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lana66690 [7]
3 years ago
5

Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the

outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Engineering
1 answer:
Alex3 years ago
6 0

Answer:

Explanation:

The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory

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Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K
MrRa [10]

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K  ,T₂ = 500 K

Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}

Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}

Q=33.34 KW/m

3 0
4 years ago
Steel bar A of 5 mm diameter is subject to a stress of 500 MPa. Aluminum bar B of 10 mm diameter is subject to a stress of 150 M
weqwewe [10]

Answer:

aluminum bar carrying a higher load than steel bar

Explanation:

Given data;

steel abr

diameter = 5 mm

stress = 500 MPa

aluminium bar

diameter = 10 mm

stress = 150 MPa

we know

stress = laod/area

for steel bar

500 = \frac{P}{\frac{\pi}{4} 5^2}

solving for P

P = 9817.47 N

for Aluminium bar

150 = \frac{P}{\frac{\pi}{4} 10^2}

solving for P

P = 11790 N

aluminum bar carrying a higher load than steel bar

6 0
3 years ago
Which of the following influenced the hair of the 1980s?
geniusboy [140]

Answer:

OB. the styles of popular musicians.

Explanation:

6 0
3 years ago
Need help I’m giving out brainlest whoever get it me correct
ludmilkaskok [199]

Answer:

(C) Prototype Model

Explanation:

I'm sure that is the answer i am very sorry if not :)

7 0
3 years ago
At a ski resort, water at 40 °F is pumped through a 3-in.-diameter, 2001-ft-long steel pipe from a pond at an elevation of 4286
deff fn [24]

Answer:

P = 24.38 hp

Explanation:

Given data:

diameter of steel pipe = 3 inc

length of pipe = 2001 ft

elevation of pond = 4286 ft

elevation of snow making machine = 4620 ft

Rate of water = 0.26 ft^3/s

Applying bernouli equation on bioth side

\frac{P}{\rho} + \frac{v_1^2}{2g} + z_1 + h_p =\frac{P_2}{\rho} + \frac{v_2^2}{2g} + z_2 + \frac{flv^2}{2gD}.....1

where P_2 = 182 psi

P_1 = 0, v_1 = 0

V =V_2 = \frac{Q}{\frac{\pi}{4} D^2}= \frac{0.26}{\frac{\pi}{4} \times (3/12)^2} = 5.30 ft/s

calculation fro friction factor

from standard table we have

\frac{\epsilon}{D} = \frac{0.00015}{(3/12)} = 6\times 10^{-4}

Re =\frac{VD}[\nu} = \frac{5.30 \times (3/12)}{1.66 \times 10^{-5}} = 7.96 \times 10^4

so for calculated R and\frac{\epsilon}{D}, F value = 0.0212

from equation 1

h_p = \frac{P_2}{\rho} + Z_2 -Z_1 +(1 + f\frac{l}{D}) \frac{V^2}{2g}

h_p = \frac{182 \times 144/ lb /ft^2}{62.4} + 4620 - 4286 + (1 + 0.0212 \frac{2001}{(3/12)}) \frac{5.30}{2\times 32.2}

h_p = 826.76ft so that

P =  \rho Q h_p = 62.4 \times 0.26 \times 826.76 = 13413.38 ft lb/s = 24.38 hp

3 0
3 years ago
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