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Nookie1986 [14]
3 years ago
15

Your sprayer has a 60-foot wide boom with 36 nozzles along this 60-foot length. Your spray speed is 4.5 miles per hour and you w

ant 25 GPA. How much should you catch per nozzle in 15 seconds to be properly calibrated?
Engineering
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

358.52 m/s

Explanation:

From the information given:

The required amount you should catch per nozzle in 15 secs. can be  determined by using the formula:

Gallons/Min  = \dfrac{(Gallons/Acre \times Miles/Hour \times  nozzle \ spacing )}{5940}

where;

Gallons/Acre  = 25 GPA

= \dfrac{(25 \times 4.5 \times 60 \times \dfrac{12}{ 36})}{5940} \times \dfrac{15}{60}

= \dfrac{(25 \times 4.5 \times 60 \times0.333)}{5940} \times 0.25

= \dfrac{(2247.75)}{5940} \times 0.25

= 0.0946 gps

= 358.52 m/s

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enot [183]

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

Maximum allowable normal stress ,σ= 150

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The normal stress for pressure vessel given as

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We always take maximum stress for safe design.

\sigma=\dfrac{Pd}{2t}

Now by putting the values

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3 years ago
A battery is an electromechanical device. a)- True b)- False
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Answer:

b)False

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2 years ago
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat
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Answer: 10.631\times 10^3\ N/m^2

Explanation:

Given

Discharge is Q=12.5\ L

Diameter of pipe d=150\ mm

Distance between two ends of pipe L=800\ m

friction factor f=0.008

Average velocity is given by

\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s

Pressure difference is given by

\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow  \Delta P=10.631\ kPa

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2 years ago
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