資源簡介
可以用來求解模擬一維水分運動方程過程,可以得到不同時刻,不同位置的土壤剖面的含水量變化
代碼片段和文件信息
function?richards
%?Solution?of?the?Richards?equation?????
%????using?MATLAB?pdepe???????????????????
%
%???$Ekkehard?Holzbecher??$Date:?2006/07/13?$
%
%?Soil?data?for?Guelph?Loam?(Hornberger?and?Wiberg?2005)
%?m-file?based?partially?on?a?first?version?by?Janek?Greskowiak?
%--------------------------------------------------------------------------
L?=?60;???????????????????%?length?[cm]
T?=?4;????????????????????%?maximum?time?[h]
qr?=?0.102;???????????????%?residual?water?content?殘余含水量
f?=?0.368;????????????????%?porosity??飽和含水量
s1=0.4;???????????????????%下滲率
s2=0;?????????????????????%底部壓力水頭
a?=?0.0335;???????????????%?van?Genuchten?parameter?[1/L]
n?=?2;????????????????????%?van?Genuchten?parameter
m?=?1-1/n;
ks?=0.00922;????????????????%?saturated?conductivity?[L/T]
N=100;
M=25;
x?=?linspace(0LN);
t?=?linspace(0TM);?
options=odeset(‘RelTol‘1e-4‘AbsTol‘1e-4‘NormControl‘‘off‘‘InitialStep‘1e-7)
u?=?pdepe(0@unsatpde@unsatic@unsatbcxtoptionss1s2qrfanks);
figure;???????????%表示壓力水頭(基質勢)分別在t=0h4h24h時刻隨土壤深度分布圖
size(u)
plot?(xu(1:)‘-‘);
xlabel(‘Depth?[L]‘);
ylabel(‘Pressure?Head?[L]‘);
legend(‘t=0h‘‘t=4h‘‘t=24h‘);
title(‘壓力水頭分別在t=0h4h24h時刻隨土壤深度分布圖‘);
figure;??????????????????????%表示壓力水頭分別在z=0cm16cm36cm60cm位置處隨時間變化分布圖
plot?(tu(:1)‘-‘);
xlabel(‘T?[h]‘);
ylabel(‘Pressure?Head?[L]‘);
legend(‘z=0cm‘‘z=16cm‘‘z=36cm‘‘z=60cm‘)
title(‘壓力水頭分別在z=0cm16cm36cm60cm位置處隨時間變化分布圖‘);
for?j=1:length(t)
????for?i=1:length(x)
?
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