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Assembling the global stiffness matrix using Eqs. (13.9.16) and (13.9.17) and the global force matrix using Eq. (13.9.18), we obtain the global equations as

\begin{array}{l} \left[ \begin{array}{c c c c c} - 0. 5 4 2 & 0. 5 7 9 & 0 & 0 & 0 \\ - 0. 5 5 4 & 0. 5 9 1 - 0. 5 4 2 & 0. 5 7 9 & 0 & 0 \\ 0 & - 0. 5 5 4 & 0. 5 9 1 - 0. 5 4 2 & 0. 5 7 9 & 0 \\ 0 & 0 & - 0. 5 5 4 & 0. 5 9 1 - 0. 5 4 2 & 0. 5 7 9 \\ 0 & 0 & 0 & - 0. 5 5 4 & 0. 5 9 1 \end{array} \right] \left\{ \begin{array}{c} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} \\ = \left\{ \begin{array}{c} F _ {1} + 7. 3 5 \\ 1 4. 7 \\ 1 4. 7 \\ 1 4. 7 \\ 7. 3 5 \end{array} \right\} \tag {13.9.19} \\ \end{array}

Applying the boundary condition t _ { 1 } = 1 0 0 ^ { \circ } \mathrm { F } , we rewrite Eq. (13.9.19) as

\left[ \begin{array}{c c c c c} 1 & 0 & 0 & 0 & 0 \\ 0 & 0. 0 4 9 & 0. 5 7 9 & 0 & 0 \\ 0 & - 0. 5 5 4 & 0. 0 4 9 & 0. 5 7 9 & 0 \\ 0 & 0 & - 0. 5 5 4 & 0. 0 4 9 & 0. 5 7 9 \\ 0 & 0 & 0 & - 0. 5 5 4 & 0. 5 9 1 \end{array} \right] \left\{ \begin{array}{l} t _ {1} \\ t _ {2} \\ t _ {3} \\ t _ {4} \\ t _ {5} \end{array} \right\} = \left\{ \begin{array}{c} 1 0 0 \\ 1 4. 7 + 5 5. 4 \\ 1 4. 7 \\ 1 4. 7 \\ 7. 3 5 \end{array} \right\} \tag {13.9.20}

Solving the second through fifth equations of Eq. (13.9.20) for the unknown temperatures, we obtain

t _ {2} = 1 0 6. 1 ^ {\circ} \mathrm{F} \quad t _ {3} = 1 1 2. 1 ^ {\circ} \mathrm{F} \quad t _ {4} = 1 1 7. 6 ^ {\circ} \mathrm{F} \quad t _ {5} = 1 2 2. 6 ^ {\circ} \mathrm{F} \tag {13.9.21}

Using Eq. (13.8.2), we obtain the heat flow into and out of the tube as

q _ {\text {in}} = \dot {m c t} _ {1} = (4. 7 2) (0. 2 4) (1 0 0) = 1 1 3. 2 8 \mathrm{Btu/h} \tag {13.9.22}

q _ {\text { out }} = \dot {m} c t _ {5} = (4. 7 2) (0. 2 4) (1 2 2. 6) = 1 3 8. 9 \mathrm{Btu/h}

where, again, the conduction contribution to q is negligible; that is, kADT is negligible. The analytical solution in Reference [7] yields

t _ {5} = 1 2 3. 0 ^ {\circ} \mathrm{F} \quad q _ {\text { out }} = 1 3 9. 3 3 \mathrm{Btu/h} \tag {13.9.23}

The finite element solution is then seen to compare quite favorably with the analytical solution.

The element with the stiffness matrix given by Eq. (13.9.13) has been used in Reference [8] to analyze heat exchangers. Both double-pipe and shell-and-tube heat exchangers were modeled to predict the length of tube needed to perform the task of proper heat exchange between two counterflowing fluids. Excellent agreement was found between the finite element solution and the analytical solutions described in Reference [9].

Finally, remember that when the variational formulation of a problem is difficult to obtain but the differential equation describing the problem is available, a residual method such as Galerkin’s method can be used to solve the problem.

13.10 Flowchart and Examples of a Heat-Transfer Program

Figure 13–30 is a flowchart of the finite element process used for the analysis of twodimensional heat-transfer problems.

Figures 13–31 and 13–32 show examples of two-dimensional temperature distribution using the two-dimensional heat transfer element of this chapter (results obtained from Algor [10]). We assume that there is no heat transfer in the direction perpendicular to the plane.

flowchart

graph TD
    A["START"] --> B["Draw the geometry and apply any heat sources, fluxes, and boundary temperatures"]
    B --> C["Define the element type and properties (here the heat-transfer element is used)"]
    C --> D["DO JE = 1,NE"]
    D --> E["Compute the element stiffness matrix k and nodal load matrix f in global coordinates (both conduction and/or convection portions of k and f)"]
    E --> F["Use the direct stiffness procedure to add k and f to the proper locations in the assemblage stiffness matrix K and load matrix F"]
    F --> G["Account for known temperature boundary conditions and modify the global stiffness matrix and force matrix accordingly"]
    G --> H["Solve Kt = F for t"]
    H --> I["Compute the element temperature gradients and heat fluxes"]
    I --> J["Output results"]
    J --> K["END"]

Figure 13–30 Flowchart of two-dimensional heat-transfer process

text_image

500°F 100°F 1 ft 100°F 1 ft 100°F

(a)

heatmap

X Range	Y Range	Temperature (deg F)
0-10	0-10	500
0-10	10-20	460
0-10	20-30	420
0-10	30-40	380
0-10	40-50	340
0-10	50-60	300
0-10	60-70	260
0-10	70-80	220
0-10	80-90	180
0-10	90-100	140
10-20	0-10	500
10-20	10-20	460
10-20	20-30	420
10-20	30-40	380
10-20	40-50	340
10-20	50-60	300
10-20	60-70	260
10-20	70-80	220
10-20	80-90	180
10-20	90-100	140
20-30	0-10	500
20-30	10-20	460
20-30	20-30	420
20-30	30-40	380
20-30	40-50	340
20-30	50-60	300
20-30	60-70	260
20-30	70-80	220
20-30	80-90	180
20-30	90-100	140
30-40	0-10	500
30-40	10-20	460
30-40	20-30	420
30-40	30-40	380
30-40	40-50	340
30-40	50-60	300
30-40	60-70	260
30-40	70-80	220
30-40	80-90	180
30-40	90-100	140
40-50	0-10	500
40-50	10-20	460
40-50	20-30	420
40-50	30-40	380
40-50	40-50	340
40-50	50-60	300
40-50	60-70	260
40-50	70-80	220
40-50	80-90	180
40-50	90-100	140
50-60	0-10	500
50-60	10-20	460
50-60	20-30	420
50-60	30-40	380
50-60	40-50	340
50-60	50-60	300
50-60	60-70	260
50-60	70-80	220
50-60	80-90	180
50-60	90-100	140
60-70	0-10	500
60-70	10-20	460
60-70	20-30	420
60-70	30-40	380
60-70	40-50	340
60-70	50-60	300
60-70	60-70	260
60-70	70-80	220
60-70	80-90	180
60-70	90-100	140
70-80	0-10	500
70-80	10-20	460
70-80	20-30	420
70-80	30-40	380
70-80	40-50	340
70-80	50-60	300
70-80	60-70	260
70-80	70-80	220
70-80	80-90	180
70-80	90-100	140
80-90	0-10	500
80-90	10-20	460
80-90	20-30	420
80-90	30-40	380
80-90	40-50	340
80-90	50-60	300
80-90	60-70	260
80-90	70-80	220
80-90	80-90	180
80-90	90-100	140
90-100	0-10	500
90-100	10-20	460
90-100	20-30	420
90-100	30-40	380
90-100	40-50	340
90-100	50-60	300
90-100	60-70	260
90-100	70-80	220
90-100	80-90	180
90-100	90-100	140

(b)
Figure 13–31 (a) Square plate subjected to temperature distribution and (b) finite element model with resulting temperature variation throughout the plate ((b) Courtesy of David Walgrave)

text_image

T∞ = 110° F h = 5 Btu/(h-ft²-°F) 570° F 4 ft Insulation (K = 0.020 Btu/(h-ft-°F)) 2 ft

(a)

heatmap

Temperature (deg f)
570
524.1503
478.3308
432.4809
366.8012
340.7814
294.9017
246.082
203.2023
187.3528
111.5029

Figure 13–32 (a) Square duct wrapped by insulation and (b) the finite element model with resulting temperature variation through the insulation

Figure 13–31(a) shows a square plate subjected to boundary temperatures. Figure 13–31(b) shows the finite element model, along with the temperature distribution throughout the plate.

Figure 13–32(a) shows a square duct that carries hot gases such that its surface temperature is 570 F. The duct is wrapped by a layer of circular fiberglass. The finite element model, along with the temperature distribution throughout the fiberglass is shown in Figure 13–32(b).

References

[1] Holman, J. P., Heat Transfer, 9th ed., McGraw-Hill, New York, 2002.
[2] Kreith, F., and Black, W. Z., Basic Heat Transfer, Harper & Row, New York, 1980.
[3] Lyness, J. F., Owen, D. R. J., and Zienkiewicz, O. C., ‘‘The Finite Element Analysis of Engineering Systems Governed by a Non-Linear Quasi-Harmonic Equation,’’ Computers and Structures, Vol. 5, pp. 65–79, 1975.
[4] Zienkiewicz, O. C., and Cheung, Y. K., ‘‘Finite Elements in the Solution of Field Problems,’’ The Engineer, pp. 507–510, Sept. 24, 1965.
[5] Wilson, E. L., and Nickell, R. E., ‘‘Application of the Finite Element Method to Heat Conduction Analysis,’’ Nuclear Engineering and Design, Vol. 4, pp. 276–286, 1966.
[6] Emery, A. F., and Carson, W. W., ‘‘An Evaluation of the Use of the Finite Element Method in the Computation of Temperature,’’ Journal of Heat Transfer, American Society of Mechanical Engineers, pp. 136–145, May 1971.
[7] Rohsenow, W. M., and Choi, H. Y., Heat, Mass, and Momentum Transfer, Prentice-Hall, Englewood Cliffs, NJ, 1963.
[8] Goncalves, L., Finite Element Analysis of Heat Exchangers, M. S. Thesis, Rose-Hulman Institute of Technology, Terre Haute, IN, 1984.
[9] Kern, D. Q., and Kraus, A. D., Extended and Surface Heat Transfer, McGraw-Hill, New York, 1972.
[10] Heat Transfer Reference Division, Docutech On-line Documentation, Algor, Inc., Pittsburgh, PA.
[11] Beasley, K. G., ‘‘Finite Element Analysis Model Development of Leadless Chip Carrier and Printed Wiring Board,’’ M. S., Thesis, Rose-Hulman Institute of Technology, Terre Haute, IN, Nov. 1992.

Problems

13.1 For the one-dimensional composite bar shown in Figure P13–1, determine the interface temperatures. For element 1, let K _ { x x } = 2 0 0 \mathrm { \ W / ( m \cdot ^ { \circ } C ) } ; for element 2, let K _ { x x } = 1 0 0 \ \mathrm { { W / ( m \cdot ^ { \circ } C ) } } ; and for element 3, let K _ { x x } = 5 0 ~ \mathrm { W / ( m \cdot ^ { \circ } C ) } . Let A = 0 . 1 \ \mathrm { m } ^ { 2 } . The left end has a constant temperature of 1 0 0 ^ { \circ } \mathrm { C } and the right end has a constant temperature of 3 0 0 ^ { \circ } \mathrm { C } .

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100°C 1 ① 2 ② 3 ③ 300°C 4 2 m 1 m 0.5 m

Figure P13–1

13.2 For the one-dimensional rod shown in Figure P13–2 (insulated except at the ends), determine the temperatures at L / 3 , 2 L / 3 , and L. Let K _ { x x } = 3 \ \mathrm { B t u } / ( \mathrm { h . { \dot { \mathrm { { ‰} } } } } ) , h = 1 . 0 \operatorname { B t u } / ( \mathrm { h } \mathrm { - i n } ^ { 2 } \mathrm { - } ^ { \circ } \mathrm { F } ) , and T _ { \infty } = 0 ^ { \circ } \mathrm { F } . The temperature at the left end is 2 0 0 ^ { \circ } \mathrm { F } .

text_image

L = 9 in. 200°F

2-in. radius

Figure P13–2

13.3 A rod with uniform cross-sectional area of 2 ~ \mathrm { i n } ^ { 2 } and thermal conductivity of 3 Btu/ ( \mathrm { h - i n . ^ { \circ } F } ) has heat flow in the x direction only (Figure P13–3). The right end is insulated. The left end is maintained at 5 0 ^ { \circ } \mathrm { F } _ { : } , and the system has the linearly distributed heat flux shown.

Use a two-element model and estimate the temperature at the node points and the heat flow at the left boundary.

text_image

q*(x) = 0.1x Btu/h-in. T(0) = 50°F Area = 2 in² k = 3 Btu/h-in.-°F 60 in.

Figure P13–3

text_image

q*(0) = 0 \quad q*(2) = 3 \quad q*(3) = 6 1 \quad ① \quad 2 \quad ② \quad 3 ← 30 in. →

13.4 The rod of 1-in. radius shown in Figure P13–4 generates heat internally at the rate of uniform Q = 1 0 , 0 0 0 { \mathrm { B t u } } / ( { \mathrm { h } } { - } { \mathrm { f } } { \mathrm { t } } ^ { 3 } ) throughout the rod. The left edge and perimeter of the rod are insulated, and the right edge is exposed to an environment of T _ { \infty } = 1 0 0 ^ { \circ } \mathrm { F } . The convection heat-transfer coefficient between the wall and the environment is h = 1 0 0 ~ \mathrm { B t u } / ( \mathrm { h } \mathrm { - f t } ^ { 2 } \mathrm { - } ^ { \circ } \mathrm { F } ) . The thermal conductivity of the rod is K _ { x x } = 1 2 ~ \mathrm { B t u } / ( \mathrm { h } \mathrm { - f t } \cdot ^ { \circ } \mathrm { F } ) . The length of the rod is 3 in. Calculate the temperature distribution in the rod. Use at least three elements in your finite element model.

text_image

L = 3 in. Q T∞ = 100°F 1-in. radius

Figure P13–4

13.5 The fin shown in Figure P13–5 is insulated on the perimeter. The left end has a constant temperature of 1 0 0 ^ { \circ } \mathrm { C } . A positive heat flux of \mathbf { \bar { \eta } } q ^ { * } = 5 0 0 0 \mathbf { \Psi } \mathbf { W } / \mathbf { m } ^ { 2 } acts on the right end. Let K _ { x x } = 6 \mathrm { \ W / ( m \cdot ^ { \circ } C ) } and cross-sectional area A = 0 . 1 \mathrm { ~ m } ^ { 2 } . Determine the temperatures at L/4, L/2, 3L/4, and L, where L = 0 . 4 ~ \mathrm { m } .

text_image

T = 100°C q = 1000 W/m² L = 0.4 m A = 0.1 m²

Figure P13–5

13.6 For the composite wall shown in Figure P13–6, determine the interface temperatures. What is the heat flux through the 8-cm portion? Use the finite element method. Use three elements with the nodes shown. 1 cm ¼ 0:01 m.

text_image

K = 5 W/m · °C K = 15 W/m · °C t₁ = 500°C t₂ ① t₃ ② ③ t₄ = 100°C 10 cm 6 cm 8 cm K = 0.8 W/m · °C

Figure P13–6

13.7 For the composite wall idealized by the one-dimensional model shown in Figure P13–7, determine the interface temperatures. For element 1, let K _ { x x } = 5 \mathrm { \ W } / ( \mathrm { m \cdot ^ { \circ } C } ) ; for element 2, K _ { x x } = 1 0 \mathrm { \ W / ( m \cdot ^ { \circ } C ) } ; and for element 3, K _ { x x } = 1 5 \ \mathrm { { W / ( m \cdot ^ { \circ } C ) } } . The left end has a constant temperature of 2 0 0 ^ { \circ } \mathrm { C } and the right end has a constant temperature of 6 0 0 ^ { \circ } \mathrm { C } .

text_image

T = 200°C 1 2 3 T = 600°C 0.1 m 0.1 m 0.1 m A = 0.1 m²

Figure P13–7

13.8 A double-pane glass window shown in Figure P13–8, consists of two 4 mm thick layers of glass with k = 0 . 8 0 \ \mathrm { W / m \cdot ^ { \circ } C } separated by a 10 mm thick stagnant air space with k = 0 . 0 2 5 \mathrm { W } / \mathrm { m } { \cdot } \mathrm { ° } . . Determine (a) the temperature at both surfaces of the inside layer of glass and the temperature at the outside surfaces of glass, and (b) the steady rate of heat transfer in Watts through the double pane. Assume the inside room temperature T _ { i \infty } = 2 0 ^ { \circ } \mathrm { C } with h _ { i } = 1 0 ~ \mathrm { \bar { W } / m ^ { 2 } { \cdot } ^ { \circ } C } and the outside temperature T _ { 0 \infty } = 0 ^ { \circ } \mathrm { C } with h _ { 0 } = 3 0 \mathrm { W } / \mathrm { m } ^ { 2 } { \cdot } ^ { \circ } \mathrm { C } . Assume one-dimensional heat flow through the glass.

text_image

Glass Glass 20°C Air T1 T2 Inside Outside 0°C T3 T4 4 mm 10 mm 4 mm

Figure P13–8

text_image

Plaster wall Fiberglass insulation Plywood Inside Outside 2.5 cm 9 cm 1.25 cm

Figure P13–9

13.9 For the composite wall of a house, shown in Figure P13–9, determine the temperatures at the inner and outer surfaces and at the interfaces. The wall is composed of 2.5 cm thick plaster wall ( k = 0 . 2 0 \mathrm { W } / \mathrm { m } ^ { \circ } \mathrm { C } ) on the inside, a 9 cm thick layer of fiber glass insulation ( k = 0 . 0 3 8 \mathrm { W } / \mathrm { m } { \cdot } ^ { \circ } \mathrm { C } ) , and a 1.25 cm plywood layer ( k = 0 . 1 2 \mathrm { W } / \mathrm { m } \mathrm { - } ^ { \circ } \mathrm { C } ) on the outside. Assume the inside room air is 2 0 ^ { \circ } \mathrm { C } with convection coefficient of 1 0 \mathrm { W } / \mathrm { m } ^ { 2 } { \cdot } ^ { \circ } \mathrm { C } and the outside air at - 1 0 ^ { \circ } \mathrm { C } with convection coefficient of 2 0 \mathrm { W } / \mathrm { m } ^ { 2 } { \cdot } ^ { \circ } \mathrm { C } . . Also, determine the rate of heat transfer through the wall in Watts. Assume one-dimensional heat flow through the wall thickness.

13.10 Condensing steam is used to maintain a room at 2 0 ^ { \circ } \mathbf { C } . The steam flows through pipes that keep the pipe surface at 1 0 0 ^ { \circ } \mathrm { C } . To increase heat transfer from the pipes, stainless steel fins ( k = 1 5 \mathrm { W } / \mathrm { m } { \cdot } ^ { \circ } \mathrm { C } ) , 20 cm long and 0.5 cm in diameter, are welded to the pipe surface as shown in Figure P13–10. A fan forces the room air over the pipe and fins, resulting in a heat transfer coefficient of 5 0 \mathrm { W } / \mathrm { m } ^ { 2 } { \cdot } ^ { \circ } \mathrm { C } at the base surface of the fin where it is welded to the pipe. However, the air flow distribution increases the heat transfer coefficient to 8 0 \mathrm { W } / \mathrm { m } ^ { 2 } { \cdot } ^ { \circ } \mathrm { C } at the fin tip. Assume the variation in heat transfer coefficient to then vary linearly from left end to right end of the fin surface. Determine the temperature distribution at L/4 locations along the fin. Also determine the rate of heat loss from each fin.

text_image

100°C h = 50 W/m²-°C T∞ = 20°C Fin diameter = df h = 80 W/m²-°C Fin tip 20 cm Pipe wall

Figure P13–10

13.11 A tapered aluminum fin ( k = 2 0 0 \mathrm { W / m { \cdot } ^ { \circ } C ) } , shown in Figure P13–11, has a circular cross section with base diameter of 1 cm and tip diameter of 0.5 cm. The base is maintained at 2 0 0 ^ { \circ } \mathrm { C } and looses heat by convection to the surroundings at T _ { \infty } = 1 0 ^ { \circ } \mathrm { C } , h = 1 5 0 \mathrm { W } / \mathrm { m } ^ { 2 } \mathrm { - } ^ { \circ } \mathrm { C } . The tip of the fin is insulated. Assume one-dimensional heat flow and determine the temperatures at the quarter points along the fin. What is the rate of heat loss in Watts through each element? Use four elements with an average cross-sectional area for each element.

text_image

T₀ = 200°C 1 2 3 4 Insulated

Figure P13–11

13.12 A wall is constructed of an outer layer of 0.5 inch thick plywood ( k = 0 . 8 0 \ \mathrm { B t u / h – f t – ^ { \circ } F } ) , an inner core of 5 inch thick fiberglass insulation ( k = 0 . 0 2 0 \ \mathrm { B t u } / \mathrm { h } \mathrm { - f t } \mathrm { - } ^ { \circ } \mathrm { F } ) , and an inner layer of 0.5 inch thick sheetrock (k ¼ 0:10 Btu/h-ft-F) (Figure P13–12). The inside temperature is 6 5 ^ { \circ } \mathrm { F } with h ¼ 1:5 Btu/h-ft2- F, while the outside temperature is 0 ^ { \circ } \mathrm { F } with h = 4 ~ \mathrm { B t u / h \mathrm { - } f t ^ { 2 } \mathrm { - } ^ { \circ } F } . Determine the temperature at the interfaces of the materials and the rate of heat flow in Btu/h through the wall.

text_image

Sheetrock Fiberglass insulation Plywood Inside Outside 0.5 in. 5 in. 0.5 in.

Figure P13–12

13.13 A large plate of stainless steel with thickness of 5 cm and thermal conductivity of k = 1 5 \mathrm { W / m ^ { \circ } C } is subjected to an internal uniform heat generation throughout the plate at constant rate of Q = 1 0 \times 1 0 ^ { 6 } \mathrm { W } / \mathrm { m } ^ { 3 } . One side of the plate is maintained at 0 ^ { \circ } \mathbf { C } by ice water, and the other side is subjected to convection to an environment at T _ { \infty } = 3 5 ^ { \circ } \mathrm { C } . , with heat transfer coefficient h = 4 0 \mathrm { W / m } ^ { 2 . \circ } \mathrm { C } , as shown in Figure P13–13. Use three elements in a finite element model to estimate the temperatures at each surface and in the middle of the plate’s thickness. Assume a one-dimensional heat transfer through the plate.

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Stainless steel 0°C → Q = 10 × 10⁶ W/m³ h T∞ 0 0 1 2 L → x 5 cm

Figure P13–13

13.14 The base plate of an iron is 0.6 cm thick. The plate is subjected to 600 W of power (provided by resistance heaters inside the iron, as shown in Figure P13–14), over a base plate cross-sectional area of 1 5 0 \mathrm { c m } ^ { 2 } , resulting in a uniform flux generated on the inside surface. The thermal conductivity of the metal base plate is k = 2 0 \mathrm { W } / \mathrm { m } { \cdot } ^ { \circ } \mathrm { C } . The outside temperature of the plate is 8 0 ^ { \circ } \mathrm { C } at steady state conditions. Assume onedimensional heat transfer through the plate thickness. Using three elements, model the plate to determine the temperatures at the inner surface and interior one-third points.

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Insulation Resistance heater 600 W Base plate 0.6 cm 80°C Figure P13-14 x 0 1 2 3 150 cm²

13.15 A hot surface is cooled by attaching fins (called pin fins) to it, as shown in Figure P13–15. The surface of the plate (left end of the pin) is 9 0 ^ { \circ } \mathbf { C } . The fins are 4 cm long and 0.25 cm in diameter. The fins are made of copper ( k = 4 0 0 \mathrm { W / m ^ { \circ } C } ) . The temperature of the surrounding air is T _ { \infty } = 2 5 ^ { \circ } \mathrm { C } with heat transfer coefficient on the surface (including the end surface) of h = 3 0 \mathrm { W / m ^ { 2 } { \cdot } ^ { \circ } C } . A model of the typical fin is also shown in Figure P13–15. Use four elements in your finite element model to determine the temperatures along the fin length.