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2 years ago

The controller determines if a(n) _________ exists by calculating the difference between the SP and the PV.

Engineering

1 answer:

shusha [124]2 years ago

7 0

The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

<h3>How does a controller work in control system?</h3>

The Control system is one where it entails if the output is one that has an effect on the input quantity.

So it uses the PV(Process Variable) set against the SP(Setpoint) to know if an error exists.

So, The controller determines if a(n) error exists by calculating the difference between the SP and the PV.

Learn more about controller from

brainly.com/question/14617664

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A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. Obtain the yield strength and ultimate str

Charra [1.4K]

Answer:

yield strength before cold work = 370 MPa

yield strength after cold work = 437.87 MPa

ultimate strength before cold work = 440 MPa

ultimate strength after cold work = 550 MPa

Explanation:

given data

AISI 1018 steel

cold work factor W = 20% = 0.20

to find out

yield strength and ultimate strength before and after the cold-work operation

solution

we know the properties of AISI 1018 steel is

yield strength σy = 370 MPa

ultimate tensile strength σu = 440 MPa

strength coefficient K = 600 MPa

strain hardness n = 0.21

so true strain is here ∈ = $ln\frac{1}{1-0.2}$ = 0.223

yield strength after cold is

yield strength = $K \varepsilon ^n$

yield strength = $600*0.223^{0.21)$

yield strength after cold work = 437.87 MPa

and

ultimate strength after cold work is

ultimate strength = $\frac{\sigma u}{1-W}$

ultimate strength = $\frac{440}{1-0.2}$

ultimate strength after cold work = 550 MPa

8 0

3 years ago

What must engineers keep in mind so that their solutions will be appropriate?

vekshin1

Answer:

Context

Explanation:

It is of great value for an engineer to keep the context of his/her experiment in mind.

7 0

3 years ago

The mass flow rate in a 4.0-m wide, 2.0-m deep channel is 4000 kg/s of water. If the velocity distribution in the channel is lin

IceJOKER [234]

Answer:

V = 0.5 m/s

Explanation:

given data:

width of channel = 4 m

depth of channel = 2 m

mass flow rate = 4000 kg/s = 4 m3/s

we know that mass flow rate is given as

$\dot{m}=\rho AV$

Putting all the value to get the velocity of the flow

$\frac{\dot{m}}{\rho A} = V$

$V = \frac{4000}{1000*4*2}$

V = 0.5 m/s

4 0

3 years ago

Multiple Choice

Ymorist [56]

Answer:

Sealing agent

Explanation:

Generally, when we have water leaks in almost any building or equipment, we use a sealant. However, this sealant could be of different types depending on the peculiarity of the leakage.

Thus, the correct answer is sealing agent.

5 0

2 years ago

Read 2 more answers

A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i

Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter $d_{o}$ = 12.8 mm

Final diameter $d_{f}$ = 10.7

Engineering stress $\alpha _{E}$ = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4( $d_{o}$ )² ; Ag = π/4( $d_{f}$ )²

% = π/4 [ ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( π/4 ( $d_{o}$ )²) ]

= ( ( $d_{f}$ )² - ( $d_{o}$ )²) / ( $d_{o}$ )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress $\alpha _{T}$ = $\alpha _{E}$ ( 1 + $E_{E}$ )

$E_{E}$ is engineering strain

$E_{E}$ = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49

$E_{E}$ = 0.431

so we substitute the value of $E_{E}$ into our initial equation;

True stress $\alpha _{T}$ = 460 ( 1 + 0.431)

True stress $\alpha _{T}$ = 460 (1.431)

True stress $\alpha _{T}$ = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0

2 years ago