1 Heat Transfer Lecturer : Dr. Rafel Hekmat Hameed University of Babylon Subject : Heat Transfer College of Engineering Year : Third B.Sc. Mechanical Engineering Dep. Drive the general conduction equation in Cartesian coordinate system Determining the basic equation that governs the transfer of heat in a solid, using Equation (1-1) as a starting point. Consider the one-dimensional system shown in Figure 1-2. If the system is in a steady state, i.e., if the temperature does not change with time, then the problem is a simple one, and we need only integrate Equation (1-1) and substitute the appropriate values to solve for the desired quantity. However, if the temperature of the solid is changing with time, or if there are heat sources or sinks within the solid, the situation is more complex. We consider the general case where the temperature may be changing with time and heat sources may be present within the body. For the element of thickness dx, the following energy balance may be made: Energy conducted in left face+ heat generated within element = change in internal energy + energy conducted out right face
Transcript
1
Heat Transfer
Lecturer : Dr. Rafel Hekmat Hameed University of Babylon
Subject : Heat Transfer College of Engineering
Year : Third B.Sc. Mechanical Engineering Dep.
Drive the general conduction equation in Cartesian coordinate system
Determining the basic equation that governs the transfer of heat in a solid, using Equation (1-1) as
a starting point. Consider the one-dimensional system shown in Figure 1-2. If the system is in a
steady state, i.e., if the temperature does not change with time, then the problem is a simple one,
and we need only integrate Equation (1-1) and substitute the appropriate values to solve for the
desired quantity. However, if the temperature of the solid is changing with time, or if there are heat
sources or sinks within the solid, the situation is more complex. We consider the general case where
the temperature may be changing with time and heat sources may be present within the body. For
the element of thickness dx, the following energy balance may be made:
Energy conducted in left face+ heat generated within element = change in internal energy +
energy conducted out right face
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These energy quantities are given as follows:
where
˙q = energy generated per unit volume, W/m3
c= specific heat of material, J/kg · ◦C
ρ= density, kg/m3
Combining the relations above gives
Or
This is the one-dimensional heat-conduction equation. To treat more than one-dimensional heat
flow, we need consider only the heat conducted in and out of a unit volume in all three coordinate
directions, as shown in Figure 1-3. The energy balance yields
and the energy quantities are given by
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so that the general three-dimensional heat-conduction equation is
For constant thermal conductivity, Equation (1-3) is written
Steady-state one-dimensional heat flow (no heat generation):
Steady-state one-dimensional heat flow with heat sources:
Fig.(1-3) Elemental volume for three-
dimensional heat-conduction analysis
in Cartesian coordinates systems
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Two-dimensional steady-state conduction without heat sources:
Heat conduction equation in cylindrical coordinate systems
Consider an element volume having the coordinates (r, ,z) for three-dimensional heat conduction
analysis, as shown in figure above
Heat generated within the element qg= qg (dr rd dz). dt
Energy stored in the element = c (dr rd dz) ∂T
∂t. dt
The volume of the element=rd dz dr
Heat flow in radial direction (z-) plane:
Heat in qr= -k (rd dz)∂T
∂rdt …..(1) ; heat out q(r+dr)= qr+
∂
∂r qr dr …..(2)
Heat accumulation in the element due to heat flow in radial direction
dqr= qr - qr+dr
= - 𝜕
𝜕𝑟 qr dr ; = -
𝜕
𝜕𝑟 (−𝑘(𝑟 𝑑 𝑑𝑧)
𝜕𝑇
𝜕𝑟 𝑑𝑡) dr ; = 𝑘(𝑑𝑟 𝑑 𝑑𝑧)
𝜕
𝜕𝑟 (𝑟
𝜕𝑇
𝜕𝑟)
Elemental volume for three-dimensional heat-conduction analysis: (cylindrical coordinates)
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=k(dr d dz) (r𝜕2𝑇
𝜕𝑟2+
𝜕𝑇
𝜕𝑟)𝜕𝑡 ; = k(dr rd dz) (
𝜕2𝑇
𝜕𝑟2 +1
𝑟
𝝏𝑻
𝝏𝒓)𝜕𝑡
By applying the same analysis in the other two planes
.
Then the general heat conduction equation in cylindrical coordinates
Heat conduction equation in spherical coordinate systems
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Solved examples
See solved examples in Holman book
where T represents the difference in surface temperatures.
Since T is the same for both walls, it follows that
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Under steady-state conditions with Est = 0, it follows that
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H.W
1) The wall of an industrial furnace is constructed from 0.2 m thick fireclay brick having a
thermal conductivity of 2.0 W/m.K. Measurements made during steady state operation reveal
temperatures of 1500 and 1250 K at the inner and outer surfaces, respectively. What is the
rate of heat loss through a wall which is 0.5 m by 4 m on a side ?
2) An uninsulated steam pipe passes through a room in which the air and walls are at 25°C.
The outside diameter of pipe is 80 mm, and its surface temperature and emissivity are 180°C
and 0.85, respectively. If the free convection coefficient from the surface to the air is 6 W/m2K,
what is the rate of heat loss from the surface per unit length of pipe ?
3) The inner and outer surfaces of a 25-cm-thick wall in summer are at 27°C and 44°C,
respectively. The outer surface of the wall exchanges heat by radiation with surrounding
surfaces at 40°C, and convection with ambient air also at 40°C with a convection heat transfer
coefficient of 8 W/m2·°C. Solar radiation is incident on the surface at a rate of 150 W/ m2. If
both the emissivity and the solar absorptivity of the outer surface are 0.8, determine the
