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3 years ago

Basic concepts surrounding electrical circuitry?

Engineering

1 answer:

Elanso [62]3 years ago

3 0

Hopefully that helps you out and is this for history or science?

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Please Help !!

Alla [95]

Answer:

how are we supposed to help?

4 0

3 years ago

Unitate de masura in SI pt F

Elena-2011 [213]

Answer:

Electrical Capacitance

Explanation:

To find - unit of measure in SI for F

Solution -

The answer is - Electrical Capacitance

Reason -

The farad (symbol: F) is the SI derived unit of electrical capacitance, the ability of a body to store an electrical charge.

Răspuns:

Capacitate electrică

Explicaţie:

Pentru a găsi - unitate de măsură în SI pentru F

Soluție -

Răspunsul este - Capacitate electrică

Motiv -

Farada (simbolul: F) este unitatea de capacitate electrică derivată din SI, capacitatea unui corp de a stoca o sarcină electrică.

7 0

3 years ago

What is a magnitute?

ValentinkaMS [17]

Answer:

Magnitude is a specific type of norm. Magnitude is what is known as the Euclidean norm. There are other norms, such as the Leibniz and the Chebychev norm.

Explanation:

5 0

3 years ago

Read 2 more answers

The rate at which velocity changes is called?

Travka [436]

The rate at which velocity changes is called acceleration

7 0

4 years ago

Read 2 more answers

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

8 0

3 years ago

Basic concepts surrounding electrical circuitry?​

Basic concepts surrounding electrical circuitry?