Question

ED - EXPERIMENTAL DESIGN RATE LIMITING STEP IN HEAT TRANSFER SYSTEM WITH SERIES RESISTANCES A water stream at a mass flow rate of 0.3 kg/s and temperature of 70°C flows through an aluminum horizontal pipe with an inner diameter of 0.025 m, wall thickness of 3 mm and length of 2 m. The ambient air is at 25°C and the outlet temperature of water is 65 °C. Assume steady-state conditions are reached and temperature of air is constant. Given data: Re 4 IND Repr 1/3 ; Re < 2300 L/D Re>10 000 Nup-1.86 Nu-0.023Re+/SP03 hair = 12 W/m2K ka - 237 W/mK kwater = 0.66 W/mK Cp water 4190 J/kg K Hwater 0.4125.10-kg/ms a) Determine the rate limiting step for this system by comparing the thermal resistances. (10 pts) b) Calculate the change in the total resistance when one of the resistances is halved while the others are held constant? Repeat your calculation for each resistance. (10 pts)

Answer 1

Solution Miller 0.4125*10 Cop de 1.90 J/kg mec Kx=237 W/mK = 0.031m m=0.3.14 di = 0.025 m * The = 65°C dec Tmi = 70°C woman L=2m # Re 4m Mao Water=o. = 0.66W тK T-25°C ho 12WhlK # Re = 4*0.3 kg/sec = 37040 > 10000 0.4125_kg X* 0.025m m sec # Nu= hidi 0.8 0.3 = 0.023 Re På Kwater Pr= Map 0.9425*10* yule * 41.907 y K – 2:62875 What 0.66 0.8 = 0.66 0.3 * hi * 0.025m 0.023 * 37040 WAK (37040 )" # hi = 3661.82 Womek Affron (2:61845

R Rz 1 bi (do/di) 1 h; Td; L hox del 2XKL # Ry= 1 3661W_** *0-02594 2017 # P = ln Coszempo ozsm) = 72225*105 K 2X* 237W *2m 1.739 * 10-3 K W mx พ 1 Rez 12W**** 0-63t yf*24 0.428 K W 씁 vate limiting # Because Rz is more as compare to Rs f k so the step is R3 because it will be slowest step. Restal = Ry+ Rx+ R3 = 1.739*10-3k_+7.2228*105* # 1 Peotal = 0.45 h W +0.428K w Alfox 0 Case -P> RE = 1:739* 10-3 ☆ 2 Riotal = R + R+ R = 2 # + 1-739*10-3 +7.2288*10-5 + 0.428) K W Restal = 0.429 Athos W

R = 2 2 2 2 # Change in total resistance = 6.43–0.429) = 0.001 Km RS = = "7.2228420 Care - I Affron Kw 2 Rotal = R +R+R=(1-739*10* + 7:2228*10-5 + 0.428) W 2 #Rotal = 0.4298 KW Alfons total resistance = (0.43–0.4298) K= 0.0002K # # Change in (0.45-0.4298) W AB Book Case-II R = Rs 0.428 K 2 W = 0.214 KW 2 Ripotal = R + R + B = R+R+ g = 2-785*10** +72228x10-5+0214) KW Pilotal = 0.2164 Alborn Change in total resistance (6.43–0216)* -0.214 KW #

In case of any doubt or you need any further help regarding this solution please comment below.

Thank you...........