enthalpy is to use it for flowing fluid andto include energy associated with pushing stream with the internal energy of the stream


10/35

Energy Transport by a MassIf is the total energy per unit mass

m is the total energy of a flowing fluid of mass m

Or if the fluid is flowing with a mass flow rate .m

Rate of energy flow with that stream is .

m

++== gz

V hmm E mass

2

Amount of energy Transport: (kJ)


++== gzV hmm E mass 2

2...

Rate of energy Transport: (kW)


11/35

Example: Steam is leaving a 4-L pressure cooker whose operating pressure is 150

kPa, as shown. It is observed that the amount of liquid in the cooker has decreasedby 0.6 L in 40 min after a steady operating conditions are established, and thecross-sectional area of the exit opening is 8 mm 2. Determine (a) the mass flow rateof the steam and exit velocity, (b) the total and flow energies of the steam per unitmass, and (c) the rate at which energy leaves the cooker by steam.

kg57.0. == mm

kPa f

liquid V m150@

=

=

L100m1

k / m0.001053L0.6 3

3 g= 0.57 kg

= 0.0142 kg/min = 2.37 10 -4 k /s


m n

= 173.9 kJ/kguhPe flow == 2.25191.2693 =

c

g

cg A

m

Am

V

..

==36

34-

m108) / m1594.1)(kg/s10(2.3

= kg

= 2693.1 kJ/kgh pekeh ++=

..

m E =

= 34.3 m/s

= 0.638 kJ/s (kW)) / kJ1.2693)(kg/s10(2.3 4- kg=


12/35

Energy Analysis for Steady-Flow Systems

Previously, we presented the 1 st low for closed systems.Now we apply the 1 st law for open systems represented by control volume

{Rate of change of internal energy within the C.V.}={Net energy transported by the flowing streams} +Heat + Work

Inspecting the C.V. shown in thedrawing, the balance equation,in this case, would be:


What are the types of energy carried by theflowing streams? IE, KE & PE = ( u +v 2 /2 + gZ ) Rate of energy transport by each stream

.2 )2 / ( m ZgV u ++=

Energy is also transported across the C.V. as heat and work

Taking into consideration the P work of the streams, then the energy balanceequation can be written as:


13/35

..

out

.2

in

.2 )2 / ()2 / (

)(W Qm ZgV uPm ZgV uP

dt

mud CV +++++++= -

For most applications, KE = PE 0

Or

..

in

.2

out

.2 )2 / ()2 / (

)(W Qm ZgV hm ZgV h

dt mud CV =+++++ -

..

in

.

out

.)()(

)(W Qmhmh

dt mud CV =+ -


Steady state flow processes: Implies no change of mass ( m = constat), &No change of conditions within C.V. u = constant

For single entrance and exit to C.V., it can be shown that

wq Z gV

h =++ 2

2(kJ/kg)

(kJ/s)..

in

.2

out

.2 )2 / ()2 / ( W Qm ZgV hm ZgV h =++++ - Remember: this is

..

out in E E =


14/35

Terms appearing in energy balance equations.

Q = rate of heat transfer between the control volume and its surrounding.Q when control volume loosing heat

0.

=Q when control volume is well insulated (i.e. adiabatic).

W = Power For steady flow devices, the control volume is constant; thus no boundary work

Work required to push mass into and out of the control volume is taken care by


this power represents remaining work done per unit time

Many steady-flow devices, such as turbines, compressors and pumps, transmitpower through a haft, and W simply becomes the shaft power for those devices

If the control surface is crossed by electric wires(as in the case of an electric water heater), W represents the electrical work done per unit time.

Otherwise 0.

=W


15/35

12 hhh = Determined by reading enthalpy values at exit and inlet stream

For ideal gases, it can be approximated by h = c p ,avg (T 2 T 1) Note that (kg/s)(kJ/kg) kW

( ) 2 / 2122 V V ke = Unit of kinetic energy is m 2 /s2, which isequivalent to J/kg; which should be divided by1000 to et kJ/k .


The amount of kJ/kg due to ke is negligiblecompared to enthalpy change when thechange in velocities is small; but becomesimportant for large change in velocities.

)( 12 z zg pe = Similar argument can be given for pe term as that of ke due to similar unit andcontribution compared to other forms of energies in energy equation.


16/35

Flow through nozzles and diffusers- Used in jet engines, rockets, space crafts, garden houses

Nozzle: increases the velocity of a fluid at the expense of pressure

Diffuser: increases the pressure of a fluid by slowing it down

Nozzle flow Diffuser flow Nozzle V 1 V 2 >> V 1


For flow through nozzles or diffusers:

0.. == W Q and PE 0

But KE 0

Diffuser V 1 V 2


17/35

Air at 10oC and 80 kPa enters the diffuser of a jet engine steadily with a velocityof 200 m/s. The inlet area of the diffuser is 0.4 m 2. The air leaves the diffuser with

a velocity that is smaller compared with the inlet velocity. Determine (a) the massflow rate of the air and (b) the temperature of the air leaving the diffuser

P 1 = 80 kPaT 1 = 10 oCV = 200 m/s

T 2 = ?Airm = ?

Example: Deceleration of Air in a Diffuser

kPa80)K283)(. / kPa.m287.0( 3 K kg=

P RT 1

= Specific volume using ideal-gas law:

= 1.015 m 3 /kg


0..

== W Q

1

11.

V m =

kg / m.0151)m4.0)(m/s002(

3

2= = 78.8 kg/s

Energy balance: 02

2=+ V h

T 2 = 303 KFrom Table A-17:

From Table A-17: kJ/kg14.283K283@1 == hh2

21

22

12V V

hh

=

=

22

2

2 / m1000kJ/kg1

2m/s)(200-0

-kJ/kg14.283s

h = 303.14 kJ/kg


18/35

Steam at 1.8 MPa and 400oC steadily enters a nozzle whose inlet area is 0.02 m

2.

The mass flow rate of steam through the nozzle is 5 kg/s. Steam leaves the nozzleat 1.4 MPa with a velocity of 275 m/s. Heat losses from the nozzle per unit massof the steam are estimated to be 2.8 kJ/kg. Determine (a) the inlet velocity and (b)the exit temperature of the stream.

P 1 = 1.8 kPaT 1 = 400 oCA1 = 0.02 m 2

P 2 = 1.4 MPaV 2 = 275 m/s

steamm = 5 kg/s

q = 2.8 kJ/kg

Example: Acceleration of Steam in a Nozzle

C400

MP8.1o

1

1

=

=

T

P

. 2

SteamTable: kJ/kg3251.6

/kgm16849.0

1

31

=

=

h


0.

== peW

1

11

m=

kg / m.168490

.

kg/s5 31=

V 1 = 42.1 m/s

Energy balance: qV

h =+2

2

T 2 = 378.6 oC

2

21

22

12V V

qhh

+=

=

22

22

2 / m1000

kJ/kg1

2

m/s)(241.1m/s)(275

kJ/kg)8.26.3251( sh = 3211.9 kJ/kg

Then from Table A-6: kJ/kg3211.9

MP4.1

2

2

=

=

h

P


19/35

Throttling ProcessThis is flow through a restriction, e.g. orifice, partially closed valve orporous plug, without any change in kinetic and potential energy.The main result of this process is a pressure drop in the fluid.Throttling produces no shaft work and is an adiabatic process:

Therefore, 1 st law reduces to: 0=h

12 hh =or Constant enthalpy process

For ideal gas: h = h (T ) only T 2 = T 1


For real gas: reduction in pressure at constant enthalpy results in T

Example: Throttling of steam

P 1 = 1000 kPa

T 1 = 300oC

Atmospheric

pressure

Determined the exit T ?

kJ/kg1.305212 == hh According to the steam table; @ P 2=101.325 kPa & h 2 = 3052.1 T 2 = 288.8 oC


20/35

Example : Throttling of wet steamP 1 = 1000 kPa

x=0.96 Atmospheric

pressure Determined the exit T ?

kJ/kg2695.71)0.96(1933.762.612 =+=+== fg xhhhh f

According to the steam table; @ P 2=101.325 kPa & h 2 = 2695.7 Superheated & T 2 = 109.8 oC

Example: Throttling of saturated liquid water some of the liquid


vaporizes due to pressure reductionP 1 = 1000 kPa

Saturated liquid Atmospheric

pressure Determined the quality of the

exit stream?

kJ/kg762.6kPa100012 === =P f hhh

kPa253.1012kPa532.1012 == += P fgP f h xhh

9.22561.4196.762 2 x+= x = 0.152


21/35

Turbine (Expanders)Expansion of gas through a nozzle:

WORKenergyKineticenergyInternal shafttoattachedblades)V&(Pnozzlea

Turbine (or Expander): a device composed of alternate set of nozzles and rotatingblades through which vapor or gas flows in a steady-state expansion process

whose overall effect is the efficient conversion of the internal energy of ahigh-pressure stream into shaft work.

Turbine : when steam is the operating fluid (Steam power plant)


xpan er : w en g -pressure gas rom a c em ca or pe roc em caplant is used as the operating fluid

Turbine sW .

1

2

Energy balance : Inlet and exit pipes aredesigned to make the velocities similar

PE = 0; Assume adiabatic

Q = 0 For steady-state flow process:

)( 12...

hhmhmW out ==


22/35

Compression ProcessesExpansion pressure reduction in a flowing fluid (Work produced)Compression Pressure augmentation (Work required)

Examples: Compressors, pump, fan, blowers and vacuum pump.Application: Transport of fluids, fluidization of particulate solids,

bringing fluid to certain desired pressure for reaction or processes, etc.Compressors: for gasesAccomplished with rotating blades, like turbine operating in reverse

direction or cylinder with reciprocating piston.


Compressor sW

.

1

2

Energy balance : Inlet and exit pipes aredesigned to make the velocities similar PE = 0; Assume adiabatic Q = 0

For steady-state flow process:

)( 12...

hhmhmW in ==


23/35

Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The massflow rate of air is 0.02 kg/s, and heat loss of 16 kJ/kg occurs during theprocess. Assuming the changes in kinetic and potential energies are negligible,determine the necessary power input to the compressor

Example: Compressing Air by Compressor

dt dE E E systemout in / .. =

massandwork,heat,bynsferenergy traNet

4 4 4 4 84 4 4 4 76

energyetc...potential,kinetic,internal,of Change

4 4 4 4 84 4 4 4 76

0

.. ....

Compressor sW

.

1

P 1 = 100 kPa

T 1 = 280 K

m = 0.02 kg/s


out in E E = 21 hmQhmW out in

+=+

= 2.74 kW

kJ/kg)13.28098.400)(kg20.0(kJ/kg)16)(kg20.0(.

+= inW

( ) 12 hhmqmW .

out

.

in

.+=

kJ/kg13.802280@1 == K hhkJ/kg98.400400@2 == K hh

2

P 2 = 600 kPaT 2 = 400 K

q = 16 kJ/kg


24/35

The power output of an adiabatic turbine is 5 MW, and the inlet and the exitconditions of the steam are as indicated in the drawing.(a) Compare the magnitudes of h , KE, and PE(b) Determine the work done per unit mass of the steam flowing through the

turbine(c) Calculate the mass flow rate of the steam

Example: Power Generation by steam turbine

P 1 = 2 MPaT 1 = 400

o CV 1 = 50 m/s

MP21 =P

Inlet:


Turbine out W .

2

1

= 5 MW

P 2 = 15 MPa

x 2 = 0.9V 2 = 180 m/s

z 2 = 6 m

Q = 0C400

o1

=T

fg f xhhh += 2

= 2361.01 kJ/kg

.1

Turbine exit is saturated liquid vapormixture at 15 MPa:

kJ/kg)4.324801.2361(12 == hhhThen,

kJ/kg)]3.2372)(9.0(94.22[ +=

= - 887.39 kJ/kg


25/35

=

22

22

/ m1000kJ/kg1

2m/s)50(m/s)180(

s

+++=

++

2

22

2

..

1

21

1

.

22gz

V hmW gz

V hm out

2

21

22 V V ke

= = - 14.95 kJ/kg

=

222

/ m1000kJ/kg1

m]10)-[(6)m/s81.9(s

( )12 z zg pe = = - 0.04 kJ/kg

0Since.

=Q

..

out in E E =

(b) Energy balance at steady state:

22 V V


kJ/kg]04.095.1439.887[ += w

v ng y e mass ow ra e: ++= 12122

z zgwout

kJ/kg72.488kJ/s0005=

wW

m out .

.=

[ ] pekeh ++=

= 87.48 kJ/kg

(c) Mass flow rate for 5 MW power output:

= 5.73 kg/s


26/35

Mixing ChambersMixing chambers: the section where the mixing process takes placeTake different shapes: T-elbow or Y-elbow

Conservation of mass principles over all streams: 0in

.

out

. = ii mm

Usually in mixing chambers: q = 0 (insulated)w = 0 (no work involved)

Kinetic and potential energies are negligible


Energy balance: 0)()(in

.

out

.

= mhmh -

Assuming steady-flow process through the chamber


27/35

Consider an ordinary shower where hot water at 60o

C is mixed with cold water at10oC. If it is desired that a steady stream of warm water at 45 oC be supplied,determine the ratio of the mass flow rates of the hot to cold water. Assume theheat losses from the mixing chamber to be negligible and the mixing to take placeat a pressure of 150 kPa.

Example: Mixing of a Hot and Cold Waters in a Shower

3

.

2

.

1

.

mmm =+Mass balance: 0in

.

out

. = ii mm

..

out in E E =Energy balance:


33

.

22

.

11

.

hmhmhm=+

0Since

..

====

pe

keW QCombining mass and energy balance:

32

.

1

.

22

.

11

.

)( hmmhmhm +=+

321 )1( h yh yh +=+ 2.

1

.

/ where mm y =2.

mDivide by :

Since T sat @ 150 kPa = 111.35 oC all streams are subcooledkJ/kg18.25160@1 = C ohhkJ/kg022.4210@1 = C ohhkJ/kg44.18845@1 = C ohh31

21

hhhh

y

= = 2.3344.18818.251022.4244.188

=


28/35

Heat ExchangersHeat Exchangers: devices where two moving fluid streams exchange heatwithout mixing

Simplest form: double-tube (tube-and-shell)

Conservation of mass:Under steady operation, the mass flow rate of each stream flowing through a heat exchanger remains constant


nergy a ance:

q depends on the selectionof control volume

w = no wor nvo ve = =


29/35

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enterswith a mass flow rate of 6 kg/min at 1 MPa and 70 oC and leaves at 35 oC. Thecooling water enters at 300 kPa and 15 oC and leaves at 25 oC. Neglecting anypressure drops, determine (a) the mass flow rate of the cooling water required and(b) the heat transfer from the refrigerant to water.

Example: Cooling of Refrigerant-134a by Water

wmmm.

2

.

1

.==

(a) Mass balance: ..

out in mm =For each fluid stream:

gmmm.

4

.

3

.==


Combining mass and energy balance:

)()( 34.

21

.

hhmhhm gw =

..

out in E E =Energy balance:44

.

22

.

33

.

11

.

hmhmhmhm +=+

0Since..

==== pe keW Q

Enthalpies


30/35

kJ/kg98.6215@1 = C f ohh kJ/kg83.10425@2 = C f ohh

Water exists as compressed liquid at both sides, thus:

Refrigerant enters as superheated vapor and leaves as compressed liquid, thus:

C70

MP1o

3

3

=

=

T

P kJ/kg303.853 =h C35

MP1o

4

4

=

=

T

P kJ/kg87.00135@4 = C f ohh

Substituting:

kJ/kg])85.30387.100([kg/min)6(kJ/kg)83.104983.62(.

=wm

.


.w

(b) Choose volume occupied by water as the control volume:

..

out in E E = 2.

1

.

,

.

hmhmQ wwinw =+

Rearranging and substituting:

kJ/kg])982.6283.104([kg/min)(29.1 =)( 12.

,

.

hhmQ winw=

kJ/min1218,.

=inwQ

Note that same result can be obtained if we choose refrigerant as control volume


31/35

Energy Analysis of Unsteady-Flow ProcessesSteady state: no change occur within the control volume

Exam les: Char in of ri id vessels from

Unsteady flow, or transient flow: processes whichinvolve changes within the control volume with time

0=dt

dmCV

- Normally starts and ends over finite time period

0)( =

dt mud CV &


supply lines

0in

.

out

.

=+ iiCV

mmdt dm

..

in

.2

out

.2 )2 / ()2 / (

)(W Qm ZgV hm ZgV h

dt mud CV = ++ +++ -

- Discharging a fluid from a pressurized vessel

Requires application of both unsteady state material and energy balances:


32/35

12 mmmmm systemout in ==

W Qm ZgV hm ZgV humum =+++++ in

2

out

2system1122 )2 / ()2 / ()( -

Multiplying by dt and integrating between state 1 and state 2 of control volumenoting that:

mdt m = . Qdt Q = . W dt W = .


systemout in E E E =

;, out ininnet QQQQ ==and that: inout out net W W W W == ,

When ke and pe changes are negligible:W Qhmmhumum =+

inoutsystem1122 )( -


33/35

Another difference between steady and unsteady

flow system:Steady flow fixed in space, size and shapeUnsteady-flow fixed in space, but might

involve moving boundary

Characterized by uniform-flow : Fluid properties at inlet and exit steams do


- A uniform flow may involveelectrical, shaft, and boundary work allat once


34/35

Example: Filling an evacuated tank with an ideal gas. m i T i

T 2

Initially m 1 = 0. Find a relation between T i and T 2.Energy balance gives:

Mass balance gives:

m 2u 2 m 1u 1 h i(m 2 m 1) = 0

= 0= 0

0)( system1122 = iihmumum

W Qhmmhumum =+ inout

system1122 )( -

12 mmmmm systemout in == 12.

mmmm systemin ==Combining:


2 i

Using defining relation for h i : u i + P i i = u 2

For ideal gas: u i + RT i = u 2or: u 2 u i = RT i C V (T 2 T i ) = RT i

12 +=V i C

RT T Solving for T 2 / T i : =

T 2 = T i Mass balance gives:


35/35

A rigid, insulated tank that is initially evacuated is connected through a valve to asupply line that carries steam at 1 MPa and 300 oC. Now the valve is opened, andsteam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, atwhich point the valve is closed. Determine the final temperature of the steam inthe tank.

Example: Charging of a Rigid Tank by Steam

Mass balance gives:12 mmmmm systemout in == 2mmm systemin ==

systemout in E E E =Energy balance:


22umhm ii=

Combining with mass balance: 2uhi =

C300

MP1o=

=

i

i

T

P kJ/kg3051.6=ih

C3051.6

MP1 @

o2

2

=

=

u

P C56.14 o2 =T

Since @

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Chapter 5 - Energy Analysis for Open Systems Compatibility Mode

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