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cen58933_ch04

# 425 the solution in such multidimensional geometries

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Unformatted text preview: onal geometry. Consider a short cylinder of height a and radius ro initially at a uniform tem0, the perature Ti. There is no heat generation in the cylinder. At time t cylinder is subjected to convection from all surfaces to a medium at temperature T with a heat transfer coefficient h. The temperature within the cylinder will change with x as well as r and time t since heat transfer will occur from the top and bottom of the cylinder as well as its side surfaces. That is, T T(r, x, t) and thus this is a two-dimensional transient heat conduction problem. When the properties are assumed to be constant, it can be shown that the solution of this two-dimensional problem can be expressed as T(r, x, t) T Ti T short cylinder T(x, t) T Ti T Plane wall FIGURE 4–27 A long solid bar of rectangular profile a b is the intersection of two plane walls of thicknesses a and b. t) cyl(r, a T(r, t) T Ti T infinite cylinder (4-25) That is, the solution for the two-dimensional short cylinder of height a and radius ro is equal to the product of the nondimensionalized solutions for the one-dimensional plane wall of thickness a and the long cylinder of radius ro, which are the two geometries whose intersection is the short cylinder, as shown in Fig. 4–26. We generalize this as follows: the solution for a multidimensional geometry is the product of the solutions of the one-dimensional geometries whose intersection is the multidimensional body. For convenience, the one-dimensional solutions are denoted by wall(x, b plane wall t) semi-inf(x, t) T(x, t) T Ti T T(r, t) T Ti T T(x, t) T Ti T plane wall infinite cylinder (4-26) semi-infinite solid For example, the solution for a long solid bar whose cross section is an a b rectangle is the intersection of the two infinite plane walls of thicknesses a and b, as shown in Fig. 4–27, and thus the transient temperature distribution for this rectangular bar can be expressed as T(x, y, t) T Ti T rectangular bar wall(x, t) wall(y, t) (4-27) The proper forms of the product solutions for some other geometries are given in Table 4–4. It is important to note that the x-coordinate is measured from the surface in a semi-infinite solid, and from the midplane in a plane wall. The radial distance r is always measured from the centerline. Note that the solution of a two-dimensional problem involves the product of two one-dimensional solutions, whereas the solution of a three-dimensional problem involves the product of three one-dimensional solutions. A modified form of the product solution can also be used to determine the total transient heat transfer to or from a multidimensional geometry by using the one-dimensional values, as shown by L. S. Langston in 1982. The cen58933_ch04.qxd 9/10/2002 9:13 AM Page 233 233 CHAPTER 4 TABLE 4–4 Multidimensional solutions expressed as products of one-dimensional solutions for bodies that are initially at a uniform temperature Ti and exposed to convection from all surfaces to a medium at T x 0 ro r r x θ (r, t) = θcyl(r, t) Infinite cylinder r θ (x,r, t) = θcyl (r, t) θwall (x,t) Short cylinder θ (x,r, t) = θcyl (r, t) θsemi-inf (x, t) Semi-infinite cylinder y x x y z θ (x, t) = θsemi-inf (x, t) Semi-infinite medium θ (x,y,t) = θsemi-inf (x, t) θsemi-inf (y, t) Quarter-infinite medium x θ (x, y, z, t) = θsemi-inf (x, t) θsemi-inf (y, t) θsemi-inf (z,t) Corner region of a large medium 2L 2L y 0 x Lx y z θ (x, t) = θwall(x, t) Infinite plate (or plane wall) θ (x, y, t) = θwall (x, t) θsemi-inf (y, t) Semi-infinite plate x θ (x,y,z, t) = θwall (x, t) θsemi-inf ( y, t) θsemi-inf (z , t) Quarter-infinite plate y x z y x z y x θ (x, y, t) = θwall (x, t) θwall ( y, t) Infinite rectangular bar θ (x,y,z, t) = θwall (x, t) θwall (y, t) θsemi-inf (z , t) Semi-infinite rectangular bar θ (x,y,z, t) = θwall (x, t) θwall (y, t) θwall (z , t) Rectangular parallelepiped cen58933_ch04.qxd 9/10/2002 9:13 AM Page 234 234 HEAT TRANSFER transient heat transfer for a two-dimensional geometry formed by the intersection of two one-dimensional geometries 1 and 2 is Q Qmax Q Qmax total, 2D Q Qmax 1 12 Q Qmax (4-28) 1 Transient heat transfer for a three-dimensional body formed by the intersection of three one-dimensional bodies 1, 2, and 3 is given by Q Qmax total, 3D Q Qmax Q Qmax 1 Q Qmax 3 12 Q 1Qmax Q Qmax 1- 1 1 Q Qmax (4-29) 2 The use of the product solution in transient two- and three-dimensional heat conduction problems is illustrated in the following examples. EXAMPLE 4–7 Cooling of a Short Brass Cylinder A short brass cylinder of diameter D 10 cm and height H 12 cm is initially at a uniform temperature Ti 120°C. The cylinder is now placed in atmospheric air at 25°C, where heat transfer takes place by convection, with a heat transfer coefficient of h 60 W/m2 · °C. Calculate the temperature at (a) the center of the cylinder and (b) the center of the top surface of the cylinder 15 min after the start of the cooling. SOLUTION A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface are to be determined. Assumptions 1...
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