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2 years ago

What are the differences between the practical and the ideal pendulum

1 answer:

4 0

lf a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called simple pendulum.

In practice, however, these requirements cannot be fulfilled. So we use a practical pendulum.

A practical pendulum consists of a small metallic solid sphere suspended by a fine silk thread from a rigid support. This is the practical simple pendulum which is nearest to the ideal simple pendulum.

Note :

The metallic sphere is called the bob.

When the bob is displaced slightly to one side from its mean position and released, it oscillates about its mean position in a vertical plane.

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Air enters the compressor of an ideal air-standard Braytoncycle at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s.The tur

Harrizon [31]

Answer:

Explanation:

Given that

Air Inlet Pressure, P1 = 100 KPa

Air Inlet temperature, T1 = 300 K

Volume flow rate, Q = 5 m³/s

Turbine inlet temperature, T₃ = 1400 K

Compressor pressure ratio, r = 6, 8, 12

Heat capacity ratio or air = 1.4

γ= 1.4

Specific heat constant pressure of air, cp = 1.005 KJ/kg.k

At r = 6,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 6 ^ (1.4 - 1)/1.4

1400/T4 = 6 ^ (1.4 - 1)/1.4

T₂ = 1.67 × 300

= 500 K

T₄ = 1400/1.67

= 839.07 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((839.07 - 300)/(1400 - 500))

= 0.40

= 40%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((500 - 300)/(1400 - 839.07))

= 0.36

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 500 - 839.07 + 300)

= 1.005 × 360.93

= 362.74 kJ/kg

Net heat, Qnet = 362.74 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 362.74

= 2103.9 kW.

At r = 8,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 8 ^ (1.4 - 1)/1.4

1400/T4 = 8 ^ (1.4 - 1)/1.4

T₂ = 1.81 × 300

= 543.4 K

T₄ = 1400/1.81

= 772.9 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((772.9 - 300)/(1400 - 543.4))

= 0.448

= 45%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((543.4 - 300)/(1400 - 772.9))

= 0.39

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 543.4 - 772.9 + 300)

= 1.005 × 383.7

= 385.62 kJ/kg

Net heat, Qnet = 385.62 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 385.62

= 2236.59 kW.

At r = 12,

For Brayton cycle,

T2/T1 = r ^ (γ - 1)/γ

T3/T4 = r ^ (γ - 1)/γ

Now by putting the values

T2/300 = 12 ^ (1.4 - 1)/1.4

1400/T4 = 12 ^ (1.4 - 1)/1.4

T₂ = 2.03 × 300

= 610 K

T₄ = 1400/2.03

= 688.32 K

Efficiency, η = 1 - ((T4 - T1)/(T3 - T2)

Inputting values,

= 1 - ((688.32 - 300)/(1400 - 610))

= 0.509

= 51%

Bwr = Wcomp/Wturb

Where,

Wcomp = workdone by compressor

Wturb = workdone by turbine

= ((T2 - T1)/(T3 - T4))

= ((610 - 300)/(1400 - 688.32))

= 0.44

Net work = Net heat

Net heat = Qa - Qr

Qr = Cp ( T₄-T₁)

Qa = Cp ( T₃-T₂)

Imputting values,

Net heat, Qnet = 1.005 (1400 - 610 - 688.32 + 300)

= 1.005 × 401.68

= 403.7 kJ/kg

Net heat, Qnet = 403.7 kJ/kg

Using the ideal gas equation,

P V = n R T

But n = mass/molar mass,

P = ρ R T

By putting the values

P = ρ R T

Inputting values,

100 = ρ x 0.287 x 300

ρ = 1.16 kg/m³

mass flow rate, m = ρ × Q

= 1.16 × 5

= 5.80 kg/s

Net power, Pnet = ms × Net heat, Qnet

= 5.8 × 403.7

= 2341.39 kW.

3 0

2 years ago

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Given that the wavelengths of visible light range from 400 nm to 700 nm, what is the highest frequency of visible light? (ccc =

enot [183]

7.5 × 10¹⁴ Hz is the highest frequency of visible light when wavelengths of visible light range from 400 nm to 700 nm.

The distance a wave travels in one unit of time is known as the wave speed (v).Taking into account that the wave travels one wavelength in one interval,

v=λ/T

Given that T = 1/f, we can write the equation above as,

V = f λ

Given data:

Minimum wavelength of visible light = 400 nm = 4 × 10⁻⁷ m

Speed of light = 3 × 10⁸ m/s

Frequency = c/λ = 3 × 10⁸ / 4 × 10⁻⁷

= 7.5 × 10¹⁴ Hz

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1 year ago

All 2023 ariya ac synchronous drive motors produce ____% torque at 0 mph for impressive off-the-line acceleration and smooth cru

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Torque = Fd

<u>Where:</u>

F is the force.

d is the distance.

Generally, torque is a rotational force which is developed by the crankshaft of an automobile vehicle and its capacity to move at a specific acceleration.

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2 years ago

What are the differences between the practical and the ideal pendulum​

What are the differences between the practical and the ideal pendulum