By Professor Ronald J. Hanks, Professor Gaylen L. Ashcroft (auth.)

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Substituting Z = Zo + ZHg into Eq. /1m = - ZHg PHg [ Pw - I ] + Zo . 4). 2 em (see Fig. 2). Find: The matric potential. 2 em + 20 em = -179 em + 20 em = -159 em. In the vacuum gage tensiometer, the mercury is replaced by a vacuum gage. /1m (weight quantity) in a way similar to that used with the mercury manometer. One should be very careful to determine the units in which the gage is calibrated. Most commercially available gage tensiometers have a dial calibrated from 0 to 100 which in weight matric potential units corresponds to a range of 0 to -1000 cm.

600 2 Irr Irr Irr Fig. 4. potential as a function of time near the top of the root zone (solid curve) and towards the bottom of the root zone (broken curve). Irrigations occurred near the end of June and July and near the middle of August Measuring Matric Potential in the Laboratorv 33 Vacuum sensing switch Fig. 5. A waterstat consists of a vacuum-sensing switching device that senses the vacuum of a tensiometer and switches the water in a sprinkling line home. In a similar way, a tensiometer can be connected to a switch.

If the soil has a uniform water content throughout the length of the tube - which is the case, for example, in the soil column just described - then the resistance to water flow is uniform throughout the tube. Now consider 1 cm increments of the flow tube. , with equal amounts of water flowing through each increment in a given time and equal flow resistance occurring in each increment - it follows that the force causing flow must be equal for each increment. [After we study Eq. 3) in the next unit, we will be able to verify quantitatively the qualitative assumption that a uniform driving force is required to cause uniform flow across uniform soil increments.