You want to know how to solve it?
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
Desired operation: A + B = C; {A,B,C) are vector quantities.
<span>Issue: {A,B} contain error (measurement or otherwise) </span>
<span>Objective: estimate the error in the vector sum. </span>
<span>Let A = u + du; where u is the nominal value of A and du is the error in A </span>
<span>Let B = v + dv; where v is the nominal value of B and dv is the error in B </span>
<span>Let C = w + dw; where w is the nominal value of C and dw is the error in C [the objective] </span>
<span>C = A + B </span>
<span>w + dw = (u + du) + (v + dv) </span>
<span>w + dw = (u + v) + (du + dv) </span>
<span>w = u+v; dw = du + dv </span>
<span>The error associated with w is the vector sum of the errors associated with the measured quantities (u,v)</span>
Answer:
Explanation:
Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field
a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.
b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.
Situations when electron is affected by an external magnetic field but not by an external electric field
There is no such situation in which electric field will not affect an electron . It will always affect an electron .
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