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2 years ago

What type of diagrams are used to present the conditions that are available to control outputs based on inputs?

1 answer:

5 0

The type of diagrams used to present the conditions that are available to control outputs based on inputs are called; B: Ladder

<h3>How to Utilize Ladder Diagrams?</h3>

Ladder diagrams are defined as specialized schematics usually used to document industrial control logic systems.

Now, these are called ladder diagrams due to the fact that they resemble a ladder, with two vertical rails that supply power and as many “rungs” as there are control circuits to represent.

Read more about Ladder Diagrams at; brainly.com/question/20519838

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A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn

mixas84 [53]

Answer:

a) | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

a) Identify the variables in the solution vector assume Bus 2 is a load bus

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 | and β1 been specified while p1 and q1 are not specified

Hence the variables in the solution

= | v2 | , β2 ( load, bus voltage at bus 2 )

p1 , q1 ( slack, bus power at bus 1 )

b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 | and β1

unspecified : p1 and q1

Hence the variables in the solution

= q2 , β2

p1 and q1 ( slack, bus power at bus 1 )

7 0

3 years ago

The underground storage of a gas station has leaked gasoline into the ground. Among the constituents of gasoline are benzene, wi

vovangra [49]

Answer:

a) benzene = 910 days

b) toluene = 1612.67 days

Explanation:

Given:

Kd = 1.8 L/kg (benzene)

Kd = 3.3 L/kg (toluene)

psolid = solids density = 2.6 kg/L

K = 2.9x10⁻⁵m/s

pores = n = 0.37

water table = 0.4 m

ground water = 15 m

u = K/n = (2.9x10⁻⁵ * (0.4/15)) / 0.37 = 2.09x10⁻⁶m/s

a) For benzene:

$R=1+\frac{\rho * K_{d} }{n}, \rho = 2.6\\ R=1+\frac{2.6*1.8}{0.37} =13.65$

The time will take will be:

$t=\frac{xR}{a} , x=12,a=0.18\\t=\frac{12*13.65}{0.18} =910days$

b) For toluene:

$R=1+\frac{2.6*3.3 }{0.37} = 24.19$

$t=\frac{12*24.19}{0.18} =1612.67days$

6 0

3 years ago

Read 2 more answers

A large building will need several different types of workmen to install and repair pipes for water, heating,

siniylev [52]

Answer:

Plumber and pipefitters

Explanation:

3 0

3 years ago

A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N

soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, $P_{t} = 8 kW = 8000 W$

Now, the force, F is given by:

$P_{t} = Force(F)\times velocity(v) = Fv$ (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, $F_{c}$ is provided by the gravitational force, $F_{G}$ :

$F_{c} = F_{G}$

$\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}$

Thus from the above, velocity comes out to be:

$v = \sqrt{\frac{GM_{e}}{R}}$

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s$

where

R = $R_{e} + H$

R = $\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}$

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

$8000 = F\times 3075.36$

F = 2.60 N

6 0

3 years ago

Unit for trigonometric functions is always "radian". 1. 10 points: Do NOT submit your MATLAB code for this problem (a) Given f(x

RoseWind [281]

Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

clear all;

clc;

x0=input('Initial guess:\n');

x=x0;

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

i=i+1;

temp=x;

x=x-(f/g);

f=exp(-x)-sin(x)-0.2;

g=-exp(-x)-cos(x);

ep=abs(x-temp);

fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

end

fprintf('The solution x = %6f \n',x);

%% End of MATLAB Program

Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

The solution x = 216.971251

Guess x=3.0:

Initial guess:

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

Guess x=4.7:

Initial guess:

4.7

Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

7 0

3 years ago