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grin007 [14]
3 years ago
7

Fluid Dynamics: How do I find gauge pressure of an air current at various points around a cylinder?A freestream air current of v

elocity 15m/s flows through a wind tunnel. The current hits a cylinder of diameter 19mm, and has a laminar boundary layer. The stagnation pressure in the test section of the wind tunnel is 1 atm. Determine the minimum and maximum gauge pressures of the flow around the cylinder.I'm given the equation Cp=2(P-Po)/(?V^2), where Cp is the pressure coefficient, Po is the upstream static pressure. My problem is largely confusion over what to do with the diameter of the cylinder and where it comes into play here.
Engineering
1 answer:
DaniilM [7]3 years ago
7 0

Answer:

The answer is as given in the explanation.

Explanation:

The 1st thing to notice is the assumptions required. Thus as the diameter of the cylinder and the wind tunnel are given such that the difference is of the orders of the magnitude thus the assumptions as given below are validated.

  1. Flow is entirely laminar, there's no boundary layer release.
  2. Flow is streamlined, ie, it follows the geometrical path imposed by the curvature.

By D'alembert's paradox, "The net pressure drag exerted on a circular cylinder that moves in an inviscid fluid of large extent is identically zero".Just in the surface of the cylinder, the velocity profile can be given in the next equation:

V=2Usin\theta

And the pressure P on the surface of cylinder is given by Bernoulli's equation along the streamline through that point:

P=P_{_{\infty }}+\frac{1}{2}\rho U^{2}(1-4sin^{2}\Theta ))

where P_∞ is  Pressure at stagnation point, U is the velocity given, ρ is the density of the fluid (in this case air) and θ is the angle measured from the center of cylinder to the adjacent point where your pressure point will be determine.

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I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal
tigry1 [53]

Answer:

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>

<em></em>

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

where, Nₐ₀-Nₐ = is the amount in moles of A used up

            Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if  3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

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A 132mm diameter solid circular section​
Ganezh [65]

Answer:

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5 0
3 years ago
For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
horrorfan [7]

Answer:

attached below

Explanation:

a) G(s) = 1 / s( s+2)(s + 4 )

Bode asymptotic magnitude and asymptotic phase plots

attached below

b) G(s) = (s+5)/(s+2)(s+4)

phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4

attached below

c) G(s)= (s+3)(s+5)/s(s+2)(s+4)

solution attached below

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Answer:

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1.) You can prevent making a fire by making sure you close and seal flammable materials.

2.) You can prevent heat burns by teaching the students how to properly use tongs,water baths, and other cooling equipment. 

3.) You can prevent chemical burns by treating the chemicals with caution, measure carefully, and use the approved containers.

4.) You can prevent cuts and scrapes by telling the students how to use the blades safely, and also when they are disposing broken or sharp items they should know how to wrap them up so no one else will get hurt. 

5.) You can prevent contamination by washing your hands, protect their clothing and skin with a lab coat or a lab apron, gloves and glasses, and cleaning your area where the germs of the chemicals were so no one will become.

6.) You can prevent inhalation by opening up windows, using ventilation fans, and using an equipment that measures the amount of gas emission in a room.

7.) Finally, you can prevent spills and breaks by telling the students what will happen if anything spills, and tell them to clean up.  

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