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labwork [276]
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
Engineering
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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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.

5 0
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
  • Electricity and magnetism that studies charges, their effects and relation with magnetism
  • Classical physics that studies laws of motion and gravity
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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

3 0
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
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