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3 years ago

9

Which career related to architecture deals with the planning of entire cities and focuses on designing and arranging buildings,

streets, and neighborhoods?

2 answers:

snow_lady [41]3 years ago

8 0

Answer:

City planner would be a good answer. Unless there are any options, this seems like the best bet. Hope this helps!

adell [148]3 years ago

4 0

Answer:

Urban Design

I got it right on the test

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A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$

Where t is given by

$1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec$

So,

$\Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2$

$$ \Delta \delta = 6.75 \: elec \: deg$

The change in torque in rpm/s is given by

$\Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ }$

$\Delta \delta =28.12 \: \: rpm/s$

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

$Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t$

Where P is the number of poles of the turbo-generator.

$Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2$

$Rotor \: speed = 1500 + 5.62$

$$ Rotor \: speed = 1505.62 \:\: rpm$

4 0

3 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

Sergeeva-Olga [200]

Answer:

HEAT LOST

polycarbonate = 252 W

soda lime glass = 1680 W

aerogel = 16.8 W

COST associated with heat loss

polycarbonate = $ 262.08

soda lime glass = $ 1,747.2

aerogel = $ 17.472

The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

Explanation:

Given that;

surface area for each window = 0.4m * 0.4m = 0.16m^2

DeltaT = 90°C, L = 12mm = 0.012m

thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER

so at 300K

KsL = 1.4 W/mK

Kag = 0.014 W/mK

Kpc = 0.21 W/mK

Now HEAT LOSS

for polycarbonate;

Qpc = -KA dt/dx

NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)

so Qag = (0.21 * 0.16 * 90) / 0.012

= 252 W

for soda lime glass;

Qsl = (1.4 * 0.16 * 90) / 0.012

= 1680 W

for aerogel

Qaq = (0.014 * 0.16 * 90) / 0.012

= 16.8 W

Now for COST associated with heat lost

for polycarbonate;

cost = Qpc * 130 * 8 * 1/1000

= 252 * 130 * 8 * 1/1000

= $ 262.08

for soda lime glass;

cost = 1680 * 130 * 8 * 1/1000

= $ 1,747.2

for aerogel

cost = 16.8 * 130 * 8 * 1/1000

= $ 17.472

Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

6 0

3 years ago

Whats 47,000 resistance converted to kilo or mega ohms.

Julli [10]

The conversion of 47,000 Ohms to kilo-ohms is equal to 47 kilo-ohms.
The conversion of 47,000 Ohms to mega-ohms is equal to 0.047 kilo-ohms.

<h3>What is resistance?</h3>

Resistance can be defined as an opposition to the flow of current in an electric circuit. Also, the standard unit of measurement of the resistance of an electric component is Ohms, which can be converted to kilo-ohms or mega-ohms.

For Ohms to kilo-ohms, we have:

1 Ohms = 0.001 kilo-ohms

47,000 Ohms = X kilo-ohms

Cross-multiplying, we have:

X = 0.001 × 47000

X = 47 kilo-ohms.

For Ohms to mega-ohms, we have:

1,000,000 ohms = 1 mega-ohms

47,000 Ohms = X mega-ohms

Cross-multiplying, we have:

X1,000,000 = 47,000

X = 47,000/1,000,000

X = 0.047 kilo-ohms.

Read more resistance here: brainly.com/question/19582164

#SPJ1

7 0

1 year ago

1 import java.util.Scanner; 3 public class EqualityAndRelational { 4 public static void main (String args) args) { int userBonus

Anastaziya [24]

Missing Part of the Question

Complete the expression so that userPoints is assigned with 0 if userBonus is greater than 20 (second branch). Otherwise, userPoints is assigned with 10 (first branch

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

( Your solution goes here)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

Answer;

Replace

( Your solution goes here)

With

if(userBonus>20).

The full program becomes

import java.util.Scanner;

public class EqualityAndRelational {

public static void main (String args) args) {

int userBonus; int userPoints;

userPoints=0;

Scanner scnr = new Scanner(System.in);

userBonus = scnr.nextInt();

// Program will be tested with values : 15, 20, 25, 30, 35. 12 13 14 15 16 17 18 19

if(userBonus>20)

{

userPoints= 10 ;

}

else {

userPoints = 0;

}

}

}

7 0

4 years ago

5. The water in an 8-m-diameter, 3-m-high above-ground swimming pool is to be emptied by unplugging a 3-cm-diameter, 25-m-long h

frosja888 [35]

Answer:

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Friction head and pressure head will cause the actual flow rate to be less.

Explanation:

Considering point 1 at the free surface of the pool, and point 2 at the exit of

pipe.

Using Bernoulli equation between

these two points simplifies to

P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2

Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),

P/(p*g) + z1 = P/(p*g) + V2²/2g

z1 = V2²/2g

Note; z1 = h

V2max = √2gh

h = 3 m

V2max = √2 * 9.81 * 3

V2max = √58.86 = 7.67 m/s

maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max

Qmax = A * V2max

Diameter d = 3 cm = 0.03 m

A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³

Qmax = 0.00071 * 7.67 = 0.00545 m³/s

Qmax = 5.45 L/s

The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.

Actual flow rate will be less because of heads such as friction head and pressure head.

7 0

3 years ago