If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

 

 

006.  `query 5

 

 

Question:    query  introset  change in pressure from velocity change.

 

Explain how to get the change in fluid pressure given the initial and final fluid velocities, assuming constant altitude

 

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Bernoulli's Equation can be written

If altitude is constant then y1 = y2, so that rho g y1 is the same as rho g y2.  Subtracting this quantity from both sides, these terms disappear, leaving us

The difference in pressure is P2 - P1.  If we subtract P1 from both sides, as well as 1/2 rho v2^2, we get

Thus

Caution:  (v1^2 - v2^2) is not the same as (v1 - v2)^2.  Convince yourself of that by just picking two unequal and nonzero numbers for v1 and v2, and evaluating both sides.

ALTERNATIVE FORMULATION

Assuming constant rho, Bernoulli's Equation can be written

1/2 rho `d(v^2) + rho g `dy + `dP = 0.

If altitude is constant, then `dy = 0 so that

1/2 rho `d(v^2) + `dP = 0

so that

`dP = - 1/2 rho `d(v^2).

Caution:  `d(v^2) means change in v^2, not the square of the change in v.  So `d(v^2) = v2^2 - v1^2, not (v2 - v1)^2.

STUDENT SOLUTION: The equation for this situation is Bernoulli's Equation, which as you note is a modified KE+PE equation. Considering ideal conditions with no losses

 

 (rho*gy)+(0.5*rho*v^2)+(P) = 0

 

g= acceleration due to gravity

y=altitude

rho=density of fluid

v=velocity

P= pressure

 

Constant altitude causes the first term to go to 0 and disappear.

(0.5*rho*v^2)+(P) = constant

 

So here is where we are:

 

Since the altitude h is constant, the two quantities .5 rho v^2 and P are the only things that can change. The sum 1/2 `rho v^2 + P must remain constant. Since fluid velocity v changes, it therefore follows that P must change by a quantity equal and opposite to the change in 1/2 `rho v^2.

 

MORE FORMAL SOLUTION: 

 

More formally we could write

 

·          1/2 `rho v1^2 + P1 = 1/2 `rho v2^2 + P2

 

and rearrange to see that the change in pressure, P2 - P1, must be equal to the change 1/2 `rho v2^2 - 1/2 `rho v1^2 in .5 rho v^2:

 

·          P2 - P1 = 1/2 `rho v2^2 - 1/2 `rho v1^2 = 1/2 rho (v2^2 - v1^2). **

 

 

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Question:    query  billiard experiment

 

Do you think that on the average there is a significant difference between the total KE in the x direction and that in the y direction?  Support your answer.

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** In almost every case the average of 30 KE readings in the x and in the y direction differs between the two directions by less than 10% of either KE.  This difference is not statistically significant, so we conclude that the total KE is statistically the same in bot directions. **

 

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Question:    What do you think are the average velocities of the 'red' and the 'blue' particles and what do you think it is about the 'blue' particle that makes is so?

 

 

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** Student answer with good analogy:  I did not actually measure the velocities. the red were much faster. I would assume that the  blue particle has much more mass a high velocity impact from the other particles made very little change in the blue particles velocity. Similar to a bycycle running into a Mack Truck.

 

INSTRUCTOR NOTE:  :  It turns out that average kinetic energies of red and blue particles are equal, but the greater mass of the blue particle implies that it needs less v to get the same KE (which is .5 mv^2) **

 

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Question:    What do you think is the most likely velocity of the 'red' particle?

 

 

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** If you watch the velocity display you will see that the red particles seem to average somewhere around 4 or 5 **

 

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Question:    If the simulation had 100 particles, how long do you think you would have to watch the simulation before a screen with all the particles on the left-hand side of the screen would occur?

 

 

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** STUDENT ANSWER:  Considering the random motion at various angles of impact.It would likely be a very rare event.

 

INSTRUCTOR COMMENT

 

This question requires a little fundamental probability but isn't too difficult to understand:

 

If particle position is regarded as random the probability of a particle being on one given side of the screen is 1/2.  The probability of 2 particles both being on a given side is 1/2 * 1/2.  For 3 particles the probability is 1/2 * 1/2 * 1/2 = 1/8.  For 100 particlles the probability is 1 / 2^100, meaning that you would expect to see this phenomenon once in 2^100 screens.  If you saw 10 screens per second this would take about 4 * 10^21 years, or just about a trillion times the age of the Earth.

 

In practical terms, then, you just wouldn't expect to see it, ever. **

 

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Question:    prin phy and gen phy problem 10.36  15 cm radius duct replentishes air in 9.2 m x 5.0 m x 4.5 m room every 16 minutes; how fast is air flowing in the duct?

 

 

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The volume of the room is 9.2 m * 5.0 m * 4.5 m = 210 m^3.

 

This air is replentished every 16 minutes, or at a rate of 210 m^3 / (16 min * 60 sec/min) = 210 m^3 / (960 sec) = .22 m^3 / second.

 

The cross-sectional area of the duct is pi r^2 = pi * (.15 m)^2 = .071 m^2.

 

The speed of the air flow and the velocity of the air flow are related by

 

rate of volume flow = cross-sectional area * speed of flow, so

 

speed of flow = rate of volume flow / cross-sectional area = .22 m^3 / s / (.071 m^2) = 3.1 m/s, approx.

 

 

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Question:    prin phy and gen phy problem 10.40  What gauge pressure is necessary to maintain a firehose stream at an altitude of 15 m?

.......................................

 

 

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** We use Bernoulli's equation. Between the water in the hose before it narrows to the nozzle and the 15m altitude there is a vertical change in position of 15 m.

 

Between the water in the hose before it narrows to the nozzle and the 15 m altitude there is a vertical change in position of 15 m. 

 

Assuming the water doesn't move all that fast before the nozzle narrows the flow, and noting that the water at the top of the stream has finally stopped moving for an instant before falling back down, we see that we know the two vertical positions and the velocities (both zero, or very nearly so) at the two points. 

 

All that is left is to calculate the pressure difference. The pressure of the water after its exit is simply atmospheric pressure, so it is fairly straightforward to calculate the pressure inside the hose using Bernoulli's equation.

 

Assuming negligible velocity inside the hose we have

 

change in rho g h from inside the hose to 15 m height:  `d(rho g h) = 1000 kg/m^3 * 9.8 m/s^2 * 15 m = 147,000 N / m^2, approx. 

 

Noting that the velocity term .5 `rho v^2 is zero at both points, the change in pressure is `dP = - `d(rho g h) = -147,000 N/m^2.

 

Since the pressure at the 15 m height is atmospheric, the pressure inside the hose must be 147,000 N/m^2 higher than atmospheric. **

 

 

 

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Question:    Openstax:  The heart of a resting adult pumps blood at a rate of 5.00 L/min.

Convert this to cm^3 /s .

What is this rate in m^3 /s ?

 

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A cube 10 cm on a side is a liter.  It takes 10 layers each of 10 rows each of 10 such cubes to fill a one-meter cube.

 

It follows that l liter is a volume of 10 cm * 10 cm * 10 cm = 1000 cm^3, and a cubic meter is a volume equivalent to 10 * 10 * 10 liters = 1000 liters.

5 L / min is therefore the same as 5 L * (1000 cm^3 / L) / min = 5000 cm^3 / min.

 

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Question:     Openstax:  Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. Approximately what is the force due to the Bernoulli effect on a roof having an area of 220 m^2 ? Typical air density in Boulder is 1.14 kg/m^3 , and the corresponding atmospheric pressure is 8.89×104 N/m^2 . (Bernoulli’s principle as stated in the text assumes laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

 

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Bernoulli's Equation says that

rho g y + 1/2 rho v^2 + P = constant.

On one side of the roof v = 0 (that would be the inside).  On the other side v = 45 m/s.

The difference in y from one side of the roof to the other is small, so that the change in rho g y from one side to the other is negligible.

Thus 1/2 rho v^2 + P is constant, and we conclude that

1/2 rho v_1^2 + P_1 = 1/2 rho v_2^2 + P_2,

where v_1 and P_1 are on one side of the roof (let's say that's the inside) and v_2 and P_2 are on the other.

The pressure change from inside to outside is thus

P_2 - P_1 = 1/2 rho v_2^2 - 1/2 rho v_1^2 = 1/2 rho ( v_2^2 - v_1^2) = 1/2 * 1.14 kg/m^3 ( (45 m/s)^2 - (0 m/s)^2) = 1400 kg m^2 / (s^2 m^3) = 1400 kg / (m s^2) = 1400 N / m^2, or 1400 Pa.

 

Exerted on a roof area of 220 m^2 this pressure will thus exert a force of

F = P * A = 1400 Pa * 220 m^2 = 1400 N/m^2 * 220 m^2 = 300 000 Newtons, very approximately.

 

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Question:    Openstax:  The Huka Falls on the Waikato River is one of New Zealand’s most visited natural tourist attractions (see Figure 12.29). On average the river has a flow rate of about 300,000 L/s. At the gorge, the river narrows to 20 m wide and averages 20 m deep. (a) What is the average speed of the river in the gorge? (b) What is the average speed of the water in the river downstream of the falls when it widens to 60 m and its depth increases to an average of 40 m?

1 m^3 is 1000 L so

300 000 L / s = 300 m^3 / s.

The cross-sectional area of the river in the gorge is 20 m * 20 m = 400 m^2.

A section of water with this cross-sectional area and length 1 meter would have volume 400 m^3, a little more than the 300 m^3 that flows in 1 second.  So the water speed is less that 1 m/s.  (A good swimmer could make headway against a 1 m/s current).

If `dx stands for the distance the water actually flows in 1 second, then the volume of flow in one second is 400 m^2 * `dx.   It follows that

400 m^2 * `dx = 300 m^3

so that

`dx = 300 m^3 / (400 m^2) = .75 m

and the water speed is therefore

.75 m/s.

 

The continuity equation can of course be applied to this situation, but perhaps without complete understanding.  This equation says that

A_cs * v = volume rate of flow,

where v is the velocity of the fluid and A_cs the cross-sectional area.  This

v = volume rate of flow / A_cs = 300 m^3/s / (400 m^2) = .75 m/s.

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Question:    Openstax:  A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.750 L/s, with an output pressure of 3.00×10^5 N/m^2 .

(a) The water enters a hose with a 3.00-cm inside diameter and rises 2.50 m above the pump. What is its pressure at this point?

(b) The hose goes over the foundation wall, losing 0.500 m in height, and widens to 4.00 cm in diameter. What is the pressure now? You may neglect frictional losses in both parts of the problem.
 

The speed of the water in the hose satisfies the continuity equation

A_cs * v = rate of volume flow.

A_cs is the cross-sectional area of the hose, which is about 7 cm^2, so

v = rate of volume flow / A_cs = .750 L/s / (7 cm^2) = .000750 m^3 / s / (.0007 m^2) = 1.1 m/s, approx..

 

The water enters at rest.  At the point 2.50 m above the pump its velocity is 1.1 m/s.  Since

change in 1/2 rho v^2 + change in rho g y + change in pressure = 0

we have

(1/2 rho v_2^2 - 1/2 rho v_1^2) + (rho g y_2 - rho g y_1) + (P_2 - P_1) = 0

with v_1 = 0, v_2 = 1.1 m/s, y_1 = 0, y_2 = 2.5 m and P_1 = 3 * 10^5 N/m^2.

We get

1/2 * 1000 kg/m^3 * (1.1 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * 2.5 m + P_2 - 3 * 10^5 N/m^2 = 0

so that

P_2 = 2.74 * 10^5 N/m^2, approx..

 

With a .500 m loss of height the quantity rho g y will change by - 5000 Pa.  With the diameter increase to 4.00 cm the velocity will reduce to about .6 m/s so 1/2 rho v^2 will reduce to about 180 Pa, which is pretty much negligible.  The pressure at this point will therefore be about 2.79 * 10^5 Pa.

 

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Question:    Gen phy:   Assuming that the water in the hose is moving much more slowly than the exiting water, so that the water in the hose is essentially moving at 0 velocity, what quantity is constant between the inside of the hose and the top of the stream?  what term therefore cancels out of Bernoulli's equation?

 

 

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** Velocity is 0 at top and bottom; pressure at top is atmospheric, and if pressure in the hose was the same the water wouldn't experience any net force and would therefore remain in the hose **

 

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Question:    query  gen phy problem 10.43  net force on 240m^2 roof from 35 m/s wind. 

What is the net force on the roof?

 

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** air with density around 1.29 kg/m^3 moves with one velocity above the roof and essentially of 0 velocity below the roof. Thus there is a difference between the two sides of Bernoulli's equation in the quantity 1/2 `rho v^2. At the density of air `rho g h isn't going to amount to anything significant between the inside and outside of the roof. So the difference in pressure is equal and opposite to the change in 1/2 `rho v^2.

 

On one side v = 0, on the other v = 35 m/s, so the difference in .5 rho v^2 from inside to out is

 

`d(.5 rho v^2) = 0.5(1.29kg/m^3)*(35m/s)^2 - 0 = 790 N/m^2.

 

The difference in the altitude term is, as mentioned above, negligible so the difference in pressure from inside to out is

 

`dP = - `d(.5 rho v^2) = -790 N/m^2.

 

The associated force is 790 N/m^2 * 240 m^2 = 190,000 N, approx. **

 

 

 

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Question:    gen phy  which term 'cancels out' of Bernoulli's equation and why?

 

 

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** because of the small density of air and the small change in y, `rho g y exhibits practically no change. **

 

 

 

 

 

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Question:    `q001.  Explain how to get the change in velocity from a change in pressure, given density and initial velocity, in a situation where altitude does not change.

Can you tell from your expressions whether the change in velocity, for a given pressure change, is greater, less or equal when the initial velocity is greater?

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Question:    `q002.  Water is moving inside a garden hose at 4 meters / second, and it exits the nozzle at 8 meters / second.  What is the change in pressure as the water moves from the end of the hose out into the surrounding air?

Neglecting the effect of air resistance:

How high would the stream be expected to rise if the hose was pointed straight upward?

How far would the stream travel in the horizontal direction before falling back to the level of the nozzle, if directed at 45 degrees above horizontal?

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Question:    univ phy problem 12.77 / 14.75 (11th edition 14.67): prove that if weight in water if f w then density of gold is 1 / (1-f).  Meaning as f -> 0, 1, infinity.  Weight of gold in water if 12.9 N in air.  What if nearly all lead and 12.9 N in air?

 

 

 

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** The tension in the rope supporting the crown in water is T = f w.

 

Tension and buoyant force are equal and opposite to the force of gravity so

 

T + dw * vol = w  or f * dg * vol + dw * vol = dg * vol.

 

Dividing through by vol we have

 

f * dg + dw = dg, which we solve for dg to obtain

 

dg  = dw / (1 - f).

 

Relative density is density as a proportion of density of water, so

 

relative density is 1 / (1-f).

 

For gold relative density is 19.3 so we have

 

1 / (1-f) = 19.3, which we solve for f to obtain

 

f = 18.3 / 19.3.

 

The weight of the 12.9 N gold crown in water will thus be

 

T = f w = 18.3 / 19.3 * 12.9 N = 12.2 N.

 

 

 

STUDENT SOLUTION:

 

After drawing a free body diagram we can see that these equations are true: 

 

Sum of Fy =m*ay ,

 

T+B-w=0,

 

T=fw,

 

B=(density of water)(Volume of crown)(gravity). 

 

Then

fw+(density of water)(Volume of crown)(gravity)-w=0.

(1-f)w=(density of water)(Volume of crown)(gravity). 

 

Use

w==(density of crown)(Volume of crown)(gravity). 

(1-f)(density of crown)(Volume of crown)(gravity) =(density of water)(Volume of crown)(gravity). 

 

Thus, (density of crown)/(density of water)=1/(1-f).  **

 

 

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Question:    univ phy  What are the meanings of the limits as f approaches 0 and 1?

 

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** GOOD STUDENT ANSWER:  f-> 0 gives (density of crown)/(density of water) = 1 and T=0.  If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero.  When f-> 1, density of crown >> density of water and T=w.  If density of crown >> density of water then B is negligible relative to the weight w of the crown and T should equal w. **

 

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Question:    `q003.  Water exits a large tank through a hole in the side of a cylindrical container with vertical walls.  The water stream falls to the level surface on which the tank is resting.  The tank is filled with water to depth y_max.  The water stream reaches the level surface at a distance x from the side of the container.

Without doing any calculations, explain why there must be at least one vertical position at which the hole could be placed to maximize the distance x.  Explain also why there must be distances x that could be achieved by at least two different vertical positions for the hole.

Give all the possible vertical levels of the hole.

What is the maximum possible distance x at which it is possible for a water stream to reach the level surface, and where would the hole have to be to achieve this?

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