The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated
1 answer:
Answer:
a)Are generally associated with factor.
Explanation:
We know that losses are two types
1.Major loss :Due to friction of pipe surface
2.Minor loss :Due to change in the direction of flow
As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and it produce losses in the energy.
Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as
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Answer:
a) 35%
b) yes it can be improved by moving the tray near the top
Tray should be located ( 1 to 2 meters below surface )
max removal efficiency ≈ 70%
c) The maximum removal will drop as the particle settling velocity = 0.5 m/h
Explanation:
Given data:
flow rate = 8000 m^3/d
Detention time = 1h
depth = 3m
Full length movable horizontal tray : 1m below surface
<u>a) Determine percent removal of particles having a settling velocity of 1m/h</u>
velocity of critical sized particle to be removed = Depth / Detention time
= 3 / 1 = 3m/h
The percent removal of particles having a settling velocity of 1m/h ≈ 35%
<u>b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency</u>
The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the
Total Maximum removal efficiency
= percent removal + percent removal
= ( d,v
) .
+ ( d
,v
) .
= 100
hence max removal efficiency ≈ 70%
<u>c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?</u>
The maximum removal will drop as the particle settling velocity = 0.5 m/h
Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2foL=1/2
foC
Solving this expression for fo yields
fo=1/2
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
