Implement each of the following functions using only two-input gates. The multi-level circuit should have AND and OR gates alter
nating at adjacent levels. a. Z = AB + C’D’E’ b. X = ABC + AD + A’C’D’ + A’E’F’ (last gate should be an AND gate) c. Solve part (b) with last gate as an OR gate.
1 answer:
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Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
<u>B) magnitude of primary and secondary winding currents </u>
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
<u>C ) Calculate the Primary voltage magnitude and the Voltage regulation</u>
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
<em>Voltage regulation </em>
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%
Answer:
A. 72.34mol/min
B. 76.0%
Explanation:
A.
We start by converting to molar flow rate. Using density and molecular weight of hexane
= 1.59L/min x 0.659g/cm³ x 1000cm³/L x 1/86.17
= 988.5/86.17
= 11.47mol/min
n1 = n2+n3
n1 = n2 + 11.47mol/min
We have a balance on hexane
n1y1C6H14 = n2y2C6H14 + n3y3C6H14
n1(0.18) = n2(0.05) + 11.47(1.00)
To get n2
(n2+11.47mol/min)0.18 = n2(0.05) + 11.47mol/min(1.00)
0.18n2 + 2.0646 = 0.05n2 + 11.47mol/min
0.18n2-0.05n2 = 11.47-2.0646
= 0.13n2 = 9.4054
n2 = 9.4054/0.13
n2 = 72.34 mol/min
This value is the flow rate of gas that is leaving the system.
B.
n1 = n2 + 11.47mol/min
72.34mol/min + 11.47mol/min
= 83.81 mol/min
Amount of hexane entering condenser
0.18(83.81)
= 15.1 mol/min
Then the percentage condensed =
11.47/15.1
= 7.59
~7.6
7.6x100
= 76.0%
Therefore the answers are a.) 72.34mol/min b.) 76.0%
Please refer to the attachment .
Same question idea but different values... I hope I helped you... Don't forget to put a heart mark
