PRESSURE/FLIGHT
UNIT SOME NOTES… ALL stuff we know already!!

Density is Mass
/Vol (in Kg/^{3}) for Physics

So Mass = Dens *
Volume, Since Weight = Mass *g, Weight=Dens*Vol*g

Pressure is
Force/Area in Newtons/ square meter

1
N/m^{2} = 1 Pascal of Pressure, Sea Level = 101320 Pascals

Since Energy is
Force * Distance = Work

Pressure = (Force/Area *
Distance/Distance)

= (Force * Distance) / (Area*Distance)

Pressure = Work / Volume = Energy/Volume

For an object in a
fluid the pressure on top of it is:

Δ
Pressure = Force/Area = Weight/Area

= (Dens * Vol*g)/ Area

=
Dens * g * Vol/Area

Δ
Pressure = Dens*g*Height

That’s
the same as saying that ΔWork/Volume = mgh /Volume !

In fact the
conservation of energy for a fluid is:

ΔWork + Δ
GPE + Δ KE = 0 (with no heat) but dividing by volume

(Mass/Vol = Dens)

F*dist/Vol + mgh/Vol + ˝ m vel^{2} / Vol =
F*dist/Vol + mgh/Vol + ˝ m vel^{2}/
Vol

Pressure_{1}
+ Dens*g*h_{1} + ˝ Dens*Vel_{1}^{2} = Pressure_{2}
+ Dens*g*h_{2} + ˝ Dens*Vel_{2}^{2}

This is called
Bernoulli’s Principle, and is just a restatement of conservation of energy that
relates Pressure (instead of work), Height and Velocity for a fluid.

If velocity stays
the same, then pressure changes depending on density and height.

If height stays the
same, then the faster the velocity, the lower the pressure!

Pg 337 #1

A large storage
tank open to the atmosphere at the top and filled with water, develops a small
hole at a point 16 m below the water. If the rate of flow of water from the
leak is .0025 m^{3} / min, determine the following:

a)
the speed at which the water leaves the hole.

b)
the diameter of the hole (hint : Area = Volume/Dist= πr^{2} )

**TOP:**

**Pressureair + Dens _{water}*g*h
+ ˝ Dens_{water}* Vel^{2}**

**101320 Pascals +
1000 kg/m ^{3}* 9.8 * 16m + 0
(large top means v≈**

** **

**BOTTOM:**

**Pressureair + Dens _{water}*g*h + ˝ Dens_{water}*
Vel^{2}**

**101320 Pascals +
0 ( h= 0!_ + ˝ 1000 kg/m ^{3}* Vel^{2}**

** **

**SO:**

** **

**Dens*g*h = ˝
Dens * Vel ^{2}**

**Vel ^{2}
= 2(9.8)*16, Vel = 17.7 m/s**

**b) VOLUME/TIME =
.0025 m ^{3}/min = .000042 m^{3}/sec,**

**AREA =
VOLUME/DISTANCE = (VOLUME/TIME) / (DISTANCE/TIME)= Flow Rate/Velocity**

**AREA = .000043 m ^{3}/sec
/ 17.7 m/s = 0.00000235 m^{2}**

**AREA (circle) = π**

** **

** r
= .000865 m so diameter = 2r= .00173 m = 0.173 cm**

#2 Skip

#3 When a person
inhales, air (Density is 1.29 kg/m^{3} ) moves down the windpipe at 15
cm/s (.15 m/s) . The average flow speed doubles when passing through a
constriction in the bronchus. Assuming incompressible flow, determine the
pressure drop in the constriction

NO HEIGHT CHANGE
(worth noting!)

SO

**Pressureairin + ˝ Dens _{air}*
Vel^{2} = Pressureairconstriction + ˝ Dens_{air}*
Vel^{2}**

**Pressureairin
– Pressureairconstriction = ˝* 1.29 * .3 ^{2} – ˝* 1.29 *
.15^{2}**

**ΔP = - .0435375 Pascals (drop in pressure)**

1) The time
required to fill a bucket with water from a certain garden hose is 30 sec. If
you cover part of the hose’s nozzle with your thumb so that the speed of water
leaving the nozzle doubles, how long does it take to fill the bucket?

*Because water
is treated as an incompressible fluid…*

*Volume stays
the same so Vol/Time=Vol/Time or Area1*Vel1=Area2*Vel2*

* So
if you double the flow speed, the area will be half*

* BUT
the VOLUME/TIME stays the same…. So 30 seconds…*

* *

2) Skip

3) The water supply
of a building is fed through a main entrance pipe that is 6 cm in DIAMETER. A 2
cm DIAMETER faucet tap positioned 2 m above the main pipe fills a 25 L (25 kg)
container in 30 seconds. Water is 1000 kg/m^{3}

a)
What is the speed at which water leaves the faucet?

b)
What is the gauge pressure in the pipe? (Pressure compared to atmosphere)

**TOP: Diameter =
2 cm = .02 m , Radius = .01 m**

**FLOW RATE =
VOL/TIME = 25 L/30 Sec = .025 m ^{3} /30 sec = .0008333 m^{3}/sec**

**AREA = π**

**VOL/TIME = AREA*
DIST/TIME = AREA * VEL**

** .0008333 m ^{3}/sec = .000314159
m^{2} * VEL**

** **

**VEL = 2.65258
m/s at top…**

** **

**At bottom
V1A1=V2A2**

** **

**Area at bottom =
AREA = π**

**At bottom
V1A1=V2A2**

**Vbottom=1/9 ^{th}
Vtop**

**2.65 (.000314) =
.0028 (Vbottom)**

** Vbottom = .2947 m/s**

** **

**We want gauge
pressure (difference in pressure)**

** **

**TOP:**

**Pressuretop + Dens _{water}*g*h
+ ˝ Dens_{water}* Vel^{2}**

**P1
+ 1000 kg/m ^{3}* 9.8 * 2m + ˝ 1000 kg/m^{3}* (2.65 m/s)^{2}**

** **

**BOTTOM:**

**Pressureair + Dens _{water}*g*h + ˝ Dens_{water}*
Vel^{2}**

**P2
+ 0 ( h= 0!_ + ˝ 1000 kg/m ^{3}*
(.2947 m/s) ^{2}**

** **

**P1-P2 = 1000 (9.8)*2 + ˝*1000*2.65 ^{2}
– ˝*1000*.2947^{2}**

** =
19600 + 3511.25 - 43.42405**

** ΔP
= 23067.83 Pascals**