MODULE 6: Worked-Out Problems

8/14/2019 MODULE 6: Worked-Out Problems

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MODULE 6: Worked-out Problems

Problem 1:

For laminar free convection from a heated vertical surface, the localconvection coefficient may be expressed as h x=Cx

-1/4 , where h x is thecoefficient at a distance x from the leading edge of the surface and thequantity C, which depends on the fluid properties, is independent of x.Obtain an expression for the ratio x x hh /

, where xh

is the average

coefficient between the leading edge (x=0) and the x location. Sketch thevariation of h

xand xh

with x.

Known : Variation of local convection coefficient with x for freeconvection from a heated vertical plate.

Find: ratio of average to local convection coefficient.

Schematic:

Boundary layer,h x=C x

-1/4 where Cis a constant

T s

x


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Analysis: It follows that average coefficient from 0 to x is given by

x1/4

x

x

0

1/4x

0xx

h34

Cx34

x3/4xC

34

h

dxxxC

dxhx1

h

===

==

Hence34

hh

x

x =

The variation with distance of the local and average convection coefficient isshown in the sketch.

x x hh

x x hh 3

4=

4/1= Cxh x

Comments: note that

hh x / x =4/3, independent of x. hence the average

coefficients for an entire plate of length L is Lh

=4/3L, where h L is the localcoefficient at x=L. note also that the average exceeds the local. Why?


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Problem 2:Experiments to determine the local convection heat transfer coefficientfor uniform flow normal to heated circular disk have yielded a radial

Nusselt number distribution of the form

+==

n

oD r

ra1

k h(r)D

Nu

Where n and a are positive. The Nusselt number at the stagnation

point is correlated in terms of the Reynolds number (Re D=VD/ ) andPrandtl

0.361/2Do Pr0.814Rek

0)Dh(rNu ===

Obtain an expression for the average Nusselt number, k Dh Nu D /

= ,corresponding to heat transfer from an isothermal disk. Typically

boundary layer development from a stagnation point yields a decaying

convection coefficient with increasing distance from the stagnation point. Provide a plausible for why the opposite trend is observed for the disk.

Known: Radial distribution of local convection coefficient for flownormal to a circular disk.

Find: Expression for average Nusselt number.

Schematic:


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Assumptions: Constant properties.

Analysis: The average convection coefficient is

o

o

s

r

0no

2n2

3o

no0

r

02

o

As

s

2)r(nar

2r

rkNuo

h

]22 r)a(r/r[1NuDk

r1

h

hdAA1

h

++=

+=

=

+

Where Nu o is the Nusselt number at the stagnation point (r=0).hence,

( )

0.361/2D

o

r

o

2n2

o

Pr2)]0.841Re2a/(n[1NuD

2)]2a/(n[1NuNuD

ror

2)(na

2r/ro

2Nuk Dh

NuDo

++=

++=

++==

+

Comments: The increase in h(r) with r may be explained in terms of the sharp turn, which the boundary layer flow must take around theedge of the disk. The boundary layer accelerates and its thicknessdecreases as it makes the turn, causing the local convection coefficientto increase.


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Problem 3:

In a flow over a surface, velocity and temperature profiles are of the forms

u(y)=Ay+By 2-Cy 3 and T(y)=D+Ey+Fy 2-Gy 3

Where the coefficients A through G are constants. Obtain expressionsfor friction coefficients C f and the convection coefficient h in terms of u, T and appropriate profile coefficients and fluid properties.

Known: form of the velocity and temperature profiles for flow over asurface.

Find: expressions for the friction and convection coefficients.

Schematic:

u,TT(y)=D+Ey+Fy 2-Gy 3

Ts=T(0)=D

U(y)=Ay-By2

-Cy3

y y

Analysis: The shear stress at the wall is

.]32[ 02

0

ACy By A yu

y

y

s =+== =

=

Hence, the friction coefficient has the form,


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Problem 4:

In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as

=

v

yuPrexp1

TTTT

s

s

Where y is the distance normal to the surface and the Prandtl number,Pr=c p/k=0.7, is a dimensionless fluid property. If T =400K, T s=300K,and u /v=5000m

-1, what is the surface heat flux?

Known: Boundary layer temperature distribution

Find: Surface heat flux.

Schematic:

Pr exp(1)(

yu

T T

T yT

s

s

=

Properties:Air ( k T S 300= ): k = 0.0263 W/m.k


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Analysis:Applying the Fouriers law at y=0, the heat flux is

0y

s

0y

"s yuPrexp uPr)Tk(TyTk q =

=

==

u)PrTk(Tq s

"s

=

1/m0.7x5000.K(100K)0.02063w/mq "s =

Comments: (1) Negligible flux implies convection heat transitedsurface (2) Note use of k at sT to evaluate from

"sq Fouriers law.


9/55

Problem 5:

Consider a lightly loaded journal bearing using oil having the constant properties =10 -2 kg/s-m and k=0.15W/m. K. if the journal and the

bearing are each mentioned at a temperature of 400C, what is themaximum temperature in the oil when the journal is rotating at 10m/s?

Known: Oil properties, journal and bearing temperature, and journalspeed for lightly loaded journal bearing.

Find: Maximum oil temperature.

Schematic:

Assumptions: (1) steady-state conditions, (2) Incompressible fluidwith constant properties, (3) Clearances is much less than journalradius and flow is Couette.

Analysis: The temperature distribution corresponds to the resultobtained in the text example on Couette flow.

+=2

2o L

yLy

U2k

TT(y)

The position of maximum temperature is obtained from

== 22

L2y

L1

U2k

0dydT


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y=L/2.

Or,The temperature is a maximum at this point since hence 0.T/dyd 22 <

22

2 U8k

To41

21

U2k

ToT(l/2)axTm, +=

+==

C40.83T0.15W/m.K 8

/s)kg/s.m(10m10C40T

max

22

max

=

+=

Comments: Note that T max increases with increasing and U, decreaseswith increasing k, and is independent of L.


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Problem 6:

Consider two large (infinite) parallel plates, 5mm apart. One plate isstationary, while the other plate is moving at a speed of 200m/s. both plates

are maintained at 27 C. Consider two cases, one for which the plates areseparated by water and the other for which the plates are separated by air.

For each of the two fluids, which is the force per unit surface arearequired to maintain the above condition? What is the correspondingrequirement?What is the viscous dissipation associated with each of the two fluids?What is the maximum temperature in each of the two fluids?

Known: conditions associated with the Couette flow of air or water.

Find: (a) Force and power requirements per unit surface area, (2) viscousdissipation,(3) maximum fluid temperature.

Schematic:

Assumptions: (1) Fully developed Couette flow, (2) Incompressible fluidwith constant properties.

Properties: Air (300K); =184.6*10-7 N.s/m2, k=26.3*10-3W/m.K; water (300K): =855*106N.s/m2,k=0.613W/m.K

Analysis: (a) the force per unit area is associated with the shear stress.Hence, with the linear velocity profile for Couette flow = (du/dy) = (U/L).


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34.2N/m0.005m200m/s

N.s/m10855 :Water

0.738N/m0.005m200m/s

N.s/m10184.6 :Air

226-water

227-air

==

==

With the required power given by P/A= .U

6840W/m200m/s)(34.2N/m(P/A) :Water

147.6W/m200m/s)(0.738N/m(P/A) :Air22

water

22air

==

==

(b) The viscous dissipation is .hence(U/L)(du/dy) 22 ==

W/m101.370.005m200m/s

N.s/m10855( :Water

W/m102.950.005m200m/s

N.s/m10184.6 ( :Air

362

26-water

342

27-air

=

=

=

=

The location of the maximum temperature corresponds to y max=L/2. HenceTmax=To+U

2/8k and

C34.0K 80.613W/m.

(200m/s)2N.s/m10855C27(T :Water

C30.5K 0.0263W/m.8

(200m/s)2_ N.s/m10184.6C27)(T :Air

26-

watermax)

2-7

airmax

=+=

=

+=

Comments: (1) the viscous dissipation associated with the entire fluid layer,),( LA must equal the power, P.

(2) Although ater w)( >> air )( , k water >>k air . Hence, T max,water Tmax,air .


13/55

Problem 7:

A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to anatmospheric air stream having a velocity of 40m/s. the air is at atemperature of T =20 C, while the plate is maintained at T s=120 C. Thesir flows over the top and bottom surfaces of the plate, and measurementof the drag force reveals a value of 0.075N. What is the rate of heattransfer from both sides of the plate to the air?

Known: Variation of h x with x for flow over a flat plate.

Find: Ratio of average Nusselt number for the entire plate to the local Nusselt number at x=L.

Schematic:

Analysis: The expressions for the local and average Nusselt number are


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2.NuLNuL

andk

2CLk

(L)2CLNuL

Hence

2CLLL

2Cdxx

LC

dxhL1

h

where

k Lh

Nu

k CL

k )L(CL

k Lh

Nu

-

1/21/2-

1/21/2L

0

1/2L

0xL

LL

1/21/2L

L

=

==

====

=

===

Comments: note the manner in which-

NuL is defined in terms of Lh

. Alsonote that

L

0xL dxNuL

1Nu


15/55

Problem 8:

For flow over a flat plate of length L, the local heat transfer coefficient hx

is known to vary as x -1/2 , where x is the distance from the leading edge of the plate. What is the ratio of the average Nusslet number for the entire

plate to the local Nusslet number at x=L (Nu L)?

Known: Drag force and air flow conditions associated with a flat plate.

Find: Rate of heat transfer from the plate.

Schematic:

Assumptions: (1) Chilton-Colburn analogy is applicable.

Properties: Air(70 C,1atm): =1.018kg/m 3, c p=1009J/kg.K, pr=0.70, =20.22*10 -6m2/s.

Analysis: the rate of heat transfer from the plate is


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)T(T(L)h2q s2

=

Whereh may be obtained from the Chilton-Colburn analogy,

240Wq

C20)(120.K)(0.3m)2(30W/mq

is rateheatThe

.K 30W/mh

2/3.70)9J/kg.K)(0)40m/s(100(1.018kg/m105.76h

Prcu2

Ch

hence,

105.76/2(40m/s)1.018kg/m

(0.2m)(0.075N/2)21

/2u

21

2

C

Prcu

htPrS

2

C j

22

2-

34-

2/3p

f -

423

2

2sf

2/3

p

2/3f H

==

=

=

=

===

===

Comments: Although the flow is laminar over the entire surface( )100.4/1022.20/2.0/40/Re 526 === smmsm Lu L , the pressure gradient is zero

and the Chilton-Colburn analogy is applicable to average, as well as local,surface conditions. Note that the only contribution to the drag force is made

by the surface shear stress.


17/55

Problem 9:

Consider atmospheric air at 25 C in parallel flow at 5m/s over both surfaces of 1-m-long flat plate maintained at 75 C. Determine the boundary layer thickness, thesurface shear stress, and the heat flux at the trailing edge. Determine the drag force onthe plate and the total heat transfer from the plate, each per unit width of the plate.

Known: Temperature, pressure, and velocity of atmospheric air in parallel flow over aPlate of prescribed length and temperature.

Find: (a) Boundary layer thickness, surface shear stress and heat flux at trailing edges,(b) drag force and total heat transfer flux per unit width of plate.

Schematic:

Fluid

=5m/sT=25C

P=1 atm L=1m

x)

Ts=75C

Assumptions: (1) Critical Reynolds number is 5*10 5, (2) flow over top and bottomsurfaces

Properties : (T f =323K, 1atm) Air: =1.085kg/m3

, =18.2*10-6m 2/s,k=0.028W/m.K,pr=0.707

Analysis: (a) calculate the Reynolds number to know the nature of flow

526L 102.75/sm1018.2

m1m/s5

LuRe =

==


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Hence the flow is laminar, and at x=L

22Ls,

1/2523

1/2L

2Ls,

1/251/2L

0.0172N/m.s0.0172kg/m

)1050.664/(2.7(5m/s)mkg

21.085

/2)0.664Re(

9.5mm)101m/(2.7555LRe

==

==

===

Using the correct correlation,

22s

2L

1/31/21/31/2L

LL

C)217W/m25C.K(754.34W/m)ThL(Ts(L)q".K 4.34W/m8W/m.K)/1m155.1(0.02h

hence,

155.1(0.707)105)0.332(2.75Pr0.332Rek

hLNu

====

====

(b) The drag force per unit area plate width is Ls L D

= 2' where the factor of two isincluded to account for both sides of the plate. Hence with

868W/mC25).K(75/m2(1m)8.68W)T(T2Lhq

.K 8.68W/m2hLhwithalso

0.0686N/m3N/m2(1m)0.034D

0.0343N/m

isdrag The

)101.328(2.75/2)(5m/s)(1.085kg/m/2)1.328Re(

2s

-

L"

2-

L

2'

2Ls

1/25231/2L

2Ls

===

==

==

=

==


19/55

Problem 10:Engine oil at 100 C and a velocity of 0.1m/s flows over both surfaces of a 1-m-longflat plate maintained at 20 C. Determine

a. The velocity and thermal boundary thickness at the trailing edge. b. The local heat flux and surface shear stress at the trailing edge.c. The total drag force and heat transfer per unit area width of the plate.

Known: Temperature and velocity of engine oil Temperature and length of flat plate.

Find: (a) velocity and thermal boundary thickness at the trailing edge, (b)Heat flux and surface shear stress at the trailing edge, (c) total drag force and heattransfer per unit plate width.

Schematic:

Assumptions: engine oil (T f =33K): =864kg/m 3, =86.1*10 6m2/s, k=0.140W/m /K,Pr=1081.

Analysis: (a) calculate the Reynolds number to know the nature of flow

1161/sm10*86.1

1m0.1m/s

LuReL 26- =

==

Hence the flow is laminar at x=L, and

0.0143m)0.147(1081Pr

0.147m)5(1m)(11615LRe1/31/3

t

1/21/2L

===

===

(b) The local convection coefficient and heat flux at x=L are


20/55

22

s

21/31/31/2LL

1300W/mC100).K(2016.25W/m)ThL(Tq"

.K 16.25W/m)(1081)0.332(11611m

0.140W/m.K Pr0.3325LRe

Lk

h

===

===

Also the local shear stress is

22Ls

1/223

1/2L

2Ls

0.0842N/m.s0.0842kg/m

)0.664(1161(0.1m/s)2

864kg/m/2)0.664Re(

==

==

(c) With the drag force per unit width given by Ls L D

= 2' where the factor of 2 isincluded to account for both sides of the plate, is follows that

5200W/mC100).K(20/m2(1m)32.5W)T(Th2Lq

thatfollowsalsoit.K,32.5W/m2hhwith

0.673N/m)1.328(1161(0.1m/s)/m2(1m)864kg/2)1.328Re(

2sL

_

2LL

_

1/2231/2L

2Ls

===

==

===

Comments: Note effect of Pr on ( /t).


21/55

Problem 11:

Consider water at 27 C in parallel flow over an isothermal, 1-m-long, flat plate with avelocity of 2m/s. Plot the variation of the local heat transfer coefficient with distancealong the plate. What is the value of the average coefficient?

Known: velocity and temperature of air in parallel flow over a flat plate of prescribedlength.

Find: (a) variation of local convection coefficient with distance along the plate, (b)Average convection coefficient.

Schematic:

Assumptions: (1) Critical Reynolds number is 5*10 5.

Properties: Water (300K): =997kg/m 3,=855*10 -6 N.s/m 2, =/=0.858*10 -6m2/s,k=0.613W/m. K, Pr=5.83

Analysis: (a) With

526L 102.33/sm100.858

1m2m/s

LuRe =

==

Boundary layer conditions are mixed and


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4055 4240 4491 4871 5514 .K)(W/mh

1.0 0.8 0.6 0.4 0.0215 x(m)

xPr

u0.0296k Pr0.0296Re

xk

hx0.215m,xfor

1206 1768 hx(W/m2.K)

0.215 0.1 x(m)

xPr

u0.332k Pr0.332Re

xk

hx 0.215m,xfor

0.215m)10/2.33101m(5)/ReL(Rex

x

0.21/34/5

1/34/5x

1/21/31/2

1/31/2x

65Lcx,c

2

==>

==

===

The Spatial variation of the local convection coefficient is shown above

(b) The average coefficient is

.K 4106W/mh

871(5.83))1033[0.0379(2.1m

0.613W/m.K 871)Pr(0.037Re

Lk

h

2L

_

1/34/561/34/5LL

_

=

==


23/55

Problem 12:

A circular cylinder of 25-mm diameter is initially at 150C and is quenched byimmersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over thecylinder. What is the initial rate of heat loss unit length of the cylinder?

Known: Diameter and initial temperature of a circular cylinder submerged in an oil bathf prescribed temperature and velocity.

Find: initial rate of heat loss unit per length.

Schematic:

Assumptions: (1) Steady-state conditions, (2) uniform surface temperature.

Properties: Engine oil (T =353K): =38.1*10 -6m2/s, k=0.138W/m. K, Pr =501;(T s=423K): Pr s=98.

Analysis: The initial heat loss per unit length is


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( ) 8.8W/mC80)0.025m(150.K 1600W/mq'

.K 1600W.mh

98501

(501)0.26(1312)0.025m

0.138W/m.K )1/4/Pr(PrPrCRe

Dk

h

hence.7.4tablefrom0.6mand 0.26CFind

1312/sm1038.1

m)2m/s(0.025

VDRe

whenrelation.Zhukauskasthefromcomputedbemay where

)TD(Thq

2

2 _

1/40.370.6

snm

D

_

26D

s

_

==

=

==

==

=

==

=

=

_

h

Comments: Evaluating properties at the film temperature, T f =388K( =14.0*10 -6m2/s,k=0.135W/m. K, Pr=196), find Re D=3517.


25/55

Problem 13:

An uninsulated steam pipe is used to transport high-temperature steam from one building to one another. The pipe is 0.5-m diameter, has a surface temperature of 150C, and is exposed to ambient air at -10C.the air moves in cross flow over the pipe

with a velocity of 5m.s.What is the heat loss per unit length of pipe?

Known: Diameter and surface temperature of uninsulated steam pipe. Velocity andtemperature of air in cross flow.

Find: Heat loss per unit length.

Schematic:

V=5m/sT=263K

Air D=0.5mm

Ts=150 C

Assumptions: (1) steady-state conditions, (2) uniform surface temperature

Properties: Air (T=263K, 1atm): =12.6*10 -6m2/s, k=0.0233W/m. K, Pr=0.72;(Ts=423K,1atm); Prs=0.649.

Analysis: the heat loss per unit length is

)TD(Thq s _

=


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.K 1600W.mh

98501

(501)0.26(1312)0.025m

0.138W/m.K )1/4/Pr(PrPrCRe

Dk

h

hence.7.4tablefrom0.6mand 0.26CFind

1312/sm1038.1

m)2m/s(0.025

VDRe

whenrelation.Zhukauskasthefromcomputedbemay where

2 _

1/40.370.6

snm

D

_

26D

=

==

==

=

==

=

_

h

Hence the heat rate is

( ) 4100W/mC10)(-0.5m(150.K 16.3W/mq' 2 ==

Comments: Note that q Dm, in which case the heat loss increases significantly withincreasing D.


27/55

Problem 14:

Atmospheric air at 25 C and velocity of 0.5m/s flows over a 50-W incandescent bulbwhose surface temperature is at 140 C. The bulb may be approximated as a sphere of 50-mm diameter. What is the rate of heat loss by convection to the air?

Known: Conditions associated with airflow over a spherical light bulb of prescribeddiameter and surface temperature.

Find: Heat loss by convection.

Schematic:

Assumptions: (1) steady-state conditions, (2) uniform surface temperature.

Properties: Air (T f =25 C, 1atm): =15.71*10 -6m2/s, k=0.0261W/m. /K.Pr=.0.71, =183.6*10 -7 N.s/m 2; Air (T s=140 C, 1atm): =235.5*10 -7 N.s/m 2

Analysis:)TD(Thq s

_

=


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10.3WC25)(140 (0.05m).K m

W11.4q

israteheattheand

.K 11.4W/mh

235.5183.6

](0.71)0.06(1591)[0.4(1591)20.05m

K 0.0261W/m.h

hence

1591/sm1015.71

0.05m0.5m/s

VDRe

Where

])(/)Pr0.06Re(0.4Re[2Dk

h

whenrelation.Whitakerthefromcomputedbemay where

22

2 _

1/40.42/31/2

_

26D

1/4s

0.42/3D

1/2D

_

==

=

++=

=

==

++=

_

h

Comments: (1) The low value of _

h suggests that heat transfer by free convection may

be significant and hence that the total loss by convection exceeds 10.3W(2) The surface of the bulb also dissipates heat to the surroundings by radiation. Further,in an actual light bulb, there is also heat loss by conduction through the socket.

(3) The Correlation has been used its range of application ( /s )


29/55


30/55

L2D

uf p

2m=

e=2.6*10 -4 m for clean cast iron; hence e/D=1.04*10 -3. With

5326

mD 102.92/1000kg/mN.s/m10855

0.25m1m/s

DuRe =

==

0.42barN/m104.2P

.mkg/s104.21000m2(0.25m)

(1m/s)(1000kg/m0.021P

,hence0.021.f that8.3figfromfind

24

2423

==

==

)

The pump power requirement is

2.06kW1m/s/4)m0.25(N/m104.2P

/4)up( DVp.P

2224

m2

.

==

==

(b) Increasing by 25% it follows that =3.25*10-4m and e/D=0.0013. WithReD unchanged, from fig 8.3, it follows that f 0.0225. Hence

2.21kW)P/f (f P bar0.45)/f (f P 1122122 ==== 1 p

Comments: (1) Note that L/D=4000>>(x fd,h/D) 10 for turbulent flow andthe assumption of fully developed conditions is justified.(2) Surface fouling results in increased surface and increases operating costs

through increasing pump power requirements.


31/55

Problem 16:

Consider flow in a circular tube. Within the test section length (between1 and 2) a constant heat flux q s is maintained.

(a) For the two cases identified, sketch, qualitatively, the surfacetemperature T s(x) and the fluid mean temperature Tm(x) as a functionof distance along the test section x. in case A flow is hydrodynamically and thermally fully developed. In case B flow is notdeveloped.

(b) Assuming that the surface flux q s and the inlet mean temperatureTm,1 are identical for both cases, will the exit mean temperature T m,2for case A be greater than, equal to , or less than T m,2 for case B?Briefly explain why?

Known: internal flow with constant surface heat flux, "sq .

Find: (a) Qualitative temperature distributions (x), under developing andfully developed flow, (b) exit mean temperature for both situations.

Schematic:

Flow

1 2

qs=constant

Assumptions: (a) Steady-state conditions, (b) constant properties, (c)incompressible flow.

Analysis: Based upon the analysis, the constant surface heat flux conditions,

constantdx

dT m =


32/55

Hence, regardless of whether the hydrodynamic or thermal boundary layer isfully developed, is follows that

Tm(x) is linear

T m,2 will be the same for all flow conditions.

The surface heat flux can be written as

(x)]Th[Tq ms"s =

Under fully developed flow and thermal conditions, h=h fd is a constant.When flow is developing h> h fd . Hence, the temperature distributions appear as below.


33/55

Problem 17:

A thick-walled, stainless steel (AISI 316) pipe of inside and outsidediameter D i=20mm and D o=40m is heated electrically to provide a

uniform heat generation rate of 36.

/10 mW q = . The outer surface of the pipe is insulated while water flows through the pipe at a rate of

skgm /1.0.

= (a) If the water inlet temperature is T m,1=20 C and the desired outlet

temperature is T m,o=40 C, what is the required pipe length

(b) What are the location and value of the maximum pipe temperature?

Known: Inner and outer diameter of a steel pipe insulated on the outside and experiencinguniform heat generation. Flow rate and inlet temperature of water flowing through the

pipe.

Find: (a) pipe length required to achieve desired outlet temperature, (b)location and value of maximum pipe temperature.

Schematic:

Assumptions: (1) steady-state conditions, (2) constant properties, (3)negligible kinetic energy, potential energy and flow work changes, (4) one-dimensional radial conduction in pipe wall, (5) outer surface is adiabatic.


34/55

Properties: Stainless steel 316 (T 400K): k=15W/m.k; water ( 303K T.

m = );c p=4178J/kg. K, k=0.617W/m. K, =803*10 -6 N.s/m 2, Pr=5.45

Analysis: (a) performing an energy balance for a control volumeabout the inner tube, it follows that

8.87mL

](0.02)4m( /4)[(0.W/m10C)20178(J/kg.K (0.1kg/s)4

)LD( /4)(q

)T(TmcL

)LD( /4)(qq)T(Tcm

22362

i

2

0

.

.

im,om,p

2i

20

.

im,om,p

.

=

=

=

==

(b) The maximum wall temperature exists at the pipe exit (x=L) and theinsulated surface (r=r o). The radial temperature distribution in the wall is of the form

Tnr2k qr

-r4k q

C Cnr2k

qrr

4k q-

T)T(r:rr

2k

qrC

rC

r2k

q0

drdT

;rr

;conditionsboundarythegconsiderin

CnrCr4k q

T(r)

si

.2

o2i

.

22i

.2

o2i

.

si1

.2

o1

o

1o

.

rro

212

.

o

+=++===

=+== =

++=

=


35/55

The temperature distribution and the maximum wall temperature (r=r o) are

)Th(T4D

)D(DqLD

)LDq( ( /4)q

fromfollowsitexit,atwalltheof temperturesurafaceinnertheTs,where

Trr

n2k

qr )r-(r

4k q

-)T(rT

Trr

n2k

qr )r-(r

4k q

T(r)

om,s

i

2i

2o

.

i

.2i

2o"

s

si

o

.2

o2i

2o

.

omaxw,

si

.2

o2i

2

.

=

=

=

++==

++=

Where h is the local convection coefficient at the exit. With

7928N.s/m103 (0.02m)80

0.1kg/s4D

m4Re 26

i

.

D ===

The flow is turbulent and, with (L/D i)=(8.87m/0.02m)=444>>(x fd/D)10,it isalso fully developed. Hence, from the Dittus-Boelter correlation,

.K 1840W/m5.45)0.023(79280.02m

0.617W/m.K )pr(0.023Re

Dk

h 20.44/50.44/5Di

===

Hence the inner surface temperature of the wall at the exit is

C52.4C48.20.010.02

n215W/m.K

(0.02)W/m10](0.01)[(0.02)

415W/m.K W/m10

T

and

C48.2C40.K(0.02m)180W/m4

](0.02m)[(0.04m)W/m10T

4D)D(Dq

23622

36

maxw,

2

2236

om,i

2i

2o

.

s

=++=

=+

=+

=

T


36/55

Comments: The physical situation corresponds to a uniform surface heatflux, and T m increases linearly with x. in the fully developed region, Ts alsoincreases linearly with x.

T

xfd

Ts(x)

Tm(x)

x


37/55

Problem 18:

The surface of a 50-mm diameter, thin walled tube is maintained thinwalled tube is maintained at 100 C. In one case air is cross flow very the

tube with a temperature of 25 C and a velocity of 30m/s. In another caseair is in fully developed flow through the tube with a temperature of 25 Cand a mean velocity of 30m/s. compare the heat flux from the tube to theair for the two cases.

Known: surface temperature and diameter of a tube. Velocity andtemperature of air in cross flow. Velocity and temperature of air in fullydeveloped internal flow.

Find: convection heat flux associated with the external and internal flows.

Schematic:

Air

Um=30m/sTm=25C

V=30m/sT =25 C

Ts=100 C

D=0.05m

Air

Assumptions: (1) steady-state conditions, (2) uniform cylinder surfacetemperature, (3) fully developed internal flow

Properties: Air (298K): =15.71*10-6m 2/s, k=0.0261W/m.K, Pr=0.71

Analysis: for the external and internal flow

426 1055.9/1071.15

05./30Re =

=== smmsm DuVD m D

From the Zhukauskas relation for the external flow, with C=0.26 and m=0.6


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223)1()71.0()1055.9(26.0)Pr Pr(Pr/Re 4/137.06.044/1 _

=== sm D D C u N

Hence, the convection coefficient and heat rate are

232"

2 _ _

/1073.8)25100(./4.116)(

./4.11622305.0

./0261.0

mW C k mW T T hq

K mW m

K mW NuD

Dk

h

s ===

===

Using the Dittus-Boelter correlation, for the internal flow, which isTurbulent,

232

_ _

4.05/4244.05/4 _

/1058.7)25100(./101)(

.2/10119305.0

./0261.0

193)71.0()1055.9(023.0Pr Re023.0

mW C K mW T T h

K mW K mW

Nu Dk

h

NU

ms

D

D D

===

===

===

"q

isfluxheattheand

Comments: Convection effects associated with the two flow conditions arecomparable.


39/55

Problem 19:

Cooling water flows through he 25.4 mm diameter thin walled tubes of astream condenser at 1m/s, and a surface temperature of 350K is maintained

by the condensing steam. If the water inlet temperature is 290 K and thetubes are 5 m long, what is the water outlet temperature? Water propertiesmay be evaluated at an assumed average temperature of 300K

Known: Diameter, length and surface temperature of condenser tubes. Water velocity and inlet temperature.

Find: Water outlet temperature.

Schematic:

L=5m

D=0.0254m

Tm,i=290KUm=1m/s

Ts=350K

Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligiblekinetic energy, potential energy and flow work changes.

Properties: Water (300K): =997kg/m 3, c p=4179J/kg.K, =855*10-

6kg/s.m,k=0.613W/m.K, Pr=5.83

Analysis:


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29,618kg/s.m10855

54m(1m/s)0.02997kg/m

DuRe

]h)cm( DL)exp[T(TTT

6

3m

D

_

p

.

im,ssom,

=

==

=

The flow is turbulent. Since L/D=197, it is reasonable to assume fullydeveloped flow throughout the tube. Hence

323K K)(4179J/kg.(0.505kg/s

.K)5m(4248W/m (0.0254m)(60K)exp350K T

0.505kg/s4m)m/s)(0.025(1/m( /4)997k /4)(um

With

.K 4248W/m4m)/m.K/0.025176(0.613WNuD(k/D)h

2

om,

232m

.

2 _

=

===

===

Comments: The accuracy of the calculations may be improved slightly by

reevaluating properties at 306.5K T _

m = .


41/55

Problem 20:

The air passage for cooling a gas turbine vane can be approximated as a tubeof 3-mm diameter and 75-mm length. If the operating temperature of the

vane is 650 C, calculate the outlet temperature of the air if it enters the tubeat 427 C and 0.18kg/h.

Known: gas turbine vane approximation as a tube of prescribed diameter andlength maintained at an known surface temperature. Air inlet temperatureand flow rate.

Find: outlet temperature of the air coolant.

Schematic:

hkgm /18.0.

=

Assumptions: (1) Steady-state conditions, (2) negligible Kinetic and

potential energy changes.

Properties: Air (assume )1,780 _

atmK T m = : cp=1094J/kg. K, k=0.0563 W/m.K, =363.7*10 -7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650 C=923K, 1atm):=404.2*10 -7 N.s/m 2.

Analysis: For constant wall temperatures heating,

=

.

_

,

, exp

pims

oms

mc

hPLT T

T T

Where P= D. for flow in circular passage,


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N.s/m103.7 (0.03m)36/3600s/h)0.18kg/h(14

Dm4

Re 7

.

D ==

The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate

correlation including combined entry length fields.

.K 87.5W/m10404.210363.7

250.706584

1.860.003m

K 0.0563W/m.h

L/DPrRe

1.86k Dh

Nu

2

0.14

7

71/3 _

0.14

s

1/3D

_ _

D

=

=

==

Hence, the air outlet temperature is

=

1094J/kg.K )kg/s(0.18/3600

.K 87.5W/m0.075m (0.003m)exp

C427)(650

T650 2om,

Tm,o=578 C

Comments: (1) based upon the calculations for T m,o=578 C, _

mT =775K whichis in good agreement with our assumption to evaluate the thermo physical

properties.


43/55

Problem 21

A household oven door of 0.5-m height and 0.7-m width reaches an averagesurface temperature of 32 C during operation. Estimate the heat loss to the

room with ambient air at 22 C. If the door has an emissivity of 1.0 and thesurroundings are also at 22 C, comment on the heat loss by free convectionrelative to that by radiation.

Known: Oven door with average surface temperature of 32 C in a roomwith ambient temperature at 22 C.

Find: Heat loss to the room. Also, find effect on heat loss if emissivity of door is unity and the surroundings are at 22 C.

Schematic:

0 .7 m

L =

0 . 5

m

Assumptions: (1) Ambient air in quiescent, (2) surface radiation effectsare negligible.

Properties: Air (T f =300K, 1atm): =15.89*10 -6 m2/s, k=0.0263W/m. K,=22.5*10 -6 m2/s, Pr=0.707, =1/T f =3.33*10-3K -1

Analysis: the heat rate from the oven door surface by convection to the

ambient air is

)T(TAhq ss _

=


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Where _

h can be estimated from the free convection correlation for avertical plate,

[ ]

2

8/279/16

1/6L

_ _`

(0.492/Pr)1

0.387Ra0.825k

LhLNu

++==

The Rayleigh number,

42626

322s

L 101.142/sm1022.5/sm1015.89

m322)K0.52(1/300k))39.8m/s

)LT(TgRa =

=

=

Substituting numerical values into equation, find

[ ]

.K 3.34W/m63.50.5m

K 0.0263W/m.LNu

Lh

63.5(0.492/Pr)1

)1020.387(1.140.825

k Lh

LNu

2 _` _

2

8/279/16

1/88 _

_`

===

=

+

+==

k L

The heat rate using equation is

11.7W22)K (320.7)m.k(0.53.34W/mq22

==

Heat loss by radiation, assuming =1 is

21.4W]k 22)(27332)[(273.K W/m5.67100.7)m1(0.5q

)T(TAq4244282

rad

4sur

4ssrad

=++=

=

Note that heat loss by radiation is nearly double that by free convection.

Comments: (1) Note the characteristics length in the Rayleigh number is theheight of the vertical plate (door).


45/55

Problem 22

An Aluminum alloy (2024) plate, heated to a uniform temperature of 227 C, is allowed to cool while vertically suspended in a room where the

ambient air and surroundings are at 27 C. The late is 0.3 m square with athickness of 15 mm and an emissivity of 0.25.a. Develop an expression for the time rate of change of the plate

temperature assuming the temperature to be uniform at anytime.

b. Determine the initial rate of cooling of the plate temperature is227 C.

c. Justify the uniform plate temperature assumption.

Known: Aluminum plate alloy (2024) at uniform temperature of 227 Csuspended in a room where the ambient air and the surroundings are at 27 C

Find: (1) expression for the time rate of change of the plate, (2) Initial

rate of cooling (K/s) when the plate temperature is 227 C. (3) justify the

uniform plate temperature assumption.

Schematic:


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Properties: Aluminium alloy 2024 (T=500K): =2270 kg/m 3,

k=186W/m.K, c=983 J/kg.K; Air (T f =400K, 1atm): =26.41*10-

6m2

/s,k=0.0338 W/m.K, =38.3*10-6

m2

/s, Pr=0.690.

Analysis :( a) from an energy balance on the plate considering free

convection and radiation exchange..

st out E E = .

)T (T)T(Th[c2

dtdT

ordtdT

A)T(T2A)T(T2Ah 4sur4ssL

.

cs4

sur4

ssss

_

L + ==

Where T s is the plate temperature assumed to be uniform at any time.

(b) To evaluate (dt/dx), estimate _

Lh . Find first the Rayleigh number

82626

322s

L 101.308/sm10.33/sm1026.41

0.3m27)K 27(1/400k)(29.8m/s

)LT(TgRa =

=

=

8

Substituting numerical values, find

[ ] [ ]5.55

) =

++=

++= 4/99/16

82

4/99/16

1/4L

_`

90)(0.492/0.61

1080.670(1.300.68

(0.492/Pr)1

0.670Ra0.68LNu

0.099K/s

)K 300.K)(500W/m100.25(5.6727)K .K(2276.25W/m983J/kg.K 0.015m2770Kg/m2

dtdT

.K 6.25W/mK/0.3m0.0338W/m.55.5k/L Nuh

442823

2 _`

L

_

=

+

=

===


47/55

(c) The uniform temperature assumption is justified if the Biot number

criterion is satisfied. With == )/A.(A)(V/AL sssc

and _ _ _ _

1.0/, =+= k h Bihhh tot rad convtot . Using the linearized radiation coefficient

relation find,

3.86W/m3002)K 300)(5002)(500.K 8W/m100.25(5.67)T)(TT (Th 3422sur2

ssursrad

_

=++=++=

Hence Bi=(6.25+3.86) W/m 2.k (0.015m)/186W/m. K=8.15*10 -4. Since

Bi


48/55

Problem 23

The ABC Evening News Report in news segment on hypothermia researchstudies at the University of Minnesota claimed that heat loss from the body

is 30 times faster in 1 0 C water than in air at the same temperature. Isthat a realistic statement?Known: Person, approximated as a cylinder, experiencing heat loss inwater or air at 10 C.

Find: Whether heat loss from body in water is 30 times that in air.

Assumptions: (1) Person can be approximated as a vertical cylinder of diameter D=0.3 m and length L=1.8m, at 25 C, (2) Loss is only from the

lateral surface.

Properties: Air ( _

T =(25+10) C/2=290K, 1atm); K=0.0293 W/m. K, ==19.91*10 -6m2/s, =28.4*10 -6 m2/s; Water (290K); k=0.598 W/m.K; =vf =1.081*10 -6m2/s, =k v f /cp =1.431*10 -7m2/s, f =174*10 -6K -1.

Analysis: in both water (wa) an air(a), the heat loss from the lateralsurface of the cylinder approximating the body is

)TDL(Thq s _

=

Where T s and T are the same for both situations. Hence,

_

a

wa

_

a

wa

h

hqq

=

Vertical cylinder in air:

92626

323

L 105.228/sm28.410/sm10*19.91

10)k(1.8m)5(1/290K)(29.8m/s

g TRa =

==


49/55

.K 2.82W/mh 173.4)100.1(5.228CRa

k

LhNu

1/3,n and 0.1Cwith

2L

_ 1/39n

L

_ _

LL =====

==

Vertical cylinder in water.

112726

3162

L 109.643/sm101.431/sm101.08110)K(1.8m)(25K 101749.8m/s

Ra ==

.K 328W/mh 978.9)100.1(9.643CRak

hLNu

1/3,n and 0.1Cwith

2L

_ 1/311n

L

_ _

L =====

==

Hence, from this analysis we find

30of claimthewithpoorlycompareswhich 117.K 2.8W/m.K 328W/m

qq

2

2

a

wa ==


50/55

Problem 24

In a study of heat losses from buildings, free convection heat transfer from

room air at 305 K to the inner surface of a 2.5-m-high wall at 295 K issimulated by performing laboratory experiments using water in a smaller testcell. In the experiments the water and the inner surface of the test cell aremaintained at 300 and 290K, respectively. To achieve similarity betweenconditions in the room and the test cell, what is the required test cell height?If the average Nusselt number for the wall may be correlated exclusively interms of the Rayleigh number, what is the ratio f the average convectioncoefficient for the room wall to the average coefficient for the test cell wall?

Known: Air temperature and wall temperature and height for a room. Water temperature and wall temperature for a simulation experiment.

Find: Required test cell height for similarity. Ratio and height convectioncoefficient for the two cases.

Schematic:

L a

= 2

. 5 m

L w

Assumptions: (1) Air and water are quiescent; (2) Flow conditionscorrespond to free convection boundary layer development on an isothermal

vertical plate, (3) constant properties.

Properties: Air (T f =300K, 1 atm); K=0.0293 W/m.K, ==15.9*10 -6m2/s,=22.5*10 -6 m2/s; =1/T f =3.33*10 -3K -1, k=0.0263W/m.K; water (T f =295K): =998kg/m

3, =959*10 -6 N.s/m 2, c p=4181 J/kg.K, =227.5*10 -6K -1, k=0.606 W/m.K; Hence =/ =9.61/s, =k / cp =1.45*10 -7m2/s.


51/55

Analysis: Similarity requires that Ra L,a=Ra L,w where

mm L

o H L

L

hence

w

a

air w

a

w

45.0)179.0(5.2

10*228.010*33.3

10*5.22*9.1510*45.1*61.9

)()(

,

)L-T(TgRa

3

3

12

143/1

2

3s

L

==

=

=

=

3

w

a

a

w

w

_

_

a

_

wL,

_

aL,wL,aL,

10*7.810.606

0.02632.5

0.45k k

LL

h

h

.hence Nu Nuthatfollowsit,RaRaif

===

==

Comments: Similitude allows us to obtain valuable information for one

system by performing experiments for a smaller system and a differentfluid.


52/55

Problem 25

A square plate of pure aluminum, 0.5 m on a side and 16 mm hick, isinitially at 300C and is suspended in a large chamber. The walls of thechamber are maintained at 27C, as is the enclosed air. If the surfaceemissivity of the plate temperature during the cooling process? Is itreasonable to assume a uniform plate temperature during the cooling

process?

Known: Initial temperature and dimensions of an aluminum plate.Conditions of the plate surroundings.

Find: (a) initial cooling rate, (2) validity of assuming negligible temperaturegradients in the plate during the cooling process.

Schematic:

L

Assumptions: (1) plate temperature is uniform; (2) chamber air is quiescent,(3) Plate surface is diffuse-gray, (4) Chamber surface is much larger than

that of plate, (5) Negligible heat transfer from edges.Properties: Aluminum (573k); k=232W/m.k, c p=1022J/kg.K, =2702 kg/m

3:Air (T f =436K, 1 atm): =30.72*10

-6m2/s,=44.7*10 -6m2/s,k=0.0363W/m.K,Pr=0.687, =0.00229K -1.

Analysis: (a) performing an energy balance on the plate,


53/55

psur ss

pst sur ss

wcT T T T h Adt dT

dt dT Vc E T T T T h Aq

/)]4()([2/

]/[)]4()([2

4 _

.4

_

+=

==+=

82626

323s

L 105.58/sm017.44/sm10*30.72)K(0.5m)721(300-0.00229K 9.8m/s)L-T(Tg

Ra ===

K mW h

Ra

Lk

h L

./8.5

])687.0/492.0(1[)1058.5(670.0

68.05.0

0363.0]Pr)/492.0(1[

670.068.0

2 _

9/416/9

4/18

9/416/9

4/1 _

=

++=

++=

Hence the initial cooling rate is

{ }

sK dt dT

k kg J mmkgK K K mW C K mW

dt dT

/136.0

./1022)016.0(/2702])300()573[(./1067.525.0)27300(./8.52

3

444282

=

+=

(b) To check the Validity of neglecting temperature gradients across the plate thickness,

4-

22"

103.832W/m.K (0.008m)/2(11W/m2.K)Bi

./0.11273//)14131583()/( /)2/(

===+===

hence

k mW K mW T T qhwherek wh Bi itot eff eff

And the assumption is excellent.

Comments: (1) Longitudinal (x) temperature gradients are likely to me moresevere than those associated with the plate thickness due to the variation of hwith x. (2) Initially q conv q rad.


54/55

Problem 26

Beer in cans 150 mm long and 60 mm in diameter is initially at 27 C and isto be cooled by placement in a refrigerator compartment at 4 C. In the

interest of maximizing the cooling rate, should the cans be laid horizontallyor vertically in the compartment? As a first approximation, neglect heattransfer from the ends.

Known: Dimensions and temperature of beer can in refrigerator compartment.

Find: orientation which maximum cooling rate.

Schematic:

Air,T=4CP=1 atmD=0.06m

Horizontal, (h) Vertical, (V)Ts=27 C

L =1 5 0 mm

Assumptions: (1) End effects are negligible, (2) Compartment air isquiescent, (3) constant properties.

Properties: Air (T f =288.5K, 1 atm): =14.87*10-6 m2/s, k=0.0254W/m.K,

=21.0*10 -6m2/s, Pr=0.71, =1/T f =3.47*10-3K -1.

Analysis: The ratio of cooling rates may be expressed as

h

v

s

s

h

v

qh

v

hh

T T T T

DL DL

hhq _

_

_

_

)()( =

==

For the vertical surface, find


55/55

639L

39326-26-

-1-323s

L

1044.8)15.0(105.2Ra

L105.2L/s)m10/s)(21m10(14.87

C)(23K 103.479.8m/s)L-T(TgRa

==

=

==

Using the correct correlation

K mW m

K mW Lk

Nuhh

NuL

Lv L ./03.515.0./0254.0

7.29 hence,

7.29])71.0/492.0(1[

)1044.8(387.0825.0

2 _ _ _

2

27/816/9

6/16

====

=

++=

Using the correct correlation,

97.018.503.5

,

./18.506.0

./0254.024.12

24.12])71.0/559.0(1[

)104.5(387.060.0

2 _ _ _

2

27/861/9

6/15 _

==

====

=

+

+=

h

v

Dh D

D

q

qhence

K mW m

K mW Dk

Nuhh

Nu

Comments: in view of the uncertainties associated with equations and theneglect of end effects, the above result is inconclusive. The cooling rates areare approximately the same.

MODULE 6: Worked-Out Problems

Documents

Transcript of MODULE 6: Worked-Out Problems