Question

Heat is uniformly generated at the rate of 2x 10W/m* in a wall of thermal conductivity 25 W/m-K and thickness 60 mm. The wall is exposed to convection on both sides, with different heat transfer coefficients and temperatures as shown. There are straight rectangular fins on the right-hand side of the wall, with dimensions as shown (L =20 mm) and thermal conductivity of 250 W/m-K. What is the maximum temperature that will occur in the wall? L tt-2 mm k=25 W/m-K h 50 W/m2-K T= 30°C h-12 W/m2-K T 2=15°C 16-2 mm 2L 60 mm k,-250 W/m-K Determine the fin area (neglecting the tip area) per unit depth, in m2/m. m2/m Af=i Determine the value of surface efficiency. Determine the maximum temperature in the wall, in °C. Tmax °C Determine the temperature at the base of the fins, in °C. °C Ts2 = i Determine the value of fin efficiency. PIC CO i

Answer 1

Consider the diagram showing a wall with straight rectangular fins attached on the right side. T, T. - L-20 mm 25 W/m K q=2 x 10 Wim t 2 mm 2 12 Wim2K T, 15C I&=2 mm h50 WimK T 30°C 250 Wim K 2L 60 mm Find the heat transfer rates at the two surfaces for a wall section of area A by using the Fourier's law at x- -L dl =-k. 9.3 3 hа(т., -т,)--qL4 - ka Ta - Ta ...(1) 2L Consider the heat transfer rates at the two surfaces for a wall section of area A by using the equation Fourier's law at x L. dT q.2-kA ХР 2=h,4 (7-T)n =-gLA-k4- Find the surface temperature at the left side and right side of the wall by solving the equations (1) and (2). Surface temperature at the left side of the wall is. .(2) 2L hhT+ KhAT, ++hA aL. 2L 21 I T.. +hhi+ khÀ 2L ... (3) kh Surface temperature at the right side of the walll is, k (k 2L T= 2L (4) kh 21 2L An. Here, the term A is a variable assigned to the ratio Calculate the parameter, m. hP m = 2x12 V250x0.002 6.9 m Calculate the efficiency of rectangular fins, n. tanh(mL, mL tanh (6.9x0.02) 6.9x0.02 =0.994

NA, by using the following equation, A Calculate the term NA, N(2wL, (8+t)Nw A 2L 8+t NA, 2x0.02 0.002 0.002 A 10 A Calculte the term by using the following equation NA, A A, A A N(5w) NA = (84t)Nw A NA, = A 0.002 4-10 0.002 0-002 = 10.5 Calculate the ratio, A. i-4 NA, (1-7,) A A = 10.5-10(1-0.094) 10-4 Calculate the value of the term, h,A h,A = 12W/m2-Kx10.4 =125 W/m2 . K k Calculate the value of the term, 21 k 25 2L 0.06 = 417W/m2 . K From equation (3), calculate the temperature T +h,A hT, + * h,T, 21 21 L T = kh A kh hh,A+ 2L 2L (417+125)x 50 x 303+417 x125x 288 +(2x417+125) x2x 10' x 0.03 T = (417x 50+50x 125 + 417x 125) = 365-75 K =92.7°C

From equation (4), obtain the temperature, T,. k k ^ +h \qL 2L T.2 2L +hh,AkhA 2L kh 2L (417)x 50x303(417+ 50) 125 x 288 +(2x417+50) 2x 10 x 0.03 T2 417x 50+50x125 + 417x 125 = 358- 89 K =85.89°C Write the equation of temperature distribution in a wall T2-T x T+T T(x)= 1- 2k L 2 2 The location of the maximum temperature in the wall, x can be found by setting the gradient of the temperature in the distribution equation to zero. dT perature дх, Т., —Т, k 2L фх Т, -Т, 0=-- k 2L Т, -Т, x= k 2La Т, — Т, A for x n the equation (1) to obtain the Substitute k- 2Lq expression for the maximum temperature in the wall, T GLk(T,-T) TT, Т 2k 8Lq mas 21 2x10 x (0.03) Tma = 25(85.8-92.7) 92.7+85.8 8x(0.03) x2x10 2x25 2 =93.7°C Therefore, the maximum temperature that will occur in the wall is 93.7°C