Please answer the above question- Heat is to be transferred through a.docx

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Please answer the above question? Heat is to be transferred through a plane wall of thickness H = 1.25 cm, which is composed of a material with thermal conductivity k = 200 W/mK. On the left of the wall is a fluid with T = 120 degree C and hi =450 W/m2K. On the right of the wall is a fluid with T,2 = 20 degree C and h2 =25 W/m2K. It is desired lo enhance heat transfer between two fluids bv using straight, rectangular fins of length L = 2.5 cm, thickness t = 0.16 cm. and 80 fins per meter. Assuming 1-D representation, find heat transfer rate between the two fluids, per unit area of the wall if There are no fins The fins are added to the right only The fins are added on both sides Note - Draw thermal circuit for each case, showing the value of all resistances. Solution a) When there are no fins. Total thermal resistance R = 1 / (h_inf1 *A) + H/(kA) + 1 / (h_inf2 *A) Taking a unit area (A = 1 m^2), we get R = 1 / (450*1) + (1.25*10^-2) / (200*1) + 1 / (25*1) R = 0.04228 K / W Heat transfer rate Q = (T_inf1 - T_inf2) / R Q = (120 - 20) / 0.04228 Q = 2364.9 W b) For rectangular Fin, perimeter P = 2(w + t) ...where w = width, t = thickness For rectangular fin, area A = w*t Thus, P / A = 2*(w+t) / (wt) = 2*(1 + 0.0016) / (1*0.0016) = 1252 L = 2.5 cm = 0.025 m t = 0.16 cm = 0.0016 m For fins, mL = sqrt (hP*L^2 / (kA)) = sqrt (25*1252*0.025^2 / 200) = 0.3127 For rectangular fin, R_fin = 1 / (sqrt (kAhP) Tanh (mL)) R_fin = 1/ sqrt (200*(1*0.0016)*25*2*(1 + 0.0016)) Tanh 0.3127 R_fin = 0.8254 K / W Covective Resistance of right side wall = 1 / (25*1) = 0.04 K/W All the 80 fins and the right side wall are in thermally connected in parallel. Their equivalent thermal resistance is R = 1 / [1 / 0.04 + (1/0.8254 + 1/ 0.8254 + .............80 times)] R = 0.0082 K/W This R and left side wall and slab are connected in series. Equivalent R = 1 / (450*1) + (1.25*10^-2) / (200*1) + 0.0082 R = 0.01048 K/W Q = (T_inf1 - T_inf2) / R Q = (120 - 20) / 0.01048 Q = 9537.7 W c) Convective resistance of left side wall = 1 / (450*1) = 0.0022 K/W Equivalent resistance for the 80 fins on left and the right side convective boundary is R = 1 / [1 / 0.0022 + (1/0.8254 + 1/ 0.8254 + .............80 times)] R = 0.001828 K / W Net resistance R = 0.001828 + (1.25*10^-2) / (200*1) + 0.0082 R = 0.01 K/W Q = (T_inf1 - T_inf2) / R Q = (120 - 20) / 0.01 Q = 9909.9 W .

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b)
For rectangular Fin, perimeter P = 2(w + t) ...where w = width, t = thickness
For rectangular fin, area A = w*t
Thus, P / A = 2*(w+t) / (wt) = 2*(1 + 0.0016) / (1*0.0016) = 1252
L = 2.5 cm = 0.025 m
t = 0.16 cm = 0.0016 m
For fins, mL = sqrt (hP*L^2 / (kA)) = sqrt (25*1252*0.025^2 / 200) = 0.3127
For rectangular fin, R_fin = 1 / (sqrt (kAhP) Tanh (mL))
R_fin = 1/ sqrt (200*(1*0.0016)*25*2*(1 + 0.0016)) Tanh 0.3127
R_fin = 0.8254 K / W
Covective Resistance of right side wall = 1 / (25*1) = 0.04 K/W
All the 80 fins and the right side wall are in thermally connected in parallel. Their equivalent
thermal resistance is
R = 1 / [1 / 0.04 + (1/0.8254 + 1/ 0.8254 + .............80 times)]
R = 0.0082 K/W

This R and left side wall and slab are connected in series.
Equivalent R = 1 / (450*1) + (1.25*10^-2) / (200*1) + 0.0082
R = 0.01048 K/W
Q = (T_inf1 - T_inf2) / R
Q = (120 - 20) / 0.01048
Q = 9537.7 W
c)
Convective resistance of left side wall = 1 / (450*1) = 0.0022 K/W
Equivalent resistance for the 80 fins on left and the right side convective boundary is
R = 1 / [1 / 0.0022 + (1/0.8254 + 1/ 0.8254 + .............80 times)]
R = 0.001828 K / W
Net resistance R = 0.001828 + (1.25*10^-2) / (200*1) + 0.0082
R = 0.01 K/W
Q = (T_inf1 - T_inf2) / R
Q = (120 - 20) / 0.01
Q = 9909.9 W

Please answer the above question- Heat is to be transferred through a.docx

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