Answer:
a. 0.9263
b. 0.4872
c. 13.83kN/m
Explanation:
moisture content (ω) = 0.32
void ratio (e) = 0.95
specific gravity () = 2.75
the degree of satruation (S) = =0.32×2.75/0.95 = 0.9263
b. porosity (n) = = 0.95/(0.95 + 1)= 0.4872
c. dry unit weight (γ) =
taking specific unit weight of water (V)= 9.81kN/m
γ = 2.75 × 1000/(1 + 0.95) = 13.83kN/m
Answer:
In summer the power available is 357.55 kW and in winter the power available is 59.59 kW
Explanation:
Given data:
height = 1500 ft = 457.2 m
30 gallon = 0.114 m³
5 gallon = 0.019 m³
In summer the power available is:
Where
μ = efficiency = 0.7
ρ = density of water = 1000 kg/m³
g = gravity = 9.8 m/s²
Q = 0.114 m³
Replacing:
In winter the power available is
Let <em>f(z)</em> = (4<em>z </em>² + 2<em>z</em>) / (2<em>z </em>² - 3<em>z</em> + 1).
First, carry out the division:
<em>f(z)</em> = 2 + (8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1)
Observe that
2<em>z </em>² - 3<em>z</em> + 1 = (2<em>z</em> - 1) (<em>z</em> - 1)
so you can separate the rational part of <em>f(z)</em> into partial fractions. We have
(8<em>z</em> - 2) / (2<em>z </em>² - 3<em>z</em> + 1) = <em>a</em> / (2<em>z</em> - 1) + <em>b</em> / (<em>z</em> - 1)
8<em>z</em> - 2 = <em>a</em> (<em>z</em> - 1) + <em>b</em> (2<em>z</em> - 1)
8<em>z</em> - 2 = (<em>a</em> + 2<em>b</em>) <em>z</em> - (<em>a</em> + <em>b</em>)
so that <em>a</em> + 2<em>b</em> = 8 and <em>a</em> + <em>b</em> = 2, yielding <em>a</em> = -4 and <em>b</em> = 6.
So we have
<em>f(z)</em> = 2 - 4 / (2<em>z</em> - 1) + 6 / (<em>z</em> - 1)
or
<em>f(z)</em> = 2 - (2/<em>z</em>) (1 / (1 - 1/(2<em>z</em>))) + (6/<em>z</em>) (1 / (1 - 1/<em>z</em>))
Recall that for |<em>z</em>| < 1, we have
Replace <em>z</em> with 1/<em>z</em> to get
so that by substitution, we can write
Now condense <em>f(z)</em> into one series:
So, the inverse <em>Z</em> transform of <em>f(z)</em> is .