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NATIONAL UNIVERSITY OF SINGAPORE Department Mechanical Engineering
ME 3122E β 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 οΏ½οΏ½ 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:
π =β ππππ₯
οΏ½ππππ₯οΏ½
Substituting the prescribed temperature distribution,
π =β ππππ₯
οΏ½π(π + ππ₯2)
ππ₯οΏ½ = π =β π
πππ₯
[2ππ₯] = β2ππ
π =β 2(β2000ππΆ/π2) Γ 50ππ.πΎ
= π.π Γ ππππΎ/ππ
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(b) The heat fluxes at the wall faces can be evaluated from Fourierβs law,
π"π₯ = βπ οΏ½
ππππ₯οΏ½π₯
Using the temperature distribution T(x) to evaluate the gradient, find
π"π₯ = βπ
π(π + ππ₯2)ππ₯
= β2πππ₯ The fluxes at x = 0 and x = L are then
π"π(π) = π
π"π₯(πΏ) = β2πππΏ = β2 Γ 50
ππ.πΎ
Γ (β2000ππΆ/π2) Γ 0.050π
π"π(π³) = ππ,πππ πΎ/ππ
From an overall energy balance on the wall, it follows that, for a unit area,
π"π₯(0) β π"
π₯(πΏ) + ποΏ½οΏ½ = 0
οΏ½οΏ½ =π"
π₯(πΏ) β π"π₯(0)
πΏ=
10,000 π/π2 β 00.050
= 2.0 Γ 105π/π3 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 Tβ = -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
ππππππ‘πππβ πππππ£ = 0
ππππππ‘πππ = πππππ£ = β Γ 2ππ2 Γ (ππ β πβ)
ππππππ‘πππ =50ππ2.πΎ
Γ 2π Γ 0.04 Γ οΏ½5 β (β15)οΏ½ = πππ.π πΎ/π (b) The inner rod is isothermal, that is T(0) = T(r1) Representing cylinder B by a thermal circuit,
πβ² =
π(π1) β ππ π β²π΅
π β²π΅ = ππ οΏ½π2π1οΏ½ /2πππ΅
Therefore,
π(π1) = ππ +(πβ² Γ π β²π΅) = 5ππΆ + οΏ½251ππ
Γππ οΏ½40
20οΏ½2π Γ 1.5
οΏ½ = 23.5ππΆ
Hence, π(0) = π(π1) = ππ.πππͺ Note that ππ΄ has no influence on the temperature π(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 650Β°C and then quenched
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in a pool of oil that has a temperature 55Β°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 200Β°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 200Β°C.The properties of steel ball bearings are as follows: k =50 W/m.K; πΌ= k/ΟCp = 1.3x10-5 m2/s Given: ππ = 650ππΆ,πβ = 55ππΆ, π = 200ππΆ (a) Using the lumped heat capacity method,
ππ οΏ½π β πβππ β πβ
οΏ½ = οΏ½ββπ΄π‘ππΆπ
οΏ½ (π)
π‘ = ππ οΏ½π β πβππ β πβ
οΏ½Γ οΏ½ππΆπββπ΄
οΏ½ = ππ οΏ½π β πβππ β πβ
οΏ½ Γ οΏ½ππ
β3 β βοΏ½ , οΏ½β΅
ππ΄
=π3οΏ½
Time required for the cooling of the bearings,
π‘ = ππ οΏ½200 β 55650 β 55
οΏ½ Γ οΏ½0.02 Γ 50
β3 Γ 1.3 Γ 10β5 Γ 300οΏ½ = πππ π
(b) Total amount of heat removed,
π»π‘ππ‘ππ = οΏ½ βπ΄(π β πβ)ππ‘π‘
0
Using equation (1) to replace (π β πβ),
π»π‘ππ‘ππ = οΏ½ βπ΄(ππ β πβ)ππ₯π οΏ½β3 β βπ‘ππ
οΏ½ ππ‘π‘
0
after the integration,
π»π‘ππ‘ππ = οΏ½πππ΄3 β
(πβ β ππ)οΏ½ οΏ½ππ₯π οΏ½β3 β βπ‘ππ
οΏ½ β 1οΏ½
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π»π‘ππ‘ππ
= οΏ½0.02 Γ 50 Γ 4π Γ 0.022 Γ (55 β 650)
3 Γ 1.3 Γ 10β5οΏ½ οΏ½ππ₯π οΏ½
β3 Γ 1.3 Γ 10β5 Γ 300 Γ 1200.02 Γ 50
οΏ½
β 1οΏ½ = π.ππ Γ ππππ±
c) Instantaneous heat transfer rate from the bearings when they are first placed in the oil and when they reach 200Β°C.
ππ‘ππππ ππππππ = βπ΄(ππ β πβ)ππ₯π οΏ½β3 β βπ‘ππ
οΏ½ Heat transfer rate from the bearings when they are first placed, that is when t = 0,
ππ‘ππππ ππππππ = βπ΄(ππ β πβ)ππ₯π(0) = βπ΄(ππ β πβ)
ππ‘ππππ ππππππ = 300 Γ 4π Γ 0.022 Γ (650 β 55) = πππ πΎ Heat transfer rate from the bearings when they reach 200Β°C, that is when t = 120 s, ππ‘ππππ ππππππ = 300 Γ 4π Γ 0.022 Γ (650 β 55)ππ₯π οΏ½β3Γ1.3Γ10β5Γ300Γ120
0.02Γ50οΏ½=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 Tβ = 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 wΒ»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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π = οΏ½βπππ΄
= οΏ½100ππ2.πΎ Γ (2π + 0.002π)180ππ.πΎ Γ (1π Γ 0.001π)
= 33.34 πβ1
π = βππ΄πβππ = οΏ½180ππ.πΎ
Γ (1π Γ 0.001π) Γ (2π + 0.002π) Γ100ππ2.πΎ
= 450 π/π
Case A: convection heat transfer from fin tip
ππ = ππ ππβππΏ + (β/ππ)πππ βππΏπ ππβππΏ + (β/ππ)πππ βππΏ
= 450 π/π0.340 + 0.0167 Γ 1.0571.057 + 0.0167 Γ 0.340
= ππππΎ/π
ππ =ππ
β(2πΏ + π‘)ππ=
151 π/π100 π/π2 Γ (2 Γ 0.01π + 0.001π) Γ 75ππΆ
= π.ππ
ππ =ππβπ‘ππ
=151 π/π
100 π/π2 Γ 0.001π Γ 75ππΆ= ππ.π
Case B: adiabatic fin tip
ππ = π π‘ππβππΏ = 450ππ
Γ 0.321 = πππ πΎ/π
ππ =π‘ππβππΏππΏ
=0.3210.333
= π.πππ
ππ =ππβπ‘ππ
=144 π/π
100 π/π2 Γ 0.001π Γ 75ππΆ= ππ.π
Case C: Very long fin (πΏ β β) ππ = π = πππ πΎ/π, ππ = π
ππ =ππβπ‘ππ
=450 π/π
100 π/π2 Γ 0.001π Γ 75ππΆ= ππ.π
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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 πΉ11 + πΉ12 + πΉ13 = π , β΄ πΉ12 = 1 β πΉ13
For the given configuration, π π = πππΏ
, π π = πππΏ
, π = 1 + 1+π π2
π π2 (1)
πΉππ =1
2 οΏ½π β οΏ½π2 β 4 οΏ½πππποΏ½2οΏ½οΏ½12
(2)
Using equations (1) and (2) with i = 1, j = 3,
πΉ13 = 12
οΏ½π β οΏ½π2 β 4 οΏ½π·3π·1οΏ½2οΏ½οΏ½
12 and π = 1 + 1+π 32
π 12, π = 2 + 1
π 2= 4π»2 + 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
πΉ12 = 1 β12
{4π»2 + 2 β [(4π»2 + 2)2 β 4]}12
πΉ12 = 2π» οΏ½(1 + π»2)1/2 β π»οΏ½ (4) b) Relation for F
22, inner lateral surface. Apply summation rule on A
2, recognizing that
F23
= F21
,
πΉ21 + πΉ22 + πΉ23 = 1 β΄ πΉ22 = 1 β 2πΉ21 (5) Apply reciprocity between A
1 and A
2,
πΉ21 = οΏ½π΄1π΄2οΏ½ πΉ12 (6)
and substituting into Eq. (5), and using area expressions
πΉ22 = 1 β 2πΉ21
πΉ22 = 1 β 2 οΏ½π΄1π΄2οΏ½ πΉ12 = 1 β 2 οΏ½
π·4πΏοΏ½πΉ12 = 1 β οΏ½
12π»
οΏ½πΉ12
where A
1 = ΟD
2/4 and A
2 = ΟDL.
Substituting from Eq. (4) for F
12, find
πΉ22 = 1 β οΏ½1
2π»οΏ½2π»οΏ½(1 + π»2)1/2 β π»οΏ½ = 1 + π» β (1 + π»2)1/2
6. A certain surface maintained at 1400 K has the following spectral emissive characteristics:
Ξ΅ Ξ» = 0.08 0 < Ξ» < 0.5ΞΌm, Ξ΅ Ξ» = 0.4 0.6 < Ξ» < 5ΞΌm Ξ΅ Ξ» = 0.7 5 < Ξ» < β
Calculate the emissive power of the surface.
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For the given spectral emissive characteristics,
ππ = οΏ½ πππΈπ,π(π)πΈπ(π)
ππ =β
0οΏ½ ππ
πΈπ,π(π)πΈπ(π)
ππ +π1
0οΏ½ ππ
πΈπ,π(π)πΈπ(π)
ππ +π2
π1οΏ½ ππ
πΈπ,π(π)πΈπ(π)
ππβ
π2
= οΏ½ π1πΈπ,π(π)πΈπ(π)
ππ +0.6ππ
0οΏ½ π2
πΈπ,π(π)πΈπ(π)
ππ +5ππ
0.6πποΏ½ π3
πΈπ,π(π)πΈπ(π)
ππβ
5ππ
= οΏ½π1οΏ½πΉ(0.6π) β πΉ(0π)οΏ½οΏ½+οΏ½π2οΏ½πΉ(5π) β πΉ(0.6π)οΏ½οΏ½ + οΏ½π3οΏ½πΉ(βπ) β πΉ(5π)οΏ½οΏ½ = οΏ½0.08οΏ½πΉ(0.6π) β πΉ(0π)οΏ½οΏ½+οΏ½0.4οΏ½πΉ(5π) β πΉ(0.6π)οΏ½οΏ½ + οΏ½0.7οΏ½πΉ(βπ) β πΉ(5π)οΏ½οΏ½
= 0.08[0.000016 β 0.000]+0.4[0.808109 β 0.000016] + 0.7[1 β 0.808109] ππ = 0.00000128+0.32323 + 0.13432 = 0.4575 πΈ = ππππ4 = 5.67 Γ 10β8 Γ 0.4575 Γ 14004 = π.πππππ Γ πππ π/π¦π 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 Ξ±Ξ» = 0, 0.8, 0 and 0.9 for 0 β€ Ξ» β€ 0.5ΞΌm, 0.5 < Ξ» β€ 1ΞΌm, 1 ΞΌm < Ξ» β€ 2ΞΌm and Ξ» > 2ΞΌm, 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
πΌπ = οΏ½πΌππΊππΊ
ππ =β
0οΏ½
πΌππΈπ,π(5800πΎ)πΈπ
ππβ
0
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= οΏ½ πΌ1πΈπ,π(π)πΈπ(π)
ππ +0.5ππ
0οΏ½ πΌ2
πΈπ,π(π)πΈπ(π)
ππ +1ππ
0.5πποΏ½ πΌ3
πΈπ,π(π)πΈπ(π)
ππ2ππ
1ππ
+ οΏ½ πΌ4πΈπ,π(π)πΈπ(π)
ππβ
2ππ
= οΏ½πΌ1οΏ½πΉ(0.5π) β πΉ(0π)οΏ½οΏ½+οΏ½πΌ2οΏ½πΉ(1π) β πΉ(0.5π)οΏ½οΏ½ + οΏ½πΌ3οΏ½πΉ(2π) β πΉ(1π)οΏ½οΏ½ + οΏ½πΌ4οΏ½πΉ(βπ) β πΉ(2π)οΏ½οΏ½ = οΏ½0.0οΏ½πΉ(0.5π) β πΉ(0π)οΏ½οΏ½+οΏ½0.8οΏ½πΉ(1π) β πΉ(0.5π)οΏ½οΏ½ + οΏ½0.0οΏ½πΉ(2π) β πΉ(1π)οΏ½οΏ½ + οΏ½0.9οΏ½πΉ(βπ) β πΉ(2π)οΏ½οΏ½
= [0.8(0.72 β 0.25)] + [0.9(1.0 β 0.94)] = 0.43
πΊππππ πππππ = πΌπ Γ πΊπ = 0.43 Γ 1200 = πππ πΎ/ππ The emissivity
π = οΏ½πππΈπ,π(400πΎ)
πΈπππ
β
0
= οΏ½ π1πΈπ,π(π)πΈπ(π)
ππ +0.5ππ
0οΏ½ π2
πΈπ,π(π)πΈπ(π)
ππ +1ππ
0.5πποΏ½ π3
πΈπ,π(π)πΈπ(π)
ππ2ππ
1ππ
+ οΏ½ π4πΈπ,π(π)πΈπ(π)
ππβ
2ππ
π = οΏ½πΌ1οΏ½πΉ(0.5 Γ 400) β πΉ(0 Γ 400)οΏ½οΏ½+οΏ½πΌ2οΏ½πΉ(1 Γ 400) β πΉ(0.5 Γ 400)οΏ½οΏ½ +οΏ½πΌ3οΏ½πΉ(2 Γ 400) β πΉ(1 Γ 400)οΏ½οΏ½ + οΏ½πΌ4οΏ½πΉ(βπ) β πΉ(2 Γ 400)οΏ½οΏ½ = 0+0 + 0 + [0.9(1.0 β 0.000016)]=0.9 πΈ = πππ4 = 5.67 Γ 10β8 Γ 0.9 Γ 4004 = ππππ π/π¦π The radiosity is : π½ = πΈ + ππ πΊπ = πΈ + (1 β πΌπ )πΊπ = 1306 + 0.57 Γ 1200 =πππππ/π¦π The net radiation transfer from the surface is: ππππ‘ = (πΈ β πΌπ πΊπ )π΄π = (1306 β 515) Γ 4 = ππππ π Note: Unless 3164 W are supplied to the surface by other means (for example, by
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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)
ππβπ =ποΏ½ππ4 β ππ4οΏ½
1 β πππππ΄π
+ 1π΄ππΉππ
+ 1 β πππππ΄π
ππβπ =5.67 Γ 10β8(2804 β 1004)
1 β 0.250.25 Γ π Γ 0.015 + 1
π Γ 0.015 Γ 1 + 1 β 0.350.35 Γ π Γ 0.025
ππβπ =342.84
63.69 + 21.23 + 23.66= π.ππ πΎ/π
Di = 15 mm Ti = 100 K Ξ΅i = 0.25
Do = 25 mm
Ξ΅o = 0.35
Ds = 20 mm Ξ΅s = 0.06 To = 280 K
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(b) Additional resistances due to the radiation shield
1 β ππ ππ π΄π
+1 β ππ ππ π΄π
+1
π΄π πΉπ π
1 β 0.06
0.06 Γ π Γ 0.02+
1 β 0.350.06 Γ π Γ 0.02
+1
π Γ 0.02 Γ 1= 514.86 πβ1
ππβπ,πππ€ =342.84
63.69 + 21.23 + 23.66 + 514.86= 0.55 π/π
% Reduction in heat gain = οΏ½3.15β0.553.15
οΏ½Γ 100 = ππ. .π %
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: Β΅ = 2.849 x 10-5, k = 0.04357 W/m.K Pr = 0.68, Ο = 0.6418 kg/m3
(i) The location of transition is determined from:
π₯π =π πππππ’β
=5 Γ 105 Γ 2.849 Γ 10β5
0.6418 Γ 2= 11.09 π
Since xc >>L=0.25 m, the air flow is laminar over the entire heater. For the first strip,
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π1 = β1οΏ½οΏ½οΏ½ Γ (βπΏ Γ π€) Γ (ππ β πβ)
β1οΏ½οΏ½οΏ½ =πβπΏ
Γ 0.664 Γ π π0.5 Γ ππ0.33
β1οΏ½οΏ½οΏ½ =0.0429
0.01Γ 0.664 Γ οΏ½
0.6418 Γ 0.01 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 53.23 π/π2.πΎ
π1 = 53.23 Γ (0.01 Γ 0.2) Γ (500 β 25) = ππ.πππΎ (ii) The fifth strip, π5 = π0β5 β π0β4 π5 = βοΏ½0β5(5βπΏ Γ π€) Γ (ππ β πβ) β βοΏ½0β4(4βπΏ Γ π€) Γ (ππ β πβ) π5 = (5βοΏ½0β5 β 4βοΏ½0β4)(βπΏ Γ π€) Γ (ππ β πβ)
βοΏ½0β5 =0.0429
0.05Γ 0.664 Γ οΏ½
0.6418 Γ 0.05 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 23.80 π/π2.πΎ
βοΏ½0β4 =0.0429
0.04Γ 0.664 Γ οΏ½
0.6418 Γ 0.04 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 26.61 π/π2.πΎ π5 = (5 Γ 23.80 β 4 Γ 26.61)(0.01 Γ 0.2) Γ (500 β 25) = ππ.ππ πΎ (iii) The tenth strip, π10 = π0β10 β π0β9 π10 = βοΏ½0β10(10βπΏ Γ π€) Γ (ππ β πβ) β βοΏ½0β9(9βπΏ Γ π€) Γ (ππ β πβ) π5 = (10βοΏ½0β5 β 9βοΏ½0β4)(βπΏ Γ π€) Γ (ππ β πβ)
βοΏ½0β10 =0.0429
0.1Γ 0.664 Γ οΏ½
0.6418 Γ 0.1 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 16.83 π/π2.πΎ
βοΏ½0β9 =0.0429
0.09Γ 0.664 Γ οΏ½
0.6418 Γ 0.09 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 17.74 π/π2.πΎ π10 = (10 Γ 16.83 β 9 Γ 17.74)(0.01 Γ 0.2) Γ (500 β 25) = π.ππ πΎ (iii) For the entire heater π5 = (10βοΏ½0β5 β 9βοΏ½0β4)(βπΏ Γ π€) Γ (ππ β πβ)
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βοΏ½0β25 =0.0429
0.25Γ 0.664 Γ οΏ½
0.6418 Γ 0.25 Γ 22.849 Γ 10β5
οΏ½0.5
Γ 0.680.33 = 10.64 π/π2.πΎ π1 = βοΏ½0β25 Γ (25βπΏ Γ π€) Γ (ππ β πβ) = 10.64 Γ 25 Γ 0.01 Γ 0.2 Γ (500 β 25)
= πππ.π πΎ 10. Consider a flat plate subject to parallel flow (top and bottom) characterized by uβ = 5 m/s, Tβ = 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: Β΅ = 1.846 x 10-5, k = 0.02624 W/m.K, πΆπ = 1.0049 ππ½/ππ.πΎ, Pr = 0.707, Ο = 1.177 kg/m3
π π2π =1.177 Γ 2 Γ 51.846 Γ 10β5
= 6.4 Γ 105 > 5 Γ 105 therefore, the flow is turbulent at the end of the plate.
βοΏ½ =0.02624
2.0Γ [0.37(6.4 Γ 105)0.8 β 871] Γ 0.7070.33 = π.ππ πΎ/ππ.π²
πππππ£ = βοΏ½π΄(ππ β πβ) = 8.85 Γ 2 Γ 2 Γ (50 β 20) = ππππ πΎ
We have information involving frictional force as follows:
ππ‘οΏ½ ππ2/3 =πΆποΏ½οΏ½οΏ½2
, πΆποΏ½οΏ½οΏ½ =π
12ππ’
2, π =
πΉ2π΄
, ππ‘οΏ½ =βοΏ½
ππΆππ’β
Ts = 50, 80oC
15
ππ‘οΏ½ =βοΏ½
ππΆππ’β=
8.851.177 Γ 1.0049 Γ 1000 Γ 5
= 0.0015
πΆποΏ½οΏ½οΏ½ = 2ππ‘οΏ½ ππ2/3 = 2 Γ 0.0015 Γ 0.7070.33 = 0.00238
π = πΆποΏ½οΏ½οΏ½ Γ12ππ’2 = 0.00238 Γ 0.5 Γ 1.177 Γ 5 Γ 5 = 0.035 π/π2
πΉ = 2ππ΄ = 2 Γ 0.035 Γ 2 Γ 2 = π.ππ π΅
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: Β΅ = 594 x 10-6, k = 0.638 W/m.K, πΆπ =4.181 ππ½/ππ.πΎ, Pr = 3.89, Ο = 990.19 kg/m3
π ππ· =4mΟDΞΌ
= 6.4 Γ 105 > 5 Γ 105
m = A Γ u Γ Ο = 4 Γ 10β4 Γ 3 Γ 10β2 Γ 990.19 = 0.01189 kg/s
D =4AP
=4 Γ 4 Γ 10β4
8 Γ 10β2= 2 Γ 10β2 m
π ππ· =4mΟDΞΌ
=4 Γ 0.01189
Ο Γ 2 Γ 10β2 Γ 594 Γ 10β6= 1274
ππ’οΏ½οΏ½οΏ½οΏ½ = 2.976 (π΅πππ: ππππ πππππππππππ ππππ ππ ππππππ ππ )
ππ’οΏ½οΏ½οΏ½οΏ½ =hοΏ½Dk
= 2.976
hοΏ½ =0.638 Γ 2.976
2 Γ 10β2= 94.93 π/π2.πΎ
Also, from an energy balance, we can write: mπΆπ(βπ)π€ππ‘ππ = hPL(βπ)πΏπππ· (1)
16
(βπ)πΏπππ· =(βπ1 β βπ2)
ln οΏ½βπ1βπ2οΏ½
=(45 β 25)
ln οΏ½4525οΏ½
= 34.02πC
Therefore from (1), mπΆπ(βπ)π€ππ‘ππ = hοΏ½PL(βπ)πΏπππ· 0.01189 Γ 4181 Γ 20 = 94.93 Γ 8 Γ 10β2 Γ L Γ 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: Β΅ = 594 x 10-6, k = 0.638 W/m.K, πΆπ =4.181 ππ½/ππ.πΎ, Pr = 3.89, Ο = 990.19 kg/m3
π ππ· =4mΟDΞΌ
= 6.4 Γ 105 > 5 Γ 105
π ππ· =4mΟDΞΌ
=4 Γ 0.3
Ο Γ 40 Γ 10β3 Γ 594 Γ 10β6= 16,084 > 2300
Therefore, the flow is turbulent, and
80 oC
35 oC
55 oC
ΞT1
ΞT2
17
βοΏ½ =ππ·
0.023π ππ·0.8ππ0.33
βοΏ½ =0.638
40 Γ 10β3Γ 0.023 Γ (16,084)0.8 Γ (3.89)0.33 = ππππ πΎ/ππ.π²
13. Check the dimensions for Grashof Number and rearrange it to express it as Gr = (Inertia force/Viscous force). (Buoyancy force/Viscous force)
πΊπππ βππ ππ’ππππ,πΊπ =gΞ²ΞTL3
Ξ½2
Checking the dimension,
πΊπ =οΏ½m
s2οΏ½ οΏ½1KοΏ½ (K)(m)3
οΏ½m2
s οΏ½2 = οΏ½
ms2οΏ½ Γ οΏ½
1KοΏ½ Γ (K) Γ (m)3 Γ οΏ½
s2
m4οΏ½ = 1
πΌππππ‘ππ πππππ = ππ’2
πππ πππ’π πππππ = πππ’ππ¦
= ππ’πΏ
π΅π’ππ¦ππππ¦ πππππ = ππβ = ππΞ²ΞTπΏ, Ξ² = 1
ΞT
πΊπππ βππ ππ’ππππ ππππππππ ππ₯ππππ π πππ,πΊπ =gΞ²ΞTL3
οΏ½πΟοΏ½2 =
gΞ²ΞTL3Ο2u2
π2u2
πΊπ =gΞ²ΞTL3Ο2
π2=
(ΟgΞ²ΞTL). (Οu2)
οΏ½π π’πΏοΏ½ . οΏ½π π’πΏοΏ½
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: Β΅ = 1.846 x 10-5, k = 0.02624 W/m.K Pr = 0.707, Ο = 1.177 kg/m3
πΊπ =gΞ²ΞTL3Ο2
π2, Ξ² =
1303
= 0.0033 Kβ1
πΊπππ =gΞ²ΞTL3Ο2Pr
π2= 109
L = οΏ½109π2
gΞ²ΞTΟ2PrοΏ½1/3
= οΏ½109 Γ (1.846 Γ 10β5)2
9.81 Γ 0.0033 Γ 20 Γ 1.1772 Γ 0.707οΏ½1/3
= π.πππ π
(b) For laminar boundary layer for natural convection heat transfer,
πΏπ₯
= 3.93Prβ1/2(0.952 + Pr)1/4Grxβ1/4
πΊππ₯ =gΞ²ΞTL3Ο2
π2=
9.81 Γ 0.0033 Γ 20 Γ 0.8143 Γ 1.1772
(1.846 Γ 10β5)2 = 1.42 Γ 109
πΏ = 3.93LPrβ1/2(0.952 + Pr)1/4Grxβ1/4
πΏ = 3.93 Γ 0.814 Γ 0.707β12 Γ (0.952 + 0.707)
14 Γ (1.42 Γ 109)β
14 = 0.0222 m
= ππ.π π¦π¦
(c) Average convection coefficient, β οΏ½ = 4
3βπ₯
ππ’π₯ =βπ₯πΏπ
= 0.508Pr1/2(0.952 + Pr)β1/4Grx1/4
β΄ βπ₯ =ππΏ
0.508Pr1/2(0.952 + Pr)β1/4Grx1/4
β΄ βπ₯ =0.02624
0.814Γ 0.508 Γ 0.707
12(0.952 + 0.707)β
14 Γ (1.42 Γ 109)
14
= 2.36 π/π2.πΎ
19
βοΏ½ =43
Γ 2.36 = π.ππ πΎ/ππ.π² 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: Β΅ = 1.846 x 10-5, k = 0.02624 W/m.K Pr = 0.707, Ο = 1.177 kg/m3
πΊπππ =gΞ²ΞTL3Ο2Pr
π2
πΊπππ =9.81 Γ 0.0033 Γ 60 Γ 43 Γ 1.1772 Γ 0.707
(1.846 Γ 10β5)2 = 3.57 Γ 1011
ππ’οΏ½οΏ½οΏ½οΏ½π =βοΏ½ππΏπ
= 0.021(πΊπππ)0.4
βοΏ½π =0.02624
4Γ 0.021 Γ (3.57 Γ 1011)0.4 = π.ππ πΎ/ππ.π²
π = βοΏ½ππ΄(ππ β πβ) = 5.76 Γ 4 Γ 1 Γ (60 β 0) = ππππ.π πΎ 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 ΞΌ = 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
1π
=1βπ
+1βπ
20
π ππ· =4 Γ mΟDΞΌ
=4 Γ (400/1000)
Ο Γ 0.025 Γ 960 Γ 10β6= 21,231 > 2300
Therefore, the flow is turbulent.
βπ =0.60
0.025Γ 0.023 Γ (21,231)0.8 Γ (6.6)0.33 = 2978 π/π2.πΎ
1π
=1βπ
+1βπ
=1
2978+
110,000
= 0.0004358 π2.πΎ/π
β΄ π = ππππ πΎ/ππ.π² (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,
οΏ½mtotal πΆπβποΏ½water = msteam Γ hfg
[400 Γ 4180 Γ (πππ’π‘ β 23)]water = 10 Γ 2305 Γ 103
πππ’π‘ = ππππ 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 π β πππ πππ‘βππ:
πΆπ = ππππ,π = 10 Γ 1.075 = 10.75 = πΆπππ
πΆβ = πβππ,β = 15 Γ 1.075 = 16.13 == πΆπππ₯
21
πΆππππΆπππ₯
=10.7516.13
= 0.666
π =ππ,π β ππ,π
πβ,π β ππ,π=
577 β 27820 β 27
= 0.694
Using π β πππ chart for one fluid mixed and the other unmixed,
πππ = 2.3 = ππ΄πΆπππ
π΄ =2.3 Γ 10.75 Γ 1000
100= πππ.ππ ππ