2.4  Calculating Volumetric Discharge

Eileen Poeter; Paul Hsieh

2.4 Calculating Volumetric Discharge

Volumetric discharge is the volumetric rate of water flowing through a system. This rate is reported in dimensions of volume (length cubed) over time (for example, liters per minute, or cubic meters per second).

For a flow net in which the equipotential lines and flow lines form curvilinear squares Equation 2 can be used to calculate the volumetric discharge through the flow net. The derivation of the equation for calculating flow through a flow net is provided in Box 3.

$\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}w$

(2)

where:

Q_total	=	volumetric flow through the system (L³/T)
K	=	hydraulic conductivity of the porous medium (L/T)
H	=	total head drop across the flow net domain (L)
n_f	=	number of flow tubes in the flow net (dimensionless)
n_d	=	number of head drops in the flow net (dimensionless)
w	=	distance that the system extends into the drawing (L)

As shown by the early steps of the derivation in Box 3, Equation 2 can be adjusted to accommodate a flow net drawn with shapes of a constant aspect ratio that differs from one, as shown in Equation 3.

$\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}wa_{r}$

(3)

where:

a_r

=

aspect ratio for one shape of constant aspect ratio in the flow net that needs to be calculated as the distance between the flow lines divided by the distance between the equipotential lines (dimensionless)

Equation 2 is applicable for the flow net of curvilinear squares shown in Figure 11. Suppose that the hydraulic conductivity of the material beneath the dam in the previous section is 0.5 m/d and the width of the dam into the image is 21 meters, then the volumetric flow rate under the dam is:

$\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}w$ = $\displaystyle \left (0.5\ \frac{\textup{m}}{\textup{d}} \right )\left ( 4\ \textup{m} \right )\left ( \frac{3}{14} \right )\left ( 21\ \textup{m} \right )$ = $\displaystyle 9\ \frac{\textup{m}^{3}}{\textup{d}}$

The formula for determining the volumetric flow through a flow net does not involve the absolute dimensions of the length and height of the system. It uses only the ratio of the number of head drops to the number of flow tubes. As mentioned earlier, when drawing a flow net, it is the ratio of the number of flow tubes to the number of head drops that is important. The ratio of flow tubes to head drops for the flow net of Figure 11 is 3/14 = 0.214. If the flow net is drawn with 2 flow tubes, 9 head drops will be needed to create curvilinear squares, for a ratio of 2/9 = 0.2222. If 5 flow tubes are used then 23 head drops will produce curvilinear squares, for a ratio of 5/28 = 0.217. These small differences in the ratio of flow tubes to head drops will yield slightly different values of Q_total, illustrating that drawing a flow net with paper and pencil yields an approximate solution. When calculating Q_total, for a practical application, these slight differences are trivial compared to the uncertainty in estimating an equivalent value of homogeneous hydraulic conductivity used to calculate Q_total.

In some cases, a person may choose to start with a round number for a contour interval for equipotential lines to start drawing a flow net. If the flow net of Figure 11, is drawn with a contour interval of 0.25 m (which produces 16 head drops), then the ratio required for a valid flow net indicates that 3.43 flow tubes are necessary (that is, 3.43/16 = 0.214. Thus, the flow net has a partial flow tube to maintain the valid ratio, so one of the flow tunes has to have 0.43 of the width of the tubes that form curvilinear squares as illustrated by the deepest flow tube in Figure 12. Calculation of the volumetric flow rate through the system yields the same result because the ratio of flow tubes to head drops has not changed:

$\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}w$ = $\displaystyle \left (0.5\ \frac{\textup{m}}{\textup{d}} \right )\left ( 4\ \textup{m} \right )\left ( \frac{3.43}{16} \right )\left ( 21\ \textup{m} \right )$ = $\displaystyle 9\ \frac{\textup{m}^{3}}{\textup{d}}$

Figure 12 – A flow net for the system illustrated in Figure 11 with a contour interval of 0.25 meters, requiring 3.43 flow tubes to achieve curvilinear squares and a valid ratio of the number of flow tubes to the number of head drops. Thus, the deepest flow tube is only ~40% of the width of the flow tubes that form curvilinear squares.

It is important to remember that drawing a flow net requires drawing the geometry of the flow system to scale. That is, the relative length and width of the system must be drawn correctly to determine the flow rate per unit length normal to the diagram. The absolute width of the system into the diagram must be known to determine the total flow through the system.

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