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.
1 answer:
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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Answer:
The heat is transferred is at the rate of 752.33 kW
Solution:
As per the question:
Temperature at inlet, = 273 + 20 = 293 K
Temperature at the outlet, = 273 + 200 = 473 K
Pressure at inlet,
Pressure at outlet,
Speed at the outlet,
Diameter of the tube,
Input power,
Now,
To calculate the heat transfer, , we make use of the steady flow eqn:
where
= specific enthalpy at inlet
= specific enthalpy at outlet
= air speed at inlet
= specific power input
H and H' = Elevation of inlet and outlet
Now, if
and H = H'
Then the above eqn reduces to:
(1)
Also,
Area of cross-section, A =
Specific Volume at outlet,
From the eqn:
Now,
Also,
Now, using these values in eqn (1):
Now, rate of heat transfer, q:
q = mQ =
Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that
hence , actual velocity will be equal to 8.39 m/s