Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
Answer:
c. V2 equals V1
Explanation:
We can answer this question by using the continuity equation, which states that:
(1)
where
A1 is the cross-sectional area in the first section of the pipe
A2 is the cross-sectional area in the second section of the pipe
v1 is the velocity of the fluid in the first section of the pipe
v2 is the velocity of the fluid in the second section of the pipe
In this problem, we are told that the pipe has a uniform cross sectional area, so:
A1 = A2
As a consequence, according to eq.(1), this means that
v1 = v2
so, the velocity of the fluid in the pipe does not change.
Answer:
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Explanation:
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Answer:
We choose PTFE
Explanation:
Attached are the modulus density and modulus strength chart.
Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.