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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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Harlamova29_29 [7]
D. All of the above
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3 years ago
-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63
Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0
3 years ago
Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3
viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

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Answer:

100 cm

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Answer:

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