Answer:
Option A
60 pH
Explanation:
For series connection of inductors, the total inductance is given by where the subcripts denote various inductors
For this case, inductance will be 10 pH+50 pH= 60 pH
Given the following phasors and the information related to the frequency of that phasor, provide the corresponding time-domain r
epresentation. (It is somewhat hard to tell the difference, but the terms on the left are the symbols for the phasor, e.g., V, while the terms in the interior provide the units, e.g., V is volts.) (a) V = 7e jπ/6 V, ω = 100 rad/s (b) V = √ 2e −j0.25 V, f = 100 Hz (c) V = 42 π/6 V, T = 10−6 s (d) I = 17 45◦ A, ω = 2π2 × 103 rad/s (e) I = 7 −π/2 A, f = 0.5 Hz
1 answer:
Answer:
Explanation:
In Engineering and Physics a Phasor That is a portmanteau of phase vector, is a complex number that represents a sinusoidal function whose Amplitude (A), Angular Frequency (ω), and Initial Phase (θ) are Time-invariant.
For the step by step solution to the question you asked, go through the attached documents.
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Answer:
a) the expected flow rate is 31.4 ft³/s
b) the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation is -8.4 ft
Explanation:
Given the data in the question;
free surface elevation = 200 ft
total length of pipe required = 1000 ft
diameter = 12 inch
Iron with relative roughness ( k/D ) = 0.0005
H = 665-0.051Q² [Qinft ]
a) the expected flow rate
given that;
k/D = 0.0005
k/2R = 0.0005
R/k = 1000
now, we determine the friction factor;
1/√f = 2log₁₀( R/k ) + 1.74
we substitute
1/√f = 2log₁₀( 1000 ) + 1.74
1/√f = 6 + 1.74
1/√f = 7.74
√f = 1/7.74
√f = 0.1291989
f = (0.1291989)²
f = 0.01669
Now, Using Bernoulli theorem between two reservoirs;
(p/ρq)₁ + (v²/2g)₁ + z₁ + H = (p/ρq)₂ + (v²/2g)₂ + z₂ + h
so
0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²
665-0.051Q² = 200 + flQ²/2gdA²
665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )
665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]
665 - 200 - 0.051Q² = 0.420138Q²
665 - 200 = 0.420138Q² + 0.051Q²
465 = 0.471138Q²
Q² = 465 / 0.471138
Q² = 986.97196
Q = √986.97196
Q = 31.4 ft³/s
Therefore, the expected flow rate is 31.4 ft³/s
b) the brake horsepower required to drive the pump (assume an efficiency of 78%).
we know that;
P = ρgHQ / η
where; H = 665 - 0.051(986.97196) = 614.7
we substitute;
P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )
P = 1204804.6236 / 429
P = 2808.4 bhp
Therefore, the required brake horsepower is 2808.4 bhp
c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).
NPSH = ( / ρg) - h
- ( P
/ ρg )
we substitute
25 = ( 2116 / 62.42 ) - h - ( 30 / 62.42 )
h = 8.4 ft
Therefore, the location of pump inlet to avoid cavitation is -8.4 ft
Answer:
1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices
b. Customers that utilize commutator-type motors
2) The primary voltage is 150 volts
Explanation:
The parameters given are;
Number of primary winding, = 50 turns
Number of secondary winding, = 150 turns
Voltage in secondary winding, = 450 volts
Voltage in primary winding =
The relation between ,
,
and
is as follows;
Which gives;
From which we have;
The primary voltage = 150 volts.
Answer:
Inductive reactance is denoted by where
from looking at the equation, you can see that as frequency increases, so does the inductive reactance.
Answer:
Detailed solution for this tank system is given in the images attached below:
