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d1i1m1o1n [39]
1 year ago
12

a) find the state-space representation of the system. b) is the system controllable? why? c) is the system observable? why

Engineering
1 answer:
mart [117]1 year ago
5 0

If a controlled input can transfer (alter) the control system's initial states to some other desired states in a finite amount of time, the control system is said to be controllable.

Using Kalman's test, we can determine whether a control system is controllable. The evolution model for the state variables (time-varying unknowns) and the observation model, which connects the observations to the state variables, make up the state space representation of a dynamical system. The capacity to move a system about in its full configuration space using just specific permitted actions is generally referred to as controllability. The precise definition changes slightly depending on the model type or framework used.

Learn more about control here-

brainly.com/question/28540307

#SPJ4

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A pressure gage at the inlet to a gas compressor indicates that the gage pressure is 40.0 kPa. Atmospheric pressure is 1.01 bar.
bonufazy [111]

Answer:

Given

inlet Pga =40kpa = 40000pa

Patm=1.01bar = 1.01 x 100000pa =101000pa

exit Pab= 6.5 (inlet Pab)

But generally, Pab = Patm + Pga

1. the absolute pressure of the gas at the inlet, inlet Pab?

inlet Pab = Patm + inlet Pga

            = 101000pa + 40000pa = 141kpa

the absolute pressure of the gas at the inlet, inlet Pab = 141kpa

2. the gage pressure of the gas at the exit? exit Pga?

exit Pab = Patm + exit Pga

exit Pga = exit Pab - Patm

             = (6.5 x 141kpa) - 101kpa

              = 815.5kpa

the gage pressure of the gas at the exit exit Pga=815.5kpa

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We can process oil into a lot of useful fuels to run our cars, trucks, and even airplanes. Oil is used for making lots of other
Ostrovityanka [42]

Answer:

Explanation:

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A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

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