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Akimi4 [234]
3 years ago
10

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier direct

ly on the line voltage with Vf = 0.7V, what will be your output DC voltage in Volts (assuming a negligible ripple)?
Engineering
1 answer:
vovangra [49]3 years ago
7 0

Answer:

V_{dc}=84.15\ V

Explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

V_{o}=\sqrt{2}\ V_{rms}

Now by putting the values

V_{o}=60\sqrt{2}\ V

The output voltage of the DC current given as

V_{dc}=V_{o}-V_f

V_{dc}=60\sqrt{2}-0.7\ V

V_{dc}=84.15\ V

Therefore output voltage of the DC current is 84.15 V.

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If the rotational speed of a pump motor is reduced by 35%, what is the effect on the pump performance in terms of capacity, head
FinnZ [79.3K]

Answer:

- the capacity of the pump reduces by 35%.

- the head gets reduced by 57%.

the power consumption by the pump is reduced by 72%

Explanation:

the pump capacity is related to the speed as speed is reduces by 35%

so new speed is (100 - 35) = 65% of orginal speed

speed Q ∝ N ⇒ Q1/Q2 = N1/N2

Q2 = (N2/N1)Q1    

Q2 = (65/100)Q1

which means that the capacity of the pump is also reduces by 35%.

the head in a pump is related by

H ∝ N² ⇒ H1/H2 = N1²/N2²

H2 = (N2N1)²H1

H2 = (65/100)²H1 = 0.4225H1

so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.

Now The power requirement of a pump is related as

P ∝ N³ ⇒ P1/P2 = N1³/N2³

P2 = (N2/N1)³P1

H2 = (65/100)²P1 = 0.274P1

So the reduction in power is 1 - 0.274 = 0.725 which is 72%

Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.  

8 0
3 years ago
What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

4 0
3 years ago
Which statement about lean manufacturing is true when you compare it to mass production?
Len [333]
Where are the statements then bbs lol
6 0
3 years ago
What financial arguments could you use to justify your proposed
Gnoma [55]

The recommendation to segregate FLTs and the workers are as follows:-1)Reputation of warehouse:- To be in the market the reputation of warehouse should be good,it can only happen when the worker of that warehouse is happy with the management looks after worker external and internal affairs. There should be two pathways one for vehicle and other for walking in which both can’t use vice versa.2)High Profitability:- When there is no incident or accident happens between the FLTs and the workers in the warehouse then off course the worker will be regular at work then there will be high profit .3)Insurance premium:- If there is zero accident happens in the ware house then there will no claim, the company will be in the profit..

7 0
3 years ago
An amplifier with 40 dB of small-signal, open-circuit voltage gain, an input resistance of 1 MO, and an output resistance of 100
Vanyuwa [196]

convert 40db to standard gain

AL=10^40/20=100

calculate total voltage gain

=AL×RL/RL+Ri

=83.33

38.41 DB

calculate power

Pi=Vi^2/Ri Po=Vo^2/RL

power gain= Po/Pi

=13.90×10^6

3 0
3 years ago
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