Chapter 1. Introduction
Chapter 1
Introduction Holman / Heat Transfer, 10th Edition
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Chapter 1. Introduction
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Heat transfer is thermal energy transfer that is induced by a temperature difference (or gradient) Modes of heat transfer Conduction heat transfer: Occurs when a temperature gradient exists through a solid or a stationary fluid (liquid or gas). Convection heat transfer: Occurs within a moving fluid, or between a solid surface and a moving fluid, when they are at different temperatures Thermal radiation: Heat transfer between two surfaces (that are not in contact), often in the absence of an intervening medium.
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Chapter 1. Introduction
1‐1. Conduction Heat Transfer
qx ∂T ∼ A ∂x
When the proportionality constant is inserted, ∂T qx = − kA ∂x
[1‐1]
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Chapter 1. Introduction
Energy conducted in left face + heat generated within element = change in internal energy + energy conducted out right face
These energy quantities are given as follows: Energy in left face = qx = −kA
∂T ∂x
Energy generated within element = qA dx ∂T Change in internal energy = ρ cA dx ∂τ Energy out right face = qx + dx
∂T ⎤ = − kA ⎥ ∂x ⎦ x + dx
⎡ ∂T ∂ ⎛ ∂T = − A ⎢k + ⎜k ⎣ ∂x ∂x ⎝ ∂x
⎞ ⎤ ⎟ dx ⎥ ⎠ ⎦
where q = energy generated per unit volume, W/m3 c = specific heat of material, J/kg ⋅ C
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Chapter 1. Introduction
ρ = density, kg/m3
Combining the relations above gives −kA
∂T ∂T ⎡ ∂T ∂ ⎛ ∂T + qA dx = ρ cA dx − A ⎢ k + ⎜k ∂x ∂τ ⎣ ∂x ∂x ⎝ ∂x
∂ ⎛ ∂T ⎞ ∂T k q c ρ + = or ∂x ⎜⎝ ∂x ⎟⎠ ∂τ qx + q y + qz + qgen
⎞ ⎤ ⎟ dx ⎥ ⎠ ⎦
[1‐2]
dE = qx + dx + q y + dy + qz + dz + dτ
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Chapter 1. Introduction
∂T qx = −k dy dz ∂x ⎡ ∂T ∂ ⎛ ∂T qx + dx = − ⎢ k + ⎜k ⎣ ∂x ∂x ⎝ ∂x q y = −k dx dz
⎞ ⎤ ⎟ dx ⎥ dy dz ⎠ ⎦
∂T ∂y
⎡ ∂T ∂ ⎛ ∂T ⎞ ⎤ q y + dy = − ⎢ k + ⎜k ⎟ dy ⎥ dx dz ⎣ ∂y ∂y ⎝ ∂y ⎠ ⎦ ∂T qz = −k dx dy ∂z ⎡ ∂T ∂ ⎛ ∂T ⎞ ⎤ qz + dz = − ⎢ k + ⎜k ⎟ dz ⎥ dx dy ⎣ ∂z ∂z ⎝ ∂z ⎠ ⎦ qgen = q dx dy dz dE ∂T = ρ c dx dy dz ∂τ dτ
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Chapter 1. Introduction
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Figure 1‐2 Elemental volume for one‐dimensional heat conduction analysis.
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Chapter 1. Introduction
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Figure 1‐3 Elemental volume for three‐dimensional heat‐conduction analysis: (a) cartesian coordinates; (b) cylindrical coordinates; (c) spherical coordinates.
Holman / Heat Transfer, 10th Edition
Chapter 1. Introduction
∂ ⎛ ∂T ⎜k ∂x ⎝ ∂x
⎞ ∂ ⎛ ∂T ⎟+ ⎜k ⎠ ∂y ⎝ ∂y
⎞ ∂ ⎛ ∂T ⎟+ ⎜k ⎠ ∂z ⎝ ∂z
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∂T ⎞ ρ + = q c ⎟ ∂τ [1‐3] ⎠
For constant thermal conductivity, Equation (1‐3) is written ∂ 2T ∂ 2T ∂ 2T q 1 ∂T + 2 + 2 + = 2 ∂x ∂y ∂z k α ∂τ [1−3a]
Equation (1‐3a) may be transformed into either cylindrical or spherical coordinates by standard calculus techniques. The results are as follows: Cylindrical coordinates: ∂ 2T 1 ∂T 1 ∂ 2T ∂ 2T q 1 ∂T + + 2 2+ 2 + = 2 r ∂r r ∂φ k α ∂τ [1−3b] ∂r ∂z
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Spherical coordinates: 1 ∂2 1 ∂ ⎛ ∂T ⎞ 1 ∂ 2T q 1 ∂T (rT ) + 2 + = ⎜ sin θ ⎟+ 2 2 2 2 ∂θ ⎠ r sin θ ∂φ r ∂r r sin θ ∂θ ⎝ k α ∂τ
[1−3c]
Steady‐state one‐dimensional heat flow (no heat generation): d 2T =0 2 dx
[1‐4]
Steady‐state one‐dimensional heat flow in cylindrical coordinates (no heat generation): d 2T 1 dT + =0 2 [1‐5] dr r dr
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Chapter 1. Introduction
Steady‐state one‐dimensional heat flow with heat sources: d 2T q + =0 2 [1‐6] dx k
Two‐dimensional steady‐state conduction without heat sources: ∂ 2T ∂ 2T + 2 =0 2 ∂x ∂y
[1‐7]
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Chapter 1. Introduction
1‐2. Thermal Conductivity
Figure 1‐4
Thermal conductivities of some typical gases . Holman / Heat Transfer, 10th Edition
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Chapter 1. Introduction
Figure 1‐5
Thermal conductivities of some typical liquids.
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Chapter 1. Introduction
Figure 1‐6
Thermal conductivities of some typical solids. Holman / Heat Transfer, 10th Edition
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1‐3. Convection Heat Transfer To express the overall effect of convection, we use Newton’s law of cooling: q = hA (Tw − T∞ ) [1‐8]
Figure 1‐7
Convection heat transfer from a plate.
Holman / Heat Transfer, 10th Edition
Chapter 1. Introduction
Convection Energy Balance on a Flow Channel
Figure 1-8 Convection in a channel. q = m(ie − ii ) q = mc p (Te − Ti ) q = mc p (Te − Ti ) = hA(Tw, avg − Tfluid , avg )
[1−8a]
m = ρ umean Ac
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Chapter 1. Introduction
1-4. Radiation Heat Transfer qemitted = σ AT 4 qnetexchange A
[1‐9]
∝ σ (T14 − T24 )
[1‐10]
q = Fε FGσ A (T14 − T24 ) [1‐11]
Radiation in an Enclosure q = ε1σ A1 (T14 − T24 )
[1‐12]
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Chapter 1. Introduction
1‐5. Dimensions and Units L = length M = mass F = force τ = time T = temperature
For example, Newton’s second law of motion may be written Force ∼ time rate of change of momentum d (mv) F =k dτ F = kma F=
[1‐13]
1 ma gc [1‐14]
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Chapter 1. Introduction
Some typical systems of units are . 1. 1‐pound force will accelerate a 1‐lb mass 32.17 ft/s 2 . 2. 1‐pound force will accelerate a 1‐slug mass 1 ft/s 2 . 3 1‐dyne force will accelerate a 1‐g mass 1 cm/s 2 . 4. 1‐newton force will accelerate a 1‐kg mass 1 m/s 2 . 5. 1‐kilogram force will accelerate a 1‐kg mass 9.806 m/s 2 . 1.
g c = 32.17lb m ⋅ ft/lb f ⋅ s 2
2.
g c = 1slug ⋅ ft/lb f ⋅ s 2
2 3. g c = 1g ⋅ cm/dyn ⋅ s
4.
g c = 1kg ⋅ m/N ⋅ s 2
5.
g c = 9.806kg m ⋅ m/kg f ⋅ s 2
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Chapter 1. Introduction
1. 2. 3. 4. 5.
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lb f ⋅ ft lb f ⋅ ft dyn ⋅ cm = 1erg N ⋅ m = 1joule(J) kg f ⋅ m = 9.806J
In addition, we may use the units of energy that are based on thermal phenomena: 1 Btu will raise 1 lb m of water 1 F at 68 F . 1 cal will raise 1 g of water 1 C at 20 C . 1 kcal will raise 1 kg of water 1 C at 20 C .
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Some conversion factors for the various units of work and energy are 1Btu = 778.16lb f ⋅ ft 1Btu = 1055J 1kcal = 4182J 1lb f ⋅ ft = 1.356J 1Btu = 252cal W=
g m gc
[1‐15]
kinW/m ⋅ C hinW/m 2 ⋅ C 1N = 1kg ⋅ m/s 2
[1‐16]
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Example 1‐1 Conduction Through Copper Plate One face of a copper plate 3 cm thick is maintained at 400 C , and the other face is maintained at 100 C . How much heat is transferred through the plate? Solution From Appendix A, the thermal conductivity for copper is 370 W/m ⋅ C at 250 C . From Fourier’s law q dT = −k A dx
Integrating gives q ΔT −(370)(100 − 400) = −k = = 3.7MW/m 2 [1.173 ×106 Btu/h ⋅ ft 2 ] −2 Δx A 3 ×10
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Example 1‐2 Convection Calculation Air at 20 C blows over a hot plate 50 by 75 cm maintained at 2 250 C . The convection heat‐transfer coefficient is 25 W/m ⋅ C . Calculate the heat transfer. Solution From Newton’s law of cooling q = hA (Tw − T∞ ) = (25)(0.50)(0.75)(250 − 20) = 2.156kW[7356Btu/h]
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Example 1‐3 Multimode Heat Transfer Assuming that the plate in Example 1‐2 is made of carbon steel (1%) 2 cm thick and that 300 W is lost from the plate surface by radiation, calculate the inside plate temperature. Solution The heat conducted through the plate must be equal to the sum of convection and radiation heat losses: qcond = qconv + qrad −kA
ΔT =2.156+0.3=2.456kW Δx (−2456)(0.02) ΔT = = −3.05 C[−5.49 F] (0.5)(0.75)(43)
where the value of k is taken from Table 1‐1. The inside plate temperature is therefore Ti = 250 + 3.05 = 253.05 C Holman / Heat Transfer, 10th Edition
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Example 1‐4 Heat Source and Convection An electric current is passed through a wire 1 mm in diameter and 10 cm long. The wire is submerged in liquid water at atmospheric pressure, and the current is increased until the water boils. For this situation h = 5000W/m ⋅ C , and the water temperature will be 100 C . How much electric power must be supplied to the wire to maintain the wire surface at 114 C ? Solution 2
The total convection loss is given by Equation (1‐8): q = hA (Tw − T∞ ) For this problem the surface area of the wire is A = π dL = π (1× 10−3 )(10 ×10−2 ) = 3.142 ×10−4 m 2
The heat transfer is therefore q = (5000W/m 2 ⋅ C)(3.142 × 10−4 m 2 )(114 − 100) = 21.99W[75.03Btu/h] and this
is equal to the electric power that must be applied. Holman / Heat Transfer, 10th Edition
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Example 1‐5 Radiation Heat Transfer Two infinite black plates at 800 C and 300 C exchange heat by radiation. Calculate the heat transfer per unit area. Solution Equation (1‐10) may be employed for this problem, so we find immediately q/ A = σ (T14 − T24 ) = (5.669 ×10−8 )(10734 − 5734 ) = 69.03kW/m 2 [21, 884Btu/h ⋅ ft 2 ]
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Example 1‐6 Total Heat Loss by Convection and Radiation A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50 C in a large room where the air and wall temperature are at 20 C . The surface emissivity of the steel may be taken as 0.8. Using the data of Table 1‐3, calculate the total heat lost by the pipe per unit length. Solution The total heat loss is the sum of convection and radiation. From Table 1‐3 we see that an estimate for the heat‐transfer coefficient 2 for free convection with this geometry and air is h = 6.5W/m ⋅ C . The surface area is π dL , so the convection loss per unit length is
q/L]conv = h(π d )(Tw − T∞ ) = (6.5)(π )(0.05)(50 − 20) = 30.63W/m
The pipe is a body surrounded by a large enclosure so the radiation heat transfer can be calculated from Equation (1‐12). Holman / Heat Transfer, 10th Edition
Chapter 1. Introduction
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With T1 = 50 C = 323 K and T2 = 20 C = 293 K , we have q/L]rad = ε1 (π d1 )σ (T14 − T24 ) = (0.8)(π )(0.05)(5.669 × 10−8 )(3234 − 2934 ) = 25.04W/m
The total heat loss is therefore q/L]tot = q/L]conv + q/L]rad =30.63+25.04=55.67W/m
In this example we see that the convection and radiation are about the same. To neglect either would be a serious mistake.
Holman / Heat Transfer, 10th Edition
Chapter 1. Introduction
1‐6. Summary dT ⎤ −kA ⎥ = hA (Tw − T∞ ) + Fε FGσ A (Tw4 − Ts4 ) dy ⎦ wall Ts = temperature of surroundings Tw = surface temperature T∞ = fluid temperature
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Chapter 1. Introduction
Figure 1‐9 Combination of conduction, convection, and radiation heat transfer.
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