425 the solution in such multidimensional geometries

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online