<span>FACTS:
|
|
It’s
that time of year again when the days are wet and cool.
The rainy
season is the best season.
Rain makes up part of Earth’s water cycle.
Water evaporates from streams, lakes, and oceans, then condensation
and
precipitation occur in the form of rain.
Precipitation in the form of
rain is better than snow.
Snow this time of year makes people gloomy.
Rain is a great boon to local farmers.
It helps their crops grow.
</span>OPINIONS:
|
<span>| It’s
that time of year again when the days are wet and cool.
The rainy
season is the best season.
Rain makes up part of Earth’s water cycle.
Water evaporates from streams, lakes, and oceans, then condensation
and
precipitation occur in the form of rain.
Precipitation in the form of
rain is better than snow.
Snow this time of year makes people gloomy.
Rain is a great boon to local farmers.
It helps their crops grow.</span>
Answer:
2.25in³
Explanation:
For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³
See attached Table 314.16(B) from NEC 2011
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:

where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)

A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr