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ser-zykov [4K]
3 years ago
10

Consider an experiment in which the number of pumps in use at each of two seven-pump gas stations was determined. Call the two g

as stations Station A and Station B. Define rv's X, Y, and U by X
Engineering
1 answer:
MakcuM [25]3 years ago
8 0

Answer: Determine the number of pumps in each of the two six-pump gas stations.

• X = the total No. of pumps that are in use at the 2 stations

• Y = the difference between the No. of pumps in utilization at station 1 and the

NO. of pumps in use at the station 2

• U = the max number of pumps in use at the 2 stations

W (observed) = (3, 4)

X (W) = 3 + 4 = 7

Y (W) = 3 − 4 = − 1

U (W) = max (3, 4) = 4

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Air is to be compressed so that its volume is reduced to half of its original volume if the initial pressure is 200 kPa and the
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Answer:

P_{2} = 527.803\,kPa

Explanation:

The politropic relationship for a isentropic process is:

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Where \gamma is the ratio of specific heats

The final pressure is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}}\right)^{\gamma}

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(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length
EastWind [94]

This question is incomplete, the complete question is;

(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.

D₃ = _____D.

{ the tolerance is +/-3% }

Answer:

the diameter of the second pipe D₃ is 1.13D

Explanation:

Given the data in the question;

Length = 2l

pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.

Now, we know that for Laminar Flow;

V' = πD⁴ΔP / 128μL

where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃

Hence,

V'₁ = πD⁴ΔP₁₋₂ / 128μL  = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL

so

D₃ = D( ΔP₁₋₂ / ΔP₂₋₃ )^{\frac{1}{4}

we substitute

D₃ = D( 1.657 )^{\frac{1}{4}

D₃ = D( 1.134568 )

D₃ = 1.13D

Therefore, the diameter of the second pipe D₃ is 1.13D

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