Question

All steps please with explanations

heat exc hange A double smooth pi 345-10* Pa-s, N 4.47) at -20 C to cool water enters at $s C (G, -4191 Jg , -0664 Wmk 389-10 Pa-s, N 245). The water flows through a copper tube of UD 19 m thickness (0.05.). I.-2.5 m, k-280 Wim K) while the refrigerant flows through uses refrigerant (C-907 J/kg K, k-0.07 winKy , annulu crn)as shown in Figure bellow. sing the Effectiveness-NTU method. Determine a) The rate of heat transter, q by the exchanger b) The outlet temperatures for both the fluids c) If the exchanger is operated with parallel flow, would the outlet temperatures of the fluids increase, decrease or stay unchanged? tnd 0.5 kg 15 cm -98 m-0.05 kg's

Answer 1

Water T^n_in, w = 98 c m^n_w = 0.05 kg/s C^n_P_w = 4191 tau/kg k k^n_w = 0.664 w/mk M^n_w = 389 times 10^- 6 Pa - s N^n_Pnw = 2.45 Refrigent T^c_in,R = -20 c m^c_R = 0.5 kg/s c^c_PR = 907 tau/kg k k^c_R = 0.07 w/mk M^c_R = 3.45 times 10^- 4 Pa - s N^c_PnR = 4.47 copper tube k_c = 280 w/mk for the net fluid (water) C^n = m^n_w C^n_p_w = (0.05 times 4191) tau/ks = 209.55 w/k For the cold fluid (refrigent) C^c = m^c_R C^c_P_R = (0.5 times 907) w/k = 453.5 w/k so, C_min = C^n = 209.55 w/k U = overall heat transfer co - efficient A = surface area (effective) in which geat in transferred 1/UA = 1/h_i A_i + lr (D_0/D_i)/2 r k_c l + 1/ /h_0 A_0 I^R_nmr = 1/h_0 A_0 IR_w = ln (D_0/D_i)/2r kl h_i, h_0 are heat - transfer co - efficient for inner (water) and