Question

A sprinkler irrigation system has been designed for a crop with the water application rate of \(1.17 \,\text{cm h}^{-1}\) and sprinkler discharge of \(1.3 \,\text{L s}^{-1}\). The coefficient of discharge and uniformity coefficient are 0.9 and 0.8, respectively. If the sprinkler spacing along the lateral is 20 m, the lateral spacing in m is

• A. 14.4
• B. 16.0
• C. 18.0
• D. 20.0

Hint:

In sprinkler design, the effective discharge must include both the discharge coefficient and uniformity coefficient. Then, spacing is simply area per sprinkler divided by row spacing.

Answer 1

The Correct Option is A

Step 1: Convert given data.
Sprinkler discharge: \[ q=1.3 \,\text{L/s}=1.3\times 10^{-3}\,\text{m}^3/\text{s}. \] Application rate: \[ 1.17\,\text{cm/h}=0.0117\,\text{m/h}=\frac{0.0117}{3600}=3.25\times 10^{-6}\,\text{m/s}. \] Effective discharge: \[ q_{\text{eff}}=q \times C_d \times C_u = 1.3\times 10^{-3}\times 0.9\times 0.8=9.36\times 10^{-4}\,\text{m}^3/s. \]

Step 2: Area served per sprinkler.
Area per sprinkler is \[ A=\frac{q_{\text{eff}}}{I_r} = \frac{9.36\times 10^{-4}}{3.25\times 10^{-6}} = 288\,\text{m}^2. \]

Step 3: Relating area to spacings.
Let sprinkler spacing along lateral = \(S_m=20\,\text{m}\). Let lateral spacing = \(S_l\). \[ A=S_m \times S_l. \] Thus, \[ S_l=\frac{288}{20}=14.4\,\text{m}. \]

Final Answer:
\[ \boxed{14.4\,\text{m}} \]