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3 years ago

7

Suppose that you are working in a home that has a 15kw oven that operates on 240 V, The oven is on a branch circuit by itself. W

hat’s the demand load for the oven circuit? Show all of your calculations.

1 answer:

krok68 [10]3 years ago

7 0

Explanation:

please see it my work solution

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Nec ________ covers selection of time-delay fuses for motor- overload protection.

Murljashka [212]

Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

<h3>What article in the NEC covers motor overloads?</h3>

Article 430 that is found in National Electrical Code (NEC) is known to be state as “Motors, Motor Circuits and Controllers.” .

Note that the article tells that it covers areas such as motors, motor branch-circuit as well as feeder conductors, motor branch-circuit and others.

Therefore, Nec Article 430 covers selection of time-delay fuses for motor- overload protection.

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1 year ago

Describe the engineering design process in your own words.

MrMuchimi

Answer:

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3 years ago

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The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

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3 years ago

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Elena L [17]

Answer:

nah

Explanation:

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Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?

Nina [5.8K]

dutile is the correct answer

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