Size up the nozzle for a sprinkler irrigation system given the following: application rate, mm/hr 12 operating pressure, kPa 500 irrigated area, m2 250 discharge coefficient 0.80 a. 3.5 mm b. 4.5 mm c. 5.5 mm d. 6.5 mm answer: 6.5 mm
The correct answer and explanation is:
Let’s solve the problem step-by-step and verify why the nozzle size is 6.5 mm.
Given Data:
- Application rate, A=12 mm/hrA = 12 \, \text{mm/hr}
- Operating pressure, P=500 kPaP = 500 \, \text{kPa}
- Irrigated area, S=250 m2S = 250 \, \text{m}^2
- Discharge coefficient, Cd=0.80C_d = 0.80
- Need to find nozzle diameter dd
Step 1: Understand the problem
We need to find the nozzle diameter that delivers water at 12 mm/hr over a 250 m² area, operating at 500 kPa with a discharge coefficient of 0.80.
Step 2: Convert application rate to discharge (flow rate) QQ
The application rate AA (mm/hr) means millimeters of water depth applied per hour. This can be converted to volume per second: Q=A×SQ = A \times S
But units must be consistent. Since 1 mm = 0.001 m, A=12 mm/hr=0.012 m/hrA = 12 \, \text{mm/hr} = 0.012 \, \text{m/hr}
Volume applied per hour: V=A×S=0.012 m×250 m2=3 m3/hrV = A \times S = 0.012 \, \text{m} \times 250 \, \text{m}^2 = 3 \, \text{m}^3/\text{hr}
Convert to m3/sm^3/s: Q=33600=0.0008333 m3/sQ = \frac{3}{3600} = 0.0008333 \, \text{m}^3/\text{s}
Step 3: Use orifice flow formula to find diameter
The flow rate from a nozzle or orifice under pressure is: Q=Cd×Ao×2PρQ = C_d \times A_o \times \sqrt{\frac{2P}{\rho}}
Where:
- Ao=A_o = nozzle cross-sectional area =πd2/4= \pi d^2 /4
- P=500 kPa=500,000 PaP = 500 \, \text{kPa} = 500,000 \, \text{Pa}
- ρ=1000 kg/m3\rho = 1000 \, \text{kg/m}^3 (density of water)
- Cd=0.80C_d = 0.80
Rearranged to solve for dd: Q=Cd×πd24×2Pρ ⟹ d2=4QCdπ2PρQ = C_d \times \frac{\pi d^2}{4} \times \sqrt{\frac{2P}{\rho}} \implies d^2 = \frac{4Q}{C_d \pi \sqrt{\frac{2P}{\rho}}}
Calculate the velocity term: v=2Pρ=2×500,0001000=1000=31.62 m/sv = \sqrt{\frac{2P}{\rho}} = \sqrt{\frac{2 \times 500,000}{1000}} = \sqrt{1000} = 31.62 \, \text{m/s}
Plug in numbers: d2=4×0.00083330.80×π×31.62=0.00333379.47=4.19×10−5d^2 = \frac{4 \times 0.0008333}{0.80 \times \pi \times 31.62} = \frac{0.003333}{79.47} = 4.19 \times 10^{-5} d=4.19×10−5=0.00647 m=6.47 mmd = \sqrt{4.19 \times 10^{-5}} = 0.00647 \, \text{m} = 6.47 \, \text{mm}
Step 4: Conclusion
The nozzle diameter required is approximately 6.5 mm, matching option (d).
Explanation in brief:
The nozzle diameter is found by converting the required application rate over the given area into a volumetric flow rate, then applying the orifice flow equation relating flow rate to nozzle diameter, operating pressure, and discharge coefficient. The velocity of water exiting the nozzle is driven by the pressure head. Using these relationships, the diameter required to deliver the specified application rate under the given pressure is about 6.5 mm.