There is a picture linked for the question I have if anyone can help me with my whole quiz I’ll try to give 60 points

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement b, no actuator movement c.

Volgvan

If a pilot-operated check valve (POC) does not check flow, you will see a. erratic actuator movement.

<h3>What is a pilot-operated check valve (POC)?</h3>

Pilot operated test valves paintings through permitting loose float from the inlet port via the opening port. Supplying a pilot strain to the pilot port permits float withinside the contrary direction. Air strain on pinnacle of the poppet meeting opens the seal permitting air to float freely.

An actuator fault is a form of failure affecting the machine inputs. Due to strange operation or fabric aging, actuator faults might also additionally arise withinside the machine. An actuator may be represented through additive and/or multiplicative fault.

Read more about the pilot-operated check valve:

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7 0

2 years ago

(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific

lina2011 [118]

Answer:

The answer to the question is 514.17 lbf

Explanation:

Volume of cylindrical tank = πr²h = 3.92699 ft³

Weight of tank = 125 lb

Specific weight of content = 66.4 lb/ft³

Mass of content = 66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

5 0

4 years ago

For goods-producing firms, at which of the following levels of resource planning does scheduling for individual subassemblies an

VashaNatasha [74]

Answer:

Disaggregation

Explanation:

In a company it is a way to create operational plans that are focused, either by time or by section.

3 0

3 years ago

) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, de

Delicious77 [7]

Answer:

$NH_4^+ = 2.5 mg/lt$

Explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

$NH_4 ^{+} ⇄ H^{+} + NH_3$

Rate constant is given as

$K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}}$ ...........1

$K_a = 5.6 \times 10^{-10}$

Total ammonia as NItrogen is given as 30 mg/l

$\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}$

= $\frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}$

$= \frac{100}{1+ \frac{NH_3^+}{NH_4^+}}$ .....2

from equation 1 we have

$\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}$ {10^{8.5}}

plug this value in equation 2 we get

$\%NH_4^{+} = 84.96 \%$

Total ammonia as N = 30 mg/lt

$NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt$

7 0

3 years ago

I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have

drek231 [11]

Answer:

(a) The mean time to fail is 9491.22 hours

The standard deviation time to fail is 9491.22 hours

(b) 0.5905

(c) 3.915 × 10⁻¹²

(d) 2.63 × 10⁻⁵

Explanation:

(a) We put time to fail = t

∴ For an exponential distribution, we have f(t) = $\lambda e^{-\lambda t}$

Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);

$P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1$

e^(1000·λ) - 0.1·e^(1000·λ) = 1

0.9·e^(1000·λ) = 1

1000·λ = ㏑(1/0.9)

λ = 1.054 × 10⁻⁴

Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours

The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours

b) Here we have to integrate from 5000 to ∞ as follows;

$p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905$

(c) The Poisson distribution is presented as follows;

$P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}$

p(x = 3) = 3.915 × 10⁻¹²

d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours

The Cumulative Distribution Function is given as follows;

p( t ≤ 1/4) $CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}$ .

4 0

3 years ago