Answer:
16.8ohms
Explanation:
According to ohm's law which states that the current passing through a metallic conductor at constant temperature is directly proportional to the potential difference across its ends.
Mathematically, V = IRt where;
V is the voltage across the circuit
I is the current
R is the effective resistance
For a series connected circuit, same current but different voltage flows through the resistors.
If the initial current in a circuit is 19.3A,
V = 19.3R... (1)
When additional resistance of 7.4-Ω is added and current drops to 13.4A, our voltage in the circuit becomes;
V = 13.4(7.4+R)... (2)
Note that the initial resistance is added to the additional resistance because they are connected in series.
Equating the two value of the voltages i.e equation 1 and 2 to get the resistance in the original circuit we will have;
19.3R = 13.4(7.4+R)
19.3R = 99.16+13.4R
19.3R-13.4R = 99.16
5.9R = 99.16
R= 99.16/5.9
R = 16.8ohms
The resistance in the original circuit will be 16.8ohms
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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
D ............................
