Assignment 6

course

I obmitted the problems dealing with the experiment because I haven't purchased my lab kit yet.

Question: query introset change in pressure from velocity change.

Explain how to get the change in fluid pressure given the change in fluid velocity, assuming constant altitude

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

According to Bernoulli’s equation, (0.5*rho*v^2)+P has to remain constant. But altitude is also constant. So if there is a change in fluid velocity, the fluid pressure must counter balance that change to keep (0.5*rho*v^2)+P at a constant value. So pretty much

(0.5*rho*V1^2+P1)=(0.5*rho*V2^2+P2)

Confidence Rating: 3

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

Given Solution:

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). **

Your Self-Critique:

ok

Your Self-Critique Rating:3

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Volume of the room= (9.2m*5.0m*4.5m)= 207m^3

Air replenishes every 16 minutes, so we can find the velocity

V= 207m^3/ (16minutes*60sec)

V= 207m^3/ 960 s

V= 0.22 m^3/s

The area of the duct= pi*r^2

Pi*(.15m)^2= .071 m^2

The speed of the air flow= rate of volume flow/cross sectional area

Speed= (0.22 m^3/s)/0.071 m^2

Speed= 3.1 m/s

Confidence Rating:3

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

Given Solution:

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.

Your Self-Critique:

ok

Your Self-Critique Rating:3

*********************************************

Question: prin phy and gen phy problem 10.40 gauge pressure to maintain firehose stream altitude 15 m ......!!!!!!!!...................................

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

0.5*rho*V2^2- 0.5*rho*V1^2= 1000 kg/m^3* 9.8 m/s^2* 15m= 147,000 N/m^2

Over altitude

Confidence Rating:1

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

Given Solution:

** 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. **

Your Self-Critique:

ok

Your Self-Critique Rating:2

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Confidence Rating:

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

Given Solution:

** 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 **

Your Self-Critique:

Your Self-Critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

P2-P1= 0.5*rho*v^2- 0.5*rho*v^2

0.5*1.29 kg/m^3* 35m/s^2-0= 790 N/m^2

This gives us the a force of 790 N/m^2

We multiply that by the volume

790 N/m^2* 240 m^2= 189,600 N

Confidence Rating:2

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

Given Solution:

** 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. **

Your Self-Critique:

ok

Your Self-Critique Rating:3

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

Altitude is 0 or close to it, so that makes rho*gy equal to 0 and cancel out.

Confidence Rating:3

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

Given Solution:

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

Your Self-Critique:

ok

Your Self-Critique Rating:

3

"

&#Very good responses. Let me know if you have questions. &#