ME 335 (HEAT TRANSFER), FALL 2007, FINAL EXAM, SOLUTIONDepartment of Mechanical Engineering, University of Michigan

PROBLEM 1 (30%)

GIVEN:The exposed human leg, in standing position, losses heat by surface convection (thermobuoyant flow)

and surface radiation. Model the leg as a uniform-temperature cylinder of diameter D and length L, FigurePr.1, in a room with air at Tf,∞ and the radiation surroundings (representing wall to wall windows in a cold

outdoor condition) at Ts. For leg, the metabolic heat generation rate is Sr,c.

Sr,c = 15 W, D = 12 cm, L = 40 cm, Tf,∞ = 25◦C, and Ts = 15◦C. Assume blackbody surfaces, use airproperties at 300 K. Assume no heat losses at both ends. Use (6.89) which is for laminar flow only, for〈Nu〉L.SKETCH:

Figure Pr.1(a) shows a human leg model [(i) without clothing (ii) with clothing] exposed to air andradiation surroundings, loses heat by the thermobuoyant flow surface convection and by surface radiation.

Surroundings, Ts Surroundings, Ts

(i) No Clothing

Leg Model

(ii) With Clothing

D

L

D l

LClothing

g

, ,0,f fu T∞ ∞=

g

, ,0,f fu T∞ ∞=

lT

'lT

Figure Pr.1(a) A human leg model [(i) without clothing (ii) with clothing] exposed to air and radiationsurroundings, loses heat by the thermobuoyant flow surface convection and by surface radiation.

OBJECTIVE:(a) Draw the thermal circuit diagram.(b) Calculate the fraction of heat losses by surface convection and surface radiation.(c) Redo (b) with a 1 mm thick fabric covering the leg skin, how much do the results in (b) change?

SOLUTION:(a) Figure Pr.1(b) shows the thermal circuit diagram.

(Rr,F)l-s

Ql Qs

lT ,b lE sT,b sE ,o( )r sq,o( )r lq

, -r l sQ,r cSɺ

ku LR

,fT∞ku L

Q

= =

Figure Pr.1(b) Thermal circuit diagram.

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(b) From Figure Pr.1(b), the energy equation in a steady-state condition is given as

〈Qku〉L + Qr,l-s = Sr,c.

We then evaluate each of the heat transfer terms as

(i) Heat Transfer By Surface Convection:For the thermobuoyant flow on the cylinder, the surface-convection heat transfer rate is given by (6.49) as

〈Qku〉L = Aku〈Nu〉LkfTs − Tf,∞

L.

From Table C.22 for air at Tδ = 300 K, kf = 0.0267 W/m-K, νf = 15.66 × 10−6 m2/s, and Pr = 0.69,and treating air as an ideal gas, from (6.77),

βf = 1/Tf = 1/〈Tf〉δ = 1/300 1/K.

The 〈Ra〉L from Table 6.5 is given as

RaL =gβf(Ts − Tf,∞)L3

νfαf

=9.807(m/s2) × 1/300( 1/K) × (T1 − 298.15)(K) × 0.43(m3)

15.66 × 10−6(m2/s) × 22.57× 10−6(m2/s)

= 5.919× 106(1/K) × (T1 − 298.15)(K)

For the 〈Nu〉D using (6.89), we have

〈Nu〉L =4

3

0.503

[1 + (0.492/Pr)9/16]4/9Ra

1/4

L

=4

3

0.503

[1 + (0.492/0.69)9/16]4/9× [5.919 × 106 × (Tl − 298.15)]1/4

= 25.31 × (Tl − 298.15)1/4.

Then, we have

〈Qku〉L =π × 0.12.4× 25.31 × (Tl − 298.15)1/4 × 0.0267

0.4× (Tl − 298.15)

= 0.2548× (Tl − 298.15)5/4.

(ii) Heat Transfer By Surface Radiation:Since the leg is modeled as a black body, and the area of the radiation surrounding is very large, (Rr,ǫ)l = 0,and (Rr,ǫ)s = 0. The view factor from the leg to the surrounding is Fl-s = 1.Then, we have

Rr,Σ =1

Ar,lFl-s

=1

1 × π × 0.12 × 0.4(m2)

= 6.631 1/m2

Qr,l-s =Eb,l − Eb,s

Rr,Σ=

σSB(T 4

l − T 4

s )

Rr,Σ

=5.67 × 10−8(W/m2-K4) × (T 4

l − 298.154)(K4)

6.631(1/m2)

= 8.551 × 10−9 × (T 4

l − 298.154).

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Then, the energy equation becomes

0.2548× (Tl − 298.15)5/4 + 8.551 × 10−9 × (T 4

l − 298.154) = 15.

Using solver, we have

Tl = 303.03 K = 29.88◦C.

Using Tl, the 〈Qku〉L, and Qr are

〈Qku〉L = 0.2548× (303.03− 298.15)5/4 = 1.848 W,

Qr,l-s = 8.551× 10−9 × (303.034 − 298.154) = 13.15 W.

The fraction of the surface radiation and the surface convection is

〈Qku〉L

Sr,c

=13.15

15= 0.8767,

〈Qr,l-s〉

Sr,c

=1.848

15= 0.1232.

(c) Figure Pr.1(c) shows the thermal circuit diagram with the fabric.

Ql

lT,r cSɺ

ku LR

,fT∞ku L

Q

lT ′, -k l lR ′

(Rr,F)l-s

Qs

,b l′E sT,b sE ,o( )r sq,o( )r l′q

, -r l sQ

= =

′

′

Figure Pr.1(c) Thermal circuit diagram with the fabric.

Even though we introduce the extra conduction resistance for the fabric Rk,l-l′ . The total heat flow fromthe surface is the same. So, the fraction does not change (except for slight increase in Aku = Ar,l). The skintemperature is replaced with fabric outside temperature. Thus, the fraction of the surface radiation and thesurface convection is same as the results in part (b).

〈Qku〉L

Sr,c

=13.15

15= 0.8767,

〈Qr,l-s〉

Sr,c

=1.848

15= 0.1232.

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PROBLEM 2 (40%)

GIVEN:Water column (circular cylinder with diameter D) flowing from the faucet at temperature T1 = 0◦C and

average velocity uw, freezes under extreme surface-convection conditions (air cross flow uf,∞ and Tf,∞), asshown in Figure Pr.2.D = 5 mm, uw = 0.01 m/s, uf,∞ = 30 m/s, Tf,∞ = -50◦C.

SKETCH:Figure Pr.2 shows the tube insulation.

D

Cold Air Stream

, ,,f fu T∞ ∞

wu

1 0 CT °=

1L

2L

Figure Pr.2(a) Cold water flowing out of facet at 0◦C and freezing by extreme surface convection heattransfer. After complete freezing, the solid matter (ice) continues to cool and approaches the ambient

temperature.

OBJECTIVE:(a) Draw the thermal circuit diagram for freezing water column.(b) Calculate the length L1 required for complete freezing of water.(c) After freezing length, L1 the ice continues to cool and gradually reaches the ambient temperature. Esti-mate (rough approximation is fine) this ice cooling length L2, using the results for fins.(d) If the ice-cooling portion was included in calculation of (b), would L1 change? Give reason.

SOLUTION:(a) Figure Pr.2(b) shows the thermal circuit diagram.

Q1

1TlsSɺ

ku DR

,fT∞

ku 1-∞Q

Figure Pr.2(b) Thermal circuit diagram.

(b) From Figure Pr.2(b), the energy equation in a steady-state condition is given as

〈Qku〉1-∞ = Ssl.

From Table C.22 for air at Tδ = (Ts + Tf,∞)/2 = 248.15 K, kf = 0.0234 W/m-K, νf = 11.28 × 10−6

m2/s, and Pr = 0.69.

4

For the force surface convection heat transfer rate, the Reynolds number is

ReD =〈uf,∞〉D

νf=

30 × 0.005

11.28× 10−6= 1.330 × 104.

From Table 6.4, we have a1 = 0.193, and a2 = 0.618. Then,

〈Nu〉D = a1Rea2

D Pr1/3

= 0.193 × (1.330× 104)0.618 × 0.691/3 = 60.31.

The surface convection resistance is

〈Rku〉D =D

Aku〈Nu〉Dkf=

0.005

π × 0.005 × L1 × 60.31× 0.0234

=0.2256

L1

.

〈Qku〉1-∞ =T1 − Tf,∞

〈Rku〉D

=0 − (−50)

0.2256/L1

= 221.6× L1.

From Table 2.1, the freezing rate is given as

Sls = −Mls∆hls = Mls∆hsl.

For the water from Table C.6, ∆hsl = 333.6 × 103 J/kg, and for liquid water from Table C.23, ρl = 1, 002kg/m3.

Ml = Mls = ρluwA = 1, 002(kg/m3) × 0.01(m/s) × π × 0.0052/4(m2)

= 1.967× 10−4 kg/s.

Then, the energy equation becomes

〈Qku〉1-∞ = Sls = −Mls∆hls = Mls∆hsl W.

221.6 × L1 = 1.967 × 10−4 × 333.6× 103 = 65.62 W.

L1 =65.62

221.6= 0.2961 m.

(c) Treating the moving ice as quasi steady-state, the frozen ice can be modeled as a fin, since we haveconduction heat transfer through the ice and cooling by the surface convection. For the ice, from TableC.15, we have ks = 2.3 W/m-K.

The fin efficiency is given by (6.147) as

ηf =tanh(mLc)

mLc

m =

(

Pku,f 〈Nu〉Dkf

AkksD

)1/2

,

where

Pku,f = πD, Ak =D2

4.

5

Then

m =

[

π × 0.005× 60.31 × 0.0234

0.25 × π × 0.0052 × 2.3 × 0.005

]1/2

= 313.3 1/m.

Note that m is sufficiently large, for low fin efficiency (the tip temperature reaching the ambient temperature)ηf → 0, from Figure 6.31(a) here we use ηf = 0.1 as an approximate value. Then, from Table 6.7 for ηf , wehave

ηf =tanh(mLc)

mLc=

1.000

10.00

Then

mLc = 10

Lc =10

m=

10

313.3= 0.03191 m = 3.191 cm.

Assuming no heat transfer through the end (at L2), we use Lc = L2.

L2 = Lc = 0.03191 m.

For ηf → 0, L2 becomes larger (for ηf = 0, L2 → ∞), as this is an approximation for L2. Note that thetemperature of the ice is non-linear along the cylinder axis. If we assume uniform distribution 〈Qku〉2-∞,then we can also use

〈Qku〉2-∞ = Mscv,s(0 − Tf,∞)

221.6 L2 ≃ Mlcp,l(0 − Tf,∞),

and this gives

L2 = 0.0.

Note that this assumes linear temperature distribution (constant 〈qku〉2-∞).

(d) There is heat conduction through the end of L1 into the L2 region. If we add this heat transfer to theenergy equation for L1, L1 is influenced by L2 region through this conduction. This additional conductionwill decrease L1. Also, the shorter L2 is, the shorter L1 becomes.

6

PROBLEM 3 (25%)

GIVEN:Trans-Alaska oil pipeline has a metallic inner tube with inside diameter D and wall thickness l1, covered

by a thick fiberglass insulation l2, and finally a thin metallic tube (wrapping) protecting the insulation fromthe environment, as shown in Figure Pr.3. The extracted oil is warm when pumped (there are a handful ofvery large pumps distributed along the long pipeline) into the pipe and remains warm by viscous heating.The pipeline is suspended, and surrounded by air. Consider a segment of the suspended pipeline, where thelength is L and the inlet temperature is 〈Tf 〉o. The volume flow rate through the pipe is V . Albeit small,there is heat loss from the oil to the ambient (air cross flow uf,∞ and Tf,∞), especially in winter.

D = 1.2 m, l1 = 1.2 cm, l2 = 10 cm, L = 50 km, uf,∞ = 3 m/s, Tf,∞ = −40◦C, 〈Tf 〉o = 50◦C, V = 3.8m3/s, and fiberglass insulation conductivity 〈k〉 = 0.02 W/m-K. Use engine oil properties at 310 K and airproperties at 270 K, and use stainless steel AISI 302 for the inner tube.

SKETCH:Figure Pr.3(a) shows warm oil losing heat by surface convection flowing through pipeline.

Thin Wrapping Tube

Inner Tube

L

Oil

Cold Air Stream

Fiberglass Insulation Thickenss,

Fiberglass Insulation

Thin Wrapping Tube

Inner Tube Wall Thickness,

Inner Tube Diameter,

2l

1l

,fu∞ ,fT

∞

D

ofT

Figure Pr.3(a) Warm oil loses heat by surface convection flowing through pipeline.

OBJECTIVE:(a) Draw the thermal circuit diagram.(b) Neglect the viscous heating and calculate the oil exit temperature 〈Tf〉L.

(c) What should be the viscous heating rate Sm,µ, so there will be no oil temperature change?

SOLUTION:(a) The thermal circuit diagram is shown in Figure Pr.3(b).(b) For the bounded flow in the pipeline, 〈Rku〉D,1 is estimated. From Table C.23 for the oil at T = 310 K,ρf = 877.8 kg/m3, kf = 0.145 W/m-K, νf = 417 × 10−6, Pr = 4,000, and cp,f = 1,950 J/kg-K.

Using V = 〈uf 〉D,1A,

〈uf〉 =V

Au,1=

3.8

π × (1.2)2/4

= 3.360 m/s.

7

(Mcp)f

Qu L-0

Qu,0 Qu,L

0fTLfT

,mS μɺ

,fT∞

1T

2T

3T

,1ku DR

,2ku DR

,2-3kR

,1-2kR

Control Volume for Part (c)

Figure Pr.3(b) Thermal circuit model.

ReD,1 =〈uf 〉DD

νf

=3.360× 1.2

417 × 10−6= 9, 669 ≃ 104 assume fully developed turbulent flow.

The entrance length using (7.39) is

Lδ = 4.4Re1/6

D D = 4.4 × (9, 669)1/6 × 1.2 = 24.37 m.

and this is much smaller than L, Lδ/L = 4.874×10−4 < 0.1. So we neglect entrance effects. Even thoughPr is lager than 160 for oil, from Table 7.2, we use only the relation for turbulent flows, i.e.,

〈Nu〉D,1 = 0.023Re4/5

D,1 Prn where n = 0.3 for cooling

= 0.023 × (9, 669)4/5 × (4, 000)0.3

= 427.2.

Then,

〈Rku〉D,1 =D

Aku,1〈Nu〉D,1kf=

1.2

π × 1.2 × 50 × 103 × 472.2 × 0.145

= 9.298 × 10−8 W/K.

For AISI 302, from Table C.16, ks = 15 W/m-K, and we have from Table 3.2,

Rk,1-2 =ln(R2/R1)

2πksL=

ln(0.5 × 1.2 + 0.012

0.5 × 1.2)

2π × 15 × 50 × 103

= 4.202 × 10−9 W/K.

8

Similarly for the insulation, we have

Rk,2-3 =ln(R3/R2)

2πkL=

ln(0.5 × 1.2 + 0.012 + 0.1

0.5 × 1.2 + 0.012)

2π × 0.02 × 50 × 103

= 2.409 × 10−5 W/K.

From Table C.22 for air at Tδ = 270 K, we have kf = 0.0249 W/m-K, νf = 13.09 × 10−6 m2/s, and Pr =0.69. Then

ReD,2 =uf,∞D2

νf=

3 × [1.2 + 2 × (0.012 + 0.1)]

13.06× 10−6

= 3.271 × 105.

Then, from Table 6.4, we have for cross flow

〈Nu〉D,2 = a1Rea2

D Pr1/3,

where a1 = 0.027, and a2 = 0.805, so 〈Nu〉D becomes

〈Nu〉D,2 = 0.027× (3.271 × 105)0.805 × (0.69)1/3

= 656.1,

and

〈Rku〉D,2 =D

Aku,2〈Nu〉D,2kf=

1.2 + 2 × (0.012 + 0.1)

π × [1.2 + 2 × (0.012 + 0.1)] × 50 × 103 × 656.1 × 0.0249(K/W)

= 3.897× 10−7 K/W.

Then,

RΣ = 〈Rku〉D,1 + Rk,1-2 + Rk,2-3 + 〈Rku〉D,2

= 9.298 × 10−8 + 4.202 × 10−9 + 2.409× 10−5 + 3.897× 10−7

= 2.458 × 10−5 W/K.

Note that the insulation resistance is dominant.

(Mcp)f = ρf V cp,f

= 877.8 × 3.8 × 1, 950 = 6.505× 106 K/W.

From (7.74), we have

NTU =1

RΣ(Mcp)f

=1

2.459 × 10−5 × 6.505 × 106= 6.254× 10−3.

For a single stream, (or from Table 7.7 for Cr = 0, since the outside flow is semi-bounded having infiniteheat capacitance), so no change in Tf,∞. We have from (7.22),

εhe = 1 − e−NTU = 1 − e−6.254×10−3

= 6.234 × 10−3.

εhe =〈Tf 〉L − 〈Tf〉0

Ts − 〈Tf 〉0

〈Tf 〉L = εhe × (Ts − 〈Tf 〉0) + 〈Tf 〉0

= 6.234 × 10−3 × (−40 − 50) + 50 = 49.44◦C.

9

(c) To have no oil temperature change, the viscous heating Sm,µ should compensate the heat loss throughthe oil pipe line. The energy equation for the entire volume gives, using 〈Tf 〉L from above,

Sm,µ = −〈Qu〉L-0 = −(Mcp)(〈Tf 〉L − 〈Tf 〉0)

= 3.336× 103 × 1, 950 × (49.44− 50)

= 3.643× 106 W.

10