MMPS Lecture Notes 2

8/3/2019 MMPS Lecture Notes 2

1/29

Tubular Reactor ( Distributed system ).

- A Tubular reactor is like a tube.

- Reactant enter the reactor from end and products leave from the other end.

- As the reactants flows through the reactor they react to form products.

- It is assumed that the cocentration at any cross section is uniform and there is

no axial mixing.

- The concentration is therefore a function of x ( and possibly ) t.i. e. CA = CA(x,

t)

- Since in this case we are dealing with cross section, Lets make a balance in

terms volume flux, F, instead of of volume flow rates. F [ m3/m

2-s]

- Consider that the reactant contains only one component .e. A that undergoes

the 1st

order reaction.

AK

B

xAxA FCeFC )()( xA dFCx )(

Ux

.AccSourceOUTIN

xx = 0

Reactant

dx

Productoqout

Flow rate = qin

Volume Flux=Fin


2/29

)()()()( dxCt

dxKCdxFCx

FCaFCaAAAxAxA

?

Assuming a is constant and by dx we get 0)(

AAA

KCFCx

Ct

P. D. E.

in time and space

If the density is constant the F is constant i. e.

inqF then

0

AAA

KCCx

FCt

constant F.

Also if CA at any x does not change with time i. e. Steady state tubular reactor

then 0

A

Ct

and

x

dx

di. e. 0

AAKCC

dx

dF

Solution : Fx

K

AA eCC

steady state tubular reactor.

Note F and are related as

LF ; where L

xK

AA eCC


3/29

Mass transport (A general form of distributed models)

In the tubular reactor we have seen that reactant were flowing through the

system as a stream.

Why does air freshner spreads in a room even if there is no fan and flow in the

room?

- The physical phenomenon responsible is diffusion governed by Fick,s law of

diffusion, that states (simple words) Mass diffuses in a direction of decreasing

concentration.

- A 1D diffusion in x- direction for A is given as:

dx

dCDdxJ A

Dx

,

gradientConc.

Decrease

D: is called diffusivity [m2/s]

Jx, D : is the diffusive mass- flux of A crossing a unit area perpendicular to x-

coordinate. [ sm

g 2

]

This equation governs the transport of A with out any bulk fluid movement. e.

g . Air fresher.


4/29

What happens if there is an imposed fluid movement e .g .

The transport will be much faster it will be transport due to diffusion + transport

due to bulk movement.

We call the 2nd thing as Convection.

- Now consider a small control volume in space. Let Jx be the net flux entering

the surface are perpendicular to dx ( including diffusion and any other flux due

to fluid movement).

Spray

Fan

y

Jx

z

xdy

dx

dzJx+ Jx/ x dx


5/29

X- faces.

Net amount of A out of control volume in x- direction.

Out - In

zyx

x

x

x

x

xddd

JdydzJdx

x

JJdydz

][

Similarly for y- direction zyxy

dddy

J

And Z- direction zyxz

dddz

J

Net out flow (Efflux) of A out of control vol.

zyxzyx dddz

J

y

J

x

J)(

Net out flow per unit volume:

z

J

y

J

x

Jzyx

J.

We have not imposed any restriction on the nature of flux

J .

Indeed it is the sum of diffusive and convective flux i. e .

CD

JJJ

The diffusive flux is given by Fick,s law.

ADCDJ

The connective flux is due to fluid movement. So if

U is the velocity of fluid the

conrective flux is given


6/29

byAC

CUJ

There fore net efflux/ or net out flow is ).(AA

CDCU

Remember that this quantity is the net out flow of A from a unit volume per

unit time.

The overall balance of A on the unit volume is:

In - Out + source = Acc.

Or Out - In - source = - Acc.

Acc. + Out - IN = source.

rCDCUVt

CAA

A

).( nKCA ) = r

Where r = rate of reaction e . g. -KCAn

Another form is: rCDCUt

CAA

A

2).( for constant D.

Remember CA is defined amount/Vol. , if concentration is defined askg

gmA

then CA is replaced by Am the transparent equation then becomes:

rmDmUt

m

AA

A

).(

)(


7/29

Energy Balance.

Law of conservation of energy

For any closed system,

Rate of energy entering the system - Rate of energy leaving the system = Rate of

Accumulation of energy

Energy can change its forms and the Law of conservation is applicable to the

sum of all types of energy. e. g. in an electric motor:

Not energy is conserved

Therefore in its full form

Rate of Rate of

Rate of Acc.

{Thermal energy + Electrical energy {-----------

of{---------

+Kinetic energy +Potential energy - ------------- =

--------

+Chemical energy +------etc} -----------}

--------}

Mechanical EnergyElectrical EnergyThermalEnergy

Acc.


8/29

Entering the system Leaving the system

In practical systems however all types of energies are not significant. e. g.

a) In an electric motor no chemical energy change is observed. Moreover same

times thermal energy may also be negligible .

b) In an electric heater mechanical energy is negligible.

Fig-----------?

c) In a gas fired boiler most chemical energy is converted in to thermal energy

and mechanical energy of steam can be neglected.

d) If steam is flowing in a nozzle and achieves speed comparable to velocity of

sound then the kinetic energy achieved may be comparable to thermal energy

converted.

Steam carries save much energy


9/29

Conclusion:

One has to use his Engineering judgement as to decide which forms must be

included in the energy balance.

To develop engineering judgement - Practice + experience is required.

Rule: Usually systems that are designed for certain form of energies do not

waste their energy in other forms.

Review : (Various forms of energy and their governing physical

laws)

Thermal Energy

Temperature: Average K. E. of the molecules.

Heat capacity / Specific Heat.

1 Btu

Cp =0.11 Cp =1.0 Cp =0.25

t =10 C t =1 t 4 C

Sensible heat /Enthalpy (h):

The thermal energy of a sample of matter due to its temperature.

1 lb

Air at const.

pressureWater

1 lb1 lb

Steel


10/29

tmCph ; Absolute h is never found, only h is of importance.

Phases of matter.

Consider a solid at a temperature less than the melting point. e. g. ice at -10

C.

When phase changes - temp is constant - Energy is absorbed as latent heat.

- Latent heat of Fusion ( slh )

Melting Solids absorbs slh to become liquid or

Heat

Temp Inc.Ice -10 C

Sub cooledsolid

Ice 0Csolid atmeltingpoint

{ Addedheat

changesthe

temp.}

Heat

Temp const.

Ice water0C

meltingpoint

{Addedheat

increaseswater

constant}

Heat

Tempconst.Water 0C

(subcooledwater)

Freezing/

meltingpoint

{Addedheat willincreasetemp.}

Temp Inc.

Heat

Temp const.

Heat

Temp const.

Heat

Temp Inc.

Water at 100 CBoiling point{saturated

water}

Steam +water,100 C{Added heatwill increase

steam%}

Steam 100 C{Saturated

steam}

Steam{Saturated

Steam}


11/29

FreezingLiquid liberates slh to become solid.

- Latent heat of vaporisation ( lgh )

MeltingSolids absorbs lgh to become gas ( vapour ).

CondensationVapour liberator lgh to become liquid

(Steady state) Mixing ( Single phase) ( Shower)

Cp = 4100 J/Kg-C

(Assume constant)

What flow rates of the two waters (hot & cold) may be mixed to produce 0.1 kg/s

of water at 35

C

Let the flow rate of hot water = H kg/s

Let the flow rate of cold water = C kg/s

Water 80 C

Water 10 C

Water 35 C

0.1Kg/s


12/29

Analysis.

1. Steady state

2. No phase change

3. Only thermal energy is involved

Assumption.

1. Cp is constant.

2. No losses.

3. Tref=10 C

( Any other temp can be taken)

Variables.

1. H kg/s

2. C kg/

Energy Balance.

0)1035(**1.0})1080(*)1010(**{00

p

OUTIN

pp CHCCC

s

kgHH 5.270


13/29

Mass Balance.

H + C = 0.1

s

kgHC 1.0

Home work

a) Hot water flow rate fixed. Outlet fixed find C at a function of hot water

temp.

Steam Heater:

Steam at 350F

and psi is use to heat water from 40F to 180 F. How much steam

is required to heat 10 lb/s of water. Assume that condensate is not sub cooled .

Analysis:

1. Steam under goes sensible heat loss as well as Condensation.

2. Water only gives steam heat.

3. Ref. Temp =30 F steam tables make this reference.

4. Cp water = 1Btu /lb F.

Steam

Water 40 F

Condensate

Water 180 F


14/29

From steam tables.

1. Sat. temperature of steam at 40 psia =

F267

Condensate ( condensed water of steam ) is at

F2.267 & 40 psia

2. Specific enthalpy ( heat content) of incoming steam = 1211.7 Btu/lb

3. Specific enthalpy of condensate = 236.1 Btu/lb.

{ Note energy lost per lb of steam = 1211.7-236.1 this includes the sensible

heat losts from

F350 to 267,2

Fand hls }

Assumption:

1. Let the flow rate of steam = x lb/s.

2. No heat loss.

3. Cp for water Heat IN = Heat OUT

1211.7x * 10*1*(40-32) = 236.1x + 10* 1* *(180-32)

(1211.7 236.1)x = 10x(180-40)

x = 1.4 lb/s ( Approx. 1/7 of water flow rate)


15/29

TTmCt

Wp

(1000 )

)( TThA

Energy Balance ( Dynamic Model)

Consider the heating of a bucket of water using an electric unmersion heater of

1000 W. Find the water temperature as a function of time if the amount of water

in bucket is 25 kg and initial temp is T .Provided

a) There is not heat loss.

b) Heat is lost as a rate proportional to the temperature difference between the

bucket temp. and surrounding temp.


16/29

Analysis:

1. Non flow - Un steady state.

2. Electric energy converting to heat.

a) Assumptions:

1. No heat loss

2. The bucket is at a uniform temp. ( lumped model)

3. Cp is constant.

Variables. T, t .

Model:

atTTCBdt

dTmC

Ckg

JCkgm

TTmCdt

d

AccOUTIN

p

p

p

:..10001

4100;25

1)}({01000

.

t = 0.

.1000)(

1000

tTTmC

tmCtTmC

p

pp

b) Assumptions:

1. Rate of heat lost x )( TT


17/29

Rate of heat lost = )(

TThA

{

hA is a constant ; we shall see what is

h & A}

2. All other as before.


18/29

Modeling:

p

pp

p

p

mC

Ah

whenprpTdt

dT

or

mC

ATh

mC

ATh

dt

dT

dt

dTmCThAThA

TTmCdt

dTThA

AccOUTIN

;

1000

1000

))(()(1000

.

B. C. = T = T at t = 0

pmC

AThr

1000

(ii ) ].[ 4 CrdteeT h

where h = pdt

Q,s : a) Is this analysis valid for ?t

b) what may be the limit on T ?


19/29

Heat Transport/ Transfer

Conventionally the transport of heat can be through three modes:

1. Conduction: which is indeed diffusion.

2. Convection: Transport with the flow of a fluid.

3. Radiation: via electromagnetic infra-red radiation.

- Unique with heat. When compared with mass transport.

Brief Description of the three modes

1. Conduction:

Different of heat.

Fourier law of heat conduction:

Heat flux - Temperature Gradient

q

dt

dTa ( 1 - D case )


20/29

q

=dt

dTk { Unit area quantity}

{ Remember that mass flux of a specie was proportional to the gradient of mass

concentration.

Here heat flux depends on Temp gradient}

2. Convection: (Solid surface Fluid hot )

Consider heat transferring from a solid surface in to cold flowing fluid.

The surface temp = TS and Bulk fluid temp = T

What happens in between TTS & - usually very difficult to determine.

Newton laws of cooling

Heat flux from surface )( TTS

Or q

= )(

TThS { Unit area quantity }

Fluid ( cold) y

x

y

T

T

Ts ( Hot )


21/29

Where

h is the average heat transfer coeff. A complicated parameter that takes

care of all possible

parameters that can influence the Convection.

h is a function of surface conditions (including geometry , roughness ) and

flow condition ( flow

velocity , fluid properties etc.)

h is usually not constant on the whole solid surface and the local quantity , h x

,is called the local heat

transfer coeff.

h ( Btu / Ffth ..

Condensing steam 1000 - 2000

Boiling water 300 - 9000

Heating / cooling of water 50 - 1000

Heating / cooling of air 0.2 - 20

Radiation Heat Transfer.

Stefan - Boltzmann law


22/29

The energy ( heat radiation ) flux from a black body is proportional to the 4th

power of absolute temp i. e .

q T4 ( for black body )

q

= T4 ( is the Stefan Boltzmann constant 5.6703 * 10-842

km

w

For non - black body radiations.

q= 4T (where is called the emissionity of the surface, it is a material as

well as surface

property)

A hot body at Abs. Temp T , loses heat to its colder infinite surroundings at

Abs. Temp T as :

q

= (T4

- T

4

)

In principal the above laws must be considered for every heat transfer situation.

However the magnitude of radiation heat transfer is significant only for

high T (say more than about 50 C )

Comparison Radiation Vs. Convection.

q


23/29

T )( KT q= )(**

44

TT Natural

Convection

Air

h 10

Forced

convection

Water.

h 500 w/m2-

K

Air

Forced

Convecti

on

H 100

31

3

293 279 400 100,000 4000

Ffth

Btu

Cm

wT

221

86.5

Concept of Thermal Resistance.


24/29

Consider a multi layered heat transfer situation as shown above.

The heat flow has an analogy with current flow i. e.

T Voltage

q Current

other terms Resistance.

So for the above figure q

is same.

11 TTT 21 TTT 32 TTT

)(23 TTT

T 1

T1

T

2T

3

( x) ( x)

x

radiationT 2

T 1 T1 T2T3 Radiation

T 2

Convection Conduction in

( x)1

Conduction in

( x)2


25/29

h

Rth

1

1

1 )(

K

xR

th

2

2 )(

K

xR

th

)(

1

5 tftRth

TTTTTTtf2233)(

q=

1

121 1;K

x

hRWhere

R

TTth

th

The thermal conductance is the reciprocal of thermal resistance.

.

1

th

thR

C ith

inetth CCU ,,

111

UU is also called over U heat transfer coeff.

Modeling of a thermometer.

Consider a mercury thermometer initially at temp. Ti placed in a fluid of

temperature. T at any given time t = 0. Find the temperature of the mercury as a

function of time.


26/29

Fig------?

Let area of bulb = A

Let overall heat transfer coef = U ( for Bulb )

Let the mass of mercury = m

Let the specific heat of mercury = Cp

)()(

))((0)(

TTUATmCdt

d

TTmC

dt

dTTUA

AccOUTIN

p

ip

First order difference Equation.

IC. : =tI at t = 0

)( TTCdt

dT

CtCTT

)ln(


27/29

pmC

UAC

Ct

eCTT

1

Ct

eCTT

As t TT

T = 0 T = TI

C

Analysis:

i- Lumped var.

ii- Unsteady

iii- Non Flow (Assumed)

Variable: T ,t .

Assumption:

i- mass of glass 0

ii- mercury out flow 0

iii- heat out flow 0 ( No condensation through thermometer stem)


28/29

iv- Cp constant.

Distributed Model.

IN - OUT =

dxdx

dT.

0)())()( Re.Re TTAUTdTTmCTTmC ffpp

0))(2( TTdxRUdtmCTmCTmC ppp

0)(2 TTURdx

dTmC

p

First order difference equation situated in form as the lumped transient . Only the

independent variable is x instead to T.

Consider steady state heat loss from a pipe carrying a heated fluid, as in the fig.

It is required to build a distributed model of fluid temp. i. e .T (x) is required.

T

U

T T + dtdA

x x

x = 0


29/29

Home work.

Work out this example for hot water entering a pipe of outer diameter

2 at 200 F with 3 ft/s velocity. The pipe is 100 ft long and outside temp. is

98 F . Assume U = 10 Btu/h-ft2- F & ignore radiation heat transfer.

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MMPS Lecture Notes 2