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2 years ago

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We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the

Venturi effect, what will happen to the velocity of the fluid flowing through this section? What will happen to the pressure?

2 answers:

Ksju [112]2 years ago

6 0

Can I get a picture? Of the question so I can understand it more?

NemiM [27]2 years ago

5 0

Can we get a Picture?

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Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:

Xelga [282]

Answer:

(a) Current density at P is $J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$ .

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

$J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\$

where $\textbf{a}_{\rho}$ and $\textbf{a}_{\phi}$ unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation. Therefore

$J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\$

(b) Total current flowing outward can be calculated by using the relation,

$I=\int {\textbf{J} \, \textbf{ds}$

where integral is calculated through the circular band given in the question. We can write the integral as below,

$I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\$

due to unit vector multiplication. Then,

$I=10\int\(\rho^3z.dz.d\phi$

where $\rho=3,\ 0$ . Therefore

$I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A$

4 0

3 years ago

The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,

sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity $= \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}$

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

$T_i$ = 25 degree C = 298 k

$T_o$ = 60 degree C = 333 k

$T_{\infty}$ = 75 degree C = 348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

$Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226$

$\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}$

$\tau = 0.036 t$

$\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}$

at Bi = 0.226

Ai = 0.982

$\lambda = 0.876$

$\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}$

$0.3 = 0.982 e^{-0.2t}$

$0.305 = e^{-0.2t}$

-1.187 = - 0.02t

t = 59.37 s

7 0

3 years ago

An article that discusses pharaohs and gives examples and explains how they look. Which text structure is that?

fgiga [73]

The correct answer to this open question is the following.

The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description.

The text structure called description allows the reader to fully know the characteristics of the people it is referring to, including some important details. That is why the author of a description text adds words like "such as" and "for example."

When describing something, the write is giving structure to the text and sequence. What comes first., what is followed, and so on.

That is why The text structure of an article that discusses pharaohs and gives examples and explains how they look is a description. It includes cause and effect sentences, and some comparisons in order to contrast an idea.

3 0

3 years ago

A tank contains initially 2500 liters of 50% solution. Water enters the tank at the rate of 25 iters per minute and the solution

sergejj [24]

Answer:

Percentage of solution=32. 96%

Explanation:

Given that initially tank have 50% solution.It means that amount of solution=0.5 x 2500 =1250 lts

Lets take the amount of solution at time t = A

So

$\dfrac{dA}{dt}=rate\ in\ solution -rate\ out\ solution$

$\dfrac{dA}{dt}=concentration\times rate\ in\ of\ water -concentration\times rate\ out\ of\ water$

$\dfrac{dA}{dt}=0\times 25-\dfrac{A}{2500}\times 50$

$\dfrac{dA}{dt}=-\dfrac{A}{50}$

Now by integration

$\ln A=-\dfrac{1}{50}t+C$

Where C is the constant

Given that at t=0 ,A=1250

So $C=\ln 1250$

$\ln A=-\dfrac{1}{50}t+\ln 1250$

When t=20 min

$\ln A=-\dfrac{1}{50}\times 20+\ln 1250$

A=837.90

So percentage of solution after 20 min

$=\dfrac{1250-837.9}{1250}\times 100$

Percentage of solution=32. 96%

4 0

3 years ago

David wants to obtain a job as a quality engineer at a manufacturing facility. Which of the following paths should he follow to

Elden [556K]

Answer:

David should obtain a bachalor's degree of training

Explanation:

It says which of the following paths should he take to obtain a higher-level metrology position.

8 0

3 years ago