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% Problem 16.06 % Problem Statement % Calculate the percentage reduction in heat loss when a layer of hair felt % is wrapped around the outside surface (see problem 16.05) format short % Input Values t_water=80; % Water temperature [C] t_outside=15; % Atmospheric temperature [C] inner_dia=0.05; % Inner diameter [m] thickness=0.006; % [m] Lambda_steel=48; % Thermal conductivity of steel [W/mK] AlfaInside=2800; % Heat transfer coefficient of inside [W/m2K] AlfaOutside=17; % Heat transfer coefficient of outside [W/m2K] % Neglect radiation % Additional layer thickness_insulation=0.012; % [m] Lambda_insulation=0.03; % Thermal conductivity of insulation [W/mK] % Output Values % Q_dot=(t_water-t_outside)/TotalResistance % TotalResistance=(1/(AlfaInside*AreaInside))+(thickness/(Lambda_steel*AreaM))+ ... (1/(AlfaOutside*AreaOutside) % Calculating the unknown terms r_i=inner_dia/2 % Inner radius of pipe [m] r_o=r_i+thickness % Outer radius of pipe [m] r_i_insulation=r_o % Inner radius of insulation [m] r_o_insulation=r_i_insulation+thickness_insulation % Outer radius of pipe [m] AreaInside=2*pi*r_i AreaOutside=2*pi*r_o AreaOutside_insulated=2*pi*r_o_insulation AreaM_pipe=(2*pi*(r_o-r_i))/log(r_o/r_i) % Logarithmic mean area for pipe AreaM_insulation=(2*pi*(r_o_insulation-r_i_insulation)) ... /log(r_o_insulation/r_i_insulation) % Logarithmic mean area for insulation TotalResistance=(1/(AlfaInside*AreaInside))+(thickness/ ... (Lambda_steel*AreaM_pipe))+(1/(AlfaOutside*AreaOutside)) TotalResistance_insulated=(1/(AlfaInside*AreaInside))+(thickness/ ... (Lambda_steel*AreaM_pipe))+(thickness_insulation/(Lambda_insulation*AreaM_insulation)) +(1/(AlfaOutside*AreaOutside_insulated)) Q_dot=(t_water-t_outside)/(TotalResistance*1000) % converting into kW Q_dot_insulated=(t_water-t_outside)/(TotalResistance_insulated*1000) % converting into kW PercentageReducttion=((Q_dot-Q_dot_insulated)/Q_dot)*100
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