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14.6.3: Emission Uniformity

  • Page ID
    44697
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    Emission uniformity can be treated like distribution uniformity, DU, in Chapter 5 and is a measure of the uniformity of emissions from all the water applicators within the entire microirrigation system. For field tests,

    \(DU=\dfrac{q_{LQ}}{q_a}100\% \) (14.6)

    where DU is the emission uniformity from a field test, %; qLQ is the average discharge for the lowest one-fourth of the field measured emitter discharges (gal/hr); and qa is the average discharge of all the emitters checked in the field (gal/hr). The efficiency of an irrigation system is the relation between gross irrigation amounts and the net addition of water to the crop root zone. Distribution uniformity and the various sources of water loss that occur during the operation of the system are the two components of microirrigation efficiency. To estimate the distribution uniformity for a proposed design, the following formula was developed (Karmeli and Keller, 1975):

    \(DU=100\left(1.0-1.27\dfrac{v}{\sqrt{n}}\right)\dfrac{q_m}{q_a} \) (14.7)

    where DU is design emission uniformity in %, v is the coefficient of manufacturing variation (Table 14.3 for typical values) and n is the number of emitters. The ratio qm/qa expresses the relationship between the minimum (qm) and the average (qa) discharges resulting from pressure variations within the system. The factor (1.0 – 1.27 v/√n) adjusts for the additional nonuniformity caused by anticipated manufacturing variations between individual emitters.

    Example 14.9

    Determine the distribution uniformity for a microirrigation system designed for an average pressure of 20 psi. The system is being designed for spiral, long path emitters with automatic flushing. The field is flat and the friction head loss is a maximum of 10 ft. The system is being designed for 2 emitters per plant and the average design emission rate is 1.0 gallon per hour.

    Given: ha = 20 psi × 2.31 ft/psi = 46.2 ft

    x = 0.75 and v = 0.06

    qa = 1.0 gal/hr

    ∆hf = change in pressure due to friction loss = 10 ft

    he = change in pressure due to elevation = 0

    n = 2

    Find: The distribution uniformity, DU, for this design

    Solution

    \(q=Kh^x\) (Eq. 14.3)

    \(DU=100\left(1.0-1.27\dfrac{v}{\sqrt{n}}\right)\dfrac{q_m}{q_a} \) (Eq. 14.7)

    \(K=\dfrac{q}{h^x}=\dfrac{1}{46.2^{0.75}}=\dfrac{1}{17.72}=0.056 \)

    \(q_m=Kh_m^x \)

    \(h_m=h_p+h_e-h_f=46+0-10 \)

    \(h_m=36\text{ ft} \)

    \(q_m=0.056(36)^{0.75}\)

    \(q_m=0.082 \text{ gal/hr}\)

    \(DU = 100\left(1.0-1.27\dfrac{0.06}{\sqrt{2}}\right)\dfrac{0.82}{1.00} \)

    \(=100(1.0-0.5)(0.82) \)


    14.6.3: Emission Uniformity is shared under a CC BY-NC-ND 4.0 license and was authored, remixed, and/or curated by LibreTexts.