In areas of low relief and/or young geologic development, the drainage network may be poorly developed. Watersheds in these areas may have many closed depressional areas, referred to as potholes. Runoff generated within these areas flows to the lowest portion of the pothole rather than contributing to flow in the main channel. Other systems that are hydrologically similar to potholes include playa lakes and fields that are artifically impounded for rice production. The algorithms reviewed in this section are used to model these types of systems.

To define an HRU as a pothole, the user must set IPOT (.hru) to the HRU number. To initiate water impoundment, a release/impound operation must be placed in the .mgt file. The water balance for a pothole is:

$V=V_{stored}+V_{flowin}-V_{flowout}+V_{pcp}-V_{evap}-V_{seep}$ 8:1.3.1

where $V$ is the volume of water in the impoundment at the end of the day (m$^3$ H$_2$O), $V_{stored}$ is the volume of water stored in the water body at the beginning of the day (m$^3$ H$_2$O), $V_{flowin}$ is the volume of water entering the water body during the day (m HO), is the volume of water flowing out of the water body during the day (m HO), is the volume of precipitation falling on the water body during the day (m HO), is the volume of water removed from the water body by evaporation during the day (m HO), and is the volume of water lost from the water body by seepage (m HO).

The surface area of the pothole is needed to calculate the amount of precipitation falling on the water body as well as the amount of evaporation and seepage. Surface area varies with change in the volume of water stored in the impoundment. For surface area calculations, the pothole is assumed to be cone-shaped. The surface area is updated daily using the equation:

$SA=\frac{\pi}{10^4}*(\frac{3*V}{\pi *slp})^{2/3}$ 8:1.3.2

where $SA$ is the surface area of the water body (ha), $V$ is the volume of water in the impoundment (m$^3$ H$_2$O), and $slp$ is the slope of the HRU (m/m).

The volume of precipitation falling on the pothole during a given day is calculated:

$V_{pcp}=10*R_{day}*SA$ 8:1.3.3

where $V_{pcp}$ is the volume of water added to the water body by precipitation during the day (m$^3$ H$_2$O), $R_{day}$ is the amount of precipitation falling on a given day (mm H$_2$O), and $SA$ is the surface area of the water body (ha).

Water entering the pothole on a given day may be contributed from any HRU in the subbasin. To route a portion of the flow from an HRU into a pothole, the variable IPOT (.hru) is set to the number of the HRU containing the pothole and POT_FR (.hru) is set to the fraction of the HRU area that drains into the pothole. This must be done for each HRU contributing flow to the pothole. Water routing from other HRUs is performed only during the period that water impoundment has been activated (release/impound operation in .mgt). Water may also be added to the pothole with an irrigation operation in the management file (.mgt). Chapter 6:2 reviews the irrigation operation.

The inflow to the pothole is calculated:

$V_{flowin}=irr+\sum_{hru=1}^n[fr_{pot,hru}*10*(Q_{surf,hru}+Q_{gw,hru}+Q_{lat,hru})*area_{hru}]$

8:1.3.4

where $V_{flowin}$ is the volume of water flowing into the pothole on a given day (m$^3$ HO), is the amount of water added through an irrigation operation on a given day (m HO), is the number of HRUs contributing water to the pothole, is the fraction of the HRU area draining into the pothole, is the surface runoff from the HRU on a given day (mm HO), is the groundwater flow generated in the HRU on a given day (mm HO), is the lateral flow generated in the HRU on a given day (mm HO), and is the HRU area (ha).

When a release operation is scheduled, all water in the pothole becomes outflow:

$V_{flowout}=V$ 8:1.3.11

where $V_{flowout}$ is the volume of water flowing out of the water body during the day (m$^3$ H$_2$O), and $V$ is the volume of water stored in the pothole (m$^3$ H$_2$O).

The volume of water lost to evaporation on a given day is calculated:

if 8:1.3.5

if 8:1.3.6

where is the volume of water removed from the water body by evaporation during the day (m HO), is the leaf area index of the plants growing in the pothole, is the leaf area index at which no evaporation occurs from the water surface, is the potential evapotranspiration for a given day (mm HO), and is the surface area of the water body (ha).

The volume of water lost by seepage through the bottom of the pothole on a given day is calculated as a function of the water content of the soil profile beneath the pothole.

$V_{seep}=240*K_{sat}*SA$ if $SW<0.5*FC$ 8:1.3.7

$V_{seep}=240*(1-\frac{SW}{FC})*K_{sat}*SA$ if $0.5*FC \le SW <FC$ 8:1.3.8

$V_{seep}=0$ if $SW \ge FC$ 8:1.3.9

where $V_{seep}$ is the volume of water lost from the water body by seepage (m$^3$ H$_2$O), $K_{sat}$ is the effective saturated hydraulic conductivity of the 1st soil layer in the profile (mm/hr), is the surface area of the water body (ha), is the soil water content of the profile on a given day (mm HO), and is the field capacity soil water content (mm HO). Water lost from the pothole by seepage is added to the soil profile.

When drainage tiles are installed in a pothole, the pothole will contribute water to the main channel through tile flow. The pothole outflow originating from tile drainage is:

if 8:1.3.12

if 8:1.3.13

where is the volume of water flowing out of the water body during the day (m HO), is the average daily tile flow rate (m/s), and is the volume of water stored in the pothole (m HO).

Water may be removed from the pothole in three different types of outflow. When the volume of water in the pothole exceeds the maximum storage, the excess water is assumed to overflow and enter the main channel in the subbasin. When the retaining wall or berm is removed (this is done with a release/impound operation in the management file), all water stored in the pothole enters the main channel. The third type of flow from the pothole is via drainage tiles installed in the pothole.

Pothole outflow caused by overflow is calculated:

$V_{flowout}=V-V_{pot,mx}$ if $V >V_{pot,mx}$ 8:1.3.10

where $V_{flowout}$ is the volume of water flowing out of the water body during the day (m$^3$ H$_2$O), $V$ is the volume of water stored in the pothole (m$^3$ H$_2$O), and $V_{pot,mx}$ is the maximum amount of water that can be stored in the pothole (m$^3$ H$_2$O).