What is meant by the acronym ISO

Answer:

Printing press , lenses etc,

Explanation:

Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.

Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.

Answer:

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Explanation:

By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:

Principle of Mass Conservation

[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)

First Law of Thermodynamics

[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)

Second Law of Thermodynamics

[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)

By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:

[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)

[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)

Where:

[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.

[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.

[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.

[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.

[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.

[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.

[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.

[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.

[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.

By property charts for steam, we get the following information:

Inlet

[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

Outlet

[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]

If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:

[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]

[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]

[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]

[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.