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Example 2.18

8 十月, 2015 - 17:29
    % 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