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USPshnik [31]
2 years ago
8

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resis

tance R.
For a power line that supplies power to 10 000 households, we can conclude that:

a) ????V < ????^2 R
b) ????^2 R = 0
c) ????V = ????^2 R
d) ????V > I^2 R
e) ???? = V/R
Engineering
1 answer:
Rufina [12.5K]2 years ago
6 0

Question:

In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.

For a power line that supplies power to 10 000 households, we can conclude that

a) IV < I²R

b) I²R = 0

c) IV = I²R

d) IV > I²R

e) I = V/R

Answer:

d) IV > I²R

Explanation:

In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.

The power delivered to households is given by

P = IV

The losses in the transmission line are given by

Ploss = I²R

Therefore, the relation IV > I²R  holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.

Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.

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agasfer [191]

Answer:

a gazebo

Explanation:

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3 years ago
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You have a motor such that if you give it 12 Volt, it will eventually reach a steady state speed of 200 rad/s. If it starts from
Aleksandr [31]

Answer:

a) \frac{Ws}{Es}  = \frac{200}{1+1.2s}

b) attached below

c) type zero system

d) k > \frac{g}{200}

e) The gain K increases above % error as the  steady state speed increases

Explanation:

Given data:

Motor voltage  = 12 v

steady state speed = 200 rad/s

time taken to reach 63.2% = 1.2 seconds

<u>a) The transfer function of the motor from voltage to speed</u>

let ; \frac{K1}{1+St} be the transfer function of a motor

when i/p = 12v then steady state speed ( k1 ) = 200 rad/s , St ( time constant ) = 1.2 sec

hence the transfer function of the motor from voltage to speed

= \frac{Ws}{Es}  = \frac{200}{1+1.2s}

<u>b) draw the block diagram of the system with plant controller and the feedback path </u>

attached below is the remaining part of the detailed solution

c) The system is a type-zero system because the pole at the origin is zero

d) ) k > \frac{g}{200}

7 0
3 years ago
Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the
zysi [14]

<u>Explanation:</u>

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = \frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600m^{2}

let sunlight hours be 8 hours

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=5.62×3600 = 20232 kJ/m^{2}/day

energy input in lawn = (600) (20232) (7)

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Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

                                            = 3398.97 mJ

Mass of the clippers  \(=(30)(20)(1 \cdot 096)^{2}(667)\)

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Removing water content,

dried grass clippings \(=95726.46\) pound

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Trash cans repaired  

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By burning the gas, total energy input = 3398.97 MJ × 0.2

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3 years ago
A 0.19-m3 rigid tank equipped with a pressure regulator contains steam at 2 MPa and 300°C. The steam in the tank is now heated.
AVprozaik [17]

Answer:

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Explanation:

#We know that:

The balance mass m_{in}+m_{out}=\bigtriangleup m_{system}

so, m_e=m_1-m_2

Energy \ Balance\\E_{in}-E_{out}=\bigtriangleup E_{system}\\\\\therefore Q_i_n+m_eh_e=m_2u_2-m_1u_1

#Also, given the properties of water as;

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#We assume constant properties for the steam at average temperatures:h_e=\approx(h_1+h_2)/2

#Replace known values in the equation above;h_e=(3024.2+3468.3)/2=3246.25kJ/kg\\\\m_1=V_1/v_1=0.19m^3/(0.12551m^3/kg)=1.5138kg\\\\m_2=V_2/v_2=0.19m^3/(0.17568m^3/kg)=1.0815kg

#Using the mass and energy balance relations;

m_e=m_1-m_2\\\\m_e=1.5138-1.0815\\\\m_e=0.4323kg

#We have Q_i_n+m_eh_e=m_2u_2-m_1u_1: we replace the known values in the equation as;

Q_i_n+m_eh_e=m_2u_2-m_1u_1\\\\Q_i_n=0.4323kg\times3246.2kJ/kg+1.0815kg\times3116.9-1.5138kg\times2773.2kJ/kg\\\\Q_i_n=573.21kJ

#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ

5 0
3 years ago
The proposed grading at a project site will consist of 25,100 m3 of cut and 23,300 m3 of fill and will be a balanced earthwork j
Anna [14]

Answer:

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Volume of water required = 18241 / 9.81 = 1859 m³

Volume of water required = 1859 kL

Therefore, the volume of water that will be required to bring these soils to the optimum moisture content is 1859 kL

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3 years ago
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