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1 year ago

what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .

1 answer:

3 0

Loaded, $H_{Loaded}$ (s) = $\frac{RR_{L} }{R+R_{L} } /(\frac{RR_{L} }{R+R_{L} }+SL)$ = $\frac{RR_{L}/L }{R+R_{L} } /(\frac{RR_{L} /L}{R+R_{L} }+S)$ is the loaded filter's transfer function.

A graded filter that, by virtue of its weight and permeability, stabilises the foot of an earth dam or other construction when it is installed at the base of that structure.

Air filters with depth loaded are made to achieve precisely that. They add particles gradually to create air passageways, reducing constriction. You may save time and money by using filters that last longer thanks to them. The bigger particles are caught at the filter's beginning, while the smaller particles are caught as it gets closer. This is intended to avoid rapid surface loading, hence facilitating more airflow. This enables longer-lasting filtration as well.

On the other hand, surface loading filters catch every particle that is on its surface. No matter how big or little the particles are, it doesn't care.

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Procurement and usage of safe tools.

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3 years ago

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Answer:

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Explanation:

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2 years ago

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Answer:

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2 years ago

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Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i

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Answer:

Gc(s) = $\frac{0.1s + 0.28727}{s}$

Explanation:

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attached is the detailed solution and the plot in Matlab

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3 years ago

An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va

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Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$