Fundamentals of Heat and Mass Transfer - CH04

Fundamentals of Heat and Mass Transfer - CH04

(Parte 1 de 10)

PROBLEM 4.1 KNOWN: Method of separation of variables (Section 4.2) for two-dimensional, steady-state conduction.

FIND: Show that negative or zero values of l2 , the separation constant, result in solutions which cannot satisfy the boundary conditions.

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties.

ANALYSIS: From Section 4.2, identification of the separation constant l2 leads to the two ordinary differential equations, 4.6 and 4.7, having the forms 2

X0Y0
l+=-=(1,2)

d X dY dx dy

Consider now the situation when l2 = 0. From Eqs. (1), (2), and (3), find that

()()()12341234XCCx,YCCy and x,yCCx CCy.q=+=+=++(4)

Evaluate the constants - C1, C2, C3 and C4 - by substitution of the boundary conditions:

x0:0,yCC0CCy0 C0
y0:x,00CXCC00 C0
xL:L,00CL0Cy0 C0
yW:x,W00x0CW1

q q

==+×+=01„

The last boundary condition leads to an impossibility (0 „ 1). We therefore conclude that a l2 value of zero will not result in a form of the temperature distribution which will satisfy the boundary conditions. Consider now the situation when l2 < 0. The solutions to Eqs. (1) and (2) will be

5678XCeCe,YCcos yCsin yllll=+=+ (5,6)

and ( ) [ ]- x +x 5678x,yCeCe Ccos yCsin y.llqllØø=++Œœºß

y0:x,0CeCe Ccos 0Csin 0 C0
x0:0,yCeCe 0Csin y0 C0

llqql

()-xL+xL58xL:L,yCee Csin y0.qlØø==-=Œœºß

From the last boundary condition, we require C5 or C8 is zero; either case leads to a trivial solution with either no x or y dependence.

PROBLEM 4.2

KNOWN: Two-dimensional rectangular plate subjected to prescribed uniform temperature boundary conditions.

FIND: Temperature at the mid-point using the exact solution considering the first five non-zero terms; assess error resulting from using only first three terms. Plot the temperature distributions T(x,0.5) and T(1,y).

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: From Section 4.2, the temperature distribution is

1 1 sinh n y LTT 2 n x, y sin

T T n L sinh n W L θ ππθ

T n 2 sinh n 2 θ ππθ

When n is even (2, 4, 6), the corresponding term is zero; hence we need only consider n = 1, 3, 5, 7

If only the first three terms of the series, Eq. (2), are considered, the result will be θ(1,0.5) = 0.46; that is, there is less than a 0.2% effect.

Using Eq. (1), and writing out the first five terms of the series, expressions for θ(x,0.5) or

T(x,0.5) and θ(1,y) or T(1,y) were keyboarded into the IHT workspace and evaluated for sweeps over the x or y variable. Note that for

T(1,1) is greater than 150°C. Upon examination of the magnitudes of terms, it becomes evident that more than 5 terms are required to provide an accurate solution.

T(x,0.5) or T(1,y), C

T(1,y) T(x,0.5)

PROBLEM 4.3 KNOWN: Temperature distribution in the two-dimensional rectangular plate of Problem 4.2.

FIND: Expression for the heat rate per unit thickness from the lower surface (0 ≤ x ≤ 2, 0) and result based on first five non-zero terms of the infinite series.

ASSUMPTIONS: (1) Two-dimensional, steady-state conduction, (2) Constant properties. ANALYSIS: The heat rate per unit thickness from the plate along the lower surface is

() () x2 x2 x2

out y 2 1 y0x0 x0 x 0 y0

Tqd q x,0 k dx k T T dx where from the solution to Problem 4.2,

1 1 sinh n y LTT 2 n xsin T T n L sinh n W L ππθ

Evaluate the gradient of θ from Eq. (2) and substitute into Eq. (1) to obtain n1x2 out 2 1 n1 y0x0

1 n L cosh n y L2n xqk T T sin dx n L sinh n W L

out 2 1 x0n1

1121 n xqk T T cos ns inh n W L Lπ ππ

1121q k T T 1 cos n ns inh n L π ππ

To evaluate the first five, non-zero terms, recognize that since cos(nπ) = 1 for n = 2, 4, 6, only the n-

odd terms will be non-zero. Hence,

Continued …..

PROBLEM 4.3 (Cont.)

COMMENTS: If the foregoing procedure were used to evaluate the heat rate into the upper surface,

()x2 in yx0

112q k T T coth n 2 1 cos n ππ π

However, with coth(nπ/2) ≥ 1, irrespective of the value of n, and with ()n1n1 divergent series, the complete series does not converge and inq′→∞. This physically untenable condition results from the temperature discontinuities imposed at the upper left and right corners.

PROBLEM 4.4

KNOWN: Rectangular plate subjected to prescribed boundary conditions. FIND: Steady-state temperature distribution. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, 2-D conduction, (2) Constant properties. ANALYSIS: The solution follows the method of Section 4.2. The product solution is

and the boundary conditions are: T(0,y) = 0, T(a,y) = 0, T(x,0) = 0, T(x.b) = Ax. Applying

BC#1, T(0,y) = 0, find C1 = 0. Applying BC#2, T(a,y) = 0, find that l = np/a with n = 1,2,…. Applying BC#3, T(x,0) = 0, find that C3 = -C4. Hence, the product solution is

Tx,yXxYyCC sinx e.a llpØø=× =- Œœºß

(Parte 1 de 10)

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