All categories

3 years ago

Plot da(t) if the output voltage of the converter pole-a is vaN(t)=Vd/2+0.85 Vd/2 sin(Ï1t), where Ï1=2Ï x 60 rad/s

Engineering

1 answer:

maw [93]3 years ago

5 0

Answer:

Explanation:

given data :

output voltage ( Van(t) ) = (Vd /2) + (0.85 Vd/2 sin ( w1 t ) )

w1 = 2 $\pi$ 60 rad/sec

find the value of da(t) by inputting the value of Van (t) into

da = Van(t) / Vd

hence: da(t) = 0.5 + 0.425 sin ((2 $\pi$ 60)t)

attached below is the plot of the da(t) against time

You might be interested in

How high of a column of sae 30 oil would be required to give the same pressure as 700 mm hg?

Rasek [7]

Hoiu-10,4000 mm.

<h3>Is positive pressure good for PC?</h3>

A balanced configuration is the most efficient way to cool your pc although it should tend towards a slight positive pressure if you can help it.
Tip: As much as it might seem important, the concept of heat rising doesn't have too much of an effect.

To learn more about it, refer

to https://brainly.in/question/413163

#SPJ4

7 0

1 year ago

Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th

Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

(1.31 kg/s)(3343.6 KJ/kg - 2442.4 KJ/kg) - 225 KW = P

5 0

3 years ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

3 years ago

with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d

Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

To know more about kVp visit:-

brainly.com/question/17095191

#SPJ4

5 0

1 year ago

Tech A says that horsepower is a measurement simply of the amount of work being performed. Tech B says that horsepower can be ca

snow_tiger [21]

Answer:

Tech B

Explanation:

Horsepower (hp) refers to a unit of measurement of power in respect of the output of engines or motors.

Horsepower is the common unit of power. It indicates the rate at which work is done.

The formula $\frac{rpm*T}{5252}$ , where rpm is the engine speed, T is the torque, and 5,252 is radians per second.

So,

Tech B is correct

6 0

3 years ago