what is the transfer function of the loaded filter? express your answer in terms of the variables r , l , rl , and s .
1 answer:
Loaded, (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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Answer:
a)
b)
c)
Explanation:
a) Let's calculate the work done by the rocket until the thrust ends.
But we know the work is equal to change of kinetic energy, so:
b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.
At the maximum height the velocity is 0, so v(f) = 0.
c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.
Using the same equation of part b)
The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.
I hope it helps you!
Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.