KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape. FIND: Sketch temperature distribution and explain shape of curve. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No internal heat generation.
ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, E in − E out = 0, itfollows that E in − E out = q x
and that q x ≠ q x x . That is, the heat rate within the object is everywhere constant. From Fourier’s law,
q x = − kA x
dT , dx
and since qx and k are both constants, it follows that
dT = Constant. dx
That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independentof distance x. It follows that since Ax increases with x, then dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above. COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2) What would the distribution be when T2 > T1? (3) How does the heat flux, q ′′ , vary with distance? x
KNOWN: Hot water pipecovered with thick layer of insulation. FIND: Sketch temperature distribution and give brief explanation to justify shape. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional(cylindrical) radial system has the form
q r = − kA r
dT dT = − k 2πr dr dr
where A r = 2πr and is the axial length of the pipe-insulation system. Recognize that for steadystate conditions with no internal heat generation, an energy balance on the system requires E in = E out since E g = E st = 0. Hence
qr = Constant.
That is, qr is independent of radius (r). Since the thermalconductivity is also constant, it follows that
dT "# = Constant. ! dr $
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r, remains constant throughout the insulation. For our situation, the temperature distribution must appear as shown in the sketch. does q ′′ r vary with r? (2) Recognize that the radial temperature gradient, dT/dr,decreases with r increasing radius. COMMENTS: (1) Note that, while qr is a constant and independent of r, q ′′ is not a constant. How r
KNOWN: A spherical shell with prescribed geometry and surface temperatures. FIND: Sketch temperature distribution and explain shape of the curve. SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction inradial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system has the form
qr = −k Ar
dT dT = − k 4π r 2 dr dr
yields E in = E out , since E g = E st = 0. Hence, qin = q out = q r ≠ q r ( r ) .
where Ar is the surface area of a sphere. Forsteady-state conditions, an energy balance on the system
That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity is constant, it follows that
dT r 2 = Constant. dr
This relation requires that the product of the radial temperature gradient, dT/dr, and the radius 2 squared, r , remains constant throughout the shell. Hence, the temperaturedistribution appears as shown in the sketch. COMMENTS: Note that, for the above conditions, q r ≠ q r ( r ) ; that is, qr is everywhere constant. How does q ′′ vary as a function of radius? r
KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution and heat rate. FIND: Expression for the thermal conductivity, k. SCHEMATIC:
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