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:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH = 
Current in the coil, i = ![1680cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=1680cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
A
where
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf, 
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:
(b) To calculate the maximum average flux,we know that:
(c) To calculate the magnitude of the induced emf at t = 0.0180 s:
An important difference between a universal and a split-phase motor is that the split-phase motor has
A. two brushes attached to the stator.
B. a single coil formed on the rotor.
<u>C. two windings on the stator. </u>
D. an armature with a commutator.
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
Answer:
1/6 del peso en la tierra.
