Answer:
International Organization For Standardization
Explanation:
This is the body within Geneva, Switzerland that promotes worldwide industrial and commercial standards
identify the unit of the electrical parameters represented by L and C and prove that the resonant frequency (fr)=1÷2π√LC
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.
Steam enters a turbine with a pressure of 30 bar, a temperature of 400 oC, and a velocity of 160 m/s. Saturated vapor at 100 oC exits with a velocity of 100 m/s. At steady state, the turbine develops work equal to 540 kJ per kg of steam flowing through the turbine. Heat transfer between the turbine and its surroundings occurs at an average outer surface temperature of 350 K. Determine the rate at which entropy is produced within the turbine per kg of steam flo
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.
Outline how the technological innovations of the Renaissance led to the Industrial Age.
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.