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Shtirlitz [24]
4 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]4 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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With thermodynamics, one cannot determine ________.
zloy xaker [14]

Answer:

Option B is correct.

Explanation:

With thermodynamics, one cannot determine the speed of the reaction.

Speed of the reaction is defined as the rate at which one component is changes to another component. With factor cannot be calculated from the thermodynamics branch of engineering.

With thermodynamics we can determine the value of the equilibrium constant,

the extent of a reaction, the temperature at which a reaction will be spontaneous & the direction of a spontaneous reaction. But we cannot determine  the speed of a reaction.

Therefore option B is correct.

4 0
3 years ago
A 1.7 cm thick bar of soap is floating in water, with 1.1 cm of the bar underwater. Bath oil with a density of 890.0 kg/m{eq}^3
PIT_PIT [208]

Answer:

The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm

Explanation:

The water level on the bar soap = 1.1 m mark

Therefore, the proportion of the bar soap that is under the water is given by the relation;

Volume of bar soap = LW1.7

Volume under water = LW1.1

Volume floating = LW0.6

The relative density of the bar soap = Density of bar soap/(Density of water)

= m/LW1.7/(m/LW1.1) = 1.1/1.7

Given that the oil density = 890 kg/m³

Relative density of the oil to water = Density of the oil/(Density of water)

Relative density of the oil to water = 890/1000 = 0.89

Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89

Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7

Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7

Where:

X  = The height of the oil on the side of the bar when the soap is floating in only the oil

Therefore;

X = 1.236 cm.

3 0
4 years ago
The town of Camp Verde has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of
Artist 52 [7]

For efficient treatment, the dissolved oxygen concentration in the aeration tank must be kept between 1-3 mg/L. The microbial biomass will die at low DO levels, and it will take time and money to rebuild it.

The depth, which normally ranges from 3 to 4.5 meters, determines how effectively air is aerated. The breadth, which is typically maintained between 5 and 10 meters, regulates the mixing. The ideal width-to-depth ratio is 1.2 to 2.2. The distance shouldn't be shorter than 30 meters or longer than 100 meters. Bacteria used in wastewater treatment and stabilization are given oxygen by aeration. The bacteria require oxygen for biodegradation to take place.

Learn more about efficient here-

brainly.com/question/13154811

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4 0
1 year ago
Over 30 day period, a lake surface area is 1260 acres. The inflow is 36 cfs, thee outflow is 30 cfs. Seepage loss is 1.5 in. The
Elis [28]

Answer:

  -0.1 inches

Explanation:

The net inflow is ...

  36 cfs -30 cfs = 6 cfs

The number of seconds in 30 days is ...

  (3600 s/h)(24 h/da)(30 da) = 2,592,000 . . . . seconds/(30 days)

Then the volume of inflow is ...

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The number of square feet in 1260 acres is ...

  (1260 ac)(43560 ft^/ac) = 54,885,600 ft^2

So, the increase in depth due to the inflow is ...

  (15,552,000 ft^3)/(54,885,600 ft^2) ≈ 0.283353 ft ≈ 3.4002 in

__

The net change in water level is then ...

  inflow - seepage + precipitation - evaporation

  3.4 in -1.5 in +4.0 in -6.0 in = -0.1 in

The water level change in the period is -0.1 inch.

5 0
3 years ago
It was found experimentally that a certain material does not change in volume when subjected to an elastic state of stress. Calc
lora16 [44]
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5 0
3 years ago
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