Question

Use the unit-gradient approach and Campbell's hydraulic conductivity model (Eq. \(4-7\) ) to: a. estimate the drainage rate for a loam soil with \(\mathrm{K}_{\mathrm{s}}=310 \mathrm{~mm}\) \(\mathrm{d}^{-1}, b=4.5, \theta_{\mathrm{s}}=0.46\), and \(\theta=0.40\), and b. estimate the volumetric water content at which the drainage rate would be \(1 \mathrm{~mm} \mathrm{~d}^{-1}\).

Short answer

Question: Estimate the drainage rate for the given loam soil using Campbell's hydraulic conductivity model and find the volumetric water content at which the drainage rate would be 1 mm/day. Answer: The drainage rate for the given loam soil is approximately 204.5 mm/day, and the volumetric water content at which the drainage rate would be 1 mm/day is approximately 0.22.

Step-by-step solution

Calculate the hydraulic conductivity for the given water content

To estimate the drainage rate for the given loam soil with the given values, we first need to find out the hydraulic conductivity K of the soil. We can use Campbell's conductivity equation \(\mathrm{K}(\theta)=\mathrm{K}_{\mathrm{s}}\left(\frac{\theta-\theta_{r}}{\theta_{\mathrm{s}}-\theta_{\mathrm{r}}}\right)^{b}\) where \(\theta_{r}\) is residual water content. In this case, we can assume the residual water content of the soil is zero. Therefore \(\mathrm{K}(\theta)=\mathrm{K}_{\mathrm{s}}\left(\frac{\theta}{\theta_{\mathrm{s}}}\right)^{b}\) Now, plug in the values for Ks, theta, and thetas into the equation: \(\mathrm{K}(0.40)=310\left(\frac{0.40}{0.46}\right)^{4.5}\)

Calculate the drainage rate

Calculate K(theta) using the equation from Step 1: \(\mathrm{K}(0.40)=310\left(\frac{0.40}{0.46}\right)^{4.5}\approx 204.5 \,\mathrm{mm}\,\mathrm{d}^{-1}\) So the drainage rate for the given loam soil is approximately 204.5 mm/day.

Estimate the volumetric water content at which the drainage rate would be 1 mm/day

To find the volumetric water content at which the drainage rate is 1 mm/day, we need to solve the equation: \(\mathrm{K}(\theta)=\mathrm{K}_{\mathrm{s}}\left(\frac{\theta}{\theta_{\mathrm{s}}}\right)^{b} = 1 \,\mathrm{mm}\,\mathrm{d}^{-1}\) We know the values of \(\mathrm{K}_{\mathrm{s}}\) and \(b\). So we can solve for \(\theta\): \(1 = 310\left(\frac{\theta}{0.46}\right)^{4.5}\) Rearrange the equation to make \(\theta\) the subject: \(\theta = 0.46\left(\frac{1}{310}\right)^{1/4.5}\)

Calculate the volumetric water content

Now, calculate the value of \(\theta\) using the formula from Step 3: \(\theta = 0.46\left(\frac{1}{310}\right)^{1/4.5}\approx 0.22\) Hence, the volumetric water content at which the drainage rate would be 1 mm/day is approximately 0.22.