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Vinil7 [7]
3 years ago
5

For an installation with a 150-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary, calculate the maximum sta

ndard size circuit breaker permitted for primary-only protection using Table 450.3(B), Note 1.
Engineering
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer: The 500 kVA, 3-phase transformer in the image has a 12,470 volt primary. To size a fuse for the primary side, use the formula in the image as seen ...

Explanation:

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For the R function shown below(Attachment):
Virty [35]

Answer:

The given function will return an integer in term of j on each iteration of while loop. However, it is noted that this program has some syntax error. The complete code executed in an online R compiler with explanation is given below

Explanation:

def<-function(j){

               k=0;

               i=1;

               while (i<j**3) #j raise to power 3

                   {

               

                   k<-k+1;#k=k+1

                   i <- i*2;#i=i*2

                   print(k);#

               }

 

    }

 def(6)# if you pass the value 6 to function def(j)

5 0
3 years ago
(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinuso
sammy [17]
Njjhvgghjjjhhhhhhhb hhh. I h. I’ve. Know
5 0
3 years ago
How do you get your drivers lisnes when your 15
prohojiy [21]

Answer:

You take a drivers test!

Explanation:

5 0
2 years ago
Can somebody help me with that
skelet666 [1.2K]

Answer:

I think it's 23 ohms.

Explanation:

Not entirely sure about it.

hope this helps

4 0
3 years ago
A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The
wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

q=\frac{Ak}{L}dT

\frac{q}{A}=\frac{k}{L}dT

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

{k_{fi}= 0.03 W/m.K

{L_{fi}= 5 cm = 0.05 m

{T_{fi}= 25 \°C

{k_{cb} = 0.5 W/m.K

{L_{cb}= 20 cm = 0.20 m

{T_{cb}= 0 \°C

{T_{m}= ? \°C = temperature at the interface

Solving for {T_{m} will give the temperature at the interface:

\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})

\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})

15 -0.6T_{m}=2.5T_{m}

3.1T_{m}=15

T_{m}=4.8

3 0
4 years ago
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