Page 478 - Design and Operation of Heat Exchangers and their Networks

Appendix 461

% density, kg/m3; viscosity, sPa; thermal conductivity, W/mK; isobaric heat
% capacity, J/kgK
Pr = mu ∗ cp / lambda; % Prandtl number
Re = G ∗ dh / mu; % Reynolds number
Nu_H1 = 8.235 ∗ (((((-0.1861 ∗ gamma + 1.0578) ∗ gamma - 2.4765) ...
∗ gamma + 3.0853) ∗ gamma - 2.0421) ∗ gamma + 1);
% Nusselt number for constant wall heat flux
alpha = Nu_H1 ∗ lambda / dh; % heat transfer coefficient, W/m2K
m_1 = sqrt(2 ∗ alpha / lambda_f / delta_f );
l_f_1 = h_fs;
m_2 = sqrt(alpha / lambda_f / delta);
l_f_2 = s_fs / 2;
eta_f = (m_1 ∗ sinh(m_1 ∗ l_f_1) ∗ cosh(m_2 ∗ l_f_2) ...
+ m_2 ∗ cosh(m_1 ∗ l_f_1) ∗ sinh(m_2 ∗ l_f_2)) / m_1 / (l_f_1 + l_f_2) ...
/ (m_1 ∗ cosh(m_1 ∗ l_f_1) ∗ cosh(m_2 ∗ l_f_2) ...
+ m_2 ∗ sinh(m_1 ∗ l_f_1) ∗ sinh(m_2 ∗ l_f_2)); % fin efficiency
beta = (2 ∗ h_fs + s_fs) / (2 ∗ (h_fs + s_fs));
% ratio of secondary heat transfer area to primary heat transfer area
eta_0 =1-(1- eta_f ) ∗ beta; % overall fin efficiency
k=1/(1 / alpha / eta_0 + delta / lambda_f );
% overall heat transfer coefficient, W/m2K
A=2 ∗ N ∗ (h_fs + s_fs) ∗ L; % total heat transfer area, m2
delta_t =Q/k /A;% temperature difference, K
t_max = t_out + delta_t;
fprintf('t_max = %f°C\n', t_max);

Example 2.4 Sizing a counterflow heat exchanger
(MatLab code)

% Example 2.4 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.