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2 years ago

10

The primary energy source for the controller in a typical control system is either brainlythe primary energy source for the cont

roller in a typical control system is either

1 answer:

inysia [295]2 years ago

5 0

Answer:

a pneumatic or electric power

Explanation:

The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."

This is because a typical control system has majorly four elements which include the following:

1. Sensor: this calculates the controlled variable

2. Controller: this receives and process inputs from the sensor to the controlled device as output

3. Controlled device: this tweak the controlled variable

4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power

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Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai

Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

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3 years ago

A simple formula to estimate the upward velocity of a rocket (neglecting the aerodynamic drag) is:

Bingel [31]

Answer:

Test code:

>>u=10;

>>g=9.8;

>>q=100;

>>m0=100;

>>vstar=10;

>>tstar=fzero_rocket_example(u, g, q, m0, vstar)

Explanation:

See attached image

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3 years ago

What does the branch of physics include

cricket20 [7]

Answer:

The branches of physics are;

Classical physics
Modern physics
Nuclear physics
Atomic physics
Geophysics
Biophysics

Explanation:

Physics is a branch of science that studies nature, properties of matter and energy. The subjects in study are; mechanics, light, heat, light, sound, electricity, properties of atoms and magnetism.

The common branches are;

Thermodynamics which that studies heat, how it is transferred and effects
Sound that studies production, properties and application of sound waves
Light that deals with properties,pysical aspects and objects that utilize light
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4 0

2 years ago

What is the current and power in a 120V circuit if the resistance 5Ω?

podryga [215]

Answer:

24A

Explanation:

trust me

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2 years ago

Read 2 more answers

Steam at 5 MPa and 400 C enters a nozzle steadily with a velocity of 80 m/s, and it leavesat 2 MPa and 300 C. The inlet area of

Korolek [52]

Answer:

a) the mass flow rate of the steam is $\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam is $\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle is $A_2$ = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

$P_1 =$ 5 MPa

$T_1$ = 400° C

Velocity V = 80 m/s

Exit:

$P_2 =$ 2 MPa

$T_2$ = 300° C

From the properties of steam tables at $P_1 =$ 5 MPa and $T_1$ = 400° C we obtain the following properties for enthalpy h and the speed v

$h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg$

From the properties of steam tables at $P_2 =$ 2 MPa and $T_1$ = 300° C we obtain the following properties for enthalpy h and the speed v

$h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg$

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

$m_1=m_2=m_3$

Thus

$m_1 =\dfrac{V_1 \times A_1}{v_1}$

$m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}$

$m_1 =\dfrac{0.4 }{0.057838 }$

$\mathbf{m_1 =6.92 \ kg/s}$

b) the exit velocity of the steam.

Using Energy Balance equation:

$\Delta E _{system} = E_{in}-E_{out}$

In a steady flow process;

$\Delta E _{system} = 0$

$E_{in} = E_{out}$

$m(h_1 + \dfrac{V_1^2}{2})$ $= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})$

$- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})$

$- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})$

$V_2^2 = 316631.29 \ m/s$

$V_2 = \sqrt{316631.29 \ m/s$

$\mathbf{V_2 = 562.7 \ m/s}$

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

$m = \dfrac{V_2A_2}{v_2}$

making $A_2$ the subject of the formula ; we have:

$A_2 = \dfrac{ m \times v_2}{V_2}$

$A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}$

$A_2$ = 0.0015435 m²

6 0

3 years ago