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PROBLEM 2.1 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, \ufffd \ufffd ,E Ein out\u2212 = 0 it follows that \ufffd \ufffdE E qin out x\u2212 = and that q q xx x\u2260 \ufffd \ufffd. That is, the heat rate within the object is everywhere constant. From Fourier\u2019s law, q kA dT dxx x = \u2212 , and since qx and k are both constants, it follows that A dT dx Constant.x = That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independent of 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, \u2032\u2032qx , vary with distance? PROBLEM 2.2 KNOWN: Hot water pipe covered 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\u2019s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form q kA dT dr k 2 r dT drr r = \u2212 = \u2212 \u3c0 \ufffd\ufffd \ufffd where A r and r = 2\u3c0 \ufffd \ufffd is the axial length of the pipe-insulation system. Recognize that for steady- state conditions with no internal heat generation, an energy balance on the system requires \ufffd \ufffd \ufffd \ufffd .E E since E Ein out g st= = = 0 Hence qr = Constant. That is, qr is independent of radius (r). Since the thermal conductivity is also constant, it follows that r dT dr Constant.\ufffd \ufffd\ufffd \ufffd \ufffd\ufffd = 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. COMMENTS: (1) Note that, while qr is a constant and independent of r, \u2032\u2032qr is not a constant. How does \u2032\u2032q rr \ufffd \ufffd vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with increasing radius. PROBLEM 2.3 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 in radial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties. ANALYSIS: Fourier\u2019s law, Eq. 2.1, for this one-dimensional, radial (spherical coordinate) system has the form ( )2r r dT dTq k A k 4 rdr dr\u3c0= \u2212 = \u2212 where Ar is the surface area of a sphere. For steady-state conditions, an energy balance on the system yields \ufffd \ufffd ,E Ein out= since \ufffd \ufffd .E Eg st= = 0 Hence, ( )in out r rq q q q r .= = \u2260 That is, qr is a constant, independent of the radial coordinate. Since the thermal conductivity is constant, it follows that 2 dTr Constant. dr \uf8ee \uf8f9 =\uf8ef \uf8fa\uf8f0 \uf8fb This relation requires that the product of the radial temperature gradient, dT/dr, and the radius squared, r2, remains constant throughout the shell. Hence, the temperature distribution appears as shown in the sketch. COMMENTS: Note that, for the above conditions, ( )r rq q r ;\u2260 that is, qr is everywhere constant. How does \u2032\u2032qr vary as a function of radius? PROBLEM 2.4 KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution and heat rate. FIND: Expression for the thermal conductivity, k. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3) No internal heat generation. ANALYSIS: Applying the energy balance, Eq. 1.11a, to the system, it follows that, since \ufffd \ufffd ,E Ein out= ( )xq Constant f x .= \u2260 Using Fourier\u2019s law, Eq. 2.1, with appropriate expressions for Ax and T, yields ( ) ( ) x x 2 3 dTq k A dx d K6000W=-k 1-x m 300 1 2x-x . dx m = \u2212 \uf8ee \uf8f9 \u22c5 \u22c5 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb Solving for k and recognizing its units are W/m\u22c5K, ( ) ( ) ( )( )22 -6000 20k= . 1 x 2 3x1-x 300 2 3x =\uf8ee \uf8f9 \u2212 + \u2212 \u2212\uf8ef \uf8fa\uf8f0 \uf8fb < COMMENTS: (1) At x = 0, k = 10W/m\u22c5K and k \u2192 \u221e as x \u2192 1. (2) Recognize that the 1-D assumption is an approximation which becomes more inappropriate as the area change with x, and hence two-dimensional effects, become more pronounced. PROBLEM 2.5 KNOWN: End-face temperatures and temperature dependence of k for a truncated cone. FIND: Variation with axial distance along the cone of q q k, and dT / dx.x x, ,\u2032\u2032 SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in x (negligible temperature gradients along y), (2) Steady-state conditions, (3) Adiabatic sides, (4) No internal heat generation. ANALYSIS: For the prescribed conditions, it follows from conservation of energy, Eq. 1.11a, that for a differential control volume, \ufffd \ufffd .E E or q qin out x x+dx= = Hence qx is independent of x. Since A(x) increases with increasing x, it follows that \u2032\u2032 =q q A xx x / \ufffd \ufffd decreases with increasing x. Since T decreases with increasing x, k increases with increasing x. Hence, from Fourier\u2019s law, Eq. 2.2, \u2032\u2032 = \u2212q k dT dxx , it follows that | dT/dx | decreases with increasing x. PROBLEM 2.6 KNOWN: Temperature dependence of the thermal conductivity, k(T), for heat transfer through a plane wall. FIND: Effect of k(T) on temperature distribution, T(x). ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3) No internal heat generation. ANALYSIS: From Fourier\u2019s law and the form of k(T), \u2032\u2032 = \u2212 = \u2212 +q k dT dx k aT dT dxx o \ufffd \ufffd . (1) The shape of the temperature distribution may be inferred from knowledge of d2T/dx2 = d(dT/dx)/dx. Since \u2032\u2032qx is independent of x for the prescribed conditions, dq dx - d dx k aT dT dx k aT d T dx a dT dx x o o 2 2 \u2032\u2032 = + \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd = \u2212 + \u2212 \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd = \ufffd \ufffd \ufffd \ufffd 0 0 2 . Hence, d T dx -a k aT dT dx where k aT = k > 0 dT dx 2 2 o o = + \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd + \ufffd \ufffd\ufffd \ufffd \ufffd\ufffd > \ufffd \ufffd \ufffd 2 2 0 from which it follows that for a > 0: d T / dx < 02 2 a = 0: d T / dx 02 2 = a < 0: d T / dx > 0.2 2 COMMENTS: The shape of the distribution could also be inferred from Eq. (1). Since T decreases with increasing x, a > 0: k decreases with increasing x = > | dT/dx | increases with increasing x a = 0: k = ko = > dT/dx is constant a < 0: k increases with increasing x = > | dT/dx | decreases with increasing x. PROBLEM 2.7 KNOWN: Thermal conductivity and thickness of a one-dimensional system with no internal heat generation and steady-state conditions. FIND: Unknown surface temperatures, temperature gradient or heat flux. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional heat flow, (2) No internal heat generation, (3) Steady-state conditions, (4) Constant properties. ANALYSIS: The rate equation and temperature gradient for this system are 1 2 x dT dT T T q k and . dx dx L \u2212 \u2032\u2032 = \u2212 = (1,2) Using Eqs. (1) and (2), the unknown quantities can be determined. (a) ( )400 300 KdT 200 K/m dx 0.5m \u2212 = = 2 x W K q 25 200 5000 W/m . m K m \u2032\u2032 = \u2212 × = \u2212 \u22c5 < (b) 2x W K q 25 250 6250 W/m m K m \u2032\u2032 = \u2212 × \u2212 = \u22c5 \uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb 2 1 dT K T T L 1000 C-0.5m -250 dx m = \u2212 = \uf8ee \uf8f9 \uf8ee