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NATIONAL UNIVERSITY OF SINGAPORE Department Mechanical Engineering

ME 3122E – Heat Transfer (2012-13)

Solution to Practice Problems

1. The steady state temperature distribution in a one dimensional wall of thermal conductivity 50 W/m.K and thickness 50 mm is observed to be T(oC) = a + bx2, where a = 200 oC, b = -2000oC/m2, and x in metres. (a) What is the heat generation rate �� in the wall? (b) Determine the heat fluxes at the two wall faces. In what manner are these heat fluxes related to the heat generation rate? (a) The appropriate form of the heat equation for steady-state, one-dimensional conditions with constant properties is:

𝑞 =− 𝑘𝑑𝑑𝑥

�𝑑𝑇𝑑𝑥�

Substituting the prescribed temperature distribution,

𝑞 =− 𝑘𝑑𝑑𝑥

�𝑑(𝑎 + 𝑏𝑥2)

𝑑𝑥� = 𝑞 =− 𝑘

𝑑𝑑𝑥

[2𝑏𝑥] = −2𝑏𝑘

𝑞 =− 2(−2000𝑜𝐶/𝑚2) × 50𝑊𝑚.𝐾

= 𝟐.𝟎 × 𝟏𝟎𝟓𝑾/𝒎𝟑

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(b) The heat fluxes at the wall faces can be evaluated from Fourier’s law,

𝑞"𝑥 = −𝑘 �

𝑑𝑇𝑑𝑥�𝑥

Using the temperature distribution T(x) to evaluate the gradient, find

𝑞"𝑥 = −𝑘

𝑑(𝑎 + 𝑏𝑥2)𝑑𝑥

= −2𝑘𝑏𝑥 The fluxes at x = 0 and x = L are then

𝒒"𝒙(𝟎) = 𝟎

𝑞"𝑥(𝐿) = −2𝑘𝑏𝐿 = −2 × 50

𝑊𝑚.𝐾

× (−2000𝑜𝐶/𝑚2) × 0.050𝑚

𝒒"𝒙(𝑳) = 𝟏𝟎,𝟎𝟎𝟎 𝑾/𝒎𝟐

From an overall energy balance on the wall, it follows that, for a unit area,

𝑞"𝑥(0) − 𝑞"

𝑥(𝐿) + 𝑞�� = 0

�� =𝑞"

𝑥(𝐿) − 𝑞"𝑥(0)

𝐿=

10,000 𝑊/𝑚2 − 00.050

= 2.0 × 105𝑊/𝑚3 2. A thin electrical heater is inserted between a long circular rod and a concentric tube with inner and outer radii of 20 and 40 mm. The rod (A) has a thermal conductivity of kA = 0.15 W/m.K, while the tube (B) has a thermal conductivity of kB = 1.5 W/m.K and its outer surface is subjected to convection with a fluid of temperature T∞ = -15oC and heat transfer coefficient 50 W/m2.K. The thermal contact resistance between the cylinder surface and the heater is negligible.

(a) Determine the electrical power per unit length of the cylinders (W/m) that is required to maintain the outer surface of the cylinder B at 5oC

(b) What is the temperature at the centre of cylinder A?

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(a) Perform an energy balance on the composite system to determine the power required to maintain T(r2) = Ts = 5oC

𝑞𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐− 𝑞𝑐𝑜𝑛𝑣 = 0

𝑞𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 = 𝑞𝑐𝑜𝑛𝑣 = ℎ × 2𝜋𝑟2 × (𝑇𝑠 − 𝑇∞)

𝑞𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 =50𝑊𝑚2.𝐾

× 2𝜋 × 0.04 × �5 − (−15)� = 𝟐𝟓𝟏.𝟐 𝑾/𝒎 (b) The inner rod is isothermal, that is T(0) = T(r1) Representing cylinder B by a thermal circuit,

𝑞′ =

𝑇(𝑟1) − 𝑇𝑠𝑅′𝐵

𝑅′𝐵 = 𝑙𝑛 �𝑟2𝑟1� /2𝜋𝑘𝐵

Therefore,

𝑇(𝑟1) = 𝑇𝑠+(𝑞′ × 𝑅′𝐵) = 5𝑜𝐶 + �251𝑊𝑚

×𝑙𝑛 �40

20�2𝜋 × 1.5

� = 23.5𝑜𝐶

Hence, 𝑇(0) = 𝑇(𝑟1) = 𝟐𝟑.𝟓𝒐𝑪 Note that 𝑘𝐴 has no influence on the temperature 𝑇(0) 3. Steel ball bearings are required to be subjected to heat treatment to obtain the desired surface characteristics. The balls are heated to a temperature of 650°C and then quenched

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in a pool of oil that has a temperature 55°C. The ball bearings have a diameter of 40 mm. The convective heat transfer coefficient between the ball bearings and oil is 300 W/m2. K Determine a) The length of time that the bearings must remain in the oil before their temperature drops to 200°C, b) Total amount of heat removed from each bearing during this time interval, and c) Instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200°C.The properties of steel ball bearings are as follows: k =50 W/m.K; 𝛼= k/ρCp = 1.3x10-5 m2/s Given: 𝑇𝑖 = 650𝑜𝐶,𝑇∞ = 55𝑜𝐶, 𝑇 = 200𝑜𝐶 (a) Using the lumped heat capacity method,

𝑙𝑛 �𝑇 − 𝑇∞𝑇𝑖 − 𝑇∞

� = �−ℎ𝐴𝑡𝜌𝐶𝑉

� (𝟏)

𝑡 = 𝑙𝑛 �𝑇 − 𝑇∞𝑇𝑖 − 𝑇∞

�× �𝜌𝐶𝑉−ℎ𝐴

� = 𝑙𝑛 �𝑇 − 𝑇∞𝑇𝑖 − 𝑇∞

� × �𝑟𝑘

−3 ∝ ℎ� , �∵

𝑉𝐴

=𝑟3�

Time required for the cooling of the bearings,

𝑡 = 𝑙𝑛 �200 − 55650 − 55

� × �0.02 × 50

−3 × 1.3 × 10−5 × 300� = 𝟏𝟐𝟎 𝒔

(b) Total amount of heat removed,

𝐻𝑡𝑜𝑡𝑎𝑙 = � ℎ𝐴(𝑇 − 𝑇∞)𝑑𝑡𝑡

0

Using equation (1) to replace (𝑇 − 𝑇∞),

𝐻𝑡𝑜𝑡𝑎𝑙 = � ℎ𝐴(𝑇𝑖 − 𝑇∞)𝑒𝑥𝑝 �−3 ∝ ℎ𝑡𝑟𝑘

� 𝑑𝑡𝑡

0

after the integration,

𝐻𝑡𝑜𝑡𝑎𝑙 = �𝑟𝑘𝐴3 ∝

(𝑇∞ − 𝑇𝑖)� �𝑒𝑥𝑝 �−3 ∝ ℎ𝑡𝑟𝑘

� − 1�

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𝐻𝑡𝑜𝑡𝑎𝑙

= �0.02 × 50 × 4𝜋 × 0.022 × (55 − 650)

3 × 1.3 × 10−5� �𝑒𝑥𝑝 �

−3 × 1.3 × 10−5 × 300 × 1200.02 × 50

�

− 1� = 𝟓.𝟕𝟖 × 𝟏𝟎𝟒𝑱

c) Instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200°C.

𝑄𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = ℎ𝐴(𝑇𝑖 − 𝑇∞)𝑒𝑥𝑝 �−3 ∝ ℎ𝑡𝑟𝑘

� Heat transfer rate from the bearings when they are first placed, that is when t = 0,

𝑄𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = ℎ𝐴(𝑇𝑖 − 𝑇∞)𝑒𝑥𝑝(0) = ℎ𝐴(𝑇𝑖 − 𝑇∞)

𝑄𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = 300 × 4𝜋 × 0.022 × (650 − 55) = 𝟖𝟗𝟕 𝑾 Heat transfer rate from the bearings when they reach 200°C, that is when t = 120 s, 𝑄𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 = 300 × 4𝜋 × 0.022 × (650 − 55)𝑒𝑥𝑝 �−3×1.3×10−5×300×120

0.02×50�=220.6 W

4. The extent to which the tip condition affects the thermal performance of a fin depends on the fin geometry and thermal conductivity, as well as the convection coefficient. Consider an alloyed aluminium (k = 180 W/m.K) rectangular fin whose base temperature is 100 oC. The fin is exposed to a fluid of temperature T∞ = 25 oC and a uniform convection coefficient of h = 100 W/m2.K may be assumed for the fin surface.

For a fin of length L = 10 mm, thickness t = 1 mm, and width w»t, determine the fin heat transfer rate per unit width, fin efficiency and effectiveness for cases of convection heat transfer and adiabatic fin tips. Contrast your results with those based on an infinite fin approximation.

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𝑚 = �ℎ𝑃𝑘𝐴

= �100𝑊𝑚2.𝐾 × (2𝑚 + 0.002𝑚)180𝑊𝑚.𝐾 × (1𝑚 × 0.001𝑚)

= 33.34 𝑚−1

𝑀 = √𝑘𝐴𝑃ℎ𝜃𝑏 = �180𝑊𝑚.𝐾

× (1𝑚 × 0.001𝑚) × (2𝑚 + 0.002𝑚) ×100𝑊𝑚2.𝐾

= 450 𝑊/𝑚

Case A: convection heat transfer from fin tip

𝑞𝑓 = 𝑀𝑠𝑖𝑛ℎ𝑚𝐿 + (ℎ/𝑚𝑘)𝑐𝑜𝑠ℎ𝑚𝐿𝑠𝑖𝑛ℎ𝑚𝐿 + (ℎ/𝑚𝑘)𝑐𝑜𝑠ℎ𝑚𝐿

= 450 𝑊/𝑚0.340 + 0.0167 × 1.0571.057 + 0.0167 × 0.340

= 𝟏𝟓𝟏𝑾/𝒎

𝜂𝑓 =𝑞𝑓

ℎ(2𝐿 + 𝑡)𝜃𝑏=

151 𝑊/𝑚100 𝑊/𝑚2 × (2 × 0.01𝑚 + 0.001𝑚) × 75𝑜𝐶

= 𝟎.𝟗𝟔

𝜀𝑓 =𝑞𝑓ℎ𝑡𝜃𝑏

=151 𝑊/𝑚

100 𝑊/𝑚2 × 0.001𝑚 × 75𝑜𝐶= 𝟐𝟎.𝟏

Case B: adiabatic fin tip

𝑞𝑓 = 𝑀 𝑡𝑎𝑛ℎ𝑚𝐿 = 450𝑊𝑚

× 0.321 = 𝟏𝟒𝟒 𝑾/𝒎

𝜂𝑓 =𝑡𝑎𝑛ℎ𝑚𝐿𝑚𝐿

=0.3210.333

= 𝟎.𝟗𝟔𝟒

𝜀𝑓 =𝑞𝑓ℎ𝑡𝜃𝑏

=144 𝑊/𝑚

100 𝑊/𝑚2 × 0.001𝑚 × 75𝑜𝐶= 𝟏𝟗.𝟐

Case C: Very long fin (𝐿 → ∞) 𝑞𝑓 = 𝑀 = 𝟒𝟓𝟎 𝑾/𝒎, 𝜂𝑓 = 𝟎

𝜀𝑓 =𝑞𝑓ℎ𝑡𝜃𝑏

=450 𝑊/𝑚

100 𝑊/𝑚2 × 0.001𝑚 × 75𝑜𝐶= 𝟔𝟎.𝟎

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5. Consider the right-circular cylinder of diameter D, length L, and the area A1, A2 and A3 representing the base, inner, and top surfaces, respectively.

(a) Show that the view factor between the base of the cylinder and the inner surface has the form F12 = 2H[(1+H2)1/2-H], where H = L/D

(b) Show that the view factor for the inner surface to itself has the form F22 = 1+H-(1+H2)1/2

Given: Right-circular cylinder of diameter D, length L and the areas A

1, A

2, and A

3 representing he base, inner lateral and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner lateral surface has he form F12 = 2H[(1+H2)1/2-H]

(a) Relation for F

12, base-to-inner lateral surface. Apply the summation rule to A

1,

noting that F11

= 0 𝐹11 + 𝐹12 + 𝐹13 = 𝟏 , ∴ 𝐹12 = 1 − 𝐹13

For the given configuration, 𝑅𝑖 = 𝑟𝑖𝐿

, 𝑅𝑗 = 𝑟𝑗𝐿

, 𝑆 = 1 + 1+𝑅𝑗2

𝑅𝑖2 (1)

𝐹𝑖𝑗 =1

2 �𝑆 − �𝑆2 − 4 �𝑟𝑗𝑟𝑖�2��12

(2)

Using equations (1) and (2) with i = 1, j = 3,

𝐹13 = 12

�𝑆 − �𝑆2 − 4 �𝐷3𝐷1�2��

12 and 𝑆 = 1 + 1+𝑅32

𝑅12, 𝑆 = 2 + 1

𝑅2= 4𝐻2 + 2 (3)

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where R1

= R3

= R = D/2L and H = L/D. Combining Eqs. (2) and (3) with Eq. (1), find after some manipulation

𝐹12 = 1 −12

{4𝐻2 + 2 − [(4𝐻2 + 2)2 − 4]}12

𝐹12 = 2𝐻 �(1 + 𝐻2)1/2 − 𝐻� (4) b) Relation for F

22, inner lateral surface. Apply summation rule on A

2, recognizing that

F23

= F21

,

𝐹21 + 𝐹22 + 𝐹23 = 1 ∴ 𝐹22 = 1 − 2𝐹21 (5) Apply reciprocity between A

1 and A

2,

𝐹21 = �𝐴1𝐴2� 𝐹12 (6)

and substituting into Eq. (5), and using area expressions

𝐹22 = 1 − 2𝐹21

𝐹22 = 1 − 2 �𝐴1𝐴2� 𝐹12 = 1 − 2 �

𝐷4𝐿�𝐹12 = 1 − �

12𝐻

�𝐹12

where A

1 = πD

2/4 and A

2 = πDL.

Substituting from Eq. (4) for F

12, find

𝐹22 = 1 − �1

2𝐻�2𝐻�(1 + 𝐻2)1/2 − 𝐻� = 1 + 𝐻 − (1 + 𝐻2)1/2

6. A certain surface maintained at 1400 K has the following spectral emissive characteristics:

ε λ = 0.08 0 < λ < 0.5μm, ε λ = 0.4 0.6 < λ < 5μm ε λ = 0.7 5 < λ < ∞

Calculate the emissive power of the surface.

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For the given spectral emissive characteristics,

𝜖𝑇 = � 𝜖𝜆𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 =∞

0� 𝜖𝜆

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +𝜆1

0� 𝜖𝜆

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +𝜆2

𝜆1� 𝜖𝜆

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆∞

𝜆2

= � 𝜖1𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +0.6𝜇𝑚

0� 𝜖2

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +5𝜇𝑚

0.6𝜇𝑚� 𝜖3

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆∞

5𝜇𝑚

= �𝜖1�𝐹(0.6𝑇) − 𝐹(0𝑇)��+�𝜖2�𝐹(5𝑇) − 𝐹(0.6𝑇)�� + �𝜖3�𝐹(∞𝑇) − 𝐹(5𝑇)�� = �0.08�𝐹(0.6𝑇) − 𝐹(0𝑇)��+�0.4�𝐹(5𝑇) − 𝐹(0.6𝑇)�� + �0.7�𝐹(∞𝑇) − 𝐹(5𝑇)��

= 0.08[0.000016 − 0.000]+0.4[0.808109 − 0.000016] + 0.7[1 − 0.808109] 𝜖𝑇 = 0.00000128+0.32323 + 0.13432 = 0.4575 𝐸 = 𝜎𝜖𝑇𝑇4 = 5.67 × 10−8 × 0.4575 × 14004 = 𝟎.𝟗𝟗𝟔𝟓𝟐 × 𝟏𝟎𝟓 𝐖/𝐦𝟐 7. An opaque surface, 2m by 2m, is maintained at 400 K and is simultaneously exposed to solar irradiation with G = 1200 W/m2. The surface is diffuse and its spectral absorptivity is αλ = 0, 0.8, 0 and 0.9 for 0 ≤ λ ≤ 0.5μm, 0.5 < λ ≤ 1μm, 1 μm < λ ≤ 2μm and λ > 2μm, respectively. Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface. Assuming: (1) Opaque, diffuse surface behavior, (2) Spectral distribution of solar radiation corresponds to emission from a blackbody at 5800 K.

The absorptivity to solar irradiation is

𝛼𝑆 = �𝛼𝜆𝐺𝜆𝐺

𝑑𝜆 =∞

0�

𝛼𝜆𝐸𝜆,𝑏(5800𝐾)𝐸𝑏

𝑑𝜆∞

0

10

= � 𝛼1𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +0.5𝜇𝑚

0� 𝛼2

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +1𝜇𝑚

0.5𝜇𝑚� 𝛼3

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆2𝜇𝑚

1𝜇𝑚

+ � 𝛼4𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆∞

2𝜇𝑚

= �𝛼1�𝐹(0.5𝑇) − 𝐹(0𝑇)��+�𝛼2�𝐹(1𝑇) − 𝐹(0.5𝑇)�� + �𝛼3�𝐹(2𝑇) − 𝐹(1𝑇)�� + �𝛼4�𝐹(∞𝑇) − 𝐹(2𝑇)�� = �0.0�𝐹(0.5𝑇) − 𝐹(0𝑇)��+�0.8�𝐹(1𝑇) − 𝐹(0.5𝑇)�� + �0.0�𝐹(2𝑇) − 𝐹(1𝑇)�� + �0.9�𝐹(∞𝑇) − 𝐹(2𝑇)��

= [0.8(0.72 − 0.25)] + [0.9(1.0 − 0.94)] = 0.43

𝐺𝑆𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 = 𝛼𝑆 × 𝐺𝑆 = 0.43 × 1200 = 𝟓𝟏𝟓 𝑾/𝒎𝟐 The emissivity

𝜖 = �𝜖𝜆𝐸𝜆,𝑏(400𝐾)

𝐸𝑏𝑑𝜆

∞

0

= � 𝜖1𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +0.5𝜇𝑚

0� 𝜖2

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆 +1𝜇𝑚

0.5𝜇𝑚� 𝜖3

𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆2𝜇𝑚

1𝜇𝑚

+ � 𝜖4𝐸𝜆,𝑏(𝑇)𝐸𝑏(𝑇)

𝑑𝜆∞

2𝜇𝑚

𝜖 = �𝛼1�𝐹(0.5 × 400) − 𝐹(0 × 400)��+�𝛼2�𝐹(1 × 400) − 𝐹(0.5 × 400)�� +�𝛼3�𝐹(2 × 400) − 𝐹(1 × 400)�� + �𝛼4�𝐹(∞𝑇) − 𝐹(2 × 400)�� = 0+0 + 0 + [0.9(1.0 − 0.000016)]=0.9 𝐸 = 𝜎𝜖𝑇4 = 5.67 × 10−8 × 0.9 × 4004 = 𝟏𝟑𝟎𝟔 𝐖/𝐦𝟐 The radiosity is : 𝐽 = 𝐸 + 𝜌𝑠𝐺𝑠= 𝐸 + (1 − 𝛼𝑠)𝐺𝑠 = 1306 + 0.57 × 1200 =𝟏𝟗𝟗𝟏𝐖/𝐦𝟐 The net radiation transfer from the surface is: 𝑞𝑛𝑒𝑡 = (𝐸 − 𝛼𝑠𝐺𝑠)𝐴𝑠 = (1306 − 515) × 4 = 𝟑𝟏𝟔𝟒 𝐖 Note: Unless 3164 W are supplied to the surface by other means (for example, by

11

8. A cryogenic fluid at 100 K is transported through a pipe 15 mm OD (outer diameter) installed in a circular duct. The emittance of the outer surface of this pipe is 0.25. (a) This 15 mm OD pipe is encased by another pipe of 25 mm OD and the space between them is evacuated. The emittance of the inner surface of this pipe is 0.35. Find the heat gain by the cryogenic fluid per unit length when outer pipe is at 280 K. (b) To reduce heat gain, a cylindrical radiation shield of 20 mm OD is inserted between the two tubes stated in (a). Both sides of this tube have an emittance of 0.06. Find the equilibrium temperature of this radiation shield and reduction of heat gain by the fluid.

(a)

𝑞𝑜−𝑖 =𝜎�𝑇𝑜4 − 𝑇𝑖4�

1 − 𝜖𝑖𝜖𝑖𝐴𝑖

+ 1𝐴𝑖𝐹𝑖𝑜

+ 1 − 𝜖𝑜𝜖𝑜𝐴𝑜

𝑞𝑜−𝑖 =5.67 × 10−8(2804 − 1004)

1 − 0.250.25 × 𝜋 × 0.015 + 1

𝜋 × 0.015 × 1 + 1 − 0.350.35 × 𝜋 × 0.025

𝑞𝑜−𝑖 =342.84

63.69 + 21.23 + 23.66= 𝟑.𝟏𝟓 𝑾/𝒎

Di = 15 mm Ti = 100 K εi = 0.25

Do = 25 mm

εo = 0.35

Ds = 20 mm εs = 0.06 To = 280 K

12

(b) Additional resistances due to the radiation shield

1 − 𝜖𝑠𝜖𝑠𝐴𝑠

+1 − 𝜖𝑠𝜖𝑠𝐴𝑠

+1

𝐴𝑠𝐹𝑠𝑜

1 − 0.06

0.06 × 𝜋 × 0.02+

1 − 0.350.06 × 𝜋 × 0.02

+1

𝜋 × 0.02 × 1= 514.86 𝑚−1

𝑞𝑜−𝑖,𝑛𝑒𝑤 =342.84

63.69 + 21.23 + 23.66 + 514.86= 0.55 𝑊/𝑚

% Reduction in heat gain = �3.15−0.553.15

�× 100 = 𝟖𝟐. .𝟓 %

9. An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an air stream that is in parallel flow over the top of the strips. Each strip is 0.2 m wide, and 25 strips are arranged side by side, forming a continuous and smooth surface over which the air flows at 2 m/s. During operation, each strip is maintained at 500oC and the air is at 25oC. What is the rate of convection heat transfer from the (i) first strip? (ii) The fifth strip? (iii) The tenth strip? and (iv) All the strips? Properties of air at at (500 + 25)/2 oC = 535.5 K: µ = 2.849 x 10-5, k = 0.04357 W/m.K Pr = 0.68, ρ = 0.6418 kg/m3

(i) The location of transition is determined from:

𝑥𝑐 =𝑅𝑒𝑐𝜇𝜌𝑢∞

=5 × 105 × 2.849 × 10−5

0.6418 × 2= 11.09 𝑚

Since xc >>L=0.25 m, the air flow is laminar over the entire heater. For the first strip,

13

𝑞1 = ℎ1��� × (∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞)

ℎ1��� =𝑘∆𝐿

× 0.664 × 𝑅𝑒0.5 × 𝑃𝑟0.33

ℎ1��� =0.0429

0.01× 0.664 × �

0.6418 × 0.01 × 22.849 × 10−5

�0.5

× 0.680.33 = 53.23 𝑊/𝑚2.𝐾

𝑞1 = 53.23 × (0.01 × 0.2) × (500 − 25) = 𝟓𝟎.𝟓𝟔𝑾 (ii) The fifth strip, 𝑞5 = 𝑞0−5 − 𝑞0−4 𝑞5 = ℎ�0−5(5∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞) − ℎ�0−4(4∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞) 𝑞5 = (5ℎ�0−5 − 4ℎ�0−4)(∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞)

ℎ�0−5 =0.0429

0.05× 0.664 × �

0.6418 × 0.05 × 22.849 × 10−5

�0.5

× 0.680.33 = 23.80 𝑊/𝑚2.𝐾

ℎ�0−4 =0.0429

0.04× 0.664 × �

0.6418 × 0.04 × 22.849 × 10−5

�0.5

× 0.680.33 = 26.61 𝑊/𝑚2.𝐾 𝑞5 = (5 × 23.80 − 4 × 26.61)(0.01 × 0.2) × (500 − 25) = 𝟏𝟏.𝟗𝟑 𝑾 (iii) The tenth strip, 𝑞10 = 𝑞0−10 − 𝑞0−9 𝑞10 = ℎ�0−10(10∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞) − ℎ�0−9(9∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞) 𝑞5 = (10ℎ�0−5 − 9ℎ�0−4)(∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞)

ℎ�0−10 =0.0429

0.1× 0.664 × �

0.6418 × 0.1 × 22.849 × 10−5

�0.5

× 0.680.33 = 16.83 𝑊/𝑚2.𝐾

ℎ�0−9 =0.0429

0.09× 0.664 × �

0.6418 × 0.09 × 22.849 × 10−5

�0.5

× 0.680.33 = 17.74 𝑊/𝑚2.𝐾 𝑞10 = (10 × 16.83 − 9 × 17.74)(0.01 × 0.2) × (500 − 25) = 𝟖.𝟐𝟎 𝑾 (iii) For the entire heater 𝑞5 = (10ℎ�0−5 − 9ℎ�0−4)(∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞)

14

ℎ�0−25 =0.0429

0.25× 0.664 × �

0.6418 × 0.25 × 22.849 × 10−5

�0.5

× 0.680.33 = 10.64 𝑊/𝑚2.𝐾 𝑞1 = ℎ�0−25 × (25∆𝐿 × 𝑤) × (𝑇𝑠 − 𝑇∞) = 10.64 × 25 × 0.01 × 0.2 × (500 − 25)

= 𝟐𝟓𝟐.𝟕 𝑾 10. Consider a flat plate subject to parallel flow (top and bottom) characterized by u∞ = 5 m/s, T∞ = 20oC. Determine the average convective heat transfer coefficient, convective heat transfer rate, and drag force associated with an L = 2 m long, w = 2 m wide flat plate for air flow and surface temperatures of Ts = 50oC and 80oC.

Properties of air at (50 + 20)/2 oC = 308 K: µ = 1.846 x 10-5, k = 0.02624 W/m.K, 𝐶𝑝 = 1.0049 𝑘𝐽/𝑘𝑔.𝐾, Pr = 0.707, ρ = 1.177 kg/m3

𝑅𝑒2𝑚 =1.177 × 2 × 51.846 × 10−5

= 6.4 × 105 > 5 × 105 therefore, the flow is turbulent at the end of the plate.

ℎ� =0.02624

2.0× [0.37(6.4 × 105)0.8 − 871] × 0.7070.33 = 𝟖.𝟖𝟓 𝑾/𝒎𝟐.𝑲

𝑞𝑐𝑜𝑛𝑣 = ℎ�𝐴(𝑇𝑠 − 𝑇∞) = 8.85 × 2 × 2 × (50 − 20) = 𝟏𝟎𝟔𝟐 𝑾

We have information involving frictional force as follows:

𝑆𝑡� 𝑃𝑟2/3 =𝐶𝑓���2

, 𝐶𝑓��� =𝜏

12𝜌𝑢

2, 𝜏 =

𝐹2𝐴

, 𝑆𝑡� =ℎ�

𝜌𝐶𝑝𝑢∞

Ts = 50, 80oC

15

𝑆𝑡� =ℎ�

𝜌𝐶𝑝𝑢∞=

8.851.177 × 1.0049 × 1000 × 5

= 0.0015

𝐶𝑓��� = 2𝑆𝑡� 𝑃𝑟2/3 = 2 × 0.0015 × 0.7070.33 = 0.00238

𝜏 = 𝐶𝑓��� ×12𝜌𝑢2 = 0.00238 × 0.5 × 1.177 × 5 × 5 = 0.035 𝑁/𝑚2

𝐹 = 2𝜏𝐴 = 2 × 0.035 × 2 × 2 = 𝟎.𝟐𝟖 𝑵

Note: Similarly, you can find out the above for the case of surface temperature 80oC 11. Water at 35oC enters a square tube of sides 2 cm with a velocity of 3 cm/s and exit with a temperature of 55oC. The wall of the square tube is maintained at a uniform temperature of 80oC. Neglecting entrance effect, determine the length of the tube required. Properties of water at (55 + 35)/2 oC = 45oC: µ = 594 x 10-6, k = 0.638 W/m.K, 𝐶𝑝 =4.181 𝑘𝐽/𝑘𝑔.𝐾, Pr = 3.89, ρ = 990.19 kg/m3

𝑅𝑒𝐷 =4mπDμ

= 6.4 × 105 > 5 × 105

m = A × u × ρ = 4 × 10−4 × 3 × 10−2 × 990.19 = 0.01189 kg/s

D =4AP

=4 × 4 × 10−4

8 × 10−2= 2 × 10−2 m

𝑅𝑒𝐷 =4mπDμ

=4 × 0.01189

π × 2 × 10−2 × 594 × 10−6= 1274

𝑁𝑢���� = 2.976 (𝑵𝒐𝒕𝒆: 𝒕𝒉𝒊𝒔 𝒊𝒏𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏 𝒘𝒊𝒍𝒍 𝒃𝒆 𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅 )

𝑁𝑢���� =h�Dk

= 2.976

h� =0.638 × 2.976

2 × 10−2= 94.93 𝑊/𝑚2.𝐾

Also, from an energy balance, we can write: m𝐶𝑝(∆𝑇)𝑤𝑎𝑡𝑒𝑟 = hPL(∆𝑇)𝐿𝑀𝑇𝐷 (1)

16

(∆𝑇)𝐿𝑀𝑇𝐷 =(∆𝑇1 − ∆𝑇2)

ln �∆𝑇1∆𝑇2�

=(45 − 25)

ln �4525�

= 34.02𝑜C

Therefore from (1), m𝐶𝑝(∆𝑇)𝑤𝑎𝑡𝑒𝑟 = h�PL(∆𝑇)𝐿𝑀𝑇𝐷 0.01189 × 4181 × 20 = 94.93 × 8 × 10−2 × L × 34.02 Required tube length, L = 3.84 m 12. Steam condensing on the outer surface of a thin walled circular tube of diameter 40 mm and length L = 6 m maintains a uniform outer surface temperature of 100oC. Water flows through a tube at a rate of 0.3 kg/s, and is heated from 25oC to 65oC. Determine the average convection coefficient associated with the water heating. Properties of water at (65 + 25)/2 oC = 45oC: µ = 594 x 10-6, k = 0.638 W/m.K, 𝐶𝑝 =4.181 𝑘𝐽/𝑘𝑔.𝐾, Pr = 3.89, ρ = 990.19 kg/m3

𝑅𝑒𝐷 =4mπDμ

= 6.4 × 105 > 5 × 105

𝑅𝑒𝐷 =4mπDμ

=4 × 0.3

π × 40 × 10−3 × 594 × 10−6= 16,084 > 2300

Therefore, the flow is turbulent, and

80 oC

35 oC

55 oC

ΔT1

ΔT2

17

ℎ� =𝑘𝐷

0.023𝑅𝑒𝐷0.8𝑃𝑟0.33

ℎ� =0.638

40 × 10−3× 0.023 × (16,084)0.8 × (3.89)0.33 = 𝟏𝟑𝟑𝟏 𝑾/𝒎𝟐.𝑲

13. Check the dimensions for Grashof Number and rearrange it to express it as Gr = (Inertia force/Viscous force). (Buoyancy force/Viscous force)

𝐺𝑟𝑎𝑠ℎ𝑜𝑓 𝑁𝑢𝑚𝑏𝑒𝑟,𝐺𝑟 =gβΔTL3

ν2

Checking the dimension,

𝐺𝑟 =�m

s2� �1K� (K)(m)3

�m2

s �2 = �

ms2� × �

1K� × (K) × (m)3 × �

s2

m4� = 1

𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒 = 𝜌𝑢2

𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒 = 𝜇𝑑𝑢𝑑𝑦

= 𝜇𝑢𝐿

𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒 = 𝜌𝑔ℎ = 𝜌𝑔βΔT𝐿, β = 1

ΔT

𝐺𝑟𝑎𝑠ℎ𝑜𝑓 𝑁𝑢𝑚𝑏𝑒𝑟 𝑚𝑜𝑑𝑖𝑓𝑖𝑒𝑑 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛,𝐺𝑟 =gβΔTL3

�𝜇ρ�2 =

gβΔTL3ρ2u2

𝜇2u2

𝐺𝑟 =gβΔTL3ρ2

𝜇2=

(ρgβΔTL). (ρu2)

�𝜇 𝑢𝐿� . �𝜇 𝑢𝐿�

Therefore, Gr = (Inertia force/Viscous force). (Buoyancy force/Viscous force) 14. A vertical plate is maintained at 40oC in 20oC still air. Determine (a) The height at which the boundary layer will turn turbulent if turbulence set in at GrPr = 109

(b) The value of boundary layer thickness (c) The value of average convection coefficient

18

(a) The height at which the boundary layer will turn turbulent if turbulence set in at GrPr = 109

Properties of air at (40 + 20)/2 oC = 303K: µ = 1.846 x 10-5, k = 0.02624 W/m.K Pr = 0.707, ρ = 1.177 kg/m3

𝐺𝑟 =gβΔTL3ρ2

𝜇2, β =

1303

= 0.0033 K−1

𝐺𝑟𝑃𝑟 =gβΔTL3ρ2Pr

𝜇2= 109

L = �109𝜇2

gβΔTρ2Pr�1/3

= �109 × (1.846 × 10−5)2

9.81 × 0.0033 × 20 × 1.1772 × 0.707�1/3

= 𝟎.𝟖𝟏𝟒 𝒎

(b) For laminar boundary layer for natural convection heat transfer,

𝛿𝑥

= 3.93Pr−1/2(0.952 + Pr)1/4Grx−1/4

𝐺𝑟𝑥 =gβΔTL3ρ2

𝜇2=

9.81 × 0.0033 × 20 × 0.8143 × 1.1772

(1.846 × 10−5)2 = 1.42 × 109

𝛿 = 3.93LPr−1/2(0.952 + Pr)1/4Grx−1/4

𝛿 = 3.93 × 0.814 × 0.707−12 × (0.952 + 0.707)

14 × (1.42 × 109)−

14 = 0.0222 m

= 𝟐𝟐.𝟐 𝐦𝐦

(c) Average convection coefficient, ℎ � = 4

3ℎ𝑥

𝑁𝑢𝑥 =ℎ𝑥𝐿𝑘

= 0.508Pr1/2(0.952 + Pr)−1/4Grx1/4

∴ ℎ𝑥 =𝑘𝐿

0.508Pr1/2(0.952 + Pr)−1/4Grx1/4

∴ ℎ𝑥 =0.02624

0.814× 0.508 × 0.707

12(0.952 + 0.707)−

14 × (1.42 × 109)

14

= 2.36 𝑊/𝑚2.𝐾

19

ℎ� =43

× 2.36 = 𝟑.𝟏𝟒 𝑾/𝒎𝟐.𝑲 15. A vertical plate 4 m high and 1 m wide is maintained at 60oC in still air at 0oC. Determine the value of average convection coefficient and heat transfer rate from the vertical plate. Properties of air at (60 + 0)/2 oC = 303 K: µ = 1.846 x 10-5, k = 0.02624 W/m.K Pr = 0.707, ρ = 1.177 kg/m3

𝐺𝑟𝑃𝑟 =gβΔTL3ρ2Pr

𝜇2

𝐺𝑟𝑃𝑟 =9.81 × 0.0033 × 60 × 43 × 1.1772 × 0.707

(1.846 × 10−5)2 = 3.57 × 1011

𝑁𝑢����𝑓 =ℎ�𝑓𝐿𝑘

= 0.021(𝐺𝑟𝑃𝑟)0.4

ℎ�𝑓 =0.02624

4× 0.021 × (3.57 × 1011)0.4 = 𝟓.𝟕𝟔 𝑾/𝒎𝟐.𝑲

𝑞 = ℎ�𝑓𝐴(𝑇𝑠 − 𝑇∞) = 5.76 × 4 × 1 × (60 − 0) = 𝟏𝟑𝟖𝟐.𝟒 𝑾 16. The condenser of a large steam power plant contains 1000 brass tubes (k = 110 W/m.K). The tubes are of thin wall construction with D = 25 mm and steam condenses on their outer surface with an associated convection coefficient of 10,000 W/m2.K. (a) If the cooling water from a large lake is pumped through the condenser tubes at 400 kg/s, what is the overall heat transfer coefficient? Properties of the water may be assumed as μ = 960 x 10-6 kg/m.s, k = 0.60 W/m.K and Pr = 6.6. (b) If water is extracted from the lake at 23oC and 10 kg/s of steam at 0.5 bars are to be condensed, what is the corresponding temperature of the water leaving the condenser? The specific heat of the water is 4180 J/kg.K (a) Outside heat transfer coefficient, ho = 10,000 W/m2.K

1𝑈

=1ℎ𝑖

+1ℎ𝑜

20

𝑅𝑒𝐷 =4 × mπDμ

=4 × (400/1000)

π × 0.025 × 960 × 10−6= 21,231 > 2300

Therefore, the flow is turbulent.

ℎ𝑖 =0.60

0.025× 0.023 × (21,231)0.8 × (6.6)0.33 = 2978 𝑊/𝑚2.𝐾

1𝑈

=1ℎ𝑖

+1ℎ𝑜

=1

2978+

110,000

= 0.0004358 𝑚2.𝐾/𝑊

∴ 𝑈 = 𝟐𝟐𝟗𝟓 𝑾/𝒎𝟐.𝑲 (b) From the steam table, for 0.5 bar, hfg = 2305 kJ/kg = 2305 x 103 J/kg Energy balance in the heat exchanger gives,

�mtotal 𝐶𝑝∆𝑇�water = msteam × hfg

[400 × 4180 × (𝑇𝑜𝑢𝑡 − 23)]water = 10 × 2305 × 103

𝑇𝑜𝑢𝑡 = 𝟑𝟕𝒐𝐂 17. Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas which has a flow rate of 15 kg/s and an inlet temperature of 820 oC, passes through a bundle of tubes, while the air, which has a flow rate of 10 kg/s and an inlet temperature of 27oC is in cross flow over the tubes. The tubes are unfinned, and the overall heat transfer coefficient is 100 W/m2.K. Determine the total tube surface area required to achieve an air outlet temperature of 577oC. The exhaust gas and the air may each be assumed to have a specific heat of 1075 J/kg.K. Using usual notations and 𝜖 − 𝑁𝑇𝑈 𝑚𝑒𝑡ℎ𝑜𝑑:

𝐶𝑐 = 𝑚𝑐𝑐𝑝,𝑐 = 10 × 1.075 = 10.75 = 𝐶𝑚𝑖𝑛

𝐶ℎ = 𝑚ℎ𝑐𝑝,ℎ = 15 × 1.075 = 16.13 == 𝐶𝑚𝑎𝑥

21

𝐶𝑚𝑖𝑛𝐶𝑚𝑎𝑥

=10.7516.13

= 0.666

𝜖 =𝑇𝑐,𝑜 − 𝑇𝑐,𝑖

𝑇ℎ,𝑖 − 𝑇𝑐,𝑖=

577 − 27820 − 27

= 0.694

Using 𝜖 − 𝑁𝑇𝑈 chart for one fluid mixed and the other unmixed,

𝑁𝑇𝑈 = 2.3 = 𝑈𝐴𝐶𝑚𝑖𝑛

𝐴 =2.3 × 10.75 × 1000

100= 𝟐𝟒𝟕.𝟐𝟓 𝒎𝟐