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zubka84 [21]
3 years ago
13

Design a plan for ""your yard"" (no matter where you live) that will capture and utilize rainwater. You can prepare the plan in

graphical form (rough sketch) on a paper. Please provide an explanation of your plan as well.

Engineering
1 answer:
Jlenok [28]3 years ago
4 0

Answer:

Explanation:

<u><em>General Considerations</em></u>

The design of the yard will affect the natural surface and subsurface drainage pattern of a watershed or individual hillslope. Yard drainage design has as its basic objective the reduction or elimination of energy generated by flowing water. The destructive power of flowing water increases exponentially as its velocity increases. Therefore, water must not be allowed to develop sufficient volume or velocity so as to cause excessive wear along ditches, below culverts, or along exposed running surfaces, cuts, or fills.

A yard drainage system must satisfy two main criteria if it is to be effective throughout its design life:

1. It must allow for a minimum of disturbance of the natural drainage pattern.  

2.It must drain surface and subsurface water away from the roadway and dissipate it in a way that prevents excessive collection of water in unstable areas and subsequent downstream erosion

The diagram below ilustrate diffrent sturcture of yard to be consider before planing to utiliza rainwater

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Determine the average power, complex power and power factor (including whether it is leading or lagging) for a load circuit whos
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Answer:

Average power = 108.25W

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Explanation:

Check attachment for step by step instructions.

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Two kilograms of air in a piston-cylinder assembly undergoes an isothermal
Lubov Fominskaja [6]

Answer:

a) 0.39795 kJ/K

b) 79.589.37 kJ

Explanation:

m = Mass of air = 2 kg

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P₂ = Final pressure = 600 kPa

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∴ Entropy is generated in the process is 0.39795 kJ/K

b)

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7 0
2 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

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Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

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Explanation:

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