Ask question

Ask question

All categories

Thepotemich [5.8K]

4 years ago

13

An engineering student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

and a voltage of 110 V, a lamp that contains a 100-W bulb, an overhead light with a 60-W bulb, and various other small devices adding up to 3.00 W.
a) Assuming the power plant that supplies 110 V electricity to the dorm is 10 km away and the two aluminum transmission cables use 0-gauge wire with a diameter of 8.252 mm, estimate the total power supplied by the power company that is lost in the transmission. (Hint: Find the power needed by the student for all devices. Use this to find the current necessary to be running through the transmission cable)
(b) What would be the result is the power company delivered the electric power at 110 kV?

1 answer:

AURORKA [14]4 years ago

3 0

Answer:

a) $P_{L}=199.075W$

b) $P_{L}=1.991x10^{-4}W$

Explanation:

1) Notation

Power on the refrigerator: $P=IV=3Ax110V=330W$

Voltage $V=110V$

$D=8.252mm$ , so then the radius would be $r=\frac{8.252}{2}=4.126mm$

$L=2x10km=20km=20000m$ , representing the length of the two wires.

$\rho=2.65x10^{-8}\Omega m$ , that represent the resistivity for the aluminum founded on a book

$P_L$ power lost in the transmission.

2) Part a

We can find the total power adding all the individual values for power:

$P_{tot}=(330+100+60+3)W=493W$

From the formula of electric power:

$P=IV$

We can solve for the current like this:

$I=\frac{P}{V}$

Since we know $P_{tot}$ and the voltage 110 V, we have:

$I=\frac{493W}{110V}=4.482A$

The next step would be find the cross sectional are for the aluminum cables with the following formula:

$A=\pi r^2 =\pi(0.004126m)^2=5.348x10^{-5}m^2$

Then with this area we can find the resistance for the material given by:

$R=\rho \frac{L}{A}=2.65x10^{-8}\Omega m\frac{20000m}{5.348x10^{-5}m^2}=9.910\Omega$

With this resistance then we can find the power dissipated with the following formula:

$P_{L}=I^2 R=(4.482A)^2 9.910\Omega=199.075W$

And if we want to find the percentage of power loss we can use this formula

$\% P_{L}=\frac{P_L}{P}x100$

3) Part b

Similar to part a we just need to change the value for V on this case to 110KV.

We can solve for the current like this:

$I=\frac{P}{V}$

Since we know $P_{tot}$ and the voltage 110 KV=110000V, we have:

$I=\frac{493W}{110000V}=4.482x10^{-3}A$

The cross sectional area is the same

The resistance for the material not changes.

With this resistance then we can find the power dissipated with the following formula:

$P_{L}=I^2 R=(4.482x10^{-3}A)^2 9.910\Omega=1.991x10^{-4}W$

You might be interested in

andrey2020 [161]

Please find attached photograph for your answer.

5 0

3 years ago

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

lukranit [14]

Explanation:

Given that,

Length of the spring, l = 50 cm = 0.5 m

Mass connected to the end, m = 330 g = 0.33 kg

The mass is released and falls, stretching the spring by 30 cm before coming to rest at its lowest point. On applying Newton's second law, 10 cm below the release point, x = 15 cm

(a) When the mass is connected, the force of gravity is balanced by the force in spring.

$kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.15}\\\\k=21.56\ N/m$

(b) The amplitude of the oscillation will be 15 cm as it is half of the total distance travelled.

(c) The frequency of the oscillation is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{21.56}{0.33}} \\\\f=1.28\ Hz$

Hence, this is the required solution.

7 0

4 years ago

Gravity causes___ objects to accelerate

kolbaska11 [484]

Gravity causes falling objects to accelerate.

4 0

4 years ago

Read 2 more answers

A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 475°C. The aluminum plate has

Anuta_ua [19.1K]

Answer:

$T_0 = 338.916 Degree\ celcius$

Explanation:

Given data:

Thickness of aluminium sheet 10 cm

initial temperature = 25 degree celcius

Assumption

Thermal properties remain constant, transfer of heat by radiation is negligible.

from the information given in the question we have

T_S ≈T_∞ , it implies we have h → ∞

from table 4.2 Biot number → ∞ the value of

$\lambda_1 = 1.5708 and A_1= 1.2732$

The fourier number is

$t = \frac{\alpha t}{l^2} = \frac{97.1\times 10^{-6} \times 15}{0.05^2} = 0.5826$

Temperature at center after 15 second of heating

$\theta _{0 wall} = \frac{T_0 - T_{\infity}}{T_i -T_{\infity}} = A_i e^{\lambda_1^2 t}$

$T_0 = T_i -T_{\infity} \times A_i e^{\lambda_1^2 t}$

$T_0 = (25 - 475) 1.2732 e^{-1.5708^2 \times 0.5826} + 475 = 356 degree celcius$

$T_0 = 338.916 Degree\ celcius$

8 0

3 years ago

Help with these please someone

Mice21 [21]

Answer:

Five: B

Ten: A

Explanation:

Five

An alpha decay looks like this.

$A^x_y====> B^{x-4}_{y - 2}+alpha^4_2$

So whatever is produced must have a mass of 4 less than 234 and a number on the periodic table of 2 less than 92. In other words, B has a mass of 230 and a number on the periodic table of 90.

The answer should be Thorium which is B.

Ten

There is actually not enough information to do 10. You get it by making an assumption and seeing if it works.

In general a beta decay can (the most common one ) look like this.

$A^x_y====> B^{x}_{y-1}+Beta^0_{-1}$

Is there anything that looks like that?

The weight stays the same (234) and the atomic number of the mother element (on the left) is 1 lower than the given chemical on the right.

$Th^{234}_{90}====> Pa^{234}_{91}+Beta^0_{-1}$

The answer is Thorium A

Beta decays are very tricky. Be very careful how you handle them. One of three items can be what is decayed. I have assumed it was an electron, but there are two other possibilities. I won't confuse you by adding them. Just be aware that they exist.

3 0

3 years ago