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arsen [322]
3 years ago
12

Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

al. The velocity varied with the radius across the flow as V = 1/r m/s where r is in meters. Find the difference in depth of the liquid from the inside to the outside radius. The inside radius of the bend is 1 m and the outside radius is 3 m.
Engineering
1 answer:
lions [1.4K]3 years ago
3 0

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

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kogti [31]

Answer:

c. 96.676 sq. Inches and 0.671 sq. Feet

Explanation:

From the list of the given option, we are told to chose the correct area in sq. inches that correspond to sq. Feet.

If we recall from the knowledge of our conversion  table that,

1 sq feet = 144 sq inches

Then, let's confirm if the option were true.

a.  966.76 sq. Inches and 8.056 sq. Feet

Assuming

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

So, 1160.064 sq. inches is equal to 8.056 sq. Feet. Then option 1 is wrong

b. 96.676 sq. Inches and 8.056 sq. Feet

if 1 sq feet = 8.056

in sq inches, we have ( 8.056 × 144 ) sq inches

= 1160.064 sq. inches

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c. 96.676 sq. Inches and 0.671 sq. Feet

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Therefore, this is the correct answer as it proves that 96.676 sq. Inches = 0.671 sq. Feet

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Based on experimental observations, the acceleration of a particle is defined by the relationa = -( 0.1 + sin(x/b) ),where a and
yKpoI14uk [10]

Answer:

a) v = +/- 0.323 m/s

b) x = -0.080134 m

c) v = +/- 1.004 m/s

Explanation:

Given:

                             a = - (0.1 + sin(x/b))

b = 0.8

v = 1 m/s @ x = 0

Find:

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(b) the position where the velocity is maximum

(c) the maximum velocity.

Solution:

- We will compute the velocity by integrating a by dt.

                           a = v*dv / dx =  - (0.1 + sin(x/0.8))

- Separate variables:

                           v*dv = - (0.1 + sin(x/0.8)) . dx

-Integrate from v = 1 m/s @ x = 0:

                          0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5

                          0.5v^2 =  0.8cos(x/0.8) - 0.1x - 0.3

- Evaluate @ x = -1

                          0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3

                          v = sqrt (0.104516)

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                             x = -0.8*0.1002

                             x = -0.080134 m

- Hence,

                            v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134

                            v = sqrt (0.504)

                            v = +/- 1.004 m/s

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Answer:

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A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

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The power drawn by the air conditioner is:

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Evaluate the performance of the proposed heat pump for three locations Using R134a. Discuss the effect of outdoor temperature on
Phoenix [80]

Answer:Table 2.2: Differences in runstitching times (standard − ergonomic).

1.03 -.04 .26 .30 -.97 .04 -.57 1.75 .01 .42

.45 -.80 .39 .25 .18 .95 -.18 .71 .42 .43

-.48 -1.08 -.57 1.10 .27 -.45 .62 .21 -.21 .82

A paired t-test is the standard procedure for testing this null hypothesis.

We use a paired t-test because each worker was measured twice, once for Paired t-test for

each workplace, so the observations on the two workplaces are dependent. paired data

Fast workers are probably fast for both workplaces, and slow workers are

slow for both. Thus what we do is compute the difference (standard − er-

gonomic) for each worker, and test the null hypothesis that the average of

these differences is zero using a one sample t-test on the differences.

Table 2.2 gives the differences between standard and ergonomic times.

Recall the setup for a one sample t-test. Let d1, d2, . . ., dn be the n differ-

ences in the sample. We assume that these differences are independent sam-

ples from a normal distribution with mean µ and variance σ

2

, both unknown.

Our null hypothesis is that the mean µ equals prespecified value µ0 = 0

(H0 : µ = µ0 = 0), and our alternative is H1 : µ > 0 because we expect the

workers to be faster in the ergonomic workplace.

The formula for a one sample t-test is

t =

¯d − µ0

s/√

n

,

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sample size, and s is the sample standard deviation (of the differences)

s =

vuut

1

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Xn

i=1

(di − ¯d )

2 .

If our null hypothesis is correct and our assumptions are true, then the t-

statistic follows a t-distribution with n − 1 degrees of freedom.

The p-value for a test is the probability, assuming that the null hypothesis

is true, of observing a test statistic as extreme or more extreme than the one The p-value

we did observe. “Extreme” means away from the the null hypothesis towards

the alternative hypothesis. Our alternative here is that the true average is

larger than the null hypothesis value, so larger values of the test statistic are

extreme. Thus the p-value is the area under the t-curve with n − 1 degrees of

freedom from the observed t-value to the right. (If the alternative had been

µ < µ0, then the p-value is the area under the curve to the left of our test

Explanation: The curve represents the sum total of the evaluation

4 0
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