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Mama L [17]
3 years ago
5

) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is inst

alled in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Engineering
1 answer:
Digiron [165]3 years ago
8 0

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

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To run. The machine one at a time

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Given an unsorted array of distinct positive integers A[1..n] in the range between 1 and 10000 and an integer i in the same rang
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Explanation:

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for j =0 to 10000 do

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7 0
3 years ago
End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf
sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

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solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

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so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

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∑ F(y) = m a(y)    .............. 4

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a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

3 0
3 years ago
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