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
Answer:
Explanation:
We know that acceleration is change in velocity over time.
v is the final velocity and u is the initial velocity.
Solve for v.
Multiply both sides by t.
Add u to both sides.
Answer:
5.1*10^3 J/m^3
Explanation:
Using E = q/A*eo
And
q =75*10^-6 C
A = 0.25
eo = 8.85*10^-12
Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]
= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]
= 5.1*10^3 J/m^3
Answer:
On real life example of a scenario that takes advantage of the inverse relationship between force and time when impulse is constant is when making a serve with a lawn tennis racket
How It is an example of impulse is that when a serve is made by moving the bat slowly, the lawn tennis player uses less force and the ball is in contact with the string for longer a period
When however, the lawn tennis player moves the racket faster, with the strings of the racket highly tensioned he uses more force and the ball also spends less time on the racket to produce the same momentum
Explanation:
The impulse of a force, ΔP is given by the following formula;
ΔP = F × Δt
Where ΔP is constant, we have;
F ∝ 1/Δt
Therefore, for the same impulse, when the force is increased, the time of contact is decreases and vice versa.
