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2 years ago

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A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. C

alculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.

1 answer:

Minchanka [31]2 years ago

5 0

Answer:

Peak to peak Ripple voltage 8.33mV

DC output voltage =19.11 V

Explanation:

Peak voltage( Vp)= 30v

Load resistance =600 ohms

Capacitor filter = 50mF

Frequency of supply = 120Hz

The peak to peak Ripple is not only dependent on the capacitor value but also on the frequency and load current.

To calculate the,

Peak to peak ripple = I (load)/f×c

I (load) = loadd current = 30/600 = 0.05 A

Peak to peak ripple = 0.05/6

= 8.33mV

The average Dc output voltage for a full wave rectifier is double that of a half wave rectifier.

The DC output voltage is equal to 0.637Vp assuming no losses.

Vdc= 0.637 × 30

Vdc =19.11V

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A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt

= $\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}$ watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = $\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}$

15999.86 $=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}$

$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

4 0

2 years ago

What does the following program segment do? Declare Count As Integer Declare Sum As Integer Set Sum = 0 For (Count = 1; Count &l

Troyanec [42]

1225

<u>Explanation:</u>

This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.

6 0

2 years ago

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

poizon [28]

Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

7 0

2 years ago

An inventor claims to have devised a cyclical engine for use in space vehicles that operates with a nuclear fuel generated energ

slava [35]

Answer:

Inventor claim is not valid.

Explanation:

Given that

Source temperature = 510 K

Sink temperature = 270 K

Power produce = 4.1 KW

Heat reject = 15,000 KJ/h

Heat reject =4.16 KW

As we know that

Heat addition = Heat rejection + Power produce

Heat addition = 4.16 + 4.1 KW

Heat addition = 8.16 KW

So efficiency of engine

$\eta =\dfrac{Power\ produces}{heat\ added}$

$\eta =\dfrac{4.1}{8.16}$

$\eta =0.502$

Now check the maximum efficiency can be possible by using Carnot heat engine

As we know that efficiency of Carnot heat engine given as

$\eta =1-\dfrac{T_L}{T_H}$

By putting the value

$\eta =1-\dfrac{T_L}{T_H}$

$\eta =1-\dfrac{270}{510}$

$\eta =0.47$

So the efficiency of Carnot cycle is less than the efficiency of above given engine.So this engine is not possible.It means that inventor claim is not valid.

6 0

2 years ago