A three-phase power of 460 MW is to the transmitted to a substation located 500 km from the source of power. With VS = 1 per unit, VR = 0.9 per unit, λ = 5000 km, ZC = 500 Ω, and δ = 36.87°, determine a nominal voltage level for the lossless transmission line.

Answers

Answer 1

The nominal voltage level for the lossless transmission line is approximately 2.585 kV.

To determine the nominal voltage level for the lossless transmission line, we can use the voltage and power equations for a transmission line.

The power equation for a transmission line is given by:

P = √3 * VL * VR * |Y| * cos(δ)

Where P is the power (460 MW), VL is the line voltage, VR is the receiving-end voltage (0.9 per unit), |Y| is the magnitude of the admittance, and δ is the phase angle.

Given that VS (sending-end voltage) is 1 per unit and |Y| = 1/ZC (where ZC is the characteristic impedance), we can rewrite the power equation as:

P = √3 * VL * VR * (1/ZC) * cos(δ)

We can rearrange this equation to solve for VL:

VL = (P * ZC) / (√3 * VR * cos(δ))

Substituting the given values into the equation:

VL = (460 MW * 500 Ω) / (√3 * 0.9 * cos(36.87°))

Simplifying the equation:

VL ≈ 2.585 kV

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pipelining increases the number of machine cycles completed per second

is  true

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A Quartz piezo-electric crystal having a thickness of 2 mm and voltage sensitivity of 0.055 V-m/N is subjected to a pressure of 1.5 MN/2. Calculate the voltage output. If the permittivity of quartz is 40.6 X 10−12 F/m, calculate its charge sensitivity.

Answers

The charge sensitivity is approximately 3.3495 × 10⁻¹² C.

How to calculate the value

First, let's convert the pressure from mega-newtons to newtons:

1 MN = 1,000,000 N.

P = 1.5 MN/2 = (1.5 * 1,000,000 N) / 2 = 750,000 N.

Now we can calculate the voltage output:

V = (0.002 m) * (0.055 V-m/N) * (750,000 N).

V = 0.0825 V.

Therefore, the voltage output is 0.0825 V.

To calculate the charge sensitivity, we can use the equation:

Q = C * V,

where:

Q is the charge sensitivity,

C is the permittivity of quartz (40.6 × 10^−12 F/m), and

V is the voltage output (0.0825 V).

Let's substitute the values into the equation:

Q = (40.6 × 10⁻¹² F/m) * (0.0825 V).

Q = 3.3495 × 10⁻¹² C.

Therefore, the charge sensitivity is approximately 3.3495 × 10⁻¹² C.

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