Answer:
a) benzene = 910 days
b) toluene = 1612.67 days
Explanation:
Given:
Kd = 1.8 L/kg (benzene)
Kd = 3.3 L/kg (toluene)
psolid = solids density = 2.6 kg/L
K = 2.9x10⁻⁵m/s
pores = n = 0.37
water table = 0.4 m
ground water = 15 m
u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s
a) For benzene:
The time will take will be:
b) For toluene:
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 .
The solution is in the attachment
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W =
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271