Module6 WorkedOut Problems

7/28/2019 Module6 WorkedOut Problems

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MODULE 6: Worked-out Problems

Problem 1:

For laminar free convection from a heated vertical surface, the local

convection coefficient may be expressed as hx=Cx-1/4, where hx is the

coefficient at a distance x from the leading edge of the surface and the

quantity C, which depends on the fluid properties, is independent of x.

Obtain an expression for the ratioxx hh /

, where

xh

is the average

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

variation of hx and xh

with x.

Schematic:

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

x

1/4

x

x

0

1/4

x

0

xx

h3

4Cx

3

4x3/4

x

C

3

4h

dxxxCdxh

x1h

===

==

Hence3

4

h

h

x

x =

Boundary layer,

hx=Cx-1/4 where C

is a constant

Ts

x


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The variation with distance of the local and average convection coefficient is

shown in the sketch.

Comments: note that

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

coefficients for an entire plate of length L isLh

=4/3L, where hL is the local

coefficient at x=L. note also that the average exceeds the local. Why?

Problem 2:

Experiments to determine the local convection heat transfer coefficient

for uniform flow normal to heated circular disk have yielded a radial

Nusselt number distribution of the form


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+==

n

o

Dr

ra1

k

h(r)DNu

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

point is correlated in terms of the Reynolds number (ReD=VD/) and

Prandtl

0.361/2

Do Pr0.814Rek

0)Dh(rNu =

==

Obtain an expression for the average Nusselt number, kDhNuD

/

= ,

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 forthe disk.

Known: Radial distribution of local convection coefficient for flow

normal to a circular disk.

Find: Expression for average Nusselt number.

Schematic:

Assumptions: Constant properties.

Analysis: The average convection coefficient is


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o

o

s

r

0

n

o

2n2

3

o

n

o0

r

0

2

o

A

s

s

2)r(n

ar

2

r

r

kNuoh

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

k

r

1h

hdAA

1h

+

+=

+=

=

+

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

( )

0.361/2

D

o

r

o

2n2

o

Pr2)]0.841Re2a/(n[1NuD

2)]2a/(n[1NuNuD

ro

r

2)(n

a

2

r/ro2Nu

k

DhNuD

o

++=

++=

++==

+

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 the

edge of the disk. The boundary layer accelerates and its thickness

decreases as it makes the turn, causing the local convection coefficient

to increase.


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Problem 3:

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

u(y)=Ay+By2-Cy3 and T(y)=D+Ey+Fy2-Gy3

Where the coefficients A through G are constants. Obtain expressions

for friction coefficients Cf 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 a

surface.

Find: expressions for the friction and convection coefficients.

Schematic:

Analysis: The shear stress at the wall is

.]32[ 02

0

ACyByAy

uy

y

s=+=

=

=

=

Hence, the friction coefficient has the form,


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2f

22

sf

u

2AC

u

2A

/2u

C

=

==

The convection coefficient is

=

=

=

+

=

=

TD

Ekh

TD

]3Gy2Fy[Ek

TT

y)T/(kh

f

0y

2

f

s

0yf

Comments: It is a simple matter to obtain the important surface

parameters from knowledge of the corresponding boundary layers

profile. However is rarely simple matter to determine the form of theprofile.

Problem 4:


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In a particular application involving airflow over a heated surface, the

boundary layer temperature distribution may be approximated as

=

v

yuPrexp1TT

TT

s

s

Where y is the distance normal to the surface and the Prandtl number,

Pr=cp/k=0.7, is a dimensionless fluid property. If T =400K, Ts=300K,

and u/v=5000m-1, what is the surface heat flux?

Known: Boundary layer temperature distribution

Find: Surface heat flux.

Schematic:

Properties:

Air ( kTS 300= ): k = 0.0263 W/m.k

Analysis:

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


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0y

"

sy

Tkq==

)PTk(Tq s"

s

=

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

s =

Comments: (1) Negligible flux implies convection heat transited

surface (2) Note use of k at sT to evaluate from"

sq Fouriers law.


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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 the

maximum temperature in the oil when the journal is rotating at 10m/s?

Known: Oil properties, journal and bearing temperature, and journal

speed for lightly loaded journal bearing.

Find: Maximum oil temperature.

Schematic:

Assumptions: (1) steady-state conditions, (2) Incompressible fluid

with constant properties, (3) Clearances is much less than journal

radius and flow is Couette.

Analysis: The temperature distribution corresponds to the result

obtained in the text example on Couette flow.

+=

2

2

oLy

LyU

2kTT(y)

The position of maximum temperature is obtained from


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==

2

2

L

2y

L

1U

2k

0

dy

dT

y=L/2.

Or,

The temperature is a maximum at this point since hence0.T/dyd 22 > air)( , kwater>>kair. Hence, Tmax,water Tmax,air .

Problem 7:

A flat plate that is 0.2m by 0.2 m on a side is orientated parallel to an

atmospheric air stream having a velocity of 40m/s. the air is at a

temperature of T=20C, while the plate is maintained at Ts=120C. The


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2.

NuL

NuL

and

k

2CL

k

(L)2CLNuL

Hence

2CLLL

2Cdxx

L

Cdxh

L

1h

where

k

LhNu

k

CL

k

)L(CL

k

LhNu

-

1/21/2-

1/21/2

L

0

1/2

L

0

xL

LL

1/21/2L

L

=

==

====

=

===

Comments: note the manner in which-

NuL is defined in terms of Lh

. Also

note that

L

0

xL dxNuL

1Nu


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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 ofthe plate. What is the ratio of the average Nusslet number for the entire

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

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(70C,1atm): =1.018kg/m3, cp=1009J/kg.K, pr=0.70,

=20.22*10-6m2/s.

Analysis: the rate of heat transfer from the plate is

)T(T(L)h2q s2

=


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Where

h may be obtained from the Chilton-Colburn analogy,

240Wq

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

israteheatThe

.K30W/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)

2

1

/2u

2

1

2

C

Prcu

htPrS

2

Cj

22

2-

34-

2/3

p

f-

4

23

2

2

sf

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 === smmsmLuL , 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.


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Problem 9:

Consider atmospheric air at 25C in parallel flow at 5m/s over both surfaces of 1-m-

long flat plate maintained at 75C. Determine the boundary layer thickness, the

surface 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:

Assumptions: (1) Critical Reynolds number is 5*105, (2) flow over top and bottomsurfaces

Properties: (Tf=323K, 1atm) Air: =1.085kg/m3,=18.2*10-6m2/s,k=0.028W/m.K,pr=0.707

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

5

26L102.75

/sm1018.2

m1m/s5

LuRe =

==


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Hence the flow is laminar, and at x=L

22

Ls,

1/252

3

1/2

L

2

Ls,

1/251/2

L

0.0172N/m.s0.0172kg/m

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

kg

2

1.085/2)0.664Re(

9.5mm)101m/(2.7555LRe

==

==

===

Using the correct correlation,

2

s

L

1/31/2

LL

L

.K(754.34W/m)ThL(Ts(L)q"

4.34W/m8W/m.K)/1m155.1(0.02h

hence,

0.332(2Pr0.332Rek

hLNu

==

==

===

(b) The drag force per unit area plate width isLs

LD

= 2' where the factor of two is

included to account for both sides of the plate. Hence with

/m2(1m)8.68W)T(T2Lhq

.K8.68W/m2hLhwithalso

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

0.0343N/m

isdragThe

(1.085k/2)1.328Re(

s

-

L

"

2-

L

2'

2

Ls

1/2

L

2

Ls

==

==

==

=

==


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Problem 10:Engine oil at 100C and a velocity of 0.1m/s flows over both surfaces of a 1-m-long

flat plate maintained at 20C. 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 (Tf=33K): =864kg/m3,=86.1*106m2/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)(11615LRe

1/31/3

t

1/21/2

L

===

===

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


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2

s

1/31/2

LL

C100).K(2016.25W/m)ThL(Tq"

0.3321m

0.140W/m.KPr0.3325LRe

L

kh

===

==

Also the local shear stress is

22

Ls

1/223

1/2

L

2

Ls

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 byLs

LD

= 2' where the factor of 2 is

included 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(

2

sL

_

2

LL

_

1/2231/2

L

2

Ls

===

==

===

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


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Problem 11:

Consider water at 27C in parallel flow over an isothermal, 1-m-long, flat plate with a

velocity of 2m/s. Plot the variation of the local heat transfer coefficient with distance

along 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*105.

Properties: Water (300K): =997kg/m3,=855*10-6N.s/m2, =/=0.858*10-6m2/s,k=0.613W/m. K, Pr=5.83

Analysis: (a) With

5

26L 102.33/sm100.858

1m2m/s

Lu

Re =

==

Boundary layer conditions are mixed and


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40554240449148715514.K)(W/mh

1.00.80.60.40.0215x(m)

xPr

u0.0296kPr0.0296Re

x

khx0.215m,xfor

12061768hx(W/m2.K)

0.2150.1x(m)

xPr

u0.332kPr0.332Re

x

khx0.215m,xfor

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

x

0.21/3

4/5

1/34/5x

1/21/3

1/2

1/31/2

x

65Lcx,c

2

==>

==

===

The Spatial variation of the local convection coefficient is shown above

(b) The average coefficient is

.K4106W/mh

871(5.83))1033[0.0379(2.1m

0.613W/m.K871)Pr(0.037Re

L

kh

2L

_

1/34/561/34/5

LL

_

=

==

Problem 12:


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A circular cylinder of 25-mm diameter is initially at 150C and is quenched by

immersion in a 80C oil bath, which moves at a velocity of 2m.s in cross flow over the

cylinder. 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;(Ts=423K): Prs=98.

Analysis: The initial heat loss per unit length is


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( ) 8.8W/mC80)0.025m(150.K1600W/mq'

.K1600W.mh

98

501(501)0.26(1312)

0.025m

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

D

kh

hence.7.4tablefrom0.6mand0.26CFind

1312/sm1038.1

m)2m/s(0.025VDRe

whenrelation.Zhukauskasthefromcomputedbemaywhere

)TD(Thq

2

2_

1/4

0.370.6

s

nm

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 ReD=3517.

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


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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 and

temperature of air in cross flow.

Find: Heat loss per unit length.

Schematic:

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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.K1600W.mh

98

501(501)0.26(1312)

0.025m

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

D

kh

hence.7.4tablefrom0.6mand0.26CFind

1312/sm1038.1

m)2m/s(0.025

VDRe

whenrelation.Zhukauskasthefromcomputedbemaywhere

2_

1/4

0.370.6s

nmD

_

26D

=

==

==

=

==

=

_

h

Hence the heat rate is

( ) 4100W/mC10)(-0.5m(150.K16.3W/mq' 2 ==

Comments: Note that qDm, in which case the heat loss increases significantly with

increasing D.

Problem 14:


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Atmospheric air at 25C and velocity of 0.5m/s flows over a 50-W incandescent bulb

whose surface temperature is at 140C. 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 prescribed

diameter and surface temperature.

Find: Heat loss by convection.

Schematic:

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

Properties: Air (Tf=25C, 1atm): =15.71*10-6m2/s, k=0.0261W/m. /K.Pr=.

0.71,=183.6*10-7N.s/m2; Air (Ts=140C, 1atm): =235.5*10-7N.s/m2

Analysis:

)TD(Thq s

_

=


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10.3WC25)(140(0.05m).Km

W11.4q

israteheattheand

.K11.4W/mh

235.5

183.6](0.71)0.06(1591)[0.4(1591)2

0.05m

K0.0261W/m.h

hence

1591/sm1015.71

0.05m0.5m/s

VD

Re

Where

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

kh

whenrelation.Whitakerthefromcomputedbemaywhere

22

2_

1/4

0.42/31/2_

26D

1/4

s

0.42/3

D

1/2

D

_

==

=

++=

=

==

++=

_

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 )


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Problem 15:

Water at 27 C flows with a mean velocity of 1m/s through a 1-km-long

cast iron pipe at 0.25 m inside diameter.

(a) Determine the pressure drop over the pipe length and the

corresponding pump power requirement, if the pipe surface is

clean.

(b) If the pipe surface roughness is increased by 25% because of

contamination, what is the new pressure drop and pump power

requirement.

Known: Temperature and velocity of water in a cast iron pipe of prescribed

dimensions.

Find: pressure drops and power requirement for (a) a clean surface and (b) a

surface with a 25% larger roughness.

Schematic:

Assumptions: (1) Steady, fully developed flow.

Properties: Water (300K):=1000 kg/m3,=855*10-6N.s/m2.

Analysis: (a) from eq.8.22, the pressure drop is


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L2D

ufp

2

m=

e=2.6*10

-4

m for clean cast iron; hence e/D=1.04*10

-3

. With

5

326

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.fthat8.3figfromfind

24

2423

==

==

)

The pump power requirement is

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

/4)up(DVp.P

2224

m

2.

==

==

(b) Increasing by 25% it follows that =3.25*10-4m and e/D=0.0013. With

ReD unchanged, from fig 8.3, it follows that f 0.0225. Hence

2.21kW)P/f(fPbar0.45)/f(fP1122122 ==== 1p

Comments: (1) Note that L/D=4000>>(xfd,h/D) 10 for turbulent flow and

the assumption of fully developed conditions is justified.

(2) Surface fouling results in increased surface and increases operating costs

through increasing pump power requirements.

Problem 16:


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32/56

Hence, regardless of whether the hydrodynamic or thermal boundary layer is

fully 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=hfd is a constant.

When flow is developing h> hfd . Hence, the temperature distributions appearas below.


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Problem 17:

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

uniform heat generation rate of 36.

/10 mWq = . 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 Tm,1=20C and the desired outlet

temperature is Tm,o=40C, 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 experiencing

uniform heat generation. Flow rate and inlet temperature of water flowing through thepipe.

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.


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

m = );

cp=4178J/kg. K, k=0.617W/m. K,=803*10-6N.s/m2, 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/m10

C)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

2

i

2

0

.

im,om,p

.

=

=

=

==

(b) The maximum wall temperature exists at the pipe exit (x=L) and the

insulated surface (r=ro). The radial temperature distribution in the wall is of

the form

Tnr2k

qr-r

4k

qCCnr

2k

qrr

4k

q-T)T(r:rr

2k

qrC

r

Cr

2k

q0

dr

dT;rr

;conditionsboundarythegconsiderin

CnrCr4k

qT(r)

si

.2

o2

i

.

22i

.2

o2

i

.

si1

.2

o

1o

1

o

.

rro

21

2

.

o

+=++===

=+==

=

++=

=

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


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LD

Dq((/4)q

innertheTs,where

4

q-)T(rT

)r-(r4k

qT(r)

i

.2

o"

s

omaxw,

2

i

2

.

=

=

=

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

7928N.s/m103(0.02m)80

0.1kg/s4

D

m4Re

26

i

.

D =

==

The flow is turbulent and, with (L/Di)=(8.87m/0.02m)=444>>(xfd/D)10,it is

also fully developed. Hence, from the Dittus-Boelter correlation,

.K1840W/m5.45)0.023(79280.02m

0.617W/m.K)pr(0.023Re

D

kh

20.44/50.44/5

D

i

===

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

C52.4C48.20.01

0.02n

215W/m.K

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

415W/m.K

W/m10T

and

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

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

4D

)D(Dq

23622

36

maxw,

2

2236

om,

i

2

i

2

o

.

s

=++=

=+

=+

=

T


36/56

Comments: The physical situation corresponds to a uniform surface heat

flux, and Tm increases linearly with x. in the fully developed region, Ts also

increases linearly with x.


37/56

Problem 18:

The surface of a 50-mm diameter, thin walled tube is maintained thin

walled tube is maintained at 100C. In one case air is cross flow very the

tube with a temperature of 25C and a velocity of 30m/s. In another caseair is in fully developed flow through the tube with a temperature of 25C

and a mean velocity of 30m/s. compare the heat flux from the tube to the

air for the two cases.

Known: surface temperature and diameter of a tube. Velocity and

temperature of air in cross flow. Velocity and temperature of air in fully

developed internal flow.

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

Schematic:

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

temperature, (3) fully developed internal flow

Properties: Air (298K): =15.71*10-6m2/s, k=0.0261W/m.K, Pr=0.71

Analysis: for the external and internal flow

4

261055.9

/1071.15

05./30Re =

===

sm

msmDuVD mD

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


38/56

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

=== sm

DD CuN

Hence, the convection coefficient and heat rate are

2"

__

./4.116)(

05.0

./0261.0

mWTThq

m

KmWNuD

D

kh

s ==

==

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

2

__

4.05/4_

100(./101)(

1905.0

./0261.0

9(023.0PrRe023.0

KmWTTh

KmWNu

D

kh

NU

ms

D

DD

==

==

==

"q

isfluxheattheand

Comments: Convection effects associated with the two flow conditions are

comparable.


39/56

Problem 19:

Cooling water flows through he 25.4 mm diameter thin walled tubes of a

stream 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 the

tubes are 5 m long, what is the water outlet temperature? Water properties

may 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:

Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligible

kinetic energy, potential energy and flow work changes.

Properties: Water (300K):=997kg/m3, cp=4179J/kg.K, =855*10-

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

Analysis:


40/56

29,618kg/s.m10855

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

DuRe

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

6

3

mD

_

p

.

im,ssom,

=

==

=

The flow is turbulent. Since L/D=197, it is reasonable to assume fully

developed flow throughout the tube. Hence

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

.K)5m(4248W/m(0.0254m)(60K)exp350KT

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

With

.K4248W/m4m)/m.K/0.025176(0.613WNuD(k/D)h

2

om,

232

m

.

2_

=

===

===

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

reevaluating properties at 306.5KT_

m = .

Problem 20:


41/56

The air passage for cooling a gas turbine vane can be approximated as a tube

of 3-mm diameter and 75-mm length. If the operating temperature of the

vane is 650C, calculate the outlet temperature of the air if it enters the tube

at 427C and 0.18kg/h.

Known: gas turbine vane approximation as a tube of prescribed diameter and

length maintained at an known surface temperature. Air inlet temperature

and flow rate.

Find: outlet temperature of the air coolant.

Schematic:

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

potential energy changes.

Properties: Air (assume )1,780_

atmKTm = : cp=1094J/kg. K, k=0.0563 W/m.

K, =363.7*10-7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650C=923K, 1atm):

=404.2*10-7N.s/m2.

Analysis: For constant wall temperatures heating,

=

.

_

,

,exp

pims

oms

mc

hPL

TT

TT

Where P=D. for flow in circular passage,

N.s/m103.7(0.03m)36

/3600s/h)0.18kg/h(14

D

m4Re

7

.

D ==


42/56

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

correlation including combined entry length fields.

.K87.5W/m10404.2

10363.7

25

0.7065841.86

0.003m

K0.0563W/m.h

L/DPrRe1.86

kDhNu

2

0.14

7

71/3_

0.14

s

1/3

D

__

D

=

=

==

Hence, the air outlet temperature is

=

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

.K87.5W/m0.075m(0.003m)expC427)(650

T650 2om,

Tm,o=578C

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

mT =775K which

is in good agreement with our assumption to evaluate the thermo physical

properties.


43/56

Problem 21

A household oven door of 0.5-m height and 0.7-m width reaches an average

surface temperature of 32C during operation. Estimate the heat loss to the

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

relative to that by radiation.

Known: Oven door with average surface temperature of 32C in a room

with ambient temperature at 22C.

Find: Heat loss to the room. Also, find effect on heat loss if emissivity of

door is unity and the surroundings are at 22C.

Schematic:

Assumptions: (1) Ambient air in quiescent, (2) surface radiation effects

are negligible.

Properties: Air (Tf=300K, 1atm): =15.89*10-6 m2/s, k=0.0263W/m. K,

=22.5*10-6 m2/s, Pr=0.707, =1/Tf=3.33*10-3K-1

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

ambient air is

)T(TAhq ss

_

=

Where_

h can be estimated from the free convection correlation for a

vertical plate,


44/56

[ ]

2

8/279/16

1/6

L

__`

(0.492/Pr)1

0.387Ra0.825

k

LhLNu

++==

The Rayleigh number,

4

2626

322

s

L 101.142/sm1022.5/sm1015.89

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

)LT(TgRa =

=

=

Substituting numerical values into equation, find

[ ].K3.34W/m63.5

0.5m

K0.0263W/m.LNu

Lh

63.5

(0.492/Pr)1

)1020.387(1.140.825

k

LhLNu

2_`_

2

8/279/16

1/88_

_`

===

=

+

+==

kL

The heat rate using equation is

11.7W22)K(320.7)m.k(0.53.34W/mq 22 ==

Heat loss by radiation, assuming =1 is

21.4W]k22)(27332)[(273.KW/m5.67100.7)m1(0.5q

)T(TAq

4244282

rad

4

sur

4

ssrad

=++=

=

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

Comments: (1) Note the characteristics length in the Rayleigh number is the

height of the vertical plate (door).

Problem 22


45/56

An Aluminum alloy (2024) plate, heated to a uniform temperature of

227C, is allowed to cool while vertically suspended in a room where the

ambient air and surroundings are at 27C. The late is 0.3 m square with a

thickness of 15 mm and an emissivity of 0.25.

a. Develop an expression for the time rate of change of the platetemperature assuming the temperature to be uniform at any

time.

b. Determine the initial rate of cooling of the plate temperature is

227C.

c. Justify the uniform plate temperature assumption.

Known: Aluminum plate alloy (2024) at uniform temperature of 227C

suspended in a room where the ambient air and the surroundings are at 27C

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

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

uniform plate temperature assumption.


46/56

Schematic:

Properties: Aluminium alloy 2024 (T=500K):=2270 kg/m3,

k=186W/m.K, c=983 J/kg.K; Air (Tf=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..

stout EE = .


47/56

)T(T)T(Th[c

2

dt

dTor

dt

dTA)T(T2A)T(T2Ah

4

sur

4

ssL

.

cs

4

sur

4

ssss

_

L +

==

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

(b) To evaluate (dt/dx), estimate_

Lh . Find first the Rayleigh number

8

2626

322

s

L101.308

/sm10.33/sm1026.41

0.3m27)K27(1/400k)(29.8m/s

)LT(TgRa =

=

=

8

Substituting numerical values, find

[ ] [ ]5.55

)=

+

+=

++=

4/99/16

82

4/99/16

1/4

L_`

90)(0.492/0.61

1080.670(1.300.68

(0.492/Pr)1

0.670Ra0.68LNu

0.099K/s

.K)(500W/m100.25(5.6727)K.K(2276.25W/m983J/kg.K0.015m2770Kg/m

2

dt

dT

.K6.25W/mK/0.3m0.0338W/m.55.5k/LNuh

282

3

2_`

L

_

=

+

=

===

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

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


48/56

____

1.0/, =+= khBihhh totradconvtot . Using the linearized radiation coefficient

relation find,

.3.86W/m3002)K300)(5002)(500.K8W/m100.25(5.67)T)(TT(Th 23422sur2

ssursrad

_

=++=++=

Hence Bi=(6.25+3.86) W/m2.k (0.015m)/186W/m. K=8.15*10-4. Since

Bi


49/56

Problem 23

The ABC Evening News Report in news segment on hypothermia research

studies at the University of Minnesota claimed that heat loss from the body

is 30 times faster in 1 0C water than in air at the same temperature. Isthat a realistic statement?

Known: Person, approximated as a cylinder, experiencing heat loss in

water or air at 10C.

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 25C, (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 vf/cp =1.431*10-7m2/s, f=174*10-6K-1.

Analysis: in both water (wa) an air(a), the heat loss from the lateral

surface of the cylinder approximating the body is

)TDL(Thq s_

=

Where Ts and Tare the same for both situations. Hence,

_

a

wa

_

a

wa

h

h

q

q=

Vertical cylinder in air:

9

2626

323

L 105.228/sm28.410/sm10*19.91

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

gTRa =

==


50/56

.K2.82W/mh173.4)100.1(5.228CRak

LhNu

1/3,nand0.1Cwith

2L

_1/39n

L

__

LL =====

==

Vertical cylinder in water.

11

2726

3162

L 109.643/sm101.431/sm101.081

10)K(1.8m)(25K101749.8m/sRa =

=

.K328W/mh978.9)100.1(9.643CRa

k

hLNu

1/3,nand0.1Cwith

2L

_1/311n

L

__

L =====

==

Hence, from this analysis we find

30ofclaimthewithpoorlycompareswhich117.K2.8W/m

.K328W/m

q

q2

2

a

wa ==


51/56

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 test

cell. In the experiments the water and the inner surface of the test cell are

maintained at 300 and 290K, respectively. To achieve similarity between

conditions 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 in

terms of the Rayleigh number, what is the ratio f the average convection

coefficient 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 convection

coefficient for the two cases.

Schematic:

Assumptions: (1) Air and water are quiescent; (2) Flow conditions

correspond to free convection boundary layer development on an isothermal

vertical plate, (3) constant properties.

Properties: Air (Tf=300K, 1 atm); K=0.0293 W/m.K, ==15.9*10-6m2/s,

=22.5*10-6 m2/s; =1/Tf=3.33*10-3K-1, k=0.0263W/m.K; water

(Tf=295K):=998kg/m3, =959*10-6N.s/m2, cp=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.


52/56

Analysis: Similarity requires that RaL,a=RaL,w where

mmL

oHL

L

hence

w

a

airw

a

w

45.0)179.0(5.2

10*228.0

10*33.3

10*5.22*9.15

10*45.1*61.9

)(

)(

,

)L-T(TgRa

3

3

12

143/1

2

3

sL

==

=

=

=

3

w

a

a

w

w

_

_

a

_

wL,

_

aL,wL,aL,

10*7.810.606

0.0263

2.5

0.45

k

k

L

L

h

h

.henceNuNuthatfollowsit,RaRaif

===

==

Comments: Similitude allows us to obtain valuable information for one

system by performing experiments for a smaller system and a different

fluid.


53/56

Problem 25

A square plate of pure aluminum, 0.5 m on a side and 16 mm hick, is

initially at 300C and is suspended in a large chamber. The walls of thechamber are maintained at 27C, as is the enclosed air. If the surface

emissivity of the plate temperature during the cooling process? Is it

reasonable 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 temperature

gradients in the plate during the cooling process.

Schematic:

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, cp=1022J/kg.K, =2702 kg/m3:

Air (Tf=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,


54/56

psurss

pstsurss

wcTTTThAdtdT

dtdTVcETTTThAq

/)]4()([2/

]/[)]4()([2

4_

.4

_

+=

==+=

8

2626

323s

L 105.58/sm017.44/sm10*30.72

)K(0.5m)721(300-0.00229K9.8m/s)L-T(TgRa =

==

KmWh

Ra

L

kh L

./8.5

])687.0/492.0(1[

)1058.5(670.068.0

5.0

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

{ }

sKdt

dT

kkgJmmkg

KKKmWCKmW

dt

dT

/136.0

./1022)016.0(/2702

])300()573[(./1067.525.0)27300(./8.523

444282

=

+=

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

plate thickness, calculate

"

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

/)2/(

=

==

hence

qhwherekwhBi toeffeff

And the assumption is excellent.

Comments: (1) Longitudinal (x) temperature gradients are likely to me more

severe than those associated with the plate thickness due to the variation of h

with x. (2) Initially qconvqrad.

Problem 26

Beer in cans 150 mm long and 60 mm in diameter is initially at 27 C and is

to be cooled by placement in a refrigerator compartment at 4C. In the

interest of maximizing the cooling rate, should the cans be laid horizontally

or vertically in the compartment? As a first approximation, neglect heat

transfer from the ends.


55/56

Known: Dimensions and temperature of beer can in refrigerator

compartment.

Find: orientation which maximum cooling rate.

Schematic:

Assumptions: (1) End effects are negligible, (2) Compartment air is

quiescent, (3) constant properties.

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

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

Analysis: The ratio of cooling rates may be expressed as

h

v

s

s

h

v

qh

v

h

h

TT

TT

DL

DL

h

hq_

_

_

_

)(

)(=

==

For the vertical surface, find

639

L

393

26-26-

-1-323

s

L

1044.8)15.0(105.2Ra

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

C)(23K103.479.8m/s)L-T(TgRa

==

=

==

Using the correct correlation


56/56

KmWm

KmW

L

kNuhh

NuL

LvL ./03.515.0

./0254.07.29hence,

7.29])71.0/492.0(1[

)1044.8(387.0825.0

2___

2

27/816/9

6/16

====

=

+

+=

Using the correct correlation,

97.018.5

03.5,

./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

DhD

D

q

qhence

KmWm

KmW

D

kNuhh

Nu

Comments: in view of the uncertainties associated with equations and the

neglect of end effects, the above result is inconclusive. The cooling rates are

are approximately the same.

Documents

Module6 WorkedOut Problems