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Shtirlitz [24]
3 years ago
5

The Reynolds number, rhoVD/μ is a very important parameter in fluid mechanics. Determine its value for ethyl alcohol flowing at

a velocity of 4 m/s through a 2-in.-diameter pipe.
Engineering
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

The Reynold Number is 146415.34

Explanation:

The density of ethyl alcohol = ρ = 789 kg/m^3

The dynamic viscosity of ethyl alcohol = μ = 0.001095 kg/m.s

Velocity = V = 4 m/s

Diameter of pipe = D = 2 in = 0.0508 m

The Reynold's number (Re), will be:

Re = ρVD/μ

Re = (789 kg/m^3)(4 m/s)(0.0508 m)/(0.001095 kg/ms)

<u>Re = 146415.34</u>

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Answer:

l=24mm

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From the question we are told that:

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3 years ago
A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27
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Answer:

Q = 62    ( since we are instructed not to include the units in the answer)

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Q = ???

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mass of He = 3 kg  × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

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Now; the amount of  Heat  Q transferred = m_{He}Cp_{He} \delta T  + m_{Ar}Cp_{Ar} \delta T

From gas table

Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar}  = 0.5203 \  kJ/Kg/K

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Q = 62.389 *10^6

Q = 62 MJ

Q = 62    ( since we are instructed not to include the units in the answer)

5 0
4 years ago
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Answer:

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