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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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Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)
Novosadov [1.4K]

Answer:

a)  ∝  and β

   The phase compositions are :

    C_{\alpha } = 5wt% Sn - 95 wt% Pb

    C_{\beta } =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C_{\alpha } = 5wt% Sn - 95 wt% Pb

C_{\beta } =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0
3 years ago
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cricket20 [7]

Answer:

A.

The power generated by a wind farm is not constant because of irregular wind patterns.

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Lelu [443]

Explanation:

See attached file

6 0
3 years ago
Type the correct answer in the box. Spell all words correctly.
Otrada [13]

Answer:

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3 years ago
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