1<br>M<br>A BLIND COOK WHO DEFEATED<br>OVER 30,000 HOME COOKS!<br>y of

Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el

exis [7]

Answer:

a) V = 20 ∠30⁰ , I = 4 ∠-210⁰ Z inductive L = 0,0125 H

b) V = 9∠-60⁰ , I = 4 ∠ 190⁰ Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰ , I = 7 ∠ 150⁰ Z Inductive L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as function cosine

we know that sin ( x + 90 ) = cos x

Then sin ( 400*t -120 ) = cos ( 400*t - 120 -90 ) = cos ( 400t - 120 - 90)

Phasor form I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I ⇒ Z = 20 ∠30⁰ / 4 ∠-210⁰ Z = 5 ∠-180⁰

We notice that voltage advances the current then we are in presence of an inductive impedance

5 = wl ⇒ 5 = 400 *L ⇒ L = 0,0125 H

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t) = 4 sin ( 900t + 280 ) ⇒ i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 ) ⇒ I = 4 ∠ 190⁰

Z = V/I ⇒ Z = 9∠-60⁰ / 4 ∠ 190⁰ Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc = 1/ 900*C 1/wc = Z ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C ⇒ 2025 =1/C ⇒ C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 ) ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90) ⇒ i(t) = 7 sin (250t + 150)

Phasor Form I = 7 ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰ ⇒ Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L 250 L = 1,86 L = 1,86/250 L = 0,0074 H

7 0

3 years ago

simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl

Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) $(\frac{783.05-778.18}{800.13-778.18} )$ = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1 ( calculated values )

= 626.57 - 295.17 = 331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4 ( calculated values )

= 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

= 541.88 - 331.4 = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

= 210.48 / ( 1324.93 - 626.57 )

= 0.3014 = 30.14%

6 0

3 years ago

What are the reasons why fine grained of alkali igneous rocks can not be used in cement

Luden [163]

Fine grained of alkali igneous rocks cannot be used in cement because of the volume expansion caused by the Alkali-silica reaction, fine-grained igneous rocks cannot be used as aggregates in cement.

<h3>What does fine grained mean in an igneous rock?</h3>

Extrusive igneous rocks have a fine-grained or aphanitic texture, with grains that are too small to see without a magnifying glass. The fine-grained texture suggests that the rapidly cooling lava did not have enough time to form large crystals. A petrographic microscope can be used to examine these tiny crystals.

The texture of an igneous rock (fine-grained vs coarse-grained) is determined by the rate at which the melt cools: slow cooling produces large crystals, while fast cooling produces small crystals.

The chemical reaction that occurs in both alkali cations and hydroxyl ions in the pore solution of hydrated cement paste and certain reactive silica phases present in concrete aggregates is known as the alkali-silica reaction (ASR).

Learn more about cement on:

brainly.com/question/14323034

#SPJ1

8 0

2 years ago

A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul

lorasvet [3.4K]

<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

$\frac{1}{R_x}$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$

Solving for Rₓ gives;

$R_{x}$ = $\frac{R_1 * R_2}{R_1 + R_2}$ ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ = resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;

$R_{x}$ = $\frac{40 * 60}{40 + 60}$

$R_{x}$ = $\frac{2400}{100}$

$R_{x}$ = 24 Ω

Therefore, the total resistance is 24Ω

4 0

3 years ago

What does a block tester do?

tatyana61 [14]

Answer:

Block design test. A block design test is a subtest on many IQ test batteries used as part of assessment of human intelligence. It is thought to tap spatial visualization ability and motor skill.

Explanation:

6 0

3 years ago

1MA BLIND COOK WHO DEFEATEDOVER 30,000 HOME COOKS!y of​

1MA BLIND COOK WHO DEFEATEDOVER 30,000 HOME COOKS!y of