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Inessa [10]
3 years ago
5

Assume a voltampere-base S3b = 100 MVA and a line-to-neutral voltage base Vb = 7.5 kV. Find the current base Ib. Also, by using

this information per-unitize the phasors V˜ as and ˜Ias, e.g., V˜ as,u = V˜ as/Vb.
Engineering
1 answer:
Rasek [7]3 years ago
6 0

<u>Explanation:</u>

Let the frequency of the operation "W"ras/s

\begin{aligned}V(t) &=7.5 \cos (\omega t+\varnothing v) k V \\&=7.5 \cos w t \quad k V \\I(t) &=\cos \left(\omega t+\phi_{i}\right) \quad k A \\&=\cos (\omega t-\pi / 6) \quad k A\end{aligned}

\(\therefore\) Power, \(P(t)=v(t) I(t)\)$$\begin{array}{l}=[7.5 \cos \omega t][\cos (\cot -\pi / 6)] \times 10^{6} w \\=7.5 \cos ^{2}(2 \omega t-\pi / 6)+7.5 \cos \left(\pi_{6}\right) MW \\=3.75+7.5 \cos (2 \omega t-\pi / 6) \quad MW\end{array}$$

Real powel, \(P=\left(7.5 \times 10^{3}\right)\left(10^{3}\right) \cos \left(\varnothing_{v}-\varnothing_{I}\right)\)$$\begin{array}{l}=7.5 \cos (-\pi / 6) MW \\=3.75 MW\end{array}$$

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Impedance is defined as the total opposition to current in an AC circuit. Question 17 options: True False
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8 0
2 years ago
A data bus can be visualized as a multilane highway
iris [78.8K]

Answer:

B. with each component having an individual address

Explanation:

Data bus is a system within a computer or device, consisting of a connector or set of wires, that provides transportation for data. Data bus needs an address unique to each component in order to deliver the right data to the right place. Every memory location has a unique binary address. A microprocessor architecture is mainly composed of two main buses: The data bus and the address bus.

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3 years ago
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe
kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c.  101.24kj/kg

Explanation:

\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }

\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. n=\frac{t2-t1}{t'2-t1}

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

0.80 =\frac{48.5}{t'2-360}

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0
3 years ago
A bar having a length of 5 in. and cross-sectional area of 0.7 i n . 2 is subjected to an axial force of 8000 lb. If the bar str
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Answer:

E=1.969 × 10¹¹ Pa

Explanation:

The formula to apply is;

E=F*L/A*ΔL

where

E=Young modulus of elasticity

F=Force in newtons

L=Original length in meters,m

A=area in square meters m²

ΔL= Change in length in meters,m

Given

F= 8000 lb = 8000*4.448 =35584 N

L= 5 in = 0.127 m

A= 0.7 in² =0.0004516 m²

ΔL = 0.002 in = 5.08e-5 m

Applying the formula

E=(35584 * 0.127)/(0.0004516*5.08e-5 )

E=1.969 × 10¹¹ Pa

8 0
3 years ago
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