TRADITIONAL PLATE EXCHANGER CALCULATION

	Number of plates 100 (101) [-] Plate Length 8.000 [m] Plate Width 0.500 [m] Plate Thickness 0.002 [m] Hot and Cold gap 0.008 [m] Hot water temperature 353.15 [K] Cold water temperature 293.15 [K] Hot and Cold fluid mass flow 400.0 [kg/s] Hot and Cold fouling resistance 0.00005 [m2W/K] Plate metal conductivity 50 [W/m/K] Water properties are taken at average temperatures. As the hot and cold inlet temperatures are 80 and 20 degrees Centigrade, respectively, the average temperature is 50 degree C. The film correction for the heat transfer coefficient is neglected for the hand calculation. It is small and of opposite influence at each process side. Heat exchanging Area A_hx = 8.000 * 0.500 * 100 = 400 [m2] Nr of hot and cold channels N_ch = 50 [-] A_flow/channel A_fch = 0.008 * 0.5 = 0.004 [m2] Channel Circumference C_fch = 2 * (0.008 + 0.5) = 1.016 [m] Hydraulic Diameter D_hyd = 4 * A_fch / C_fch = 0.015748 [m] Flow Area per fluid A_flow = N_ch * A_fch = 0.2 [m2] Fluid Mass Velocity G = M_flow / A_flow = 400.0 / 0.2 = 2000.0 [kg/m2/s] Water viscosity @ 50 deg.C u_w = 0.000525 [Pa.s] Water conductivity @ 50 deg C k_w = 0.6435 [W/m/K] Reynolds Re = G * D / u_w = 59993 [-] Prandtl water @ 50 deg C Pr = 3.555 [-] Hot & Cold  heat transfer coefficient U_w = 0.023 * k_w/D_hydr * Re^0.8 * Pr^0.4 =10372 [W/m2/K] Plate resistance / m2 R_pl = thickness/cond = 0.002 / 50 = 0.00004 [m2W/K] Total heat transfer resistance / m2 R_t = 2/U_w + 2 * R_foul + R_pl = R_t = 2/10372 + 2*0.00005 + 0.00004 = 0.0003328 [m2W/K] Overall Heat transfer Coefficient U_oa = 1 / R_t = 3004.6 [W/m2/K]
	Now, the overall heat transfer coefficient is calculated. We have the following equations:           Q_transferred = delta_T_mean * U_oa * A_hx (eq.1) Q_fluid = delta_T_fluid * M_flow * Cp_fluid (eq.2) Because the fluid and the fluid mass flow are identical on both sides, delta_T_mean equals the initial temperature difference (ITD=T_hot,in-T_cold,in) minus the delta_T_fluid, or: delta_T_mean = ITD - delta_T_fluid (eq.3) Inserting this into (eq.1), equating (eq.1) and (eq.2), we get: (ITD - delta_T_fluid) * U_oa * A_hx = delta_T_fluid * M_flow * Cp_fluid (eq.4)
	Solving for delta_T_fluid : delta_T_fluid = ITD * U_oa*A_hx / (U_oa*A_hx  +  M_flow*Cp_fluid) (eq.5) delta_T_fluid = 60.0 * 3004.6*400.0 / (3004.6*400.0 + 400.0*4035) = 25.61 [K] Q_fluid = M_flow * Cp * delta_T_fluid = 400.0 * 4035 * 25.61 = 41334540 [W], or 41.33 [MW] This results in: outlet temperature of 80 - 25.61 = 54.39 degree Centigrade (hot side) outlet temperature of 20 - 25.61 = 45.61 degree Centigrade (cold side) The results of AHTL are 55.41 degree C (hot side) 44.64 degree C (cold side) Heat transferred: 41.16 [MW]. The hand calculation is 0.4% in error. It is interesting to inspect the output of AHTL. The effect of varying fluid properties on heat transfer is significant. Overall effects on heat transfer cancel out to great extent. However, the effect on metal temperatures is much greater. The hot inlet side plate temperatures should by at around the average of hot flow inlet- and cold flow outlet temperatures, i.e. 0.5 * (80.0 + 44.64) = 62.32 degree C. Average plate temperature calculated by AHTL at this location is 63.197 degree C. This example is exceedingly simple, but in more complex cases, with phase transitions and for technically difficult problems it pays to have a solution that is as accurate as possible. source: www.heattransferconsult.nl read more on:  Double pipe heat exchanger