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Juliette [100K]
1 year ago
15

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

Engineering
1 answer:
NISA [10]1 year ago
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.

Learn more about Loaded here:

brainly.com/question/20039214

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T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

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Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

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Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

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