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Artist 52 [7]
3 years ago
12

Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe annulus of inner radius Ri a

nd outer radius Ro Ignore the effects of gravity. A constant negative pressure gradient _P/_x is applied in the x-direction, (∂P/∂x)=(P1-P2)/(x2-x1) where x1 and x2 are two arbitrary locations along the x-axis, and P1 and P2 are the pressures at those two locations. The pressure gradient may be caused by a pump and/or gravity. Note that we adopt a modified cylindricaloordinate system here with x instead of z for the axial component, namely, ( r, θ ,x) and (Ur, Uθ ,Ux) Derive an expression for the velocity field in the annular space in the pipe.
Engineering
1 answer:
svetlana [45]3 years ago
8 0

Answer:

is it multiple choice?

Explanation:

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anygoal [31]

In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.

<h3>What is a thrust angle?</h3>

A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.

This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.

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5 0
1 year ago
A brake caliper is considered a suspension item.<br> True<br> False
user100 [1]
True


Suspension is the system of tires, tire air, springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two.[1] Suspension systems must support both road holding/handling and ride quality
5 0
2 years ago
For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
2 years ago
Technician A says that mechanical shifting controls can wear out over time. Technician B says that vacuum control rubber diaphra
diamong [38]

Based on the information, both technician A and technician B are correct.

<h3>How to depict the information?</h3>

From the information given, Technician A says that mechanical shifting controls can wear out over time.

Technician B says that vacuum control rubber diaphragms can deteriorate over time.

In this case, both technicians are correct as the information depicted is true.

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8 0
1 year ago
The melting point of Pb (lead) is 327°C, is the processing at 20°C hot working or cold working?
bonufazy [111]

Answer:

Explained

Explanation:

Cold working: It is plastic deformation of material at temperature below   recrystallization temperature. whereas hot working is deforming material above the recrystallization temperature.

Given melting point temp of lead is 327° C and lead recrystallizes at about

0.3 to 0.5 times melting temperature which will be higher that 20°C. Hence we can conclude that at 20°C lead will under go cold working only.

6 0
3 years ago
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