# 2.3 Components of Hydraulic Head

Figure 7 is a modification of the experimental setup described previously and illustrated in Figure 1. If water was no longer introduced into the left vessel, the water level in the left vessel would gradually decline, the hydraulic gradient would lessen, and flow through the cylinder (*Q*_{out}) would also gradually decline in accordance with Darcy’s law. These time-dependent flow conditions are referred to as *transient conditions*, which are relevant to groundwater modeling and pumping tests, for example. In the transient scenario depicted in Figure 7, both *h*_{1} and *Q*_{out} can be expressed as a function of time, *f(t)*, often denoted as *h*_{1}*=f(t) *and *Q*_{out}*=f(t)*. Eventually, the water levels on both sides will equilibrate, the hydraulic gradient will equal zero, and thus water will no longer flow through the cylinder. The water is static everywhere.

Referring to the hydraulic equilibrium condition (*t*_{∞}) in Figure 7, imagine inserting piezometers into the vessel to various depths (Figure 8). The water in all the piezometers will rise to the same elevation that is equal to the water level elevation in the vessel. This may be intuitively obvious, since it is analogous to inserting straws into a glass of water: no matter to what depth a straw is inserted, the water level inside the straw will be equal to the water elevation in the glass. Note that if the straws were sufficiently small in diameter, water may be drawn upward above the surrounding water level due to capillary action; however, in practice the diameter of wells (field-scale piezometers) are not small enough to create a significant capillary effect. Since the *point of measurement is the open end of the piezometer in the water*, this exercise shows that the hydraulic head is the same everywhere and that the hydraulic head measured at the open end of the piezometer is equal to the elevation to which the water rises. Figure 8 shows the resulting head distribution; the hydraulic head is 5 cm everywhere, regardless of the depth of the measurement. Hydraulic head distribution is discussed in more detail in Section 3 and 4 in the context of developing hydraulic head contour maps and potentiometric cross sections, which are used to infer the direction and magnitude of groundwater flow.

Figure 9 shows the components of hydraulic head in a static vessel of water. The hydraulic head (*h*) at each location is the sum of the elevation of the point of measurement and the height of the water column above that point. Since the latter is proportional to the pressure of the water column, it is often referred to as the pressure head (*Ψ*_{i}), whereas the elevation of the point of measurement is referred to as the elevation head (*z*_{i}):

(8) |

With regard to flow in the saturated zone, the elevation of the water level in the piezometer is what is of interest, and that is what the reader should be sure to understand: the *hydraulic head in a saturated formation is equal to the elevation of the water that rises in a well, which is effectively a piezometer*.

The hydraulic head in Figure 9 is the same everywhere. Accordingly, the hydraulic gradient is zero everywhere such that the water elevation in piezometers will be equal. However, the pressure head (*Ψ*) in the piezometers is different (they are proportional to the height of the water column in each piezometer). This simple setup illustrates that pressure head should not be used to infer flow.

The same principle applies to flowing conditions. For example, Figure 10 shows that the direction of decreasing pressure head (*Ψ*) in each configuration is the opposite of one another, but the direction of flow and hydraulic gradient is the same for each scenario. Hence, Figure 10 shows quite clearly that the flow direction cannot be based on the pressure head alone, but requires evaluation of the hydraulic head, which is defined by water elevation in the piezometers.

# Example Problem 2

What is the hydraulic head at the point in the column shown below?

Click here for solution to Example Problem 2