Calculate total heat loss by conduction given temperature and depth profile vectors
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Hi all,
I need to solve this problem in Matlab. I have a hot body placed over cold ground, this having a thermal conductivity k = 1.
I need to estimate the total heat loss by conduction into the substratum at each timestep.
For each of these timesteps I have a temperature profile and the associated vertical profile at 1cm intervals (see .mat files attached).
How can I estimate the total heat loss (w/m2) at each time interval (i.e., the heat loss for every profile)?
Any help woud be grately appreciated!
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1707156/image.png)
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采纳的回答
Torsten
2024-5-31
移动:Torsten
2024-5-31
Here is another way to determine the temporal change of heat content in the substratum.
rho*cp*dT/dt = d/dx (k*dT/dx)
Integrating with respect to x gives
d/dt (integral_{x=-20}^{x=0} (rho*cp*T) dx ) = k*dT/dx @h=0 - k*dT/dx @h=-20
Above, we tried to approximate the right-hand side. Here, we try to approximate the left-hand side.
close all
clc
load("time_in_days.mat")
load("Depth_Profiles.mat")
load("Temperature_Profiles.mat")
num_intervals = size(Temperature_Profiles, 2);
depth_index = find(Depth_Profiles <= -20, 1, 'last');
rhocp = 1.0; % Specify rho*cp of the substratum [J/(m^3*K)]
for i = 1:num_intervals
temperature_profile = Temperature_Profiles(:, i);
energy_content(i) = trapz(Depth_Profiles(depth_index:end),temperature_profile(depth_index:end))*rhocp; %[J/m^2]
end
figure(1)
plot(time_in_days,energy_content) %[J/m^2]
change_of_energy_content = gradient(energy_content)./gradient(time_in_days*24*3600);%[W/m^2]
figure(2)
plot(time_in_days,change_of_energy_content*1e3) %[mW/m^2]
8 个评论
Torsten
2024-6-1
Here is a second-order approximation of the temperature gradients at the boundaries, but it doesn't seem to change that much:
dT_dh_surface = (1.5*temperature_profile(end) - 2*temperature_profile(end-1) + 0.5*temperature_profile(end-2)) / 0.01;
dT_dh_Depth = (-1.5*temperature_profile(2+depth_index) + 2*temperature_profile(1+depth_index) - 0.5*temperature_profile(1+depth_index - 1)) / 0.01;
For reference:
更多回答(1 个)
Torsten
2024-5-31
编辑:Torsten
2024-5-31
How can I estimate the total heat loss (w/m2) at each time interval (i.e., the heat loss for every profile)?
Total heat flow into the domain is (k*dT/dh @h=0) - (k*dT/dh @h=-20). Maybe you only want to evaluate this at one end.
To estimate the gradient, use a finite difference approximation.
2 个评论
Torsten
2024-5-31
编辑:Torsten
2024-5-31
Maybe you only want to evaluate this at one end. Sorry, can you provide more details on this? What does it mean?
It means that at h = 0, you have a heat flow into the domain and at h = -20, you have a heat flow out of the domain. Maybe you only want to compute the flow into the domain (k*dT/dh @h=0) or only the flow out of the domain (-k*dT/dh @h=-20) and not the net flow (k*dT/dh @h=0) - (k*dT/dh @h=-20).
To estimate the gradient, use a finite difference approximation: I came up with this code, but the output does not make any sense... :
You could use difference quotients of higher order that involve more h-points than only two to approximate the temperature gradient. But if it's correct that the distance between two adjacent h-points is 0.01 m, your code looks fine in principle. But you have to use
dT_dh_surface = (temperature_profile(end) - temperature_profile(end-1)) / 0.01;
instead of
dT_dh_surface = (temperature_profile(2) - temperature_profile(1)) / 0.01;
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