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2 years ago

In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?

Engineering

1 answer:

taurus [48]2 years ago

3 0

Answer:

No, the velocity profile does not change in the flow direction.

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.

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Certain pieces made by an acoustic lathe are subject to three kinds of defects X,Y,Z. A sample of 100 pieces was inspected with

Sonja [21]

Explanation:

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2 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of Pressure is $M^{1}L^{-1}T^{-2}$
Dimensional formula of length is L
Dimensional formula of mass is M
Dimensional formula of velocity is $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

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6 0

2 years ago

An air-conditioning system operating on the reversed Carnot cycle is required to transfer heat from a house at a rate of 755 kJ/

Lyrx [107]

Answer:

There is 0.466 KW required to operate this air-conditioning system

Explanation:

Step 1: Data given

Heat transfer rate of the house = Ql = 755 kJ/min

House temperature = Th = 24°C = 24 +273 = 297 Kelvin

Outdoor temperature = To = 35 °C = 35 + 273 = 308 Kelvin

Step 2: Calculate the coefficient of performance o reversed carnot air-conditioner working between the specified temperature limits.

COPr,c = 1 / ((To/Th) - 1)

COPr,c = 1 /(( 308/297) - 1)

COPr,c = 1/ 0.037

COPr,c = 27

Step 3: The power input cna be given as followed:

Wnet,in = Ql / COPr,max

Wnet, in = 755 / 27

Wnet,in = 27.963 kJ/min

Win = 27.963 * 1 KW/60kJ/min = 0.466 KW

There is 0.466 KW required to operate this air-conditioning system

3 0

2 years ago

In an orthogonal cutting operation, the tool has a rake angle = 12°. The chip thickness before the cut = 0.32 mm and the cut yie

Snezhnost [94]

Answer:

The shear plane angle and shear strain are 28.21° and 2.155 respectively.

Explanation:

(a)

Orthogonal cutting is the cutting process in which cutting direction or cutting velocity is perpendicular to the cutting edge of the part surface.

Given:

Rake angle is 12°.

Chip thickness before cut is 0.32 mm.

Chip thickness is 0.65 mm.

Calculation:

Step1

Chip reduction ratio is calculated as follows:

$r=\frac{t}{t_{c}}$

$r=\frac{0.32}{0.65}$

r = 0.4923

Step2

Shear angle is calculated as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

Here, $\phi$ is shear plane angle, r is chip reduction ratio and $\alpha$ is rake angle.

Substitute all the values in the above equation as follows:

$tan\phi=\frac{rcos\alpha}{1-rsin\alpha}$

$tan\phi=\frac{0.4923cos12^{\circ}}{1-0.4923sin12^{\circ}}$

$tan\phi=\frac{0.48155}{0.8976}$

$\phi=28.21^{\circ}$

Thus, the shear plane angle is 28.21°.

(b)

Step3

Shears train is calculated as follows:

$\gamma=cot\phi+tan(\phi-\alpha)$

$\gamma=cot28.21^{\circ}+tan(28.21^{\circ}-12^{\circ})$ $\gamma = 2.155$ .

Thus, the shear strain rate is 2.155.

6 0

2 years ago

Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press

Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

Given data:

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0

2 years ago