Investigation of various volume–balance methods in surface irrigation

Abstract

Prediction of the water advance phase and infiltration rate is of crucial importance for the design of surface irrigation. The volume–balance model is applied to specify the parameters of infiltration and advance rate. The present research aims to present a volume–balance equation for the Philip infiltration equation using the new two-point method and to compare the proposed method with seven other ones including Elliott-Walker, Ebrahimian-Shepard, Shepard, Valiantaz, Infilt, Mailapalli, and scaling method among others. To satisfy this end, the measured data for six furrows and six borders were considered. The methods described in border irrigation were more accurate than that of furrow irrigation. The results showed that the accuracy of the new two points, Elliott-Walker and Infilt methods, was higher than the other ones. The proposed new two-point’s method has been found as the most appropriate method because it does not require to calculate base infiltration. Also, it requires less input data than Infilt and Elliott-Walker. The accuracy of scaling method was less than other methods but its simplicity is superior to other methods. According to the results, if the furrow end point is used to calculate the scaling factor in the scaling method, the accuracy of this method will be significantly increased.

Appendix. Solving the first part of the integral of Eq. (12)

$$ {V}_{\mathrm{Z}1}={\int}_0^xS{\left({\left[\frac{x}{p}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}-{\left[\frac{s}{p}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds={\int}_0^xS{\left(\frac{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}-{s}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds==\underset{0}{\overset{x}{\int }}\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}{\left({x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}-{s}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\underset{0}{\overset{x}{\int }}{x}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}{\left(1-\frac{s^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}{x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}}\underset{0}{\overset{x}{\int }}{\left(1-{\left(\frac{s}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds $$

(32)

Then using the new variable:

$$ \left\{\begin{array}{c}u=\frac{s}{x}\Longrightarrow \frac{du}{ds}=\frac{1}{x}\Longrightarrow du=\frac{ds}{x}\\ {}\left\{\begin{array}{c}s=0\to u=0\\ {}s=x\to u=1\end{array}\right.\end{array}\right. $$

(33)

By putting $u=\frac{s}{x}$ value in answer of relation (32):

$$ {V}_{\mathrm{Z}1}=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}}\underset{0}{\overset{x}{\int }}{\left(1-{\left(\frac{s}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times ds=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}}\underset{0}{\overset{x}{\int }}{\left(1-{\left(\frac{s}{x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \frac{x}{x} ds=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}}\underset{0}{\overset{x}{\int }}{\left(1-{(u)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}x\frac{ds}{x}=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}+1}\underset{0}{\overset{1}{\int }}{\left(1-{(u)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times du $$

(34)

It can be inferred that (Kreyszig 1979):

$$ \underset{0}{\overset{1}{\int }}{\left(1-{(u)}^b\right)}^{\alpha}\times du=\frac{1}{b}\times \frac{\Gamma \left(\frac{1}{b}\right).\Gamma \left(1+\upalpha \right)}{\Gamma \left(r+\frac{1}{b}+1\right)} $$

(35)

where Γ is gamma function, hence:

$$ \frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}+1}\underset{0}{\overset{1}{\int }}{\left(1-{(u)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times du=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}+1}\times \left[\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\times \frac{\varGamma \left(\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right).\varGamma \left(1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\varGamma \left(\frac{1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$r$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1}\right)}\right]=\frac{S}{p^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.}}\times {x}^{\frac{1}{2r}+1}\times {\sigma}_{{\mathrm{z}}_1} $$

(36)