Final Exam Solutions-1

8/3/2019 Final Exam Solutions-1

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NAME _______________________________________________________________________

CHEMICAL ENGINEERING 150A

Final Exam

Comments:This is a closed book, closed notes exam. Graphing calculators are allowed. The appendices

will NOT be collected for grading.

Problem Score

1 / 50

2 / 35

3 / 50

4 / 45

5 / 60

6 / 60

Total / 300


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Question 1 (50 points)

The reservoir system shown below has the following properties:

Q = 0.012 m3/s D = 0.1 m

! = 950 kg/m3

Patm = 101325 Pa

" = 0.0019 Pas g = 9.8 m/s

2

The liquid is incompressible and has a negligible vapor pressure. The globe valve is half open;

the 90o

elbow is square. The tanks are very large and the pipe is smooth. Note that the figure isNOT drawn to scale.

Reservoir system:

(a) Compute the power required for the pump.

(b) Determine the maximum distance the pump can be place from the first tank.

10 m

1200 m

100 m 50 m

20 m

Globe valve Pump

Tank #1

Tank #2

Patm

Patm


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Solution to Question 1

a) Use the engineering Bernoulli equation to size the pump:!"# $%

&'(""

) *+",!-# $%

&'(-"

) *+-./

012

34

35 ) 678

9 . :;

:; , :;=? ) :;AA>BC8 , D #$%&'(>"E>F>G>H?BCAI8J@=>=? ) D K# $%&'(L"M@L@>AA>BC8 The fluid is incompressible and exposed to atmospheric pressure:/ 0123435 , 1" . 1-2 , 1NAO . 1NAO2 , PThe tanks are very large:

$%&'(- , $%&'(" , P'QRS

The engineering Bernoulli equation between the two tanks reduces to the following:*+" , *+- ) 6789 . :;T 6789 , *U+" . +-V ) :;Calculate the mass flow rate and the velocity in the pipe:W , 2X , UYZP'[\RQ]VUP^PK#'Q]RSV , KK^_'[\RS

$%&'(= , X` , _XaG" , _UP^PK#'Q]RSVaUP^K'QV" , K^Z]'QRSThe Reynolds number in the pipe is:

bc , 2$%&'(=Gd , UYZP'[\RQ]VUK^Z]'QRSVUP^K'QVUP^PPKY'efgSV , hi_PP T jklmkncojThe friction factor can determined from the f-Re chart or calculated from either the Blasius or

von Krmn-Nikuradse equations. All results are roughly equivalent:

Graph: bc , hi_PP T F , P^PP_Z4000 < Re < 10

5(Blasius):

'F , P^PhY'bcp-Rq , P^PhY'Uhi_PPVp-Rq , P^PP_hZ4000 < Re (von Krmn-Nikuradse):' KrF , _^Pns\-tubcrFv. P^_ , KrF , _^P ns\-tuhi_PPrFv. P^_ T F , P^PP_hi The flow experiences four fittings. First, there is a sudden contraction out of tank #1:


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`" w -T `"`- x PK . yU`"R`-VK . U`"R`-V" , z

yK^#{

"

TK. PK . P , z

yK^#{

"

T y , K^#

M@- , |#y .`"`- . K}" , | #K^# .P.K }" , P^__Second, there is a half-open globe valve: M@" , Y^ZThird, there is a square 90

oelbow: M@~ , K^]

Fourth, there is a sudden expansion into tank #2:

`- w "T `-`" x PM@q , |K . `-`"}" , UK . PV" , K

Substitute the values into the viscous loss term::; , :;=? ) :;AA>BC8 , D #$%&'(>"E>F>

G>

H?BCAI8J@=>=?

) D K

#$%&'(L"M@L

@>AA>BC8

:;=? , D #$%&'(>"E>F>G>H?BCAI8J@=>=? ,#UK^Z]'QRSV"UK#PP'Q)ZP'QVUP^PP_hZVUP^K'QV , #h^#'Q#RS#

:;AA>BC8 , D K# $%&'(L"M@L@>AA>BC8 , K# UK^Z]'QRSV"UP ^__)Y^Z)K^])KV , K_^]'Q#RS#

:;

, :;=?

) :;AA>BC8

, #h'Q#

RS#

)K_^]'Q#

RS#

, #YK^Z'Q#

RS#

The pump size is:6789 , *U+" . +-V ) :; , UY^'QRS#VUhP'Q.KP'QV )#YK^Z'Q#RS# , hY^Z'Q#RS#789 , W6789 , UKK^_'[\RSVUhY^Z'Q#RS#V , KPP#i' , KP'[ , K]^_'(b) Use the engineering Bernoulli equation to determine the maximum distance the pump can be

place from the first tank.


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!=# $%&'(=" ) *+= , !-# $%&'(-" ) *+- ./ 012335 ) 6789 . :;:; , :;=? ) :;AA>BC8 , D #$%&'(>"E>F>G>H?BCAI8J@=>=?

) D K#$%&'(L"M@L

@>AA>BC8

Applying the same assumptions as before with # = 1, the engineering Bernoulli equation between

the first tanks and the pump reduces to the following:K# $%&'(=" , *+- . 1= . 1NAO2 . :;Since the vapor pressure is negligible Set the pressure at the pump equal to zero to find the

maximum distance the pump can be placed from the first tank:K# $%

&'

(="

, *+-

.1NAO2 . :

;

, *+-

)1NAO2 . :

;=?

. :;AA>BC8

Assume the pump is placed after the globe valve, but before the 90o

elbow. The flow experience

two fittings. First, there is a sudden contraction out of tank#1:M@- , P^__Second, there is a half-open globe valve: M@" , Y^ZSubstitute the values into the fittings viscous loss term:

:;AA>BC8 , DK# $%&'(L

"

M@L@>AA>BC8 ,K# UK^Z]'QRSV

"

UP^__)Y^ZV , KK^i'Q#

RS#

Solve for the piping viscous loss term::;=? , *+- . K# $%&'(=" ) 1NAO2 . :;AA>BC8:;=? , UY^'QRS#VUKP'QV . K# UK^Z]'QRSV" ) KPK]#Z'efYZP'[\RQ] .KK^i'Q#RS# , KYK^Y'Q#RS#

From the piping viscous loss term, determine the length:

:;=? , D #$%&'(>"

E>F>G>H?BCAI8J@=>=? , #UK^Z]'QRSV"EUP^PP_hZVUP^K'QV , KYK^Y'Q#RS# E , i]'Q

The maximum distance is after the globe valve (100 m) and before the 90o

elbow; therefore theprevious assumption is correct and the answer is valid.


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An incompressible, Newtonian fluid steadily flows past an eccentrically placed cylinder asshown in the figure below. Far from the cylinder, the fluid velocity field is fully developed and

described by a parabolic flow:

UV , . 'U . V"The cylinder is very long with a radius a. The cylinder rotates about its axis with a constantangular velocity $ in the direction of increasing !. For simplicity, wall effects are not explicitly

considered and the Reynolds number is assumed to be low.

Parabolic flow past an eccentrically placed rotating cylinder:

The goal of this problem is to setup, but NOT solve for, the velocity field of the fluid aroundthe

cylinder: List all assumptions in the problem, clearly expressing what the assumptions mean

mathematically. Postulate the form of the velocity field aroundthe cylinder in cylindrical

coordinates (r, !,z). Simplify the continuity and Navier-Stokes equations. Give boundary

conditions.

Conversion between Cartesian and cylindrical coordinates:

, sS , sS ) S o , So , . So ) sS , ,


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Governing Equations for Question 2:

Continuity Equation for an Incompressible Fluid in Cylindrical Coordinates

, K UV ) K ) , P

Navier-Stokes Equations for an Incompressible, Newtonian Fluid in Cylindrical Coordinates

r component:

2 ) ) . " ) , . ''

''''''''''''''''''''') K UV ) K" "

" . #"

) "

"

% component:

2 | ) ) ) ) } , .K ''''''''''''''''''''''') K UV ) K" "" ) #" ) "" z component:

2 | ) ) ) } , . ''''''''''''''''''''''')K | } ) K" "" ) "" Energy Equation for a Pure Incompressible, Newtonian Fluid with

Constant Thermal Conductivity in Cylindrical Coordinates

2= | ) ) ) } ,' '''''''''''''''''''''' , K | } ) K# ## ) "" ) 9 ) d''


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The following assumptions are used:

Newtonian => viscosity constant, no change in &

Incompressible => density constant, no change in !

Steady state => no time dependence, '/'t = 0Very long => no z dependence, '/'z = 0

Low Re => creeping flow, Re


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Simplify the r component of the Navier-Stokes equation:P , . ) K UV ) K" "" . #" Simplify the % component of the Navier-Stokes equation:

P , .K ) K UV ) K" "" ) #" Simplify the z component of the Navier-Stokes equation:P , .


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A sphere fixed in space has a diameter of 0.1 m. The sphere, initially at a uniform temperatureof 0C, is suddenly surrounded by water at 20C.

(a) Determine how long it will take for the center of the sphere to reach 5C if the water is

flowing with a velocity of 10 m/s and sphere is made of asbestos. At that moment, give thetemperature of the sphere at a radius of 0.04 m.

(b) Determine how long it will take for the center of the sphere to reach 5C if the water is stilland the sphere is made of copper. At that moment, give the temperature of the sphere at a radius

of 0.04 m.

Properties of the spheres:

Thermal conductivity: kasbestos = 2.07 W/(mK) kcopper= 401 W/(mK)Density: !asbestos = 0.35 gcm

-3 !copper= 8.94 gcm

-3

Specific heat capacity: Cp,asbestos = 0.84 kJ/(kgK) Cp,copper= 0.39 kJ/(kgK)

Properties of water:Thermal conductivity: kwater= 0.597 W/(mK)

Density: !water= 0.9982 gcm-3

Specific heat capacity: Cp,water= 4.182 kJ/(kgK)Viscosity: "water= 9.93 x 10

-4Pas

Buoyancy factors: *g/+2

= 2.035 x 109

m-3

K-1


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Continuity Equation for an Incompressible Fluid in Spherical Coordinates

, K" U"V ) KSo U SoV) KSo , P

Navier-Stokes Equations for an Incompressible, Newtonian Fluid in Spherical Coordinates

r component:

2 ) ) ) So . " ) " , . ''

')d K" "" U"V ) K" So |So } ) K" So"

""

% component:

2 ) ) ) So ) . " 'sj' , .K '' ')d K" |" } ) K" KSo U SoV ) K" So" "" ) #" . #'sj''" 'So

, component:

2 ) ) ) So ) )'sj' , . KSo '' ')d K" " ) K" KSo u Sov) K" So" "" ) #" So ) #'sj''" 'So

Energy Equation for a Pure Incompressible, Newtonian Fluid with

Constant Thermal Conductivity in Spherical Coordinates

2= | ) ) ) So } ,' '', K" |" } ) K" So |So } ) K" So" "" ) 9 ) d;''


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(a) Find h from the Nusselt number. Start by calculating the Prandtl number of water:

el ,! ,

d=


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, P^PY Read off the chart for a sphere:

, . . t ,

#P.#P.P , P^#

, Ki(b) Find h from the Nusselt number. The Prandtl number is the same as before. The Grashof

number is: l , *G~" , U#^P]Z''KPVUKh^ZVUP^KV , ]^Zi''KPNote that the -T is based on the liquid temperature and average surface temperature. Using the

Grashof number, calculate the Rayleigh number:

bf , lel , U]^Zi''KPVUi^YiV , #^_''KP

Use a correlation to find h from the Nusselt number:k , # ) P^_]'bf-Rq , k , #) P^_]U#^_''KPV-Rq , Zi^Pk , +G , +UP^K'QVP^ZYh'RUQV , Zi^P + , ]]_'RUQ"V

Check the Biot number: , +%`

, +Gi

, U]]_VUP^KViU_PKV

, P^PK_ w KSince the Biot number is less than one, use lump capacitance analysis:2%=


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An atomizer nozzle generates a fine spray of liquid and is commonly used to dispense paints andperfumes. To evaluate the performance of an atomizer nozzle, it is proposed to atomize a

nonvolatile liquid wax into a stream of cool air. The atomized wax particles are expected to

solidify in the air, from which they may later be collected and examined. The wax droplets leave

the atomizer at their freezing point To with the surrounding air at T# and

#. Assume that thereis no volume change in the solidification process and that the heat transfer coefficient h between

the drop and the air is known. In the solid phase, $ is the density, kis the thermal conductivity,and Cp is the heat capacity; these properties are constant. For this problem, determine an

analytical solution rather than using Biot number simplifications.

Freezing spherical drop:

(a) Solve the pseudo-steady state energy equation in the solid phase betweenRf andR todetermine Tas a function of position. This is effectively assumingRf moves very slowly in

time.

(b) Write an unsteady-state energy balance equating the heat liberation at r=Rf(t) resulting

from the freezing of the liquid to the heat flow q out the spherical surface at r =R. Estimate the

time tf required for a spherical drop of radiusR to completely freeze. Let -Hfbe the latent heatof fusion per unit mass.

#T#, h

$, k, Cp


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r component:

2 ) ) ) So . " ) " , . ''

')d K" "" U"V ) K" So |So } ) K" So"

""

% component:


, component:




2= | ) ) ) So } ,' '', K" |" } ) K" So |So } ) K" So" "" ) 9 ) d;''


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(a) The following assumptions are used:

Constant properties conductivity => no change in !, k, Cp

Pseudo steady state => no time dependence, '/'t = 0

Axisymmetric => no % or, dependence, '/'% = '/', = 0No generation => no heat generation in solid phase, q = 0

No velocity => no velocity in solid phase, vr= v% = v, = .v = 0

From assumptions, T is independent of t, %, and ,, such that T = T(r).

Simplify the energy equation:P' , K" |" } , ' " 00 |" 00}'Integrate the energy equation:

" 00 , - , .- ) "List the boundary conditions:

T= To @ r=Rfby known freezing point temperaturek dT/dr= h (TT#) @ r=R by heat flux at a solid-fluid interface

Apply the boundary conditions to the integrated energy equation:

J , . -@ ) "

" , J ) -@. 00 , +UUV . V

.-" , + |.- ) " . } , +.- ) J ) -@ . . -+" , - K@ . K ) J . - K . K@ . +" , J .


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- , J . K . K@ . +"

" , J ) -@ , J ) K@ J . K . K@ . +"Therefore, the temperature profile is:

, .K J . K . K@ . +") J )K@ J . K . K@ . +"

, J . K . K@ . +"

K@. K ) J

(b) Write an unsteady-state energy balance equating the heat liberation at r=Rf(t) resulting

from the freezing of the liquid to. Estimate the time tf required for a spherical drop of radiusR

to completely freeze. Let -Hfbe the latent heat of fusion per unit mass.

The heat liberation at r=Rf(t) results from the freezing of the liquid: , 2@ . 00 |_]a@~} , ._]a2@ 0@~0 , ._a2@@" 0@0 The heat flow q out the spherical surface at r =R is given by Fouriers law of conduction:

, .`00 , ._a# J . K . K@ . +" | K"}

, ._a J . K . K@ . +" , ._aUJ . VK . K@ . +"

The heat flow q out the spherical surface at r =R is also given by Newtons law of cooling:

, +`U . V , _a+# J . K . K@ . +" K@ . K ) J . , _a+#UJ . V K@ . KK . K@ . +" ) K , _a+

#UJ . V . +"K . K@ . +"


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, ._aUJ . VK . K@ . +" Equate the two relationships and integrate in time:

_a2@@" 0@0 , _aUJ . VK . K@ . +" 2@@" 0@0 , UJ . VK . K@ . +"

@" K . K@ . +"0@0 , UJ . V2@

@" . @"+" . @0@0 , UJ . V2@ / @" . @"+" . @0@t , / UJ . V2@At 0.~] ) ~]+" ) "# , UJ . V2@ @

"# .

"] )

]+ ,

"i )

]+ ,

UJ . V2@

@

@ , 2@UJ . V "i ) ]+


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A system with two concentric porous spherical shells of radii aR andR is shown in the figure

below. The inner surface of the outer shell is at a temperature To and equivalent pressure o; theouter surface of the inner shell is at a lower temperature Ta. Dry air at Ta is blown outward

radially from the inner shell into the intervening space and then through the outer shell at a mass

flow rate ofw. Assume steady, laminar flow with negligible viscous dissipation. The fluid isNewtonian and incompressible.

Transpiration cooling:

The inner sphere is being cooled by means of a refrigeration coil to maintain its temperature atTa. When air is blown outward, as shown, less refrigeration is required. Note that the tube thatsupplies the air does not disrupt the symmetry to an appreciable extent.

(a) List all assumptions in the problem, clearly expressing what the assumptions meanmathematically. Postulate the form of the velocity and temperature fields between the two

shells; simplify the governing equations. Determine the velocity field along with the pressure

distribution. Solve for the temperature field to derive the following:

. JN . J , c |. W=_a} .c |. W=_a}c |. W=_a} .c |. W=_a}(b) Assuming that the gas velocity is low such that conduction dominates, determine the rate of

heat flow out of the inner sphere. In contrast, determine the rate of heat flow out of the inner

sphere when the air is completely stagnant.


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r component:

2 ) ) ) So . " ) " , . ''

')d K" "" U"V ) K" So |So } ) K" So"

""

% component:


, component:




2= | ) ) ) So } ,' '', K" |" } ) K" So |So } ) K" So" "" ) 9 ) d;''


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, J'' , J , . W"]#a"2q ) " " , J ) W"]#a"2q

, . W"]#a"2q ) J ) W"]#a"2q , W"]#a"2q K.|}q ) J

Use the energy equation to find the temperature field, which reduces to the following:2= , " |" }After substituting in the expression for vrand using the form of the temperature field, integrate:2= | W_a2"} 00 , " 00 |" 00}W=_a 00 , 00 |" 00}W=_a " , 00

noUV , .W=_a K ) ~%

, ~ c|. W=_a} , " 00 00 , ~" c|. W=_a} , _a~W= c |. W=_a} ) q

Use the known temperatures at the edge of the shells to solve for the constants of integration: W=_a , ~ cU.RV ) q , J'' , , N'' ,


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J , ~ cU.RV ) q

N,~ cU.RV )

q

N . J , ~ cU.RV . ~ cU.RVN . J , ~ cU.RV .cU.RV~ , N . JK

cU.RV .cU.RV

J , K UN . JV cU.RVK cU.RV .cU.RV ) qJ , UN . JV cU.RVcU.RV .cU.RV ) qq , J . UN . JV cU.RV

cU.RV .cU.RV

, UN . JV cU.RVcU.RV .cU.RV ) J . UN . JV cU.RVcU.RV .cU.RV . JN . J , cU.RV .cU.RVcU.RV .cU.RV . JN . J , c |.

W=_a} .c |. W=_a}

c |.W=_a} .c |.

W=_a}

(b) When the gas velocity is low such that conduction dominates, the rate of heat flow out of the

inner sphere can be calculated from Fouriers law:

, UN . JV c|. W=_a} .c|. W=_a}c|. W=_a} .c |. W=_a} ) J


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, .`00N , ._a" UN . JVW=_a" c|. W=_a}c|. W=_a}.c|. W=_a}N

, .UN . JVW= c |. W=_a}c |. W=_a} .c |. W=_a}N , .UN . JVW= c |. W=_a}c |. W=_a} .c|. W=_a}

When the air is completely stagnant, the rate of heat flow out of the inner sphere can be

calculated by using a thermal circuit: , JBA>JB , - . "UKR-V . UKR"V

_a , _aU- . "VUKR-V . UKR"V

, _aUN . JVUKRV . UKRV , _aUN . JVK .


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In a heat exchanger at steady state, waste water from a chemical process flows inside a pipewhile cooling water boils outside the pipe. The waste water enters the heat exchanger with a

mean inlet temperature Tm,i of 630 K and is cooled to mean outlet temperature Tm,o of 580 K.

The cooling water boils at a saturation temperature Tsatof 373 K. The pipe has an outside

diameter of 450 mm with a thickness of 20 mm and a thermal conductivity of 111 W/mK. Thewaste water has fouled the pipe over time, creating a layer of salt 12 mm thick with a thermal

conductivity of 143 W/mK, giving the inside surface of the pipe a roughness of 0.46 mm.

The waste water has the following properties:

V) = 47.5 mm/s Cp = 60270 J/kgK! = 1200 kg/m

3kw = 0.051 W/mK

" = 1.1 x 10-6

Ns/m2

g = 9.8 m/s2

Assume axial convection dominates over axial conduction. Be careful when writing the

relationship between diameter and thickness (versus radius and thickness).

Heat exchanger:

(a) Using the given variables above along with the overall heat transfer coefficient (based upon

the inner area of the fouled pipe) Ui and the inner diameter of the fouled pipeDi, perform anenergy balance over a length, -x, of the pipe. Then, let -x ( 0 and integrate across the lengthL

of the heat exchanger.

(b) Solve for the power input required to pump the waste water through the heat exchanger.

Boiling Water

Pipe

Salt

Waste Water

-x


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(a) Begin with a basic energy balance on the pipe: Accumulation = In Out

P , W=O . W=O- . >-UO . 8NAVP , 2 za_ G>"{ %=O . 2 za_ G>"{ %=O- . >UaG>V-UO . 8NAVK_ 2G>%=O- . K_2G>%=O

- , . >UO . 8NAVLet -x approach zero:

K_2G>%= 0O0 , . JUO . 8NAVIntegrate with the BCs that T = Tm,i @ x = 0 and T = Tm,o @ x = L:/ KUO . 8NAV 0O%= 0

t

no O2G>%= EO2G>%= E

This is the design equation for simple single-tube evaporator heat exchanger. (This can be

confirmed by rearranging the generalized heat exchanger equation.)

wCpTm|x wCpTm|x+-x

UiP-x (Tm Tsat)


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(b) Evaluate the overall heat transfer coefficient. The system can be thought of as a thermal

circuit with three resistances in series:

>

,K

`> ,K

`>uJB;?A>JB ) JBA>JB=? ) JBA>JB=?v

> , K`> | K+>`> ) K#a8NHAE no zG> ) #8NHAG> { ) K#a=>=?E no |G> ) #8NHA ) #=>=?G> ) #8NHA }}> , KaG>E | K+>aG>E ) K#a8NHAE no zG> ) #8NHAG> { ) K#a=>=?E no |G> ) #8NHA ) #=>=?G> ) #8NHA }}

> , KK+> ) G>#8NHA no zG> ) #8NHAG> { ) G>#=>=? no |G> ) #8NHA ) #=>=?G> ) #8NHA }

Calculate the inner diameter:G> , GJ . #=>=? . #8NHA , P^_ZP m. #UP^P#P mV . #UP^PK# mV , 'P^]i mIn order to estimate the heat transfer coefficient inside the pipe, use an analogy between

momentum and heat transfer. Evaluate the Reynolds number and then calculate the frictionfactor:

Re , 2%G>d , UK#PP'kg/m3VUP^P_hZ'm/sVUP^]i mVK^KKPp'Ns/m2 , #KP jklmkncojKrF , ._^P ns\ G> ) _^ihRe'rF)#^#KrF , ._^P ns\ P^_i]i ) _^ih2xKP'rF ) #^# F , Z^K#_KPp~

Evaluate the Prandtl number and then use the Prandtl analogy, since the flow is turbulent withoutform drag when Pr > 1, to calculate the heat transfer coefficient from the friction factor:

Pr

,! ,

=d ,

UiP#hP'J/kgKVUK^KKPp'Ns/m2VP^PZK'

W/mK

, K^]

@ , F , Z^K#_KPp~St , Nu

Re Pr, @R#K ) Z@# UPr. KV


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NuU#KPVUK^]V , Z^K#_KPp~R#K ) ZZ^K#_KPp~# U1.3. KV TNu , iKYKP ,+>G>

+> , iKYKPG> , iKYKPP^PZK'W/mK

P^]i'm , KhY'W/m2

K

Calculate the overall heat transfer coefficient:

> , KKKhY'W/m2K) P^]i m#UK_]'W/mKV no zP^_KP mP^]i m{ ) P^]i m#UKKK'W/mKV no zP^_ZP mP^_KP m{ > , #h]Z'W/m2K

Substituting into the derived heat exchanger design equation, solve for the length:

ZP K.]h] Ki]P K.]h] K, c. _U#h]Z'W/m2KVUK#PP'kg/m3VUP^]i mVUP^P_hZ'm/sVUiP#hP'J/kgKV E E , #i^## mNow, calculate the pressure drop from the friction factor:F , #2%" GE

F , #UK#PP'kg/m3VUP^P_hZ'm/sV" P^]i'm

#i^##'m , Z^K#_KPp~ , K^Pa

Solve for the power input required to pump the waste water through the heat exchanger:1W , X , a_ G>"% , a_ UP^]i mV"UP^P_hZ'm/sVUK^ PaV , P^PKP_'

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Final Exam Solutions-1