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3 years ago

How to draw the output voltage waveform rectifier

Engineering

1 answer:

tatyana61 [14]3 years ago

7 0

Answer:

Half-wave rectifier converts an AC signal into a DC signal. It's called a half-wave because it only rectify the positive part of an AC signal.

AC Signal = An electrical signal that alternates between positive and negative voltage.

DC Signal = An electrical signal that only has positive voltage.

Rectify = A fancy word for converting something.

Adding a capacitor helps the positive part of the signal stay on longer. This work because the capacitor stores energy kinda like a battery. During the negative part of the AC signal, the energy stored in the capacitor will be drained and used, then the cycle repeats.

The load resistor is just there to prevent a short circuit from happening.

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A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit

alexdok [17]

Answer:

$5.31\frac{kg}{m^3}$

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to $\frac{m}{M}$ , where m is the mass and M the molar mass of the gas, and the density is $\frac{m}{V}$ .

For air $M=28.66*10^{-3}\frac{kg}{mol}$ and $\frac{5}{9}R=K$

So, $598.59 R*\frac{5}{9}=332.55K$

$pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}$

7 0

3 years ago

PLEASE HURRY TIMED TEST

ladessa [460]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

Fire andwater ______(do/does) not agree

Elina [12.6K]

Answer:

Hii Dear !!! 

How are you ....

I hope That you are Fine .....

Explanation:

So the answer is __does_____

3 0

3 years ago

A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t

maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

4 0

3 years ago

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subs

NeTakaya

Answer:

For detailed answer of "

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of S°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of O.S m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 2SO W/m2.K. The subsea surrounding has a temperature of soc and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be ISO W /m2 .K. If the pipeline is made of material with thermal conductivity of 60 W/m.K, by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, (b) determine the inner surface temperature of the pipeline wall, (c) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and (d) determine the heat flux through the outer pipeline surface."

see attachment.

Explanation:

Download pdf

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3 years ago