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Lilit [14]
4 years ago
8

A proposed piping and pumping system has 20-psig static pressure, and the piping discharges to atmosphere 160 ft above the pump.

If the piping friction loss is 20 ft head, the minimum pressure rating (psi) of the piping system is most nearly:
(A) 50
(B) 100
(C) 150
(D) 250
Engineering
1 answer:
larisa86 [58]4 years ago
7 0

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

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frosja888 [35]

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

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for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

T=(\frac{Cv}{d^{2} })t

where Cv is the coefficient of consolidation

(\frac{Cvt}{d^{2} })_{1}=  (\frac{Cvt}{d^{2} })_{2}

if Cv is constant, we have:

(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2})  }=\frac{t2}{2^{2} }

Clearing t2:

t2=32 min

3 0
3 years ago
Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo
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Z = 3 + 0.23t

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Please see attachment for the equation

8 0
3 years ago
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How should the system administrator resolve the issue?
Hunter-Best [27]

Answer:

Option B

Select the Marketing User checkbox in the user record

Explanation:

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4 0
4 years ago
I will mark as brainliest !
Sliva [168]

Answer:

7.8 Mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel =  \pi d=26\piπd=26π

Speed would be

\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s

Converting to mph

1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph

134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph

The Speed of the bicycle is 7.8 mph

3 0
3 years ago
1.
garik1379 [7]

Answer:

The answer is below

Explanation:

1)

R_{AB}=(500+500)||(500+500)\\\\R_{AB}=1000||1000\\\\R_{AB}=\frac{1000*1000}{1000+1000} \\\\R_{AB}=500\ \Omega

2)

R_{AB}=1000\|(1000+1000+1000)\\\\R_{AB}=1000||3000\\\\R_{AB}=\frac{1000*3000}{1000+3000} \\\\R_{AB}=750\ \Omega

3)

Because of the short, the resistance is zero.

R_{AB}=0

4)

R_{AB}=940\ \Omega

5)

R_{AB}=2200||2200||(2200+2200)\\\\R_{AB}=2200||2200||4400\\\\\frac{1}{R_{AB}}=\frac{1}{2200}  +\frac{1}{2200}  +\frac{1}{4400}  \\\\\frac{1}{R_{AB}}=\frac{5}{4400}\\\\R_{AB}=880

6)

R_{AB}=(220+100)||470||330\\\\R_{AB}=320||470||330\\\\\frac{1}{R_{AB}}=\frac{1}{320}  +\frac{1}{470} +\frac{1}{330} \\\\R_{AB}=120.7\ \Omega

5 0
3 years ago
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