5. Switch a in the circuit has been open for a long time and switch b has been closed for a long time. Switch a is closed at t = 0. After remaining closed for 1s, switches a and b are opened simultaneously and remain open indefinitely. Determine the expression for the inductor current i that is valid when (a) 0 ≤ t ≤ 1s and (b) t ≥ 1s

Answers

Answer 1
Answer:

Answer:

(a) 1/L∫Vdt; integral t [0,1]

(b) 1/L∫Vdt; integral t [ 1, infinity]

Explanation:

An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t

The expression for inductor current is given as:

I= 1/L∫Vdt,

where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.

The formula can also be written as:

I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch


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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment with a row of mailboxes in the post office. These mailboxes were numbered 1 through 150, and beginning with mailbox 2, he opened the doors of all the even-numbered mailboxes, leaving the others closed. Next, beginning with mailbox 3, he went to every third mail box, opening its door if it were closed, and closing it if it were open. Then he repeated this procedure with every fourth mailbox, then every fifth mailbox, and so on. When he finished, he was surprised at the distribution of closed mailboxes. Write a program to determine which mailboxes these were.

Answers

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Answers

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa    ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = ((1,0,0) \cdot (1,1,0))/(1 * \sqrt2)$

       $=(1)/(\sqrt2 )$

$\cos \lambda = ((1,0,0) \cdot (1,-1,1))/(1 * \sqrt3)$

       $=(1)/(\sqrt3 )$

  τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

ii. Slip plane  --- (1 1 0)

   Slip direction -- [1 1 1]

  $\cos \phi = ((1, 0, 0) \cdot (1, -1, 0))/(1 * \sqrt2) =(1)/(\sqrt2)$

   $\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

 τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

iii. Slip plane  --- (1 0 1)

    Slip direction --- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, 1))/(1 * \sqrt2) =(1)/(\sqrt2)$

   $\cos \lambda = ((1, 0, 0) \cdot (1, 1, -1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

    Slip direction  ---- [1 1 1]

$\cos \phi = ((1, 0, 0) \cdot (1, 0, -1))/(1 * \sqrt2)=(1)/(\sqrt2)$

$\cos \lambda = ((1, 0, 0) \cdot (1, -1, 1))/(1 * \sqrt3) =(1)/(\sqrt3)$

τ = σ cos Φ cos λ

$50= \sigma * (1)/(\sqrt2) * (1)/(\sqrt3) $

  σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

Benzene vapor at 480°C is cooled and converted to a liquid at 25°C in a continuous condenser. The condensate is drained into 1.75-m3 drums, each of which takes 2.0 minutes to fill. Calculate the rate (kW) at which heat is transferred from the benzene in the condenser.

Answers

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

A power plant burns natural gas to supply heat to a heat engine which rejects heat to the adjacent river. The power plant produces 800 MW of electrical power and has a thermal efficiency of 38%. Determine the heat transfer rates from the natural gas and to the river, in MW.

Answers

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

  • Efficiency = 38%
  • Power output = 800 MW
  • Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

  • Total heat = 2105.26 MW
  • Heat used by plant = 800 MW
  • Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

brainly.com/question/2009210

Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

\eta = (Wn)/(Qin)    ......................1

put here value we get

38% = (800)/(Qin)  

Qin  = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.

Answers

Answer:

critical stress required is  18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × 10^(-3) m

so we get here critical stress that is

\sigma _c = \sqrt{(2E \gamma )/(\pi a)}     ...............1

put here value we get

\sigma _c =   \sqrt{(2* 225* 10^9 * 1 )/(\pi * 0.4* 10^(-3))}

\sigma _c =  18923493.9151 N/m²

\sigma _c =   18.92 MPa

Where should the Q point of an amplifier be on a DC load line to have the LARGEST linear output?

Answers

Answer:

At the middle of the DC load

Explanation:

For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region

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