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 ratio xx 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.

Known: Variation of local convection coefficient with x for free convection from a heated vertical plate. Find: ratio of average to local convection coefficient.

Schematic:

Boundary layer,hx=C x

-1/4 w here C is a constant

Ts

x

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

x

1/4x

x

0

1/4x

0xx

h34Cx

34x3/4

xC

34h

dxxxCdxh

x1h

===

==

−−

−−

∫∫

Hence 34

hh

x

x =−

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

xx hh−

xx hh34

=−

4/1−=Cxhx

Comments: note that

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

coefficients for an entire plate of length L is Lh−

=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

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+==

n

oD r

ra1k

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/ν) 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 for the 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

o

o

s

r

0no

2n2

3o

no0

r

02o

As

s

2)r(nar

2r

rkNuoh

]2π2πr)a(r/r[1NuDk

πr1h

hdAA1h

⎥⎦

⎤⎢⎣

⎡+

+=

+=

=

+−

−

−

∫

∫

Where Nuo 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/ro2Nu

kDhNuD

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.

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:

u∞,T∞

T(y)=D+Ey+Fy2-Gy3

Ts=T(0)=D

U(y)=Ay-By2-Cy

3

y y

Analysis: The shear stress at the wall is

.]32[ 02

0

µµµτ ACyByAyu

yy

s =−+=∂∂= =

=

Hence, the friction coefficient has the form,

2f

22s

f

u2AνC

ρu2Aµ

/2ρuτ

C

∞

∞∞

=

==

The convection coefficient is

∞

∞

=

∞

=

−−

=

−−+−

=−

∂∂−=

TDEkh

TD]3Gy2Fy[Ek

TTy)T/(k

h

f

0y2

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 the profile.

Problem 4: In a particular application involving airflow over a heated surface, the boundary layer temperature distribution may be approximated as

⎟⎠⎞

⎜⎝⎛−−=

−− ∞

∞ vyuPrexp1

TTTT

s

s

Where y is the distance normal to the surface and the Prandtl number, Pr=cpµ/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:

)Prexp(1)(

νyu

TTTyT

s

s ∞

∞−−=

−−

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

Analysis: Applying the Fourier’s law at y=0, the heat flux is

0ys

0y

"s ν

yuPrexpν

uPr)Tk(TyTkq

=

∞∞∞

=⎥⎦⎤

⎢⎣⎡−⎥⎦

⎤⎢⎣⎡−−=

∂∂−=

νu)PrTk(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 Fourier’s law.

Problem 5:

Consider a lightly loaded journal bearing using oil having the constant properties µ=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.

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+=

22

o Ly

LyU

2kµTT(y)

The position of maximum temperature is obtained from

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−== 2

2

L2y

L1U

2kµ0

dydT

y=L/2. Or, The temperature is a maximum at this point since hence 0.T/dyd 22 <

22

2 U8kµTo

41

21U

2kµToT(l/2)axTm, +=⎥

⎦

⎤⎢⎣

⎡−+==

C40.83T

0.15W/m.K8/s)kg/s.m(10m10C40T

max

22

max

°

−°

=×

+=

Comments: Note that Tmax increases with increasing µ and U, decreases with increasing k, and is independent of L.

Problem 6: Consider two large (infinite) parallel plates, 5mm apart. One plate is stationary, 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 are separated 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 area required to maintain the above condition? What is the corresponding requirement? 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) viscous dissipation,(3) maximum fluid temperature. Schematic:

Assumptions: (1) Fully developed Couette flow, (2) Incompressible fluid with constant properties. Properties: Air (300K); µ=184.6*10-7 N.s/m2, k=26.3*10-3W/m.K; water (300K): µ=855*10—6N.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 τ=µ (du/dy) = µ (U/L).

34.2N/m0.005m200m/sN.s/m10855 τ : Water

0.738N/m0.005m200m/sN.s/m10184.6 τ : Air

226-water

227-air

=××=

=××=

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

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 .henceµ(U/L)µ(du/dy)µ 22 ==ϕ

W/m101.370.005m200m/sN.s/m10855 (µµφ : Water

W/m102.950.005m200m/sN.s/m10184.6 (µµφ : Air

362

26-water

342

27-air

×=⎥⎦⎤

⎢⎣⎡××=

×=⎥⎦⎤

⎢⎣⎡××=

The location of the maximum temperature corresponds to ymax=L/2. Hence Tmax=To+µU2/8k and

C34.0K80.613W/m.

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

C30.5K0.0263W/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, ),(LAµφ must equal the power, P.

(2) Although aterw )( µφ >> air )( µφ , 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∞=20°C, while the plate is maintained at Ts=120°C. The sir flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075N. What is the rate of heat transfer from both sides of the plate to the air?

Known: Variation of hx 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

2.NuL

NuL

andk

2CLk

(L)2CLNuL

Hence

2CLLL

2CdxxLCdxh

L1h

wherek

LhNu

kCL

k)L(CL

kLhNu

-

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

−. Also

note that

∫≠− L

0xL dxNu

L1Nu

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 (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(70°C,1atm): ρ=1.018kg/m3, cp=1009J/kg.K, pr=0.70, ν=20.22*10-6m2/s. Analysis: the rate of heat transfer from the plate is

)T(T(L)h2q s2

∞

−−=

Where

−h may be obtained from the Chilton-Colburn analogy,

240WqC20)(120.K)(0.3m)2(30W/mq

is rateheat The.K30W/mh

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

Prcρu2

Ch

hence,

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

(0.2m)(0.075N/2)21

/2ρuτ

21

2C

Prcρu

htPrS2

Cj

22

2-

34-

2/3p

f-

423

2

2sf

2/3

p

2/3fH

=°−=

=

−×=

=

×===

===

−

−∞

−

∞

−−

∞

−−

−−

Comments: Although the flow is laminar over the entire surface ( )100.4/1022.20/2.0/40/Re 526 ×=××== −

∞ smmsmLuL ν , 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.

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, the surface shear stress, and the heat flux at the trailing edge. Determine the drag force on the 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 a Plate 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=75°C

Assumptions: (1) Critical Reynolds number is 5*105, (2) flow over top and bottom surfaces Properties: (Tf=323K, 1atm) Air: ρ=1.085kg/m3,ν=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

526L 102.75/sm1018.2

m1m/s5νLu

Re ×=×

×== −∞

Hence the flow is laminar, and at x=L

22Ls,

1/2523

1/2L

2Ls,

1/251/2L

0.0172N/m.s0.0172kg/mτ

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

21.085/2)0.664Re(ρρτ

9.5mm)101m/(2.7555LReδ

==

×==

=××==−

−

Using the correct correlation,

22s

2L

1/31/21/31/2L

LL

C)217W/m25C.K(754.34W/m)ThL(Ts(L)q"

.K4.34W/m8W/m.K)/1m155.1(0.02hhence,

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

hLNu

°−°=−=

==

====

∞

(b) The drag force per unit area plate width is LsLD−

= τ2' where the factor of two is included to account for both sides of the plate. Hence with

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

.K8.68W/m2hLh with also

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

0.0343N/mτ

is drag The

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

2s

-

L"

2-

L

2'

2Ls

1/25231/2L

2Ls

=°−=−=

==

==

=

×==

∞

−

−−∞

−

Problem 10: Engine oil at 100°C and a velocity of 0.1m/s flows over both surfaces of a 1-m-long flat 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 heat transfer per unit plate width.

Schematic:

Assumptions: engine oil (Tf=33K): ρ=864kg/m3,ν=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/sνLu

ReL 26- =×== ∞

Hence the flow is laminar at x=L, and

0.0143m)0.147(1081δPrδ

0.147m)5(1m)(11615LReδ1/31/3

t

1/21/2L

===

===−−

−−

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

22s

21/31/31/2LL

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

.K16.25W/m)(1081)0.332(11611m

0.140W/m.KPr0.3325LReLkh

−=°−=−=

===

∞

−

Also the local shear stress is

22Ls

1/223

1/2L

2Ls

0.0842N/m.s0.0842kg/mτ

)0.664(1161(0.1m/s)2

864kg/m/2)0.664Re(ρρτ

==

==

−

−−∞

−

(c) With the drag force per unit width given by LsLD−

= τ2' 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

that follows alsoit .K,32.5W/m2hh with

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

2sL

_

2LL

_

1/2231/2L

2Ls

−=−=−=

==

===

°∞

−−∞

−

Comments: Note effect of Pr on (δ/δt).

Problem 11:

Consider water at 27°C 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 prescribed length.

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):ρ =997kg/m3,µ=855*10-6N.s/m2, ν=µ/ρ=0.858*10-6m2/s, k=0.613W/m. K, Pr=5.83 Analysis: (a) With

526L 102.33/sm100.858

1m2m/sνLu

Re ×=×

×== −∞

Boundary layer conditions are mixed and

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

1.0 0.8 0.6 0.4 0.0215 x(m)

xPrν

u0.0296kPr0.0296Rexkhx 0.215m,xfor

1206 1768 hx(W/m2.K) 0.215 0.1 x(m)

xPrν

u0.332kPr0.332Rexkhx 0.215m,x for

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

.K4106W/mh

871(5.83))1033[0.0379(2.1m

0.613W/m.K871)Pr(0.037ReLkh

2L

_

1/34/561/34/5LL

_

=

−×=−=

Problem 12: 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 bath f 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): ν=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

( ) 8.8W/mC80)0.025m(150π.K1600W/mq'

.K1600W.mh

98501(501)0.26(1312)

0.025m0.138W/m.K)1/4/Pr(PrPrCRe

Dkh

hence . 7.4 table from 0.6m and 0.26C Find

1312/sm1038.1

m)2m/s(0.025ν

VDRe

when relation. Zhukauskas the from computed be may where

)TπD(Thq

2

2_

1/40.370.6

snm

D

_

26D

s

_

=°−=

=

⎟⎠⎞

⎜⎝⎛==

==

=×

==

−=

∞

=

∞

_

h

Comments: Evaluating properties at the film temperature, Tf=388K(ν=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 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:

V=5m/sT∞=263K

Air D=0.5mmTs=150 C

Assumptions: (1) steady-state conditions, (2) uniform surface temperature Properties: Air (T=263K, 1atm):ν=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

)TπD(Thq s

_

∞−=

.K1600W.mh

98501(501)0.26(1312)

0.025m0.138W/m.K)1/4/Pr(PrPrCRe

Dkh

hence . 7.4 table from 0.6m and 0.26C Find

1312/sm1038.1

m)2m/s(0.025ν

VDRe

when relation. Zhukauskas the from computed be may where

2_

1/40.370.6

snm

D

_

26D

=

⎟⎠⎞

⎜⎝⎛==

==

=×

==

∞

=

_

h

Hence the heat rate is

( ) 4100W/mC10)(-0.5m(150π.K16.3W/mq' 2 =°−= Comments: Note that q’αDm, in which case the heat loss increases significantly with increasing D.

Problem 14:

Atmospheric air at 25°C and velocity of 0.5m/s flows over a 50-W incandescent bulb whose 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 prescribed diameter and surface temperature.

Find: Heat loss by convection.

Schematic:

Assumptions: (1) steady-state conditions, (2) uniform surface temperature. Properties: Air (Tf=25°C, 1atm): ν=15.71*10-6m2/s, k=0.0261W/m. /K.Pr=. 0.71,µ=183.6*10-7N.s/m2; Air (Ts=140°C, 1atm): µ=235.5*10-7N.s/m2

Analysis:

)TπD(Thq s

_

∞−=

10.3WC25)(140π(0.05m).KmW11.4q

is rateheat the and

.K11.4W/mh

235.5183.6](0.71)0.06(1591)[0.4(1591)2

0.05mK0.0261W/m.h

hence

1591/sm1015.71

0.05m0.5m/sν

VDRe

Where

])(µµ/)Pr0.06Re(0.4Re[2Dkh

when relation. Whitaker the from computed be may 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 (µ/µs )<1.

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):ρ=1000 kg/m3, µ=855*10-6 N.s/m2. Analysis: (a) from eq.8.22, the pressure drop is

L2Dρuf∆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/m108550.25m1m/s

νDuRe ×=

××== −

0.42barN/m104.2P

.mkg/s104.21000m2(0.25m)

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

, hence 0.021.fthat 8.3 fig from find

24

2423

=×=∆

×==∆

≈

)

The pump power requirement is

2.06kW1m/s/4)m0.25(πN/m104.2P

/4)u∆p(πDV∆p.P

2224

m2

.

=×××=

==

(b) Increasing ε by 25% it follows that ε=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(fP bar 0.45)∆/f(f∆P 1122122 ==== 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:

Consider flow in a circular tube. Within the test section length (between 1 and 2) a constant heat flux q”s is maintained.

(a) For the two cases identified, sketch, qualitatively, the surface

temperature Ts(x) and the fluid mean temperature Tm(x) as a function of distance along the test section x. in case A flow is hydro dynamically and thermally fully developed. In case B flow is not developed.

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

Known: internal flow with constant surface heat flux, "

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

Flow

1 2

q”s=constant

Assumptions: (a) Steady-state conditions, (b) constant properties, (c) incompressible flow. Analysis: Based upon the analysis, the constant surface heat flux conditions,

constantdx

dTm =

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 appear as below.

Problem 17:

A thick-walled, stainless steel (AISI 316) pipe of inside and outside diameter 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=20°C and the desired outlet temperature is Tm,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 experiencing uniform 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.

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

m = ); cp=4178J/kg. K, k=0.617W/m. K,ρ=803*10-6N.s/m2, Pr=5.45

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

8.87mL

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

)LD(ππ/4)(q

)T(TmcL

)LD(ππ/4)(qq)T(Tcm

22362i

20

.

.

im,om,p

2i

20

.

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

Tnr 2kqr

-r4kqC Cnr

2kqr

r4k

q-T)T(r :rr

2kqr

C rCr

2kq0

drdT ;rr

; conditions boundary the gconsiderin

CnrCr4kqT(r)

si

.2o2

i

.

22i

.2o2

i

.

si1

.2o

1o

1o

.

rro

212

.

o

+=++===

=+−−==⎟⎠⎞=

++−=

=

λλ

λ

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

)Th(T4D

)D(DqLπD

)LDq(π(π/4)q

from followsit exit,at wall the of temperture surafaceinner the Ts, where

Trr

n 2kqr

)r-(r4kq-)T(rT

Trrn

2kqr

)r-(r4kqT(r)

om,si

2i

2o

.

i

.2i

2o"

s

si

o

.2o2

i2o

.

omaxw,

si

.2o2

i2

.

−=−

=−

=

++==

++=

λ

λ

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

7928N.s/m103π(0.02m)80

0.1kg/s4µπD

m4Re 26i

.

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(0.023ReDkh 20.44/50.44/5

Di

===

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

C52.4C48.20.010.02n

215W/m.K(0.02)W/m10](0.01)[(0.02)

415W/m.KW/m10T

and

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

](0.02m)[(0.04m)W/m10T4D

)D(Dq

23622

36

maxw,

2

2236

om,i

2i

2o

.

s

°=°++−−=

°=°+×

−=+−

=

λ

T

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.

T

xfd

Ts(x)

Tm(x)

x

Problem 18: The surface of a 50-mm diameter, thin walled tube is maintained thin walled 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 case air is in fully developed flow through the tube with a temperature of 25°C 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:

Air

Um=30m/sTm=25°C

V=30m/sT =25 C

Ts=100 C

D=0.05m

Air

Assumptions: (1) steady-state conditions, (2) uniform cylinder surface temperature, (3) fully developed internal flow Properties: Air (298K): ν=15.71*10-6m2/s, k=0.0261W/m.K, Pr=0.71 Analysis: for the external and internal flow

426 1055.9

/1071.1505./30Re ×=

××=== − sm

msmDuVD mD νν

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

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

=×== smDD CuN

Hence, the convection coefficient and heat rate are

232"

2__

/1073.8)25100(./4.116)(

./4.11622305.0

./0261.0

mWCkmWTThq

KmWm

KmWNuDDkh

s ×=−=−=

=×==

°∞

Using the Dittus-Boelter correlation, for the internal flow, which is Turbulent,

232

__

4.05/4244.05/4_

/1058.7)25100(./101)(

.2/10119305.0

./0261.0

193)71.0()1055.9(023.0PrRe023.0

mWCKmWTTh

KmWKmWNuDkh

NU

ms

D

DD

×=°−=−=

=×==

=×==

"q

isflux heat the and

Comments: Convection effects associated with the two flow conditions are comparable.

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:

L=5m

D=0.0254m

Tm,i=290KUm=1m/s

Ts=350K

Assumptions: (1) Negligible tube wall conduction resistance, (2) Negligible kinetic energy, potential energy and flow work changes. Properties: Water (300K):ρ=997kg/m3, cp=4179J/kg.K, µ=855*10-6kg/s.m,k=0.613W/m.K, Pr=5.83 Analysis:

29,618kg/s.m10855

54m(1m/s)0.02997kg/mµ

DρuRe

]h)cm(ππDL)exp[T(TTT

6

3m

D

_

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/mπ(0.0254m)(60K)exp350KT

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

With

.K4248W/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 306.5KT

_

m = .

Problem 20: 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 650°C, calculate the outlet temperature of the air if it enters the tube at 427°C 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:

hkgm /18.0.

=

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, µ=363.7*10-7 N.s/m2, Pr=0.706; Pr=0.706;(Ts=650°C=923K, 1atm): µ=404.2*10-7N.s/m2. Analysis: For constant wall temperatures heating,

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−=

−−

.

_

,

, expp

ims

oms

mc

hPLTTTT

Where P=πD. for flow in circular passage,

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

πDµm4Re 7

.

D −××==

The flow is laminar, and since L/D=75mm/3mm=25, the Sieder-Tate correlation including combined entry length fields.

.K87.5W/m10404.210363.7

250.7065841.86

0.003mK0.0563W/m.h

µµ

L/DPrRe1.86

kDhNu

20.14

7

71/3_

0.14

s

1/3D

__

D

=⎟⎟⎠

⎞⎜⎜⎝

⎛××

⎟⎠⎞

⎜⎝⎛ ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛==

−

−

Hence, the air outlet temperature is

⎟⎟⎠

⎞⎜⎜⎝

⎛×

××−=°−

−1094J/kg.K)kg/s(0.18/3600

.K87.5W/m0.075mπ(0.003m)expC427)(650

T650 2om,

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

_

mT =775K which is in good agreement with our assumption to evaluate the thermo physical properties.

Problem 21 A household oven door of 0.5-m height and 0.7-m width reaches an average surface 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 the surroundings are also at 22°C, comment on the heat loss by free convection relative to that by radiation.

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

0.7m

L=0.

5m

Assumptions: (1) Ambient air in quiescent, (2) surface radiation effects are negligible. Properties: Air (Tf=300K, 1atm): ν=15.89*10-6 m2/s, k=0.0263W/m. K, α=22.5*10-6 m2/s, Pr=0.707, β=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,

[ ]2

8/279/16

1/6L

__`

(0.492/Pr)1

0.387Ra0.825kLhLNu

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++==

The Rayleigh number,

4

2626

322s

L 101.142/sm1022.5/sm1015.89

m322)K0.52(1/300k))39.8m/sνα

)LT(TgRa ×=

×××−=

−= −−

∞β

Substituting numerical values into equation, find

[ ].K3.34W/m63.5

0.5mK0.0263W/m.LNu

Lh

63.5(0.492/Pr)1

)1020.387(1.140.825

kLhLNu

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 ε=1 is

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

)Tσ(TεAq4244282

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 the height of the vertical plate (door).

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 a thickness 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 any time.

b. Determine the initial rate of cooling of the plate temperature is 227°C.

c. Justify the uniform plate temperature assumption. Known: Aluminum plate alloy (2024) at uniform temperature of 227°C suspended 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:

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

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

6m2/s,k=0.0338 W/m.K,α=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 =− .

)Tεσ(T)T(Th[c2

dtdTor

dtdTρA)Tσ(Tε2A)T(T2Ah 4

sur4ssL

.

cs4sur

4ssss

_

L −+−∫−==−−−− ∞∞ λ

λ

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

82626

322s

L 101.308/sm10.33/sm1026.41

0.3m27)K27(1/400k)(29.8m/sνα

)LT(TgRa ×=

××××−=

−= −−

∞

8β

Substituting numerical values, find

[ ] [ ] 5.55) =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+×+=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++= 4/99/16

82

4/99/16

1/4L

_`

90)(0.492/0.611080.670(1.300.68

(0.492/Pr)10.670Ra0.68LNu

0.099K/s

)K300.K)(500W/m100.25(5.6727)K.K(2276.25W/m983J/kg.K0.015m2770Kg/m

2dtdT

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

442823

2_`

L

_

=

−×+−×××

−=

=×==

−

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

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

and____

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

relation find,

.K3.86W/m3002)K300)(5002)(500.K8W/m100.25(5.67)T)(TTεσ(Th 23422sur

2ssursrad

_

=++−×=++=

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

Bi<<0.1, the assumption is appropriate.

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 0°C water than in air at the same temperature. Is that a realistic statement?

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

ν==19.91*10-6m2/s, α=28.4*10-6 m2/s; Water (290K); k=0.598 W/m.K; ν=µvf=1.081*10-6m2/s, α=k vf /cp =1.431*10-7m2/s, βf=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

_

∞−= π

Where Ts 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/sνα

gββ∆TRa ×=×

−== −−

.K2.82W/mh 173.4)100.1(5.228CRak

LhNu

1/3,n and 0.1C with

2L

_1/39n

L

__

LL ==×===

==

Vertical cylinder in water.

112726

3162

L 109.643/sm101.431/sm101.081

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

−−

.K328W/mh 978.9)100.1(9.643CRak

hLNu

1/3,n and 0.1C with

2L

_1/311n

L

__

L ==×===

==

Hence, from this analysis we find

30 of claim the with poorly compares which 117.K2.8W/m.K328W/m

qq

2

2

a

wa ==

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 is simulated 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:

L a=2

.5m L

w

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, ν==15.9*10-6m2/s, α=22.5*10-6 m2/s; β=1/Tf=3.33*10-3K-1, k=0.0263W/m.K; water (Tf=295K):ρ=998kg/m3, µ=959*10-6N.s/m2, cp=4181 J/kg.K, β=227.5*10-

6K-1, k=0.606 W/m.K; Hence ν=µ/ ρ =9.61/s, α=k / ρ cp =1.45*10-7m2/s.

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

mmLoHL

L

hence

w

a

airw

a

w

45.0)179.0(5.210*228.0

10*33.310*5.22*9.1510*45.1*61.9

)()(

,

)L-T(TgRa

3

3

12

143/1

2

3s

L

==

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

=

−

−

−

−

∞

βανβαν

ναβ

3

w

a

a

w

w

_

_

a

_

wL,

_

aL,wL,aL,

10*7.810.6060.0263

2.50.45

kk

LL

h

h

.henceNuNu that followsit ,RaRa if

−===

==

Comments: Similitude allows us to obtain valuable information for one system by performing experiments for a smaller system and a different fluid.

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 the chamber 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:

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, cp=1022J/kg.K, ρ=2702 kg/m3: Air (Tf=436K, 1 atm):υ=30.72*10-6m2/s,α=44.7*10-6m2/s,k=0.0363W/m.K, Pr=0.687, β=0.00229K-1. Analysis: (a) performing an energy balance on the plate,

psurss

pstsurss

wcTTTThAdtdT

dtdTVcETTTThAq

ρεσ

ρεσ

/)]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 ×=

××−×== −−

∞

ναβ

KmWh

RaLkh L

./8.5

])687.0/492.0(1[)1058.5(670.068.0

5.00363.0

]Pr)/492.0(1[670.068.0

2_

9/416/9

4/18

9/416/9

4/1_

=

⎭⎬⎫

⎩⎨⎧

+×+=

⎭⎬⎫

⎩⎨⎧

++=

Hence the initial cooling rate is

{ }

sKdtdT

kkgJmmkgKKKmWCKmW

dtdT

/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, calculate

4-

22"

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

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

×==

=+=−== ∞

hence

kmWKmWTTqhwherekwhBi itoteffeff

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 q”conv ≈q”rad.

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 4°C. 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. Known: Dimensions and temperature of beer can in refrigerator compartment. Find: orientation which maximum cooling rate.

Schematic:

Air,T∞=4°CP=1 atmD=0.06m

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

L=150mm

Assumptions: (1) End effects are negligible, (2) Compartment air is quiescent, (3) constant properties. Properties: Air (Tf=288.5K, 1 atm): υ=14.87*10-6 m2/s, k=0.0254W/m.K, α=21.0*10-6m2/s, Pr=0.71, β=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

hTTTT

DLDL

h

hq_

_

_

_

)()(

=−−

==∞

∞

ππ

For the vertical surface, find

639L

39326-26-

-1-323s

L

1044.8)15.0(105.2Ra

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

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

×=×=

×=××

°××== ∞

ναβ

Using the correct correlation

KmWm

KmWLkNuhh

NuL

LvL ./03.515.0

./0254.07.29 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

DhD

D

qq

hence

KmWm

KmWDkNuhh

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.