course Phy 202 Mr. Smith,I tried to do the best I could on this assignment. This was the day I left early so I don't have all the notes. 100203 Everything in today's notes is related to Bernoulli's Equation, which expresses the work-energy theorem in the context of a continuous fluid. We look at different forms of the equation, the details of the equation, how to apply the equation to systems we have observed in class, and at some other applications. Bernoulli's Equation says that if rho is the density, v the velocity, y the vertical position and P the pressure in a continuous fluid, the quantity 1/2 rho v^2 + rho g y + P is constant. This is an expression of energy conservation, assuming no dissipative forces. It provides important information on how changes in rho, v, y and P, from one point to another in a continuous fluid, are related. Bernoulli's Equation can be applied to any two points within a continuous fluid. We will see what 'continuous fluid' means in the examples we consider below. Since 1/2 rho v^2 + rho g y + P is constant, the sum of the changes in these three terms must be zero. That is, `d( 1/2 rho v^2) + `d ( rho g y) + `d P = 0. It is often easiest to think of the equation in this form. Another form of the equation is sometimes useful. If two points are denoted A and B, then the values of rho, v, y and P at point A will be denoted rho_A, v_A, y_A and P_A; the values of the same quantities at point B will be denoted rho_B, v_B, y_B and P_B. To say that 1/2 rho v^2 + rho g y + P is constant is to say that the value of 1/2 rho v^2 + rho g y + P at point A is equal to its value at point B. Thus we can write an equivalent form of Bernoulli's Equation as follows: 1/2 rho_A * v_A^2 + rho_A * g * y_A + P_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P_B There are a lot of symbols in the equation and it might look intimidating, but in many cases it simplifies a great deal. It will be helpful if you understand that 1/2 rho v^2 is the KE per unit volume, rho g y the gravitational PE per unit volume, and P is a form of energy per unit volume. Our main focus in this class will be the application of Bernoulli's equation to systems we have already observed. Outflow velocity and water depth You were given a bottle with a hole in its side and asked to observe the horizontal range of the outflow stream vs. the depth of water in the bottle relative to the hole, as the water fell to the ground at an observed distance below the hole. `q001. Give your data. **** Height of Water In Column Vertical Height of Bottle Horizontal Distance Travelled 100 cm 15 cm 43 cm 60 cm 14 cm 28 cm 80 cm 13 cm 30 cm #$&* `q002. Assume that the horizontal velocity of each drop of water remained constant as it fell, and that its vertical acceleration was equal to the acceleration of gravity. For each horizontal range determine the horizontal velocity of the stream. **** V_ave = ds/ dt It is estimated that the fall of the horizontal range was about 1 to 2 seconds. V_ave = 100 cm/ 2 seconds V_ave = 50 cm/s V_ave = 60 cm/ 2 seconds V_ave = 30 cm/s V_ave = 80 cm/ 2 seconds V_ave = 40 cm/s #$&*
100203
Everything in today's notes is related to Bernoulli's Equation, which expresses the work-energy theorem in the context of a continuous fluid. We look at different forms of the equation, the details of the equation, how to apply the equation to systems we have observed in class, and at some other applications.
Bernoulli's Equation says that if rho is the density, v the velocity, y the vertical position and P the pressure in a continuous fluid, the quantity
is constant. This is an expression of energy conservation, assuming no dissipative forces. It provides important information on how changes in rho, v, y and P, from one point to another in a continuous fluid, are related.
Bernoulli's Equation can be applied to any two points within a continuous fluid. We will see what 'continuous fluid' means in the examples we consider below.
Since 1/2 rho v^2 + rho g y + P is constant, the sum of the changes in these three terms must be zero. That is,
It is often easiest to think of the equation in this form.
Another form of the equation is sometimes useful.
If two points are denoted A and B, then
- the values of rho, v, y and P at point A will be denoted rho_A, v_A, y_A and P_A;
- the values of the same quantities at point B will be denoted rho_B, v_B, y_B and P_B.
To say that 1/2 rho v^2 + rho g y + P is constant is to say that the value of 1/2 rho v^2 + rho g y + P at point A is equal to its value at point B. Thus we can write an equivalent form of Bernoulli's Equation as follows:
- 1/2 rho_A * v_A^2 + rho_A * g * y_A + P_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P_B
There are a lot of symbols in the equation and it might look intimidating, but in many cases it simplifies a great deal.
It will be helpful if you understand that 1/2 rho v^2 is the KE per unit volume, rho g y the gravitational PE per unit volume, and P is a form of energy per unit volume.
Our main focus in this class will be the application of Bernoulli's equation to systems we have already observed.
Outflow velocity and water depth
You were given a bottle with a hole in its side and asked to observe the horizontal range of the outflow stream vs. the depth of water in the bottle relative to the hole, as the water fell to the ground at an observed distance below the hole.
`q001. Give your data.
****
Your data will consist of the horizontal range, vertical fall and water depth.
For sample data let's assume the following:
water depth 15 cm
vertical fall 100 cm and
horizontal range 40 cm.
#$&*
`q002. Assume that the horizontal velocity of each drop of water remained constant as it fell, and that its vertical acceleration was equal to the acceleration of gravity. For each horizontal range determine the horizontal velocity of the stream.
****
The water fell freely 100 cm. It initially exited the container in a direction perpendicular to the sides of the container. Assuming the container was held with its sides vertical, the initial vertical velocity is zero. The acceleration in the vertical direction is that of gravity, 980 cm/s^2. Using the fourth equation of uniformly accelerated motion the final velocity is
vf = +- sqrt( v0^2 + 2 a `ds ) = +- sqrt( 0 ^2 + 2 * 980 cm/s^2 * 100 cm) = +-470 cm/s, approx..
Assuming downward to be the positive vertical direction we conclude that vf = 470 cm/s.
Its average vertical velocity is therefore (0 cm/s + 430 cm/s) / 2 = 215 cm/s. To fall the 100 cm the time required is therefore
`dt = 100 cm / (235 cm/s) = .43 sec, approx..
The water therefore travels 40 cm in the horizontal direction in the .43 sec required to reach the ground. Its average horizontal velocity is therefore
40 cm / (.43 sec) = 90 cm/s, very approximately.
#$&*
... expt stream vs. projectile off ramp ...
`q003. Construct a graph of horizontal stream velocity vs. water depth, based on your results. Sketch the smooth curve you believe best fits your results. Give a good description of your graph.
****
You will have observed the system for at least three water depths. For each depth you will have a different horizontal range. Suppose the depths were 15 cm (as in the sample data), 10 cm and 5 cm, and the horizontal velocities 90 cm/s (as in the preceding), 70 cm/s and 50 cm/s.
Then the graph of stream velocity vs. water depth would be a straight line with slope 4 (cm/s) / cm.
Your graph might or might not be a straight line. If your graph tends to curve you should include how it curves when you give your description.
#$&*
`q004. On the same set of coordinate axes graph the function sqrt( 2 g h) vs. h, where h is the water depth. Describe how your graph compares to the graph of your data.
****
You could get a reasonable plot of sqrt( 2 g h) vs. h by evaluating this quantity at h values consistent with your data.
For example h = 5 cm yields sqrt( 2 g h) = sqrt( 2 * 980 cm/s^2 * 5 cm) = sqrt(9800 cm^2 / s^2) = 100 cm/s, approx..
h = 10 cm and h = 15 cm will yield velocities of about 140 cm/s and 170 cm/s, respectively.
These points appear to define a curve which is rising as its slope decreases.
This curve would lie higher than the three points (5 cm, 50 cm/s), (10 cm, 70 cm/s) and (15 cm, 90 cm/s) plotted in the preceding analysis.
#$&*
If water flows out of the bottle for a short time interval `dt, the center of mass of the water in the bottle gets lower, so the gravitational PE of the water remaining in the bottle decreases.
The speed of the water in the bottle changes by a small amount, and the water exiting the bottle by a much greater amount. This will be the case as long as the cross-sectional area of the outflow hole is much less than the cross-sectional area of the bottle.
Suppose that in the short time interval `dt, the mass of water exiting through the hole is `dm.
If the water depth relative to the hole is h, then the 'missing mass', equal to `dm, started out at altitude h relative to the hole, and ended up at the level of the hole, at altitude 0 relative to the hole. (The water might subsequently have fallen to the ground, at some negative altitude relative to the hole, but we assume a short time interval `dt in which the water doesn't have time to fall to any significant distance below the hole).
The PE of the mass `dm therefore changed. Relative to the hole, its PE at the beginning of the interval was
and at the end of the interval, being at altitude 0 with respect to the hole, is PE is zero:
The change in its gravitational PE is therefore
If no other PE is involved, and if no work is done against nonconservative forces, then
The original KE of the mass `dm was nearly negligible (provided the c-s area of the container is much less than that of the outflow hole), so `dKE = KE_f - KE_0 = KE_f = 0 = KE_f (i.e., the initial KE was essentially zero so the final KE is equal to the change in KE). Thus
The final KE is that of the exiting water, which has mass `dm. We let v_exit stand for the velocity of the exiting water, and recall that KE = 1/2 m v^2. We can therefore write the final KE as
Since KE_f = `dm * g * h, we set our two expressions for KE_f equal to obtain the equation
We easily solve this for v_exit, obtaining
`q005. When you compare the graph of your experimental results with the graph of sqrt(2 g y), is there evidence that the system is doing work against nonconservative forces? If so, what is this evidence?
****
The velocities indicated by the observed water stream are less than those predicted by the ideal equation.
So the KE of the exiting water is less than would be predicted by energy conservation with the assumption that no nonconservative forces are present, and that as a result all PE loss is manifested as KE gain.
We conclude that nonconservative forces must be acting in such a way as to reduce the total mechanical energy of the system.
#$&*
(challenge question): Can you estimate the rate at which work is being done against nonconservative forces, for each observed height?
****
Hints:
You could base an estimate on how fast water is falling in the container, and the cross-sectional area of the container. This would give you the rate at which PE is being lost from the system.
The observed velocity of the exiting water can be used with the same estimates to determine the rate at which KE is exiting the container.
The difference would be the rate at which nonconservative forces work on the system.
#$&*
Now let's consider the same problem we just solved. This time we're going to use Bernoulli's Equation to get our result.
We start by choosing two points within the continuous fluid.
It was said earlier that Bernoulli's Equation usually simplifies.
In this case the density rho remains constant, or very nearly so.
This reduces the number of variable quantities from four to three.
So the pressure at both points is the same, P_atm.
This leaves only two variable quantities, v and y.
Thus, for this particular system, rho and P are constant. Using the form `d( 1/2 rho v^2) + `d ( rho g y) + `d P = 0, we have
and, since P is also constant, `dP is zero and we conclude that
You might see that rho is constant so it can actually be divided out of the equation. This will be very useful. However we won't do that at this stage. Right now we want to associate the expression 1/2 rho v^2 with KE (in this equation, to be specific, it is KE per unit volume) and the expression rho g y with gravitational PE (in this case, gravitational PE per unit volume).
Now let's again suppose that the water level relative to the outflow hole is h, and that the velocity of the water surface is essentially zero.
`q006. Consider first the term `d(rho g y), the change in the value of rho g y.
Measuring y relative to the hole, what is the expression for rho g y at the water surface (the point we have designated as A), and what is the expression for rho g y just outside the hole (point B)? (hint: measuring relative to the hole, what is y at the hole, and what is y at the water surface? Just plug these expressions in for y).
What therefore is the change in rho g y as we move from point A to point B?
****
Using our previous result that water exits at 90 cm/s when the level is 15 cm above the hole:
rho and g don't change. The change in y is -15 cm. So the change in rho g y is
`d(rho g y) = rho g `dy = 1000 kg/m^3 * 9.8 m/s^2 * (-.15 m) = -1500 kg / (m s^2) = -1500 N / m^2.
#$&*
`q007. Now consider the term `d(1/2 rho v^2), the change in 1/2 rho v^2.
At the water surface (point A) we assume that v is zero, or close enough to zero that it can be neglected. So at the water surface, 1/2 rho v^2 is zero.
Letting v_exit be the velocity at the point of exit (point B), the value of 1/2 rho v^2 is 1/2 rho v_exit^2.
From point A to point B, then, what is the expression for the change in 1/2 rho v^2?
****
At point A, we have 1/2 rho v^2 = 0.
At point B, we have 1/2 rho v^2 = 1/2 rho v_exit^2.
So the change in 1/2 rho v^2 is
`d(1/2 rho v^2) = 1/2 rho v_exit^2 - 0 = 1/2 rho v_exit^2.
#$&*
`q008. Substitute your expressions for the changes in 1/2 rho v^2 and rho g y into the equation
and solve for v_exit. What is your result?
****
`d(1/2 rho v^2) = 1/2 rho v_exit^2 and `d(rho g y) = -1500 N/m^2. Thus
1/2 rho v_exit^2 + (-1500 N/m^2) = 0 and
1/2 rho v_exit^2 = 1500 N/m^2. Thus
v_exit = +- sqrt( 2 * 1500 N/m^2 / rho) = +- sqrt( 3000 N/m^2 / (1000 kg/m^3) ) = +- sqrt( 3 m^2/s^2) = +- 1.73 m/s, approx..
#$&*
How does your result compare with the result obtained for v_exit in the analysis of your experimental data?
****
The result is significantly higher than the .90 m/s result from our experiment.
Note that this result is consistent with our approximate estimate of 170 cm/s or 1.70 m/s, based on energy conservation in the absence of nonconservative forces.
#$&*
Alternatively we could have solved this problem using the form 1/2 rho_A * v_A^2 + rho_A * g * y_A + P_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P_B of Bernoulli's Equation, as outlined in the next few questions:
`q009. Since rho is very nearly constant for water, it is the same at point A as at point B, and we can just replace the symbols rho_A and rho_B both by rho.
Write out the equation 1/2 rho_A * v_A^2 + rho_A * g * y_A + P_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P_B with both rho_A and rho_B replaced by the symbol rho. What is your equation?
****
The equation would be
1/2 rho * v_A^2 + rho * g * y_A + P_A = 1/2 rho * v_B^2 + rho * g * y_B + P_B
#$&*
`q010. Using the same notation as in the preceding solution,
Substitute these values into the equation. What is the resulting equation?
****
Using the form
1/2 rho * v_A^2 + rho * g * y_A + P_A = 1/2 rho * v_B^2 + rho * g * y_B + P_B
we make the indicated substitutions to get
#$&*
Solve this equation for v_exit, in terms of the other symbols. What is your solution?
****
Starting with
1/2 rho * 0^2 + rho * g * h + P_atm = 1/2 rho * v_exit^2 + rho * g * 0 + P_atm.
we subtract P_atm from both sides and leave off 1/2 rho * 0^2, which is 0 and contributes nothing to the left-hand side, as well as rho * g * 0, to get
rho * g * h = 1/2 rho * v_exit^2 .
#$&*
`q011. In general, when P_A = P_B we could just use the same symbol P for the pressure at both points. Write the equation 1/2 rho_A * v_A^2 + rho_A * g * y_A + P_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P_B with the symbol P substituting for P_A and for P_B.
****
The equation is
1/2 rho_A * v_A^2 + rho_A * g * y_A + P = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P.
#$&*
What do you get if you subtract P from both sides of this equation? How does this simplify the equation?
****
Starting with
1/2 rho_A * v_A^2 + rho_A * g * y_A + P = 1/2 rho_B * v_B^2 + rho_B * g * y_B + P
we subtract P from both sides to get
1/2 rho_A * v_A^2 + rho_A * g * y_A = 1/2 rho_B * v_B^2 + rho_B * g * y_B.
#$&*
Now suppose that in addition rho is constant, so both rho_A and rho_B can be replaced by the symbol rho. What is the resulting equation?
****
We get the equation
1/2 rho * v_A^2 + rho * g * y_A = 1/2 rho * v_B^2 + rho * g * y_B.
#$&*
What do you get when you divide both sides by rho? How does this simplify the equation?
****
Starting with
1/2 rho * v_A^2 + rho * g * y_A = 1/2 rho * v_B^2 + rho * g * y_B
we divide both sides by rho to get
1/2 v_A^2 + g * y_A = 1/2 v_B^2 + g * y_B.
#$&*
Solve the resulting equation for v_B. Give your solution below, including a brief synopsis of the steps you took to get the solution.
****
Starting with
1/2 v_A^2 + g * y_A = 1/2 v_B^2 + g * y_B
we subtract g * y_B from both sides to get
1/2 v_A^2 + g * y_A - g * y_B = 1/2 v_B^2
then we multiply both sides by 2 to get
v_B^2 = v_A^2 + 2 g * y_A - 2 g * y_B
Thus
v_B = +-sqrt( v_A^2 + 2 g * y_A - 2 g * y_B ), which we might simplify as
v_B = +- sqrt( v_A^2 + 2 g * (y_A - y_B) ).
#$&*
For the special case in which v_A = 0, what is the expression for your solution?
****
v_B = +-sqrt( v_A^2 + 2 g * y_A - 2 g * y_B ).
If y_A = 0 then
v_B = +-sqrt( v_A^2 - 2 g * y_B )
#$&*
Now let's consider the two-bottle system.
`q012. Can you trace a path from position A to position B by following a path that doesn't leave water? (hint: if this is done as indicated the siphon tube remains full throughout, and stays that way)
****
You can trace a path from the water surface at A down into the water, into the tube, through the tube into the second bottle, out the bottom into the water in that bottle, and to the water surface at B. You never leave the water, so we say there is a continuous path through the fluid from point A to point B.
#$&*
What is the change `dP from the pressure at point A and point B? (hint: recall that both caps are loose)
****
Both caps are loose so the air inside each bottle is at atmospheric pressure. The air is in contact with both water surfaces, which are therefore also at atmospheric pressure.
#$&*
What is the are the velocities v_A and v_B of the water surface at points A and B? (hint: how fast is each water surface moving, once the two levels are equal?)
****
The two levels are equal and the system is just sitting there, so the velocities are both zero.
#$&*
Using Bernoulli's Equation, what do we conclude is therefore the difference between the values of rho g y at points A and B?
****
P_A = P_B (both at atmospheric pressure), so let P stand for the common pressure.
v_A = v_B (both surfaces are stationary)
Therefore Bernoulli's equation
1/2 rho * v_A^2 + rho * g * y_A + P_A = 1/2 rho * v_B^2 + rho * g * y_B + P_B
becomes
1/2 rho * 0 ^2 + rho * g * y_A + P = 1/2 rho * 0^2 + rho * g * y_B + P.
Subtracting P from both sides and leaving off the terms equal to 0 we end up with
rho * g * y_A = rho * g * y_B.
Dividing both sides by rho * g we have
y_A = y_B.
The only way we can have these conditions (equal pressure, surface velocities both zero) is for both y_A to equal y_B. That is, both surfaces are at the same vertical level.
#$&*
What does this tell you about the values of y at the two points?
****
y is the same at both points
#$&*
Why was it important that you could trace a path from one point to the other that didn't leave the water? (hint: read the first statements of Bernoulli's Equation at the beginning of this document)
****
Bernoulli's Equation applies to a continuous fluid. If you can't trace that path it's because the fluid isn't continuous, in which case Bernoulli's Equation doesn't hold.
#$&*
Now suppose you seal both bottles and raise one of them. You did this in class. The water level in one bottle will increase, the level in the other will decrease. If you raise the bottle to a height of 1 meter or so, one of the bottles will partially collapse.
`q013. Which bottle will partially collapse?
****
Your experience should tell you that the top bottle collapses.
#$&*
In which bottle will the water level rise?
****
From your experience you should know that the water in the lower bottle rises.
#$&*
In which bottle will the volume of the air decrease?
****
If the water level rises, the air volume decreases. If the water level decreases the air volume increases (and/or the bottle collapses).
#$&*
What will happen to the volume of air in the other bottle?
****
The bottle might partially collapse, but not until air volume has decreased to reduce the pressure inside. The collapse results from air pressure outside exceeding air pressure inside.
#$&*
If you had 'pressure tubes' in both bottles, how would the length of the air column in the higher bottle change?
****
The pressure in the higher bottle decreases, so the pressure on the air column decreases. This allows it to expand. Its length therefore increases.
#$&*
How would the length of the air column in the lower bottle change?
****
The pressure in the lower bottle increases, so the pressure on the air column increases. This compresses it. Its length therefore decreases.
#$&*
`q014. Let point A be at the water surface in the higher bottle, point B at the water surface in the lower. Suppose the system is given time to reach an equilibrium position, in which water ceases to flow from one bottle to the other.
We have already observed that for water, rho remains constant. Of the other three possibly varying quantities v, y and P, which will be the same at point A as at point B?
****
y differs, since one bottle is higher than the other
It appears that P differs from one bottle to the other.
However if the system has reached equilibrium, the speed of the water will be zero everywhere, so v is the same at point A as at point B.
#$&*
What are the two remaining variable quantities?
****
y and P are variable, in the sense that they change as we move from point A to point B.
#$&*
Of these two quantities, which is most obviously greater at point A?
****
Since the first bottle is higher it is most obvious that the y is greater at point A.
#$&*
What can be concluded about the other of the two variable quantities?
****
The other variable quantity is P.
1/2 rho v^2 doesn't change; since by Bernoulli's equation 1/2 rho v^2 + rho g y + P is constant, it follows that in this case the quantity rho g y + P is constant
Since rho g y + P is constant, it follows that a decrease in y is 'balanced' by an increase in P.
As we go from point A to point B, the fact that y decreases implies that P increases.
Thus the pressure at point B is higher than the pressure at point A, so the pressure at point A is lower than at point B.
#$&*
`q015. This is a continuation of the preceding question. Let y_A and y_B stand for the water levels at A and B, relative to the tabletop. Let P_A and P_B be the pressures at A and B. Let v_A and v_B be the water velocities at A and B. Let rho stand for the density of water.
As we move from point A to point B, is `d(rho g y) positive or negative? In terms of the given symbols, what is `d(rho g y)?
****
Since rho and g are constant, `d( rho g y) = rho g * `dy.
From point A to point B, `dy = y_B - y_A.
Since point B is lower than point A, y_B < y_A so that y_B - y_A is negative.
Thus
#$&*
As we move from point A to point B, is `d(1/2 rho v^2) positive or negative? In terms of the given symbols, what is `d(1/2 rho v^2)?
****
The system is stationary and in equilibrium so v = 0 at both points; thus 1/2 rhg v^2 is the same at A and B. We could write
#$&*
As we move from point A to point B, is `dP positive or negative? In terms of the symbols rho, y_A, y_B, v_A and v_B, what is `dP?
****
Since `d(rho g y) is negative and `d(1/2 rho v^2) is zero, and since `d(rho g y) + `d(1/2 rho v^2) + `dP = 0, it follows that `dP must be positive.
#$&*
`q016. Is the pressure at point B higher or lower than the pressure at point A?
How can you use Bernoulli's equation to confirm your conclusion?
****
In the preceding we found that as we go from A to B the pressure change `dP is positive.
Thus the pressure at B is higher than the pressure at A.
#$&*
`q017. Suppose y_A = 120 cm and y_B = 10 cm. Using Bernoulli's Equation, find `dP.
****
This could be done using Bernoulli's equation in any of three forms:
Using 1/2 rho v^2 + rho g y + P = constant:
v = 0 at both A and B
rho g y takes value rho g y_A at point A, and value rho g y_B at point B. So rho g y changes by rho g ( y_B - y_A) = 1000 kg / m^3 * 9.8 m/s^2 * (.10 m - 1.20 m) = -11 000 N / m^2, approx..
To keep 1/2 rho v^2 + rho g y + P constant while the first term remains unchanged and the second changes by -11 000 N / m^2, the third term P must increase by 11 000 N / m^2.
Using `d(1/2 rho v^2) + `d(rho g y) + `dP = 0
v = 0 at both A and B so `d(1/2 rho v^2) = 0
rho g y takes value rho g y_A at point A, and value rho g y_B at point B.
- `d(rho g y) = rho g ( y_B - y_A) = 1000 kg / m^3 * 9.8 m/s^2 * (.10 m - 1.20 m) = -11 000 N / m^2, approx..
Our equation `d(1/2 rho v^2) + `d(rho g y) + `dP = 0 becomes
0 + (-11 000 N/m^2) + `dP = 0, which we easily solve for `dP to obtain
`dP = +11 000 N/m^2.
Using the form 1/2 rho v_A^2 + rho g y_A + P_A = 1/2 rho v_B^2 + rho g y_B + P_B:
We substitute v_A = 0, v_B = 0, y_A = 1.20 m, y_B = -10 m to get
1/2 rho * (0^2) + rho g * 1.20 m + P_A = 1/2 rho * 0^2 + rho g * .10 m + P_B
which simplifies to
rho g * 1.20 m + P_A = rho g * .10 m + P_B
Subtracting P_A + rho g * .10 m from both sides
rho g * 1.20 m - rho g * .10 m = P_B - P_A so that
P_B - P_A = rho g * 1.20 m - rho g * .10 m = 1000 kg/m^3 * 9.8 m/s^2 * 1.20 m - 1000 kg/m^3 * 9.8 m/s^2 + .10 m = 12 000 N/m^2 - 1000 N/m^2 = 11 000 N/m^2.
The analysis looks slightly different with each form of the equation, but it should be clear how each form of the analysis is equivalent to each of the others.
#$&*
`q018. Continuing the preceding question, use Bernoulli's Equation to answer the following::
Suppose the top of the tube is 30 cm higher than the water surface in the higher bottle. Let the top of the tube be point C.
What is the pressure change if you move from point A to point C?
****
Point C is 30 cm higher than point A. Using the vertical positions from the preceding, y_A = 120 cm so y_C, being 30 cm higher, is y_C = 150 cm.
v = 0 at both points.
Thus 1/2 rho v^2 = 0, so that rho g y + P is constant.
From point A to point C, rho g y changes by
`d(rho g y) = rho g ( y_C - y_A) = 1000 kg/m^3 * 9.8 m/s^2 * (150 cm - 120 cm)
= 1000 kg/m^3 * 9.8 m/s^2 * .30 m
= 3000 N/m^2, approx..
Thus from point A to point C, the pressure P must change by about 3000 N/m^2.
#$&*
What is the pressure change if you move from point B to point C?
****
The analysis is identical to the preceding, except that y_B = 10 cm. Thus from point B to point C
`d(rho g y) = rho g ( y_C - y_B) = 1000 kg/m^3 * 9.8 m/s^2 * (150 cm - 10 cm)
= 1000 kg/m^3 * 9.8 m/s^2 * 1.40 m
= 14 000 N/m^2, approx..
#$&*
How are your results consistent with the result you obtained in the preceding question for `dP?
****
From point B to point A, we pass through point C so the path from B to A can be broken into a path from B to C, and a path from C to A.
From point B to point C the pressure change is 14 000 N/m^2.
From point A to point C the pressure change is 3 000 N/m^2, so from point C to point A the change in -3 000 N/m^2.
Thus the change from B to A must be +14 000 N/m^2 + (-3000 N / m^2) = 11 000 N/m^2.
This is consistent with the pressure change calculated previously from B to A.
#$&*
`q019. Suppose a horizontal water pipe narrows. Water flowing through the pipe must speed up as it goes through the narrow part of the pipe. Suppose that the speed increases from 80 cm/s at a point before the narrowing to 200 cm/s at the narrowest point.
Let A be one point and B the other. Which letter do you wish to apply to which point?
****
For the purposes of this solution let's let A be the point before the tube narrows, B the point just after the tube has narrowed.
#$&*
What is the change in rho g y from point A to point B (hint: the pipe is horizontal).
****
The tube is horizontal, so the altitude y is unchanged.
#$&*
What is the change in 1/2 rho v^2 from point A to point B?
****
At point A we have 1/2 rho v^A^2 = 1/2 * 1000 kg/m^3 * (.80 m/s)^2 = 320 N / m^2.
At point B we have 1/2 rho v^B^2 = 1/2 * 1000 kg/m^3 * (2.00 m/s)^2 = 2000 N / m^2.
The change is therefore
1/2 rho v_B^2 - 1/2 rho v_A^2 = 2000 N/m^2 - 320 N/m^2 = 1700 N/m^2, approx..
#$&*
Use Bernoulli's equation to find the change in pressure from point A to point B.
****
Using the form `d(rho g y) + `d(1/2 rho v^2) + `dP = 0, where the change is from A to B, we note again that y is unchanged so `d(rho g y) = 0 and `d(1/2 rho v^2) = +1700 N/m^2 so our equation becomes
0 + 1700 N/m^2 + `dP = 0, which gives us
`dP = -1700 N/m^2.
From point A to point B, `dP = -1700 N/m^2 so the pressure decreases by 1700 N/m^2.
#$&*
At which point is the water at the higher pressure, the point at which the pipe is wider or the point at which it is narrower?
****
The pressure decreases as we go from point A to point B, so the pressure at point A is higher.
#$&*
`q020. In a hurricane, where is the air pressure higher, on the 'inner wall' bounded by the 'eye' of the hurricane, where winds are fastest, or near the outside of the hurricane, where winds are slowest?
****
Let point A be on the 'inner wall' and point B on the outside of the hurricane.
The density of air varies with pressure, so rho isn't necessarily constant.
Let's assume that A and B are at the same altitude, so that rho g y is the same at both points.
The velocity v_A is considerably greater than v_B (winds are fastest near the 'inner wall').
From point A to point B, therefore, rho g y remains constant while 1/2 rho v^2 decreases. Thus `d(rho g y) = 0 and `d(1/2 rho v^2) is negative.
Since `d(1/2 rho v^2) + `d(rho g y) + `dP = 0, it follows that from point A to point B the pressure must increase.
The pressure is lowest near the 'eye' of the hurricane.
#$&*
`q021. When hurricane winds are blowing, where is air moving faster, above the roof of an intact dwelling or below the roof? Where therefore would you expect air pressure to be higher? If the pressure difference gets great enough, what will happen to the roof?
****
y differs by less than a meter between inside and outside. In this case the fluid is air. If rho is the density of air, the change in rho g y for around 1 meter is (very roughly) around 10 Pa, which will prove to be insignificant compared to the other quantities involved here.
v is greater above the roof than below. By Bernoulli's equation, assume rho g y to have negligible change, 1/2 rho v^2 and pressure P have equal and opposite changes. This implies that the pressure above the roof will be lower.
#$&*