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1 year ago

As resistors are added in series to a circuit, the total resistance will

Engineering

1 answer:

olganol [36]1 year ago

8 0

The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

<h3>
What is resistance?</h3>

Resistance is the obstruction offered whenever the current is flowing through the circuit.

So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.

V eq = V₁ + V₂ + V₃

IReq = IR₁ + IR₂ + IR₃

Req = R₁ + R₂ + R₃

Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

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Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr

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Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

$P _1= 15 psia$

$P _1= 100 psia$

We know that work for isothermal process

$W=mRT\ln \dfrac{P_1}{P_2}$

Lets take mass is 1 kg.

So work per unit mass

$W=RT\ln \dfrac{P_1}{P_2}$

We know that for air R=0.287KJ/kg.K

$W=RT\ln \dfrac{P_1}{P_2}$

$W=0.287\times 310.92\ln \dfrac{15}{100}$

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

$\Delta S=-R\ln \dfrac{P_2}{P_1}$

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3 years ago

The time factor for a doubly drained clay layer

Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

$T_v=\frac{\pi }{4}(\frac{U}{100})^2$

Solving for 'U' we get

$\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%$

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i) $\frac{z}{H}=0.25=U=0.71$ = 71% consolidation

ii) $\frac{z}{H}=0.5=U=0.45$ = 45% consolidation

iii) $\frac{z}{H}=0.75U=0.3$ = 30% consolidation

Part b)

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm$

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

$T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59$

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

$\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm$

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2 years ago

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Answer:

The convective coefficient is 37.3 W/m²K.

Explanation:

Use Newton’s law of cooling to determine the heat transfer coefficient. Assume there is no heat transfer from the ends of electric resistor. Heat is transferred from the resistor curved surface.

Step1

Given:

Diameter of the resistor is 2 cm.

Length of the resistor is 16 cm.

Current is 5 amp.

Voltage is 6 volts.

Resistor temperature is 100°C.

Room air temperature is 20°C.

Step2

Electric power from the resistor is transferred to heat and this heat is transferred to the environment by means of convection.

Power of resistor is calculated as follows:

P=VI

$P=6\times5$

P= 30 watts.

Step3

Newton’s law of cooling is expressed as follows:

$Q=h\times \pi DL(T_{r}-T_{\infty})$

Here, h is the convection heat coefficient and $\pi DL$ is the exposed surface area of the resistor.

Substitute the values as follows:

$30=h\times \pi (2cm)(\frac{1m}{100cm})(16cm)(\frac{1m}{100cm})(100-20)$

$h=\frac{30}{0.8042}$

h = 37.3 W/m²K.

Thus, the convective coefficient is 37.3 W/m²K.

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