Page 482 - Design and Operation of Heat Exchangers and their Networks
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Appendix  465


                   s = (1 - Q_cal / Q) ∗ (t_h_in - t_h_out);
                   % outlet temperature deviation of hot water, K
                   t_h_out = t_h_out + s;
                   if (abs(s) < 1E-6)
                        break;
                   end
              end
              m_h = Q / cp_h / (t_h_in - t_h_out);
              fprintf('m_h = %fkg/s\n', m_h);


              Example 2.6 Sizing a counterflow heat exchanger
              (MatLab code)

              % Example 2.6 Sizing a counterflow heat exchanger
              % Consider a counterflow shell-and-tube heat exchanger with one shell pass
              % and one tube pass. Hot water enters the tube at 100°C and leaves at 80°C.
              % In the shellside, cold water is heated from 20°Cto 70°C. The heat duty
              % is expected to be 350 kW. There are totally 53 tubes with the inner
              % diameter of 16 mm and wall thickness of 1 mm. The thermal conductivity of
              % the tube wall is 40 W/mK. The shellside heat transfer coefficient can be
              % established as 1500 W/m2K. Calculate the tube length of the heat
              % exchanger with the consideration of variable tubeside heat transfer
              % coefficient depending on the fluid temperature.
              clear;
              t_h_in = 100; % inlet temperature of hot water (tubeside), °C
              t_h_out = 80; % outlet temperature of hot water (tubeside), °C
              t_c_in = 20; % inlet temperature of cold water (shellside), °C
              t_c_out = 70; % outlet temperature of cold water (shellside), °C
              d_i = 0.016; % tube inner diameter, m
              delta_w = 0.001; % tube wall thickness, m
              d_o = d_i + 2 ∗ delta_w; % tube outside diameter, m
              N_tube = 53; % number of tubes
              Q = 350; % heat duty, kW
              alpha_c = 1500; % shellside heat transfer coefficient, W/m2K
              lambda_w = 40; % thermal conductivity of the tube wall, 40 W/mK
              t_h_m = (t_h_in + t_h_out) / 2;
              [cp_h, lambda_h, mu_h] = water_properties(t_h_m);
              % isobaric heat capacity, J/kg; thermal conductivity, W/mK; K; viscosity,
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