http://www.k-wave.org/forum/topic/volume-rate-of-heat-deposition Support for the k-Wave MATLAB toolbox en-US Tue, 12 May 2026 23:32:09 +0000 http://bbpress.org/?v=1.0.2 q http://www.k-wave.org/forum/search.php http://www.k-wave.org/forum/topic/volume-rate-of-heat-deposition#post-7071 Tue, 08 Oct 2019 20:25:36 +0000 vich 7071@http://www.k-wave.org/forum/ <p>The <code>source.T0</code> you provided is just an example. Any initial temperature field will do, including a constant one.</p> <p>Since you are simulating an active heat source, you are most likely interested in <code>source.Q</code>, the actual volume rate of heat deposition (in W/m^3) that you mentioned in the title of this thread. Running <code>help kWaveDiffusion</code> should give you more details. </p> http://www.k-wave.org/forum/topic/volume-rate-of-heat-deposition#post-7067 Sat, 05 Oct 2019 09:33:38 +0000 Cheng-You Lee 7067@http://www.k-wave.org/forum/ <p>I want to simulate 2D temperature distribution in water, the heating source is circle and always turn on (time-step is 3600 and dt is 1). When I set Gaussian initial temperature distribution, I confused about the equation in the example-Heat Diffusion In A Homogeneous Medium below,</p> <p>% set Gaussian initial temperature distribution [degC]<br /> width = 4 *dx;<br /> source.T0 = 37 + 5.*exp( -(kgrid.x./width).^2 - (kgrid.y./width).^2);</p> <p>My understanding is width can change the size of heating source in the figure, 37 is initial temperature, but how does '5' come from? I think it's related to Volume rate of heat deposition, and I want to know how to calculate when simulating in water. Is there any principle? Below is my whole programming, </p> <p>% create the computational grid<br /> PML_size = 5; % size of the PML in grid points<br /> Nx = 235 - 2*PML_size; % number of grid points in the x (row) direction<br /> Ny = 579 - 2*PML_size; % number of grid points in the y (column) direction<br /> dx = 1e-3; % grid point spacing in the x direction [m]<br /> dy = 1e-3; % grid point spacing in the y direction [m]<br /> kgrid = kWaveGrid(Nx, dx, Ny, dy);</p> <p>% define medium properties<br /> medium.density = 1000; % [kg/m^3]<br /> medium.thermal_conductivity = 0.6; % [W/(m.K)]<br /> medium.specific_heat = 4200; % [J/(kg.K)]</p> <p>% set Gaussian initial temperature distribution [degC]<br /> width = 25*dx;<br /> source.T0 = 25 + 5.*exp( -(kgrid.x./width).^2 - (kgrid.y./width).^2);</p> <p>% set input args<br /> input_args = {'PlotScale','auto'};</p> <p>% create kWaveDiffusion object<br /> kdiff = kWaveDiffusion(kgrid, medium, source, [], input_args{:});</p> <p>% % take time steps (temperature can be accessed as kdiff.T)<br /> Nt = 60*60;<br /> dt = 1;<br /> kdiff.takeTimeStep(Nt, dt);</p> <p>% plot the current temperature field<br /> figure;<br /> kdiff.plotTemp;</p> <p>% =========================================================================<br /> % VISUALISATION<br /> % =========================================================================</p> <p>figure;<br /> imagesc(kgrid.y_vec * 1e3, kgrid.x_vec * 1e3, kdiff.T);<br /> h = colorbar;<br /> xlabel(h, '[^\circC]');<br /> ylabel('x-position [mm]');<br /> xlabel('y-position [mm]');<br /> axis image;<br /> title('Final Temperature Distribution');<br /> colormap(jet);</p> <p>Thanks! </p>