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Bogdan [553]
3 years ago
7

A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor

is
Engineering
1 answer:
Oksanka [162]3 years ago
8 0

Answer: the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

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Cutting and abrasive machining are the two major material processes. List the differences between Cutting tool and Abrasive mach
STatiana [176]

Answer:

Explained

Explanation:

Cutting tools:

 1. Cutting tools can either be single point or multi point.

2. Cutting tools can have variety of material depending on use like ceramics, diamonds, metals, CBN, etc.

3.Cutting tools have definite shapes and geometry.

Abrasive machining tools

1. Abrasive tools are always multi point tools.

2. Abrasive tools composed of abrasives bounded in medium of resin or metal.

3. They do not have definite geometry of shape

7 0
4 years ago
What type of engineer is responsible for the protection of wetlands and beaches?
vovikov84 [41]
Environment engineer
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3 years ago
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Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pres
LekaFEV [45]

Answer:

44.59°c

Explanation:

Given data :

Total pressure = 105 kpa

complete combustion

A) Determine air-fuel ratio

A-F = \frac{N_{air} }{N_{fuel} }  = \frac{(Nm)_{air} }{(Nm)_{c} (Nm)_{n} }

N = number of mole

m = molar mass

A-F = \frac{(6.75*4.76)kmole * ( 29kg/mol)}{(3kmole)* 12kg/mol + (6kmol)*(1kg/mol)}  =  22.2 kg air/fuel

hence the ratio of Fuel-air = 1 : 22.2

B) Determine the temperature at which water vapor in the products start condensing

First we determine the partial  pressure of water vapor before using the steam table to determine the corresponding saturation temp

partial pressure of water vapor

Pv = \frac{(N_{water vapor}) }{N_{pro} } * ( P_{ro} )

N watervapor ( number of mole of water vapor ) = 3

N pro ( total number of mole of product = 3 + 3 + 2.25 + 25.28 = 33.53 kmol

Pro = 105

hence Pv = ( 3/33.53 ) * 105 =  9.39kPa

from the steam pressure table the corresponding saturation temperature to 9.39kPa =  44.59° c

Temperature at which condensing will start = 44.59°c

An equation showing the products of propylene with their mole numbers is attached below

8 0
3 years ago
A sheet of steel 2.5 mm thick has nitrogen atmosphere on both sides at 900 oC and is permitted to achieve a steady-state diffusi
DanielleElmas [232]

Answer:

1.8 mm

Explanation:

given data

thick = 2.5 mm

flux = 1 × 10^{-7} kg/m²

high pressure surface is 2 kg/m³

solution

we use fick first law for steady state diffusion

J = D × \frac{Ca - Cb}{Xa - Xb}   ..........1

we take here Ca to point which concentration of nitrogen is 2 kg/m³

so we solve Xb

Xb = Xa + D × \frac{Ca - Cb}{J}

assume Xa = 0 at surface

Xb = 0 + ( 12 × 10^{-11} ) × \frac{2 - 0.5}{1*10^{-7}}

Xb = 1.8 × 10^{-3}

Xb = 1.8 mm

8 0
3 years ago
Pls help me it’s due today
hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

6 0
3 years ago
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