Ask question

Ask question

All categories

3 years ago

11

A three-phase Y-connected 50-Hz two-pole synchronous machine has a stator with 2000 turns of wire per phase. What rotor flux wou

ld be required to produce a terminal (line-to-line) voltage of 6 kV (rms value)?

1 answer:

mamaluj [8]3 years ago

5 0

Answer:

The answer is "0.0045 Wb"

Explanation:

Using formula:

$\to E_A= \sqrt{2} \pi N_c f \phi\\\\\to \phi = \frac{E_A}{\sqrt{2} \pi N_c f}\\\\$ .......(a)

Calculating the phase voltage of the machine:

$V_\phi = \frac{V_L}{\sqrt{3}}$

$= \frac{4.16 \times 10^3}{\sqrt{3}}\\\\=2401.7$

Substitute the value from equation (a):

$\phi= \frac{2401}{\sqrt{2} \times \pi \times 2000 \times 60}$

$=4.503 \times 10^{-3}\\\\= 0.0045 \ wb$

The rotor flux = 0.0045 Wb

You might be interested in

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

<u>Check One-Sample T-Interval Conditions</u>

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

<u>One-Sample T-Interval Information</u>

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

<u>Problem 1</u>

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago

The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

7 0

3 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

You want to find all files on your server that have either the SGID or SUID permission set. Which command should you use to obta

aalyn [17]

Answer:

For SGID you type this

$ find . -perm /4000

For SUID you type this

$ find . -perm /2000

Explanation:

Auxiliary file permissions, that are commonly referred to as “special permissions” in Linux are needed in order to easily find files which have SUID (Setuid) and SGID (Setgid) set.

After typing

$ find directory -perm /permissions

Then type the commands in the attachment below to obtain a list of these files with SGID and SUID.

3 0

3 years ago

It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)

ycow [4]

Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

for example ; bronze and brass

so from above discussion it is clear that option (d) will be the correct option

8 0

3 years ago

Read 2 more answers